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THE ELEMENTS
OF
PLANE TRIGONOMETRY.
THE ELEMENTS
OF
PLANE TRIGONOMETRY
BY
^i--^
R
LEVETT,
SECOND MASTER
AND
M.A.
C.
DAVISON,
M.A.
MATHEMATICAL MASTER
KING Edward's high school, Birmingham
MACMILLAN AND CO.
AND NEW YORK
1892
All 7'ighfs reserved
LA
^^.
/X.
PREFACE.
In this treatise on the Elements of Plane Trigonometry
the subject is divided into three parts, dealing respectively
with arithmetical, real algebraical, and complex, quantity.
Such an arrangement appears to be a natural one, and
has the advantage of introducing the new names and
formulae that belong to the subject before the student
encounters the difficulty of the application of signs to
denote the sense and direction of lines. Part I. is
further simplified by the postponement of the treatment
of the circular measurement of angles. For many
practical purposes, e.g. in surveying and in the applica-
tion of trigonometry to elementary mechanics, the short
introduction to the subject comprised in Part I. will be
found useful.
The immediate substitution of the real for the tabular
logarithm of the trigonometrical ratios was recommended
by Professor De Morgan, and seems likely to be generally
adopted as the simpler method in teaching and the more
expeditious and more accurate one in working.
In Part II. the theory of the Circular and Hyperbolic
Functions, for real variables, is presented in some
detail, the analogy between the circular and hyperbolic
797949
vi PREFACE.
functions is exhibited by similarity of method of treat-
ment, and an essay has been made to lead the student to
deal with infinite series with due caution. The chapter
on the Solution of Triangles and the applications to
Surveying has been made as practical as possible, in
order to add to the interest the student will find in this
part of the subject. The chapter on Factors is a de-
velopement of the consequences of the elegant theorem
given by Professor Adams in the Transactions of the
Carnhndge Philosophical Society.
No apology is needed for the introduction of geometrical
methods in Part TIL The methods are essentially general,
and the student who learns to think of complex numbers
as lines will gain a clearness of conception and a means
of testing results he can acquire in no other way.
Abundant examples for exercise have been collected
from University and other examination papers. The
student is advised to work through the shorter sets in
order to gain skill and readiness in using formulae, and
to select from the longer sets such examples as may
appear interesting or useful. The sets marked A and B
are alternative and may be used when the same portion
of the subject is read in class in consecutive terms.
Some of the longer sets of examples have been divided
into sections, arranged in order of difficulty.
The repetition of matter the student will have read
elsewhere has, so far as possible, been avoided. In
Chapters I.-X. a knowledge of Euclid and a few
well-known additional theorems in Geometry and of
elementary Algebra, including simple properties of
logarithms, is assumed ; Chapters XL-XXII. will probably
be read in connection with Geometrical Conies and the
PREFACE. vii
more advanced Algebra given in such treatises as those
of Ur. Todhunter, Mr. Charles Smith, or Messrs. Hall
and Knight.
A text-book of Plane Trigonometry, intended for use
in schools, can, from its nature, contain little original
matter. The present work differs mainly from those
most generally read in the extent to which the treatment
adopted by Professor De Morgan has been followed. The
influence of De Morgan's writings will be seen throughout
the book, and, in particular, in the use of the negative
hypotenuse in defining the ratios (necessary if the proofs
of some of the fundamental theorems are to be general),
in the more definite meaning assigned to the notation for
inverse functions, the manner in which the addition
formulae are extended to any number of variables, the
geometrical treatment of the hyperbolic functions and of
complex numbers, and in the two-fold generalisation of a
logarithm to a given base.
Our acknowledgements are due to Professor Chrystal
for the aid we have derived from his masterly and ex-
haustive work on Algebraical Analysis. In the classifi-
cation and terminology of convergent and divergent
series, in the use made of the continuity of a series up
to the limit of convergence, and in the geometrical form
given to the proof of the Binomial Theorem we have
followed his treatment. The chapters dealing with
imaginary quantities and infinite series must, of
necessity, contain much that is due directly or indirectly
to Cauchy's Analyse Algehrique. Arts. 183, 239 are
derived from recent numbers of Mathesis ; Arts. 210, 211
from Schlomilch's Homdhnch der algehraischen Analysis.
The short chapters on the direct and inverse exponential
viii PREFACE.
functions follow the lines laid down in the early para-
graphs of Professor Cay ley's Article on Functions in the
Encyclopaedia Britannica.
We are under great obligations to Mr. R. Tucker, of
University College School, for the very valuable assist-
ance he has given while the work has passed through
the press, and for his kindness in testing the results of
examples ; and our best thanks are also offered to Mr.
E. M. Langley, of the Bedford Modern School, and to our
past and present colleagues, Mr. C. H. P. Mayo, Mr. W. H.
Wagstaff and Mr. F. O. Lane for the kindly interest they
have taken in the book, and the pains they have given to
make it as free from error as possible. Our thanks are
due to the publishers for permission to make use of a
portion of the map of the Mer de Glace of Chamouni
given in the Life and Letters of Professor Forbes.
R LEVETT.
C. DAVISON.
King Edward's School,
Birmingham, January, 1892.
CONTENTS.
PART L— ARITHMETICAL QUANTITY,
CHAPTER I.
Measurement of Angles.
ARTS. PAGE
1. Sexagesimal Measure, , . 1
Examples I. a, b, . . . . . . . . . 2
CHAPTER II.
Trigonometrical Ratios of Acdte Angles.
2-5. Definitions, 4
6-14. Relations between the Ratios, ...... 5
15-20. Ratios of 0°, 18°, 30°, 45°, 60°, 90°, ..... 10
Viv^ Voce Examples, • . . 13
21, 22. Variations in the Ratios, . . . . . . .14
23. Solution of Equations, 15
Examples II. A, B, .16
Examples III. , 19
CHAPTER III.
Trigonometrical Ratios of Compound Angles.
24-29. Ratios of a ±/3, 21
30-34. Ratios of 2a, 3a, etc., 25
35. Products expressed as the Sum or Difference of Ratios, . 29
Viva Voce Examples, .29
36. Sums or DiflFerences expressed as Products of Ratios, . . 30
Viva Voce Examples, . 33
ix
X CONTENTS.
AKT8. PAOB
37. Transformations and Solution of Equations, .... 34
Examples IV. a, B, 35
Examples V., 42
CHAPTER IV.
Use of Mathematical Tables.
38-43. Logarithms of Numbers, 44
44-47. Logarithms of Trigonometrical Ratios, .... 48
Examples VI. a, b, 61
CHAPTER V.
)LUTioN OF Right- Angled Tbllngles and Practical
Applications.
48-57. Solution of Right-Angled Triangles, .... 54
Examples VII. a, b, 59
58-63. Heights and Distances, 60
64. Dip of the Horizon, 62
65. Dip of a Stratum, 63
Examples VIII. a, b, 63
Miscellaneous Examples I., 70
PART II.— REAL ALGEBRAICAL QUANTITY.
CHAPTER VL
CracuLAR Measure of Angles.
66-76. Definitions and Fundamental Propositions, ... 76
77-79. Change of Units of Angular Measurement, ... 83
Viva Voce Examples, 85
Examples IX. a, b, 86
Examples X., 88
CHAPTER VII.
General Definitigns of the Circular Functions. Formulae
INVOLVING One Variable Angle.
§ 1. Definitions.
80-82. Sense of Lines and Angles, 91
83. Projection of Point and Line, 92
84. Extended Definitions of the Ciicular Functions, ... 93
CONTENTS. xi
§ 2. Fundamental Properties of the Circular Functions.
ARTS. PAGE
85. Circular Functions one-valued, 94
86, 87. Signs of the Functions, 95
88-90. Periodicity and Continuity of the Functions. Formulae, 97
Viv4 Voce Examples, 101
§ 3. Reduction of Functions of nZ^O.
2
91,92. Even and Odd Functions, 102
93-97. Functions of n'^±d expressed as Functions of ^, . . 103
Viva Voce Examples, 107
98. Geometrical Proofs, . . . ^ 108
§ 4. Inverse Functions.
99. Definitions of COS" ^ a, Cos" ^a, etc., 109
100-106. Expressions for all Angles which have a given Circular
Function, . .111
Viva Voce Examples, 116
§ 5. Curves of the Circular Functions.
107, 108. Curves, 117
Examples XI. A, b, 122
CHAPTER VIII.
Circular Functions of Two or More Variable Angles.
109-112. General Proofs of the Addition Formulae, . . .124
113-115. cos a and sin a in terms of cos 2a or sin 2a, . . .129
Viv4 Voce Examples, 133
116. Ex. 1, 2. If J +5 + (7= 180°, then 2 sin ^= 411 cos :|,
Scos2^+2ncos^ = l, 133
Ex.4. tan"^a; + tan"^v = ?i7r + tan"^ ^^ , where 7i = 0 or +1, 134
i-xy
Ex. 5. Euler's and Machin's Values of - ,
4
Ex.8. Solution of a cos ^ + 6 sin ^ = c.
Examples XII. A, B,
117. Illustrative Examples, ....
Examples XIII. , . . . .
135
137
139
148
150
xii CONTENTS.
CHAPTER IX.
Relations between the Elements of a Triangle. Solution
OF Triangles and Practical Api'lications.
§ 1. Relations between the Elements of a Triangle.
ARTS. PA6>
118-121. a = 6co8(7+ccos5, 186
a/sm A = 6/sin B = r/sin 0.
a2 = 62 + c2-2&ccos^.
122-125. cos^=^«"(^^ sin ^=^£1^15, sm A =2Slbc, 167
126. tan^Z^=J^cot^-, 170
2 b + e 2
127. Illustrative Examples, 171
Ex. 4. Relation between the six lines joining four points in
a plane 172
Examples XIV. a, b, 172
Examples XV., 175
l__ 128-140. § 2. Solution of Triangles. 179
Examples XVI. A, b, 188
Examples XVII., 189
/ § 3. Practical Applications.
141-147. Use of Chain and Theodolite, 191
148-151. Survey of Mer de Glace by Prof. Forbes, . . .199
152-154. Measurement of Heights, 204
155. Dip of a Stratum, 207
Examples XVni., 208
CHAPTER X.
Applications to the Geometry of Triangles, Polygons
AND Circles.
156. Enunciations of Geometrical Theorems,
157. *S' = ^bc sin A = sfsis - a)(8 - 6)(.s - c),
158. Q = ^^/{{s - a){s - b){s - c)(« -d)- abed cos-w}
/ 159, 160. Area of Polygons, ....
^ 161. JR = al2 8UiA=abcl4JS, ....
219
220
221
222
223
CONTENTS, xiii
ARTS. PA.GE
162. r = >S'/s = 4i?sin^sin:|sin^, 224
163. ?-i = *9/(,9-a) = 4i?sin^cos:|cos^, 225
Ji "Ji z
164. SO"^ = R'^{1- 8 COS A cos Bco^C), .226
SP = B^~-2Rr.
OP ^2r^-4R^-cos A COS B COB C.
165. Feuerbach's Theorem, ........ 228
166. 167. Area of Circle, 228
168. Illustrative Examples, 229
Examples XIX., 231
Miscellaneous Examples II., . 247
CHAPTER XI.
Hyperbolic Functions.
169, 170. Definitions, 257
171-173. Elementary Relations between the Hyperbolic Functions, 260
174-176. Geometrical Properties of the Rectangular Hyperbola, . 262
177-179. Addition Formulae, 265
180. Gudermannian Function, 267
181. Curves of the Hyperbolic Fvmctions, 268
Examples XX., . . , 271
CHAPTER XII.
Inequalities and Limits.
§ 1. Inequalities.
182-184. sin d>e- 6^6, cos ^ < 1 - 6^2 + 6^/24, tan ^ > ^ + d^/-i, . 274
185. sinh X > a: > tanh x% cosh a; > 1 + a;-/2, 277
186. Ex. 1. Jsin^>^> Jvers^, 277
Examples XXI. , 279
§2. limits.
187. 188. Fundamental Propositions, 281
189-191. i.4^"'^'^\ i.^. fcos-Y%tc., 283
a:=0\ X I n=a)\ ul
Examples XXII., .286
xiv CONTENTS.
CHAPTER XIII.
Series.
arts. page
192-199. § 1. The Addition Formulae Extended. 288
§ 2. Series of Powers of a Cosine or Sine.
200-204. 2cosn^ = (2cos^)"-'i(2co8^)"-2 + ^(n-3)i(2co8^)«-''-...
+ (-l)'-*V-r-l)^_i(2co8^)"-2'-+...,etc., . 296
r
§ 3. Summation of Series.
205, 206. '^ S cos(a + rj8) = co8(a + (7i-l)^}8in^/8in^, . . 306
r=0 *• ^-' ^ I ^
207. Table of Diflference-forms, 309
208. Illustrative Examples, 310
209-214 § 4. Convergency and Continuity of Series. 312
§ 5. Infinite Series for Cosines and Sines.
215-217. cosa; = S(-lf'J'*, cosh a: = 2^ 321
[2r I2r
sin r = S( - 1 Y^ — -, sinh x = S-^- - , .
^ ' 2r+l \2r+l
'218-221. cosha; = ^(e* + e-*), sinha: = i(ef-e-'), . . . .326
Examples XXm., ........ 329
CHAPTER XIV.
Factors.
§ 1. Fundamental Theorem on Trigonometrical Factors.
222-228. If v„ = 2 cos nx, 2 cosh nx, or a;" + — , then will
v„-2coana = ~ll^(vi-2cos(a + r—\\ . . .339
§2. Products for cos w^, =^, coshmt, ~^' .
wsin^ wsinhw
229-232. cos nd = cos"^ II 1 1+ tan ^/tan — ^ ir j ; etc. , . .348
CONTENTS. XV
§ 3. Infinite Products for the Cosines and Sines of x.
ARTS, PAGE
where l>i^^>l ^^^;etc., . . . .354
(r-- l)7r-^
Examples XXIV., .363
CHAPTER XV.
Approximations.
237-241. § 1. Approximations and Errors, 369
Examples XXV., 376
242-249. §2. Theory of Proportional Parts, 380
Miscellaneous Examples III. , 388
PART III. -COMPLEX QUANTITY.
CHAPTER XVI.
Complex Numbers.
250, 251, Representation of Numbers by Straight Lines, . . 398
252-258. Addition and Multiplication of Complex Numbers, . 401
259. Conjugate Complex Numbers, 405
260. Powers and Roots, 405
261. Resolution of Complex Numbers. Demoivre's Theorem, . 409
262. 263. Some Applications of Complex Numbers, . . . 412
Examples XXVI., 419
CHAPTER XVII.
Series of Complex Numbers.
264-270. Convergency and Continuity of Series of Complex Numbers, 424
271,272. Summation of Series, 431
Examples XXVII., 434
CHAPTER XVIII.
273-277. The Binomial Theorem, 437
Examples XXVIII., 441
xvi CONTENTS.
CHAPTER XIX.
The Exponential Series,
artb. page
278, 279. exp(a;)xexp(y) = exp(a: + y), 444
280. exp(arO = cos a: + » sin x, 445
281,282. exp(a; + yi) = exp(a;)(co8 y + e sin y), .... 446
Examples XXIX 447
CHAPTER XX.
Logarithms of Complex Numbers.
283-285. Log(r, ^) = log r + (^ + 27i7r)i, 449
286. When a: is real, ((a))* = exp{a:Loga), 451
287-290. Logarithmic Series, .452
291, 292. Gregory's Series. Numerical Value of tt, . . . 456
293. Some Trigonometrical Series, 457
Examples XXX., 463
CHAPTER XXI.
295-300. Complex Indices, 467
Examples XXXI. , 472
CHAPTER XXII.
CiRCCJLAR AND HYPERBOLIC FUNCTIONS OF COMPLEX NUMBERS.
301-305. Definition and Fundamental Properties of Circular and
Hyperbolic Functions of a Complex Variable, , . 474
306. Formulae of Interchange of Circular and Hyperbolic Functions, 479
307-309. Inverse Functions, 481
Examples XXXII., 486
Miscellaneous Examples IV. , , . . . . . 489
Mathematical Tables, ....... 498
Answers to Examples, 501
PART I.
AKITHMETICAL QUANTITY.
''And to fuch as delight in matter feruifable for the State J hope
this Introduction fhal not he umvelcome : meaning as I fee the fame
gratefully accepted, hereafter to impart the reft, leaving at this time
farther to tvade in the large Sea 0/ Algebra d: numbers Cofsical."—
Stratioticos.
CHAPTER I.
MEASUEEMENT OF ANGLES.
1. In selecting a unit of angular measurement for
practical purposes, it is necessary that the unit should
be : (1) constant, (2) easily obtained, and (3) of such a
magnitude that the angles most frequently measured may
be expressed by integers that are as a rule not very
great.
A right angle satisfies the first two of these conditions,
and, by sub-division, the third also. It has therefore
been adopted as the primary unit in the only system now
in use for the practical measurement of angles.
In this system, a right angle is divided into 90 equal
parts called degrees, a degree into 60 equal parts called
minutes, and a minute into sixty equal parts called
seconds. An angle containing 47 degrees, 39 minutes,
17 seconds is written 47° 39' 17".
(g A
2 l^EA&UREMENT OF ANGLES.
Examples I a.
1. Reduce 57° 14' 46" to seconds, and 121475" to degrees
etc.
2. Express 69° 47' 42" and 58° 12' 18'' as decimals of a
right angle.
3. Find the number of degrees in the angle of a regular
octagon.
4. Find the number of sides in the regular polygon each
angle of which contains lo7|°.
5. The angles of a triangle are in arithmetical progression,
and the greatest angle is double of the least. Find
the number of degrees in each angle.
6. The angles of a triangle are such that the first contains
a certain number of degrees, the second 10 times as
many minutes, and the third 120 times as many
seconds. Find the angles.
7. The numerical measures of the angles of a quadrilateral
when referred to units containing 1°, 2°, 3°, 4°
respectively, are in arithmetical progression, and
the difference between the second and fourth is
equal to a right angle. Find the angles.
Examples I. b.
1. Reduce 35° 18' 47" to seconds, and 210501" to degrees
etc.
2. Express 8° 15' 81" and 85° 3' 2" as decimals of a right
angle.
3. Find the number of degrees in the angle of a regular
quindecagon.
4. Find the number of sides in the regular polygon each
angle of which contains 162°.
MEASUREMENT OF ANGLES. 3
5. An isosceles triangle has each of the angles at the base
double of the third angle. Find the number of
degrees in each angle.
6. The angles of a quadrilateral are in arithmetical pro-
gression, and the difference between the greatest
and least is a right angle. Find the number of
degrees in each angle.
7. One regular polygon contains twice as many sides as
another, and an angle of the first is double an
angle of the second. Find the number of sides in
each polygon.
CHAPTER II.
TRIGONOMETRICAL RATIOS OF AN ACUTE ANGLE.
2. Let POM be an acute angle, and let it be denoted
^ by a. From P, any point in either
of the bounding lines, draw PM
perpendicular to the other.
The following are called the trig-
^ j^^— ooiometrical ratios of the angle a :
Base Oif /hypotenuse OP is the cosine of a.
Perpendicular Pi//hypotenuse OP is the sine of a.
Perpendicular PM /h&se DM is the tangent of a.
H3^potenuse OP/base OM is the secant of a.
Hypotenuse OP/perpendicular PM is the cosecant of a.
Base Oif /perpendicular PM is the cotangent of a.
These ratios are written as follows : cos a, sin a, tan a,
sec a, cosec a and cot a.
3. Powers of trigonometrical ratios may be denoted in
the usual way, as (cos a)^ (tan a)^ etc ; but positive inte-
gral powers are generally written thus : cos^a, sec^a, etc.
4. Inverse Notation. — A notation similar in form is
used to denote angles having a given cosine, etc. The
angle whose cosine is J is written cos~^|, the angle whose
tangent is 3 is written tan-^S, etc. It should be borne
TRIGONOMETRICAL RATIOS. ' 5
in mind that these expressions are entirely different from
the first negative powers of the ratios, which are written
in the usual way, as (cos a)"^ (tan a)"\ etc.
5. The cosine of an angle depends on the angle only.
Let POM be the given angle, and let it be denoted by
a. Let P, P' be any „
points on one bounding r^
line of the angle, and P" p^
any point on the other.
From P, P\ F' draw
perpendiculars Pif , P'M\ q
P"M" to the other bounding line.
Now, the angle POM is common to the three triangles
POM, PVM\ P'VM") and the right angles PMO,
FM'O, P"M"0 are equal.
.-. the triangles POM, FOM\ F'OM" are similar
(Eucl. VI. 4).
OMIOP = OMjOP' = OM"IOP",
i.e. the cosine of the angle a is the same wherever the
point P be taken on either bounding line.
.•. the cosine of an angle depends on the angle only.
Cor. — Similarly, it may be shewn that the other trigon-
ometrical ratios of an angle depend on the angle only.
Relations between the Trigonometrical Ratios of an
Acute Angle.
6. To shew that the cosine and secant of an angle are
reci'procal, and likewise the sine and cosecant, and the
tangent and cotangent.
Let POM (see figure of art. 2) be the given angle, and
let it be denoted by a. Then
TRIGONOMETRICAL RATIOS
OM OP ,
cosa.seca = ^.^ = l,
cosa = l/seca and seca = l/cosa \
PM OP
Also, sin a . cosec OL—TTp ' -pTf = 1,
sin a = 1/cosec a and cosec a = 1/sin a . . . .
. , , . PM OM ^
And, tana.cota=^p^-p^=l,
tan a = 1/cot a and cot a — 1/tan a.
(A)'
7. If the sum of two angles is equal to a right angle,
each angle is called the complement of the other.
8. To shew that the cosine of an angle is equal to the
sine of its complement, the tangent of an angle to the
cotangent of its complement, and the secant of an angle
to the cosecant of its complement
Let POM (see fig. of art. 2) be the given angle, and
let it be denoted by a. The angle 0PM is the comple-
ment of a, since the angle OMP is a right angle. Now,
cos a = OMjOP = sin 0PM = sin(90° - a),\
sin a=^PM/OP= cos 0PM = cos(90°-a),
t3ina = PM/0M= cot 0PM = cot(90°-a),
sec a = OP/OM= cosec 0PM = cosec(90° - a),
cosec a =OP/Pif= sec OPif= sec(90°-a),
cot a =Oilf/Pif= tan OPif= tan(90°-a),>
9. To prove that tan a = » and cot a =
'■ cos a sm a
Let POM (see fig. of art. 2) be the given angle a. Then
* Formulae that should be remembered are denoted by capital letters
at the end of the lines in which they occur.
•(B).
OF AN ACUTE ANGLE.
sin a _PM OM_PM_
cosa~ OP ' OP~OM~^'''''
cos a _0M PM_OM_
sin a OP ' OP~PM~^'^^'''
...(C).
10. To prove that cos^a4-sin2a = l, l-\-tein^a = sec^a,
and cot^a + 1 = cosec^a.
Let POM (see figure of art. 2) be the given angle a.
Since OMP is a right angle,
03P + PM' = 0P\ (End I. 47.)
Dividing both sides by OP^, we have
0M\ PM^_
cos^a + sin^a = 1 (D) .
Again, dividing both sides of the first equation by Oilf ^
we have
PM''_ OP^
1 + tan^a = sec^a, (D).
Lastly, dividing both sides of the same equation by
PM\ we have
01/2
+ 1
OP^
PM^ ' PM^
cot^a+l =cosec%, (D).
V 11. Example.— Prove that
2(cos'5a + sin^a) - 3(cos*a + sin^a) +1=0.
2(cos<'a + sin'^a) - 3(cos*a + sin''a) + 1
= 2(cos-a + sin2a)(cos^a ~ cos^a sin^a + sin^a)
— 3{(cos2a + sin^a)^ - 2 cos^a sin^a} + 1
= 2(cos*a - cos^a sin^a + sin-^a) - 3( 1 - 2 cos^a sin^a) + 1
= 2{(cos% + sin^a)^ - 2 cos^a sin^a - cos^a sin^a} -3(1-2 cos-a sin^a) + 1
= 2(1-3 cos^a sin^a) - 3(1 - 2 cos2a sin^a) + 1
= 0.
8
TRIGONOMETRICAL RATIOS
12. The cosine of an angle being given, to express the
other trigonometrical ratios in terms of it.
Let a denote the angle whose cosine is given.
(1) Algebraical method. — By art. 10, we have
cos^a + sin^a^l,
sin a = V(l — cos^a).
Again, tan a = sin a/cos a (art. 9),
= ;^(1— cos2a)/cosa.
Also, sec a = 1/cos a,
cosec a = l//v/(l — cos^a),
cot a = cos al^{\ — cos^a).
(2) Geometrical method. — Let POM be the angle a,
and let its cosine be denoted by c.
Regarding the hypotenuse OP as
the unit of length, the base OM
contains c of these units, and there-
fore the perpendicular PM contains
^(l-c2) units (Eucl. I. 47).
.-. sin a = PM/OP = V(l - o2)/l = ^(l _ cos^a),
tan a = PMJOM = s/{\- c^)/c = J(l - cos^aVcos a,
sec a = OP/OM = l/c = 1/cos a,
cosec a = OP/PM = 1/^(1 - c^) = 1/^^(1 - cos^a),
cot a = OM/PM= cj JO - c2) = cos aj J{1- cos^a).
Cor. — Similarly, the trigonometrical ratios may be ex-
pressed in terms of the sine, secant, or cosecant of the angle.
13. Example. — If cos a =f, find the other trigonometrical ratios
of a. . P Let POM be the angle a. Regarding the
hypotenuse OP as containing 5 units of
length, the base OM contains 4 such units,
and therefore the perpendicular PM con-
tains 3. (Eucl. I. 47.)
.'. sina = f, tana = f, seca = f, cosec a = |,
and cot a =|.
OF AN ACUTE ANGLE. 9
14. The tangent of an angle being given, to express the
other trigonometrical ratios in terms of it
Let a denote the angle whose tangent is given.
(1) Algebraical method. — By art. 10, we have
sec^a = 1 + tan^a,
cos a = 1/sec a = 1/^(1 +tan2c().
Also, sin a/cos a — tan a,
sin a = tan a . cos a
= tan aU{ 1 + tan'^a) ;
and sec a = ^/O- + tan^a),
cosec a = jj{ 1 + tan^a)/ tan a,
cot a — 1/tan a.
(2) Geometrical method. — Let POM be the angle a,
and let its tangent be denoted by t.
Regarding the base OM as the unit
of length, the perpendicular PM
contains t of these units, and there-
fore the hypotenuse OP contains
s/il+f^) units (Eucl. I 47).
.-. cos a=OM/ OP = l/J{l-ht^) = l/^(i+ta.n^a),
sin a = PM/OP = tlJ{l + f") = tan «/ V(l + tan^a),
sec a = OP 1 031 = J(l+ 1^)/! = J(l + tan^a),
cosec a = OPIPM= ^(1 + t^)lt = ^(1 + tan2a)/tan a,
cot a = 1/tan a.
Cor. — Similarly, if the cotangent of an angle be given,
the other trigonometrical ratios may be expressed in
terms of it.
10
TRIGONOMETRICAL RATIOS
Trigonometrical Ratios of Particular Acute Angles.
15. To find the trigonometrical ratios of angles of G0°
and 30^
Let ABC be an equilateral triangle. Draw BD per-
j3 pendicular to AG. Then BD bisects
both the angle ABC and the base AG
(Eucl. I 26).
Then, angle 5^D = 60°,
and angle ABD = ^0\
Now, BD^=AB^-AD^
' =AB^-iAB^=iAB\
BD^^.AB.
COS 60° = sin 30° = AB\AB=\AB\AB = \,
sin 60°= cos^O° = BDIAB=^^ABIAB = '4
tan 60°= cot30° = J5i)/^i)=^^5/J^j5 = V3,
sec60° = cosec30° = 2,
2
cosec 60°
cot 60°
sec 30° =
tan 30'
x/3'
1
73'
The values of these ratios may be remembered by
the aid of the accompanying figure.
16. To find the trigonometrical ratios of an angle of 4)5°.
B Let ABG be an isosceles triangle, right-
angled at (7, so that each of the angles A
and B is 45°.
Now, AB'' = AG''^BG'' = ^AG'' = 2BG\
AG=BG=^.AB.
v2
OF AN ACUTE ANGLE.
11
cos 45° = sin 45° = ^(7/^5
^^ABIAB.
1
tan45° = cot45° = 50/Aa=l,
sec 45° = cosec 45° = ^2.
The values of these ratios may be remembered
by aid of the accompanying figure.
17. Definition of a Limit. — If two quantities, A and
B, be so connected that, when any change is made in B,
a corresponding change is consequently made in A, the
limit of A for a given value of B is that value towards
which A (from and after a certain value) continually
approaches, and from which it can be made to differ as
little as we please by making B approach near enough to
its given value.
For example, let APB be a circular arc, OA and OB
radii perpendicular to one an-
other, and OP any other radius. "
Draw PM perpendicular to OA.
Then the lengths of PM and
OM depend upon the magni-
tude of the angle AOP. As
this angle diminishes, the length
of PM (from and after the value
OB) continually diminishes, and 6 M A
may be made to diflfer from zero as little as we please by
making the angle AOP small enough. Thus, the limit
of PM, when the angle AOP vanishes, is zero; or, the
length of PM is ultimately zero. In like manner, the
limit of OM, when the angle AOP vanishes, is OA. And,
when the angle A OP is a right angle, the limit of PM is
OB or OA, and the limit of OM is zero.
12
TRIGONOMETRICAL RATIOS
18. To find the tHgonometrical ratios of an angle of 0°.
Let AOP be a very small angle, APsl circular arc with
centre 0. Draw PM perpendicular
to OA.
O M Then, by the preceding article,
when the angle AOP vanishes, the limit of OM is OA,
and the limit of PM is zero.
Hence, when the angle AOP vanishes, the limit of
OMjOP is unity, i.e. the limit of cos ^ OP is unity.
This is usually written, for brevity, cosO° = l.
Similarly, sin 0" = 0, tan 0° = 0 and sec 0° = 1.
Again, when the angle AOP vanishes, the limit oi PM
is zero. Hence, the limit of OP/PM is infinitely great,
cosec 0° = 00 , and, similarly, cot 0° = x .
19. To find the trigonometrical ratios of an angle
of 90°.
Let AOP be an angle very nearly equal to 90°, APB
a quadrant, centre 0. Draw PM
perpendicular to OA.
Then, by art. 17, when the
angle AOP is a right angle, the
limit of OM is zero, and the limit
ofPif is 05 or OP.
Hence, when the angle AOP
is a right angle, the limit of
OM/OP is zero, i.e. the limit of
cos J. OP is zero.
This is usually written, for brevity, cos 90° = 0.
Similarly, sin 90° = 1, tan 90° = oo , sec 90°= oo ,
cosec 90° = 1 and cot 90° = 0.
O M
OF AN ACUTE ANGLE.
13
20. To find the sine of 18°.
Let ABC be a triangle having each of the angles at
the base BG double of the third angle A, D
a point in AB such that the rect. AB . BD
is equal to the square on AD \ then AD is
equal to BG (Eucl. IV. 10). '> " . 0.\j "
Draw AE bisecting the angle BAG, and
therefore bisecting the base BG at right
angles (Eucl. I. 4).
The angles A, B, C are in the propor- B
tion of 1 : 2 : 2, and therefore the angle BAG=\ of 2
right angles = 86°; therefore angle 5^^=18°.
Let AB = a, AD = x; 80 thsit BE =x/2.
Then, a{a — x) = x\
x^ + ax — a^ = 0,
X- 2 - 2 •
The + sign must be taken in this expression, for the —
sign would give a value numerically greater than a.
sin 18 =-i-^=— ^^
AB '.
■^a
V5-1
Viva Voce Examples.
Express in degrees the following angles :
1. cos-ii 7. tan-il.
2. tan-V^- 8. sin-^O.
3. cosec"^l. 9. cos"^0.
4. cot-V3. 10. sec-V2.
5. sec~^l.
6. sec-^x
11. cosec"^
V3-
12.
tan'^oo.
13.
cot-n.
14.
15.
• 1 1
16.
sin-ij.
17.
tan-iQ.'
18.
19.
cosec-^2.
20.
cos-^1.
21.
^-.f.
14 TRIGONOMETRICAL RATIOS
22. sec -12.
23. cosec-^oo.
24. cot"ioo.
25. sin'U.
26. tan-i-io-
27. cosec'V^.
2
28. sec-i-7K-
29. cos-i^o-
30. cot- 10.
Variations in the Trigonometrical Ratios when the
Angle changes.
21. To find the limits between which the trigonometrical
ratios of an acute angle lie.
Let POM (see figure of art. 2) be an acute angle ; from
any point P, in either bounding line, draw PM perpen-
dicular to the other.
Since the angle PMO is a right angle, it is not less
than either of the angles 0PM or POM,
.-. the side OP is never less than either of the sides OM
or PM,
.'. OMjOP and PM/OP are never greater than unity,
while OP/OM and OP/PM are never less than unity,
i.e. the cosine and sine of an angle are never greater
than unity, and the secant and cosecant are never less
than unity.
Again, when the angle POM is zero, the limit of OM
OF AN ACUTE ANGLE. 16
is OP, that of PM is zero ; and, when the angle POM is
a right angle, the limit of OM is zero, and that of PM is
OP. Hence, the values of the cosine and sine of an acute
angle lie between 0 and 1, those of the tangent and
cotangent lie between 0 and oo , and those of the secant
and cosecant between 1 and oo .
22. To trace the changes in the trigonometrical ratios
of an angle as the angle increases from 0° to 90°.
Let AOB be a quadrant (see the figure of art. 17), OA
and OB its bounding radii ; and let the radius OP re-
volve from the position OA to the position OB, so that
the angle AOP increases from 0° to 90°. Draw PM
perpendicular to OA.
As the angle AOP increases from 0° to 90°, OM
diminishes from OA to zero, and PM increases from zero
to OB or OA. Hence, as the angle AOP increases from
0° to 90°,
cos A OP decreases from 1 to 0,
sin ^ OP increases from 0 to 1,
tan A OP increases from 0 to oo,
sec -4 OP increases from 1 to oo,
cosec -4 OP decreases from oo to 1,
cot -4 OP decreases from oo to 0.
23. Example 1. — Solve the equation *
6cos2(9+17sin(9=13.
Since cos2^=l-sin2^,
6-6sin2^+17sin^=13,'
6sin2(9-17sia(9 + 7 = 0,
(2sin6'-l)(3sin(9-7)=--0,
sin^ = ^or|.
*In Part I. the phrase " Solve the equation" must be understood to
mean ' ' Find the angle or angles between 0° and 90° inclusive, which
satisfy the equation."
IC TRIGONOMETRICAL RATIOS
Of these roots, only the first is admissible, since the sine of an
angle cannot be greater than unity. Now, the sine of 30° is ^,
^=30''.
Example 2. — Solve the equation
6 tan ^ + 5 cot ^=11.
Since cot 6= 1/tan 6,
6tan^+-A-.=ll,
tan^
6tan2^-lltan^+5 = 0,
(tan^-l)(6tan^-5)=0,
tan ^=1 or f .
Both of these roots are admissible, since the tangent of an angle
can have any value between 0 and c» ;
^=45° or tan~^f.
Example 3. — To eliminate 6 between the equations
acos ^ + 6sin ^=c and 6cos ^-asin^=o?.
Squaring both sides of each equation, we have
a''^cos-^+ 2a6 cos 6 sin 6-\-hHin-$=c'^,
\ h\os''6 - 2ab cos ^ sin ^ + a^sin"^ = d\
Vdding a-(cos-^ + sin^^) + b-{cos^e + sin^^) ^(r+d',
Examples II. a.
I Prove the following identities -
1. (cos a + sin a)^ = 1 + 2 cos a sin a.
tan^a + l^cot^a + l
.« o cot^a
^4. tan a + cot a = sec a cosec a.
5. sin^a sec^^S + tan^^ cos^a = sin^a + tan-^.
6. { ;y/(sec a + tan a) + ^/(sec a — tan a) }^ = 2( 1 + sec a).
7. cos^^a + sin^a = 1-3 sin^a + 3 sin^a.
„ cosec a — sec a _ cot a — tan a
cot a + tan a cosec a + sec a
OF AN ACUTE ANGLE. 17
9. tan^a sec^a + cot-acosec^a = sec^acosec^a-Ssec^acosec^a.
^_ l+sin« — cosa , l+sina + cosa _
1 0. r--^ , h , ,— ^ = 2 cosec a.
1 + sin a + cos a 1 + sin a — cos a
11. If cos a = ^, iind sin a, tan a and cosec a.
12. If sin a = fy, find cot a, sec a and cosec a.
13. If tan a = Yy find cos a, cot a and sec a.
*yi3
14. If sec a = "—- , find cos a, sin a and tan a.
2
15. If cot a = —7^, find sin a, sec a and cosec a.
16. If tan a = -?, — Thy find cos a and sin a.
17. Find cos 18°.
Prove that :
18. cosec 60°cot 30° = sec245°.
19. tan260°- 2 tan245° = cot^SO"- 2 sin^SO"- f cosec245°.
20. cos 30°sin 30° + cos 45°sin 45° + cos 60°sin 60°
= sin 30° + sin 60°.
21. sin 90° + cos260° = (2 sin 18° + sin 30°)^.
Find the value of:
22. tan 60°cos 80° - cos 0°tan 45° + 4 sin 18°.
23. cosec "^x + cos -^0 — sec "^2.
24. 2cos-il+cot-i0-3sec-ii
25. 2 cosec -12 -cos -1-^ + 3 cos-i^-sin-^l.
26. tan-^Go — cot'^— T^ + sin-^^ — cot'^l.
Solve the equations :
27. tan 0 = cot a
28. cos20 + sin0 = l.
29. V3(tan0 + cot0) = 4.
18 TRIGONOMETIUCAL RATIOS
30. sec20=^3taue + l.
31. 2cos3e + sm2a-l=0.
32. If cos 0 = tan ft prove that sin 0 = 2 sin 18°.
Eliminate 0 between the following equations :
33. cot^ = a, sec 0 = 6.
34. a sec 0 — c tan Q = dy 6 sec 0+ cZ tan Q = c.
35. a tan20 + & tan 0 + c = 0, acoi'^O + 6'cot 0 + c' = 0.
Examples II. b.
Prove the following identities :
1. cos^a tan^a + sin^a cot2a = 1.
2. sec^a + cosec^a = sec^a cosec^a.
3. l+TH = seca.
1+seca
J. cos g + cos ^ sing + sin ;g
sin g — sin (3 cos g — cos/3 ~ *
5. (cot0 + 2)(2cot0+l) = 2cosec2e + 5cota
6. sin2g(l + n cot^g) + cos2g(l + n tan^a)
= sin2g(7i + cot^g) + co^\{n + tan^g \
• 7. sec^g — tan^g = 1 + 3 tan^g sec^g.
8. (4 cos^g — 1 )2tan^g + (3 — 4 cos^g)^ = sec^g.
9. (tan^a -f tan2/3)cos2g cos"^^ = cos^g + cos2;5-2 cos^a cos^/?.
10. (1 + secg + tan g)(l + cosecg4-cotg)
= 2(1 + tan g + cot g + sec g + cosec g).
11. If cos g = W, find sin g, tan g and cot g.
12. If sin g = y^^, find cos g, cot g and sec g.
13. If tan g = if, find cos g, sin g and cosec g.
14. If sec g = ||-, find sin g, tan g and cosec g.
15. If cosec g = f¥» ^^^^ ^^^ «> *^^ « and cot g.
2771/
16. If cosg = 3-- — 9, find sing and tang.
1+m^
17. Find tan 18°.
OF AN ACUTE ANGLE. 19
Prove that
18. tan230° + 3 sin245° = sec245° - J cot^GO^.
l + cot60°^/l + cos30Y
l-cotG0°""Vl-cos30V*
20. sin 90°cot 30° - cot 45'tan 60° = cosec245'^ - 8 siii230°.
21. 2 cos218° - sec245° = cos 72° - siii245°.
Find the value of:
22. (2 cos 0°sin 30°tan 45°
+COS 30°sin 45°tan 60°-cosec 30''cos245°)2
23. sin-iO + 2 sec-loo -3tan-V3.
24. 4tan-i0+3sec-V2-2cosec-i-^.
25. cos-ii + sin-i-y^-cosec-il+tan-U-2cot-V3.
26. 2cosec-V2 + sin-i^-3sec-n-5tan-ii
. Solve the equations :
27. 2 sin 0 = tan a
28. 2cos20 + llsin0-7 = O.
29. 3tan20-7sece+5 = O.
30. cot20(2 cosec 0 - 3) + 3(cosec 0 - 1) = 0.
31. sec 0 cosec 0 — cot 0 = 2.
32. tan 0+ sec 0 = 2.
Eliminate 0 between the following equations :
33. sec 0 = (X, cosec 0 = 6.
34. ^ cosec 0 +g' cot 0 = r, s cosec 0 — r cot 0 = g.
35. m cos20 + -71 cos 0 = p, m'sec20 + Ti'sec 0 =_p'.
Examples III
1. Simplify ( — ^— ^ -\ ^ r-^ ) x cos^a sin^a.
•^ Vsec^a — cos^a cosec^a — sin^a/
20 TRIGONOMETRICAL RATIOS.
" ^ Vcos a + tan'^a sin a cos a cot^a + sin a/
sec a cosec a — 1
cosec a — sec a
3. Express sec^0 in terms of tan Q.
7— a;
4. If cos a= ^ :r> and a? be positive, show that x can-
not be less than 2.
2 4- cc
5. If cosec a = ^ «> show that x cannot be greater than 5.
G. If ^ = ^2, and ^^ = ^3, find a and ^.
sinj8 ^ tan^ ^ ' '^
7. If sin a = m sin ^8 and cos a = ti cos /3, find tan a and
tan ;8.
Vtan ^/ Vtana/ • Vtan^/'
/ 1 Y ^ /cos ey /sin^Y
Vsin cf)) Vsin a/ Vsin /3/
9. If a tan a = 6 tan /3, and a^^^ = a^ — 6-, show that
(1 - aj2sin2^)(i _ x^co^^a) = l-x\
10. Eliminate Q between
cos 0 — sin 0 = a and tan Q=c sec^0.
11. Eliminate 0 between
cosec ^ — sin 0 = a, sec 0 — cos 0 = 6.
12. Find the least value of a^sec^^ + ^^cos^^, where a and h
are constant quantities.
<9^
CHAPTER III.
TEIGONOMETEICAL EATIOS OF COMPOUND ANGLES.
24. The principal object of the present chapter is to
express the trigonometrical ratios of the sum or difference
of two or more angles in terms of the ratios of the com-
ponent angles, and those of the multiples or sub-multiples
of a given angle in terms of the ratios of that angle. In
the latter part of the chapter, we shall shew how these
relations may be used for effecting the transformation of
trigonometrical expressions.
Throughout the chapter every angle, both component
and compound, is supposed to be acute ; but it should be
remarked that this limitation only applies to the first
two propositions. These propositions are shewn to be
true for all real values of the angles in Chapter VIII.
. 25. To exijress the cosine and sine of the sum of tiuo
angles in terms of the cosines and sines of the angles
themselves.
Let the given angles be denoted by a and /3. Draw-
the angle AOB equal to a and BOG equal to ^, so that
AOG is equal to the compound angle a + j3.
In OC, one of the bounding lines of the compound
angle, take any point P. Draw PM and PK perpen-
21
22
TRIGONOMETRICAL RATIOS
dicular to OA and OB, KL perpendicular to OA arid
KR to PM. Then, in the triangle PRK, the angle KPR
is equal to a, for
angle KPR = complement of PKR
= angle RKO
= angle AOB.
Now,
, , ^, OM OL--ML
OP
OL-RK
OP
PK
OP
_qL^ OK RK
~OK'OP PK
= cos a cos ^ — sin a sin /3
KL+RP
.(A)
., . , ^ r.. PM MR+RP
Also, 8in(a + i8) = ^p= ^ —
OP
^KL OK RP PK
OK' OP"^ PK' OP
= sin a cos /3 + cos a sin /3.,
(B)
26. To express the cosine and sine of the difference of
two angles in terms of the cosines and sines of the angles
themselves.
Let the given angles be denoted by a and ^. Draw
the angle AOB equal to a, and
BOD equal to ft so that AOD is
equal to ' the compound angle
a-/3.
In OD, one of the bounding
lines of the compound angle,
take any point Q. Draw QiV
and QK perpendicular to OA
and OB, KL perpendicular to OA and KS to NQ pro-
Also, sin(a — /3)
OF COMPOUND ANGLES. 23
duced. Then, in the triangle QSK, the angle KQS is
equal to a, for
angle iiTQ/Sf = complement of QKS = bx\^q SKB
= angle AOB.
, ^, ON OL + LN OL+KS
Now, cos(a-/5) = -^^=— ^g— =— ^g—
_0L OK KS QK
~0K' OQ^QK'OQ
= cosaCOs/3 + sin a sin/3 (C)
QN_JNS-QS_LK-Q8
0Q~ OQ OQ
_LK OKQS QK
~OK'OQ QK'OQ
= sin a COS ^ - cos a sin fi (D)
27. To express the tangent of the sum and difference
of two angles in terms of the tangents of the angles
themselves.
Let the given angles be denoted by a and ^.
(1) Algebraical proof
( I Q\ _ ^^"(<^ + /^) _ ^^^ g cos ^ + cos a sin /3
tan(a + ^)- ^^^^^ _^ ^^ - ^^^ ^ ^os ^ - sin a sin /3'
Dividing the numerator and denominator of this frac-
tion by cos a cos /3, we have
sin g sin ^
^ , , ^. cos a cos/3 tan a + tan 8 .j..
^°(" + ^^ = ^-iWihr^ = l-tanatan/3 (^>
COS a cos /3
. . ^ . r,. sin(a-/3) sinacos^-cosasin^
Again, tan(a--/3)= — 7 ^ = 15-, — — o
^ ' ^ ^^ cos(a — j5) cosaCOs/5 + smasm/5
_ tan a — tan /3 /-px
"r+tana'tajT/S ^
24 TRIGONOMETRIC A L RA TIOS
(2) Geometrical Proof. — (See figure of art. 25.)
, . ^ox P^^ MR+RP
LK,RP LK RP
OL^OL OL^OL
RK ._^PP
OL RP'OL
Now, the triangles KPR, KOL are similar, for the
angles KPR and KOL are equal, and the angles KRP
and KLO are right angles.
RP_PK_. ^
OL~OK~^^^f^'
^ '^^ 1 — tan a tan /5
Similarly, making use of the figure in Art. 26, we may
show that t8Lu(a — 8) = ^ . J^ r — ^.
^ '^^ 1+tanatanp
Govs. — Similarly, it may be shewn, both algebraically
and geometrically, that
., , Q. cotacotiS — 1
C0t(a + /3) = — 7 — , ^. o '
^ ^^ cota + cot/3
J x/ m cot a cot -5+ 1
and cot(a-/3) = — -^ — ^^^ — •
^ '^^ cot p — cot a
28. The formulae of the preceding articles may be used
to obtain the trigonometrical ratios of angles which are
the sums or diflferences of angles whose ratios are known.
For example,
cos 75° = cos(45° + 30°) = cos 45°cos 80° - sin 45°sin 30°
_J JS 1 i_V-^-i
J2' 2 J~2'2~ 2\/2 '
OF COMPOUND ANGLES. 25
sin 75° = sin (45° + 30°) = sin 45°cos 30° + cos 45°sin 30°
= _L x/? , J 1_V3 + 1
V2' 2 "^V2'2~ 2V2 '
, ^j,o , //i-o , oAox tan 45° + tan 30°
tan 75 = tan(4a + 30 ) = r — r — — ^r — ^7^
^ ^ 1 — tan4o tan 30
^+73 V3+t 4 + 2^/3
Similarly, or by art. 8, it may be shewn that
cosl5°=^^±i, sin 15°-^^2^,andtanl5° = 2-V3.
29. The product of the sines of the sum and difference
of two angles is equal to the difference of the squares of
the sines of the component angles.
Let the angles be denoted by a and /5.
Then, sin (a + /5)sin(a - ^)
= (sin a cos |8 + cos a sin /3)(sin a cos ^ — cos a sin /3)
= sin^a cos^yg — cos^a sin^/^
= sin^a — sin^^ (G)
Cor. 1. — sin(a + |8)sin (a — /3) = cos^/3 — cos^a.
Cor. 2. — Similarly it may be shewn that
cos(a+/8)cos(a — P) = cos^a — sin^/^ = cos"/3 — sin^a.
30. To express the cosine of 2a in terms of the cosine
and sine of a.
(1) Algebraical proof
cos 2a = cos(a + a) = cos a cos a — sin a sin a
= cos^a — sin^a (H)
Putting sin^a = 1 — cos^a in this equation, we obtain an
expression for cos 2a in terms of cos a, namely,
cos 2a = cos^a — (1 — cos^a) = 2 cos^a — 1 (I)
26
TRIGONOMETRICAL RATIOS
Again, putting cos^a = 1 — sin^a in (H), we obtain an
expression for cos 2a in terms of sin a, namely,
cos 2a = 1 — sin-a — sin^a = 1 — 2 sin^a (J)
(2) Geometrical proof. — Let AG he the diameter of a
circle, centre 0. Draw the angle GAP equal to a : then
the angle- GOP is equal to 2a. Draw PM perpendicular
to AG, and join PG. Then APG is a right angle, and
angle (7Pif= complement of angle PGA = angle PAG= a.
T., „ OM 2.0M {2.0G-2.MG)
Now, cos2a = ^ = 2-^ = ^^jj^
_AM-MG_AM AP^MG PG^
~ AG ~AP'AG PG'AG
= cos^a — sin^a.
The formulae (I) and (J) may be obtained in a similar
manner.
Govs. — From equations (I) and (J) we have
cos2a = Kl+cos2a) (K)
and sin2a = J(l — cos2a) ; (L)
equations which give the cosine and sine of an angle in
terms of the cosine of double the given angle.
OF COMPOUND ANGLES. 27
31. To express the sine of 2a in terms of the cosine and
sine of a.
(1) Algebraical jpr oof
sin 2a = sinfa + a) = sin a cos a + cos a sin a
= 2 cos a sin a (M)
(2) Geometrical jproof — (See figure of Art. 30.)
• 9 _PM_2.PM_2PM AP
^'"^ ''~0P~2.0P~ AP 'AG
= 2 sin a cos a.
32. To express the tangent of 2a in terms of the tan-
gent of a.
(1) Algebraical proof
tan 2a = tan (a + a)
tan a + tan a
1 — tan a . tan a
2 tana
1 — tan^a *'"
(2) Geometrical proof — (See figure of art. 30.)
PM 2PM 2PM
m
tan 2a =
OM 20M~AM-MG
2PM
AM 2 tan a
. MG PM 1-tanV
pm'am
Gor. — 2 cot 2a = cot a — tan a, for
o .o 2(l-tan2a) 1
2cot2a = -^T7T -=7 tana
Z tan a tan a
= cot a — tan a.
28 TRIGONOMETRICAL RATIOS
33. To express the tAgonometrical ratios of 3a in
tervis of those of a.
cos 3a = cos(2a + a) = cos 2a cos a — sin 2a sin a
= (2 cos^a — l)cosa — 2 sin a cos a . sin a
= 2 cos^a — cos a — 2 cos a(l— cos^a)
= 4 cos^a — 3 cos a (O)
sin 3a = sin(2a + a) = sin 2a cos a + cos 2a sin a
= 2 sin a cos a . cos a + (1 — 2 sin2a)sin a
= 2 sin a(l — sin^a) + sin a — 2 sin^a
= 3 sin a — 4 sin^a (P)
tan 3a = tan(2a + a)
_ tan 2a + tan a
~1— tan 2a . tan a
2 tan a
1 — tan^a
+ tana
^ 2tana .
1 — ^ — 77 — o- . tan a
1 — tan'^a
_ 2 tan a + tan a — tan^a
1 — tan^a — 2 tan^a
_ 3 tan a — tan^a
l-3tan2a *
34. The cosine and sine of any multiple of a may thus
be expressed in terms of the cosine and sine of a. The
general formulae for obtaining them are
cos(n + l)a = 2 cos na cos a — cos{n — l)a,
sm(n + 1 )a = 2 sin na cos a — ^m{n — l)a,
formulae which may be easily proved by means of t])e
results obtained in Arts. 25, 26. Putting 7i = 3, 4, 5, etc.,
successively in the formulae, we find the cosines and
sines of 4a, ba, 6a, etc.
OF COMPOUND ANGLES. 29
Trigonometrical Transformations.
35. We have already proved that ' ::i
cos(a + /3) = cos a cos P — sin a sin ^,
cos(a — j8) = cos a cos /3 + sin a sin ^, . ^
sin(a + /3) = sin a cos ^ + cos a sin ^,
sin(a — ^) — sin a cos /3 — cos a sin /3.
By addition and subtraction of the first and second of
these formulae, and also of the third and fourth, we obtain,
2 cos a cos /3 = cos(a + /3) + cos(a — P);
2 sin asin/5 = cos(a — /3) — cos(a + ^),l /qx
2 sin acos P = sin(a + /3) + sin(a — ^),
2 cos asin/3 =sin(a4-/3) — sin(a — /3).-
By means of these formulae, we can express twice the
product of two cosines, or of two sines, or of a sine and
cosine, as the sum or difference of two cosines or sines.
It should be noticed that the third formula is used
when we have the sine of the greater angle, and the
fourth when we have the cosine of the greater angle ;
also, that, in the second formula, the cosine of the greater
angle, a + ^, is subtracted from the cosine of the lesser,
a — /3: for the cosine of an angle diminishes as the angle
increases.
YiVA Voce Examples.
Transform the following expressions into the sums or
differences of two cosines or sines :
1. 2 cos 4a cos 2a. 7. 2 sin 7a sin 4a.
2. 2 cos 5a cos a. 8. 2 sin 10a sin 3a.
3. 2 cos 10a cos 7a. 9. 2 sin 4a sin 3a.
4. 2 cos a cos 3a. 10. 2 sin 5a sin 5a.
5. 2 cos 7a cos 6a. 11. 2 sin 3a sin 18a.
6. 2 cos 3a cos 13a. 12. 2 sin a sin 12a.
80
TRIGONOMETRICAL RATIOS
13. 2 sin 6a cos 4a.
14. 2 sin 5a cos a.
15. 2 sin 11a cos 9a.
16. 2 sin 3a cos 2a.
17. 2 sin 7a cos 4a.
18. 2 sin 10a cos a.
19. 2 cos Hasina.
20. 2 cos 14a sin 3a.
21. 2 cos 10a sin 5a.
22. 2 cos 2a sin a.
23. 2 cos a sin a.
24. 2 cos 7a sin 2a.
25. 2 cos a sin 15a.
26. J sin 3a cos 8a.
27. cos 3a . cos 7a.
28. sin 3a. sin 11a.
29. 2sin(2a + 3/3)cos(a-i8).
30. cos(2a+;8)cosa.
31. cos(a-)8)cos(2a + 4^).
32. sin3asin(2/3-3a).
33. cos(5a-2/3)sin(a-4^).
34. sin(8a-3/3)sin(5/3-3a).
35. sin(45° + a)sin(45° — a).
36. cos(30° + 2a)sin(30°-a).
36. The four formulae (Q) of the preceding article serve
also for expressing the sum or difference of two cosines
or sines as the product of sines or cosines. But they
may be put into a more convenient form by writing
a+/3 = o- and a — /8 = <5,
so that
+ S
and /? = '
The formulae thus become
COSO-+COS 8
cos ^ — coso-:
sin (T + sin 6
2 cos — y- cos -^
2 sin --r— sin —^
2 sin ^7^- cos
sin 0- — sm d = 2 cos — 7^— sin
2
.(R)
2 "" 2
They may be thus expressed :
The sum of the cosines of two angles is equal to twice
the product of the cosine of half their sum by the
cosine of half their difference.
OF COMPOUND ANGLES.
31
The difference of the cosines of two angles is equal to
twice the product of the sine of half their sum by
the sine of half their inverted difference.
The sum of the sines of two angles is equal to twice the
product of the sine of half their sum by the cosine
of half their difference.
The difference of the sines of two angles is equal to twice
the product of the cosine of half their sum by the
sine of half their direct difference.
Geometrical 'proof . — Draw the angle AOG equal to o-,
and the angle AOD equal to S,
so that the angle GOD is equal
to o- — ^. Bisect the angle GOD
by OB, making the angles BOD,
GOD each equal to ^{cr — S),
and, consequently, the angle
AOB equal to ^ + i(a--^), i.e. to
In OB take any point K, and o
through K draw PKQ perpendicular to OB, meeting OG
OD in P, Q. Draw KL, PM, QN perpendicular to OA,
and through K draw RKS perpendicular to PM and NQ
produced. Then, by elementary geometry, we have
OP=OQ, PM+ QF= 2KL, PM- QN= 2PE,
0M+0N=20L, and ON-OM=2ML = 2RK.
Also, in the triangle PRK, we have
angle ZPJS = complement of angle PKR
angle EKO =
cr + (5
82 TRIGONOMETRICAL RATIOS
„ ^ , OM ON OM+ON
Hence, cos ^+^^^^ = 'qp+-qq = — Qp —
201 _qL OK
~ OP ~ OK OP
= 2 cos — ^— cos — o— •
ON-OM 2RK
cos 6-C03(r = ^p =-Qp-
RK PK
~ PK OP
= 2 Sin ^-sin-g-. .
. , . , PM+QN 2KL
sin o-+sin S= QjT— = ^TT
_^ OK
~ OK' OP
= 2 sin ^— cos
sin a — sin S =
2 2
PM-QN_2PR
OP ~ OP
PR PZ
"^PZ" OP
„ cr + S . (T — S
= 2 cos ^ sm
2 2
Oo7's. — The cosine of an angle being equal to the sine
of its complement, we have
coso-+sin ^ = 2 cosf 45°+^-s~) cos f 45° —
2
+ S'
cos 0- — sin (5 = 2 sin (45° +^-2—) sin f45° — ^^^
OF COMPOUND ANGLES.
83
Viva Voce Examples.
Transform the following expressions into the products
of sines and cosines : —
1. cos 5a + cos a.
2. cos 7a + cos 3a.
3. cos 10a + cos 2a.
4. cos 1 5a + cos 3a.
5. cos 9a + cos 8a.
6. cos 2a + cos a.
7. cos a — cos 3a.
8. cos3a — cos9a.
9. cos 5a — cos 6a.
10. cos 8a — cos 11a.
11. cos 5a — cos 10a.
12. cos 6a — cos 16a.
13.
sin 6a + sin 4a.
14.
sin 4a + sin2a.
15.
sin 9a + sin a.
16.
sinl3a + sinlla.
17.
sin 1 2a + sin 7a.
18.
sin 8a + sin 5a.
19.
sin 9a — sin 3a.
20.
sin 11a — sin a.
21.
sin 10a — sin 2a.
22.
sin 5a — sin 4a.
23.
sin a- sin |.
24.
sin 3a — sin a.
25.
sin 3a — sin 5a.
26.
sin 7a + sin 13a.
27.
cos 3a + cos 9a.
28.
29.
30.
31.
32.
33.
cos 9a — cos 2a.
sin 11a — sin 15a.
cos 1 la + cos 2a.
• a .3a
sin ^ — sin— .
5a a
cos y- cos 2-
. K .5a
sin 7a — sin -^.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
sin a + sin
3a
cos 9a + cos 10a.
sin 11a — sin 7a.
. 7a . 11a
o ^a
cos 8a — cos -^.
. 5a , . 3a
sm-^ + sin-.
3a
cos-j — cos
4
sin. -sin
.a .a
sm^-sin^.
11a
4 '
15a
sin(2a+3^)+sin(2a+/3).
cos(3a-j8) + cos(a + 5/3).
cos(2a-/3)-cos3/3.
34 TRIOONOMETRWAL RATIOS
46. 8in(3a+8/3)-sin(7a-3/3). 49. sin 63° -- sin 27°.
47. cos 12° -cos 48°. 50. cos 60° + cos 20".
48. sin 75° + sin 15°.
51. cos{a + (r-J)/3}-cos{a + (r+J)^}.
52. sin{a + (r + i)^}-sin{a + (r-J)^}.
37. The following examples are given to shew the use
of the formulae proved in this chapter : —
Example 1. — Prove that
tana=-«Hl2^=lz^^, and tan^a-^ "^^'^a
1 + cos 2a sin 2a 1 + cos 2a
sina_2sinacosa_ sin 2a
cos a 2 cos^a 1 + cos 2a'
2 sin^a 1 - cos 2a
2 sin 2a 1 - cos 2a 1 — cos 2a
tana=
Also, tana=^ . . «
2 sin a cos a sm 2a
Hence, by multiplication,
tan^a . .
1 + cos 2a sin 2a 1 + cos 2a
Example 2. — Prove that
sin a + 2 sin 3a + sin 5a _ , o„
cos a + 2 cos 3a + cos 5a
sin a + 2 sin 3a + sin 5a _ (sin 5a + sin a) + 2 sin 3a
cos a 4- 2 cos 3a + cos 5a (cos 5a + cos a) + 2 cos 3a
_2 sin 3a cos 2a + 2 sin 3a
2 cos 3a cos 2a + 2 cos 3a
_ 2 sin 3a(cos 2a + 1 )
2cos3a(cos2a+l)
= tan 3a.
Example 3. — Prove that
3 sin 3a ,. 3 cos 3a sin 4a
cos^a . — - — + siira . ^ — = — - —
3 3 4
«"^^ + sin^a . 2^^
3 3
= ^[cos*a(3 sin a - 4 sin^a) + sin'a(4 cos^a - 3 cos a)]
= ^(3 cos^a sin a - 3 sin^a cos a)
= cos a sin a(cos^a — sin'^a)
=^ sin 2a. cos 2a
=1 sin 4a.
OF COMPOUND ANGLES. 85
Example 4. — If tana=| and tail ^8 = ^, find tan(2a-/3).
l-tan^a 1-^ ^'
tan(2a- ^)^ tan 2a-tan/? _i-i ^ ^
^ '^^ l+tan2atan/? l+l-i
Example 5. — Express 4 cos a cos /? cos y as the sum of four
cosines.
4 cos a cos /8 cosy = 2 cos a . 2cos /3cosy = 2cosa{cos(/8 + y) + cos(/8-y)}
= 2 cos a cos(/3 + y) + 2 cos a cos(^ - y )
= cos(a + /? + y) +cos(^ + y - a) + cos(y + a - /S) + cos(a + /S - y).
Example 6. — Solve the equation
2sin^sin3^=sin22(9.
2sin6'sin36'-sin22l9=0,
cos 2(9 - cos 4^ - sin22(9 = 0,
cos W - (2 cos22^ - 1 ) - ( 1 - cos22^) - 0,
cos2^-cos22(9=0,
cos 2(9(1 -cos 2(9) = 0,
cos 2^=0 or 1,
2^=90° or 0°,
(9=45° or 0°.
Examples IV. a.
Find the values of :
1. cos(a+j8) and sin(a — /?), if sma = xT ^^^ mi\^ = -^^.
2. cos(a — /8) and sin(a + /3), if sin a = f and cos/3 = ^|.
3. cos(a + /3) and sin(a + /5), if tana = ^f and cot|8 = ff.
4. tan(a — /5), if tan a = a_/^ and tan /5 = xt^q'
5. cot(a + 18), if cot a = 7 and cot |8 = f .
6. Shew that tan 75° + cot 75' = 4.
7. The value of
cos(7i + l)a cos(7i — l)a4-sin(7i + l)a sin(?i — l)a
is independent of n.
8. sinacos(/3+y) — sin /5cos(a + y) = sin(a — /9)cosy.
36 TRIGONOMETRICAL RATIOS
A L ^ I o sin(/5±a)
9. cota±cot/8 = -^-^^^-^ — 5.
' sin a sin p
iA / _L /o\ sec a sec /3
10. sec(a±/3) = 7TrT r^-^-
'^^ 1 + tan a tan ^
11. cos^a - cos a cos(60° + a) + sin2(80° - a) = f .
12. Find cos(a+/3+y) in terms of the cosines and sines
of a, /5, y; and hence shew that, if a+/3+y = 90°,
tan |8 tan y + tan y tan a + tan a tan ^ = 1 .
13. Find tan(a + j8 + y) in terms of the tangents of a, /3,
y ; and hence shew that, if a+/3+y = 90°,
tan |8 tan y + tan y tan a + tan a tan ^ = 1.
14. cos^a + cos2/3 — 2 cos a cos /3 cos(a + /3) = sin2(a + ^8).
1 5. tan^a — tan^/S = sin(a + /5)sin(a — ^)sec^a sec^/?.
16. sin44"cos74° = l-sinn4°.
Find the values of :
17. cos 2a, when (1) cos a = f, (2) sin a = i, (3) cos a = f .
18. sin 2a, when (1) cosa = -^f, (2) tana = ^V
19. tan 2a, when (1) tan a = A, (2) cot a = 4.
20. cos a, sin a, and tan a, when cos 2a = f i.
21. tan a, when (1) tan 2a = ^8, (2) cos | = i-
22. tan(a + 2/3), when tan a = ^3 and tan /3 = 2 - ^3.
/3 / 3 — 1
23. cos 3a, when cos a = '^- ; sin 3a, when sin a= -^ /o >
and tan 3a, when tan a = i-
24. Find the cosine, sine and tangent of 22^°, and shew
that 2 cos 11° 15'= V {2+ V(2+ V2)}.
25. sin 18° and sin 54° are roots of the equation
4»2_2^5i»+l=0.
o^ n 1 " tan^a
26. cos2a = =-rT — 2-
l + tan^a
27. 2 cosec 2a = sec a cosec a.
28. sec 2a —
OF COMPOUND ANGLES. 37
cot a + tan a
cot a — tan a
29. tan(45° + a)=sec2a + tan2a.
30. Simplify Z^^^ + ^^^y.
^ *" l + cosa + cos2a
Q^ c<- i-r 2(l + tanatan2a)
31. Simplify y, , . 7 — ^.
^ ^ 2 + tan a tan 2a
32. tan(45°-|) + tan(45°+|) = 2seca.
33. sin2(22r + 1) - sin2(22i° - 1) = -1 sin a.
l-tan2(45°-|)
34. Simplify ^-•
l + tan2U5°-|j
35. Simplify cos(36°+a)cos(36°-a)+cos(54°+a)cos(54°-a).
36. cos^a + sin^a cos 2(3 = cos^^ + sin^^ cos 2a.
37. {cos2a + cos2^ + 2cos(a + |8)}2
+ {sin2a + sin 2/3 + 2 sin(a + iS)}2 = 16 cos*^^.
38. tan 3a tan 2a tan a = tan 3a — tan 2a — tan a.
39. cos4a = 8cos*a — 8cos2a + l.
40. Find the value of 4 tan-i^
41. Express cos Qa in terms of cos a.
42. sin 3a = 4 sin(60° - a)sin a sin(60° + a).
43. tan 3a = tan(60° - a)tan a tan(60° + a).
44. ^i^«^ = tan4^.
cosa + cos/5 2
sin a — Sin |8 2 2
46. cos(4a + /5) = 2 cos a cos(3a + /5) - cos(2a + /5).
47. sin 1 1 a sin a + sin 7a sin 3a = sin 8a sin 4a.
48. sin a(cos 2a + cos 4a + cos 6a) = sin 3a cos 4a.
38 TRlGONOMEriUOAL RATIOS
49. Simplify '^'°"+""f"+^'"f^.
cos a + cos 2a + cos 3a
50. Simplify si" «-«'" 4«+8in 7«-sin 10a
COS a — COS 4a + COS 7a — COS 1 Oa
51. sin 10°+ sin 50° = sin 70°.
52. cos 55° + sin 25° = sin 85°.
53. sina+sin(72°+a)+sin(36°- a) = sin(72°- a)+sin(36°+a).
54. Prove geometrically that cot(a+j8)= ^ ^ , To •
55. Also that tana = =— ; =r--
1 + cos 2a
Solve the equations :
56. sin(e+a)-sin(0-a) = ^2sina.
5/. -^sm0+^cose = ^2-
58. 2cos20-2sine-l = O.
59. (1 + V3)tan20(l~tan0) = 2tana
60. sin 7^ + sin 30 = cos 2a
61. sin 20 + 1= cos 0 + 2 sin a
62. cos 60 + cos 40 + cos 20 + 1 = 2 cos 20 cos 0.
63. cos 0 + ^3 sin 0 = ^3 cos 0 - sin 0 = 2 cos(0 + <p).
Eliminate 0 between the following equations :
64. cos 0 = a, cos 20 = b.
65. cos 0 + sin 0 = a, cos 20 = 6.
66. acos0 + 6cos20 = c, asin0 + 6sin 26 = d.
67. a cos 0 + 6 sin 0 = J(^^^^)> tan 20 = c.
Examples IV. b.
Find the values of :
1. cos(a + ^) and cos(a — /3), if cos a = f f and cos /9 = f f .
2. sin(a + /8) and am(^ — a), if sin a = H ^^^ cos ^ = |f .
OF COMPOUND ANGLES. 39
12
~5~-
3. cos(a + ^) and sin(a - ^), if tan a = V- and cot /S ^
4. tan(a + P) and tan(a — /3), if tan a = J and tan ^ = J.
5. cot(a - B). if sin a = tVt and cos (3 = f f.
6. Shew that 4 sin 75° sin 15° = 1.
^ ^. yn tan('yi + l)a — tsin(n — l)a
7. bimplity i_|_tan(7i+l)atan('^-l)a'
8. cos(a + /3)sin ^ — cos(a + y)sin y
= sin(a + /5)cos ft — sin(a + y)cos y.
f^ i. _L.j. /o sin(a±/8)
9. tana±tan/3= ^ '^i-
'^ cos a cos /5
, . ^. cosec a cosec /3
10. cosec(a±/5) = -^^-^^^^-.
11. sin2a-sinasin(60° + a) + cos2(30°-a) = f.
12. Find sin(a + /3 + y) in terms of the cosines and sines of
a, A y.
1 3. Find tan(a + /5 + y), when tan a = J, tan /5 = f , tan y = | .
14. cos^a + cos2/3 — 2 cos a cos ft cos(a — ft) = sm\a — ft).
1 5. sin(a + /5)sin(a — ft) = (sin a + sin /3)(sin a — sin ft),
16. sin55°cos25° = f — sin25°.
Find the values of:
17. cos 2a, when (1) cosa = H» (2) sin a= iV» (3) sin a = 4-
18. sin 2a, when (1) sin a = f, (2) tan a = ^^!^.
19. tan 2a, when (1) tana = f, (2) sin a= — ^ — r-
20. cos a, sin a and tau a, when cos 2a = 2^.
21. tan a, when sec 2a = 3; and tan ^, when sin a = ff.
22. tan(2a + iS), when tana = l and tan^ = J.
23. cos 3a, when cos a=\^',^ ; sin 3a, when sina = J;
and tan 3a, when tana = -y-.
40 TRIGONOMETRICAL RATIOS
Shew that :
24. seen 5° = 4 tan 1 5°, and tan 7° 30' = ^6 + v^2- 2- ^3.
25. 4sml8°cos36° = l.
2 tan a
26. sin 2a
27. sec 2a =
1 + tanV
sec^a
2 — sec^a'
28. 2 cosec 2a = tan a + cot a.
29. c»««+8in« = sec2a+tau2a.
COS a — sin a
30. cot§— cota = coseca.
31. l±sina = 2sin2('45^±«Y
32. cot(45" + a) + cot(45° - a) = 2 sec 2a.
33. cot<45° + ^) = |^^^^^|^^«.
\ 2/ 2 cosec 2a 4- sec a
34. 224|5;±2_) = sec2a-tan2a.
cos(4o —a)
„^ cos 2a sin 2a _ 1
cos a + sin a cos a — sin a ~~ v^2 cos(a + 45°)'
36. sin^a — cos^a cos 2/3 = sin^^ — cos^/S cos 2a.
2
.37. seca-^^2_^^^2 + 2cos4a)r
38. From the expressions for cos(a + /3 + y), sin(a + iS + y)
and tan(a + i5 + y), deduce the values of cos 3a,
sin 3a and tan 3a.
39. 4 cos^a — 4 sin^a = 4 cos 2a — sin 2a sin 4a.
40. Express tan 4a in terms of tan a.
41. Express sin 5a in terms of sin a.
42. sin 3(a - 15°) = 4 cos(a - 45°)cos(a + 15°)sin(a - 15°).
43. tan 3a = tan a cot(30° - a)cot(30° + a).
OF COMPOUND ANGLES. 41
COS p — COS a '1
45. 52i^±£^=cot^cot^«.
COS a — cos/3 2 2
46. sec(45° + a)sec(45° — a) = 2sec2a.
47. sin 2a sin 5a + sin 3a sin 10a = sin 5a sin 8a.
48. sin 2a(cos 3a + cos 7a + cos 1 la) = sin 6a cos 7a.
An o- Ti* sin a + 2 sin 3a + sin 5a
49. Simplify -.— - — , ^ . „ — ;— r— ^_-.
^ "^ sin 3a + 2 sm 5a + sin 7a
-A o- vr cosa — cos 3a + cos 5a — cos 7a
oO. bimplity ^ — ; -. ^ =— .
'^ "^ cos 2a + cos 4a — cos 6a — 1
51. cos 25° — sin 5° = cos 35°.
52. sin 33° + cos 63° = cos 3°.
53. 4sin20°sin40°sin80° = sin60°.
54. Prove geometrically that cot(a — /3) = ^ — ^— ^ .
^ ^ V A-/ cot/3-cota
55. Also that 2 cot 2a = cot a — tana.
Solve the equations :
56. isin0+^cos^ = -^.
57. tan(45° + ^) = 4tan(45°-e).
58. sill 20 cos 0 = sin 0.
59. coslie + cos50 = cos3a
60. cos 0- cos 90 = ^2 sin 4a
61. sin 20 + sin 0 = cos 20 + cos a
62. sin 30 + sin 50-sin 7a- sin 0 = 2^2 sin 20 sin 6.
63. 4 cos(0 + 60°) -J2 = jQ-4! cos(0 + 30°).
Eliminate 0 between the following equations :
64. sin0 — cos 0 = (X, sin 20 = 6.
65. cos20 + cos0 = a, sin20 + sin0 = 6.
66. atan(0 + a) = c, 6tan(0 + /3) = d
67. cos 30 + sin 30 = a, cos 0- sin 0 = 6.
42 TRIGONOMETRICAL RATIOS
Examples V.
1. Prove geometrically, by means of Eucl. VI. 8, that
cos 75°=^^^ and cos 15° = ^5^^.
2. If tan Q = b/a, then a cos 20 +b sin 26 = a.
o. Express 4 sin a cos ^ cos y as the sum of four sines.
4. Find the maximum value of cos a + sin a.
5. Shew that cos 6 cos(a --0) is a maximum when 6 = Ja,
and that its value is then cos^^.
6. If a sin(e +a) = b am(d + fi), then
^g^asiDa-6sin^
6 COS p — a cos a
7. Shew that acos(0+a)+6cos(0+iS) may be written
in the form ccos(0 + y), and find c and y in terms
of a, b, a and /3.
8. V3 + tan 40° + tan 80° = ^3 tan 40°tan 80°.
9. sin^a + sin*2a = f(l — cos a cos 3a) — J sin 3a sin 5a.
10. If tan a = tm% then 2 tan 2a = tan 2/3 sin 2/3.
11. If cos(a — y)cos /3 = cos(a — /3+y), then tan a, tan /3,
tan y are in harmonical progression.
12. If cos(a-^)sin(y-^) = cos(a + /3)sin(y+^), then
cot S = cot a cot ^ cot y.
13. If tan I = ^^7^^, then sin 0 = ^.
14. If tan /3 = ^£_«.^, then tan(a - iS) = (1 - ^)tan a.
1 5. If cot ^(tan^a - 1) = 2 tan /3, then
sin(a + |5)sec a = cos a cosec(a — /3).
16. Shew that {2«+icosec 2'^'ri^f
= (2^cosec 2^a)2 + (2^sec 2^a)\
OF COMPOUND ANGLES. 43
irr Ti* COS a — COS i8 COS y ,,
17. If cos a = — —.—>,-. -^^ \ then
sin p sm y
1 — sec^a — sec2/3 — sec^y + 2 sec a sec /3 sec y = 0.
18. If tan0+tan0 = secO, find the relation between 6
and (p. .
19. If 0 — a, 0, 0 + a be three angles whose cosines are in
harmonical progression, then cos 0 = ^2 cos J
20. If the angle ^J5(7 Ls divided by the straight line 5Q
into two angles, a and /?, and the circle ^P(7,
which touches AB and 5(7, is cut by BQ, in P, Q :
shew that sin a sin ^8 varies as the square of the
intercepted chord PQ of the circle.
CHAPTER IV.
USE OF MATHEMATICAL TABLES.
38. The principal tables required in trigonometrical
calculations are: (1) a table of the logarithms of all
numbers from 1 to 100,000; (2) a table of the trigono-
metrical ratios of angles for every minute from 0° to 90° ;
and (3) a table of the logarithms of these ratios.
In the tables at the end of this book are given all
the logarithms of numbers, trigonometrical ratios and
logarithms of trigonometrical ratios that will be required
for the working of the examples given in the text. The
logarithms of numbers are printed in nearly the same
form as in books of Mathematical Tables. The reader
is, however, strongly recommended to obtain a collection
of tables, for example, Chambers' Mathematical Tables,
and to work exclusively with them.
(For the fundamental properties of logarithms and
their proofs, see C. Smith's Elementary Algebra, second
edition, or any other treatise on Algebra.)
Logarithms of Numbers.
39. The table of logarithms of numbers given at the
end of this book contains eleven columns. In the first
column, are the first four digits of the number, the fifth
44
USE OF MATHEMATICAL TABLES. 45
and last digit being given at the head of each of the
remaining ten columns. In these columns, the mantissse
of the logarithms of the numbers are given to seven
places of decimals, the first three being given in the
second column only. The characteristic of the logarithm
is omitted, since it is easily determined by inspection of
the number.
For example, consider a number whose first four digits
are 2296. These digits are given in the first column.
In a line with them, and in the second column (headed
0), are the figures 3609719 ; these forming the mantissa
of the logarithm of 22960. In the same line, and in the
third column (headed 1), are the figures 9908, which form
the last four figures of the mantissa of the logarithm
of 22961, the three figures 360 being omitted, so that the
complete mantissa is 3609908. Again, in the same line,
and in the fourth column (headed 2), are the figures
0097, the bar above the figures * indicating that the last
of the three figures 360 is to be increased by 1 ; so that
the mantissa of the logarithm of 22962 is 3610097.
Thus, log 22960 = 4-3609719,
log 229610 =5-3609908,
log 2-2962 =0-3610097,
log 0-22963 =1-3610286,
log 0-00022964 = 4-3610475.
40. In the examples just given, the logarithms can be
found at once from the tables ; but, frequently, this is not
the case, the number whose logarithm is required lying
between two numbers in the tables.
The logarithm of 229613, for example, must lie be-
* Sometimes the bar is placed over the first figure only, thus : 0097.
46 USE OF MA THEM A TICAL TABLES.
tween the logarithms of 229610 and 229620, i.e., between
5-3609908 and 5-3610097. To determine log 229613, we
make use of the Principle of Proportional Parts, which
may be thus stated :
If there be three numbers such that the difference
between any two of them is small in comparison with
either, then the difference between any two of the
numbers is proportional to the difference between their
logarithms.
This theorem, it should be mentioned, is only approxi-
mately true. The limits between which it is applicable
are investigated in a later chapter.
Applying this principle to find log 229613, we have
the difference between 229610 and 229620 is to the
difference between 229610 and 229613 as the difference
between log 229610 and log 229620 is to the difference
between log 229610 and log 229613. The difference be-
tween log 229610 and log 229620 is '0000189: let the
difference between log 229610 and log 229613 be denoted
by 5; then 10 :3 :: -0000189 : 5,
from which we find ^ = '0000057
to seven places of decimals.
log 229613 = log 229610+^
= 5-6309908 + 00000057
= 5-6309965.
In practice, this example should be written out as
in the first of the following article.
41. Example 1.— To find log 229613.
log 229620 = 5-3610097,
log 229613=5-3609908 + 8,
log 229610=5-3609908,
USE OF MA THEM A TWA L TA BLES. 47
10: 3:: -0000189: 8,
8= -0000057.
log 229613= 5-3609908
+0-000()057
= 5-3609965.
Example 2.— To find log 0-0356124
log 0-035613 =2-5516086,
log 0-0356124 = 2-5515964 + 8,
log 0 035612 =2-5515964,
10 : 4 :: 0-0000122 : 8,
8 = 0-0000049,
log 0-0356124= 2-5515964
+ 0-0000049
= 2-5516013.
42. In a similar manner, we find a number whose
logarithm is known from the two numbers (given in
the table) whose logarithms have mantissse respectively
just greater and less than that of the given logarithm.
Example 3. — To find the number whose logarithm is 3*5970721.
3-5970806 = log 0-0039544,
3-5970721 = log (0-0039543 + 8),
3-5970696 = log 0*0039543,
110 : 25 :: 22 : 5 :: 00000001 : 8,
8 = 0-000^005-7-22 = 0-00000002, nearly,
3-5970721 = log 0*00395432, nearly.
43. Example 4.— To find the 61st root of 329 x 3079-i-425130 to
six places of decimals.
The logarithm of the required root
= 6\(log 329 + log 3079 -log 451230)
2-5171959
-^1 + 3-.
4884097-5-6543980/
^1 / 6-0056056 \
^V- 5-6543980/
= ^^ of 0-351 2076 = 0-0057575.
48 USE OF MA THEM A TIGAL TA BLES.
Now, 0-0057809 = log 1-0134,
0-0057575 = log( 1-0 133 + 8),
0-0057380 = log 1-0133,
429: 195:: -0001 : 8,
8=0-000045,
the required root = 1 '01 3345.
Logarithms of Trigonometrical Ratios.
44. It is unnecessary to give examples of finding by-
means of tables the trigonometrical ratios of angles;
the method being, in every respect but one, the same as
that for finding the logarithms of the same ratios. It
differs only on account of the form in which the ratios
and their logarithms are printed in the tables.
The cosines and sines of all acute angles, the tangents
of angles less than 45°, and the cotangents of angles
greater than 45°, being less than unity, the logarithms of
all such ratios have negative characteristics. To avoid
printing such characteristics, it is usual to add 10 to the
logarithm, forming the tabular logarithm of the ratio,
written L cos a, etc. The letter L is used as an abbrevi-
ation of the words ' tabular logarithm ' ; thus,
Xcos a = log cos a + 10.
In working with the tables, however, the tabular loga-
rithm should not be used : the 10 should be subtracted
from it whilst copying the logarithm. For example, we
should write at once,
logcos 57° 10' = 1-7341572,
log sin 1° =2-2418553,
W tan 49° 13' = 00641 556.
USE OF MA TEEM A TICAL TABLES. 49
45. In finding the logarithms of trigonometrical ratios
which are not given exactly in the tables, it is necessary
to distinguish the cases in which the ratio increases or
decreases as the angle increases. The sine, secant, and
tangent of an angle increase as the angle increases, while
the cosine, cosecant, and cotangent decrease as the angle
increases (Art. 22), Taking, first, an example of the
former, let us find the logarithm of sin 16° 23' 27".
From the table, we find log sin 16° 23' = 1-4503452, and
log sin 16° 24' = 1-4507747. The required logarithm must
therefore lie between these values. To obtain it, we
make use of the Principle of Proportional Parts, as
enunciated above (Art. 40), with the requisite verbal
changes.
The limits between which the principle is applicable
in this case are investigated in chapter xv. : for the
present, it may be stated that the result will not be
accurate to the seventh place of decimals if the angles
differ by less than a few degrees from 0° or 90°.
Now, the difference between 16° 23' and 16° 24' is 60";
and that between 16° 23' and 16° 23' 27" is 27". Again,
the difference between log sin 16° 23' and log sin 16° 24' is
0-0004295. Let the difference between log sin 16° 23' and
log sin 16° 23' 27" be denoted by S. Then
60 : 27:: 00004295 :^,
from which we find S = 00001933.
log sin 16° 23' 27" = log sin 16° 23'+^
= T'4503452 + 00001933
= 1-4505385.
In practice, we should work this example as follows :
D
51) USE OF MATHEMATICAL TABLES.
Example 5.— To find log sin 16° 23'_27".
log sin 16° 24' =1-4507747,
log sin 1 6° 23' 27" == 1 '4503452 + 8,
logsinl6°23' =14503452,
60 : 27 : : 0-0004295 : 8.
•0004295
9
20 ) -0038655
•0001933
log sin 16° 23' 27"= 1-4503452
+ 0-0001933
= T-4505385
46. Similarly, we find an angle the logarithm of whose
sine is given, from the two angles (given in the table) the
logarithm of whose sines are respectively just greater and
just less than the given logarithm.
Example 6. — To find the angle the logarithm of whose sine is
i -9658931.
1 -9659285 = log sin 67° 36',
1-9658931 =log sin 67° 35' S",
1 -9658764 = log sin 67° 35',
521:167::60:S.
167
60
521 ) 10020 ( 19", nearly.
521
4810
4689
121
r9658931 =log sin 67° 35' 19".
47. Example 7.— To find log cos 53° 14' 51".
log cos 53° 14' =1-7771060,
log cos 53° 14'51"=T-7771000-S,
log cos 53° 1 5' =1-7769369,
USE OF MATHEMATICAL TABLES. 51
.'.
-60 :-51 :: -0001691: 8.
•0001691
51
1691
8455
60 ) -0086241
•0001437 = 8.
• •
log cos 53° 14' 51"= 1-7771060
- 0-0001437
= 1-7769623.
Example 8.-
1-9447435.
—To find the angle the logarithm of whose cosine is
1-9447862= log cos 23° 17',
1-9447435 = log cos 23° 17' 8",
1-9447182 = log cos 23° 18',
,*.
680: 427:: -60 -.-8.
427
3
34 ) 1281 ( 38", nearly.
102
261
1-9447435 =log cos 23° 17' 38", nearly.
Examples VLa.
Find the logarithms of the following numbers :
1. 249317. 4. 9152-06.
2. 0-751204. 5. 539-005.
3. 39172500.
Find the numbers whose logarithms are :
6. 5-2471903. 9^ 2-3058769.
7. 8-7192855. 10. 21572693.
8. 4-2145028.
52 VSE OF MATHEMATICAL TABLES.
Find the values of :
_, 72x5301 , „ , CA ' ^
619x70025' P ^^^^ decimals.
12. 4/(3914-26 X 130-72), to 4 places of decimals.
,^ (1-0012)2 X (-059)^
13. ^ '7a.qi9q\4 — > ^° ' places oi decimals.
Find the logarithms of:
14. sin 17° 13' 28". 17. sin 54° 15' 2".
15. tan 19° 57' 12''. 18. tan 17° 41' 39".
16. cos 78° 24' 50". 19. cos 11° 3' 12".
Find the values of a, when :
20. log sin a = 1-5127043. 23. log sin a = 1-8318514.
21. log tan a = 1-4587148. 24. log tan a = 0-7974903.
22. log cos a = 1-9849418. 25. logcos a = r9447285.
Find:
26. sin 12° 3' 12". 28. cot 52° 41' 21".
27. cos 71° 14' 39".
Find the values of a, when :
29. sin a = -37 14051. 30. cos a = "2.
Examples VI. b.
Find the logarithms of the following numbers :
1. 632503. 4. 0-000531058.
2. 00513517. 5. 7912-02.
3. 73-1459.
Find the numbers whose logarithms are :
6. 0-2175004. 9. 5-3971142.
7. 1-3721153. 10. 4-2456173.
8. 1-3190725.
USE OF MATHEMATICAL TABLES. 53
Find the values of:
.. 035 X 1-2502 , ^ , rA ' ^
11. .,-, -, o — ,, ^r,m> to 9 places or decimals.
12. a/L^^3 ,^J, to 5 places of decimals.
13. (0001528x517)7, to 6 places of decimals.
Find the logarithms of:
14. sin 52° 15' 17''. 17. sin 5° 51' 22".
15. tan 74° 11' 9". 18. tan 14° 19' 38".
16. cos 32° 30' 14". 19. cos 12° 43' 18".
Find the values of a, when :
20. log sin a = 1-9703652. 23. log sina = ro012301.
21. log tan a = 0-5175023. 24. log tan a = 1-8115891.
22. log cos a = 1-6838142. 25. log cos a = 1-5354010.
Find :
26. sin 19° 12' 37". 28. cosec 24° 14' 9".
27. cos 43° 15' 5".
Find the values of a, when :
29. sin a = -5502471. 30. cos a = "8026140.
CHAPTER V.
SOLUTION OF EIGHT- ANGLED TEIANGLES AND
PEACTICAL APPLICATIONS.
Solution of Right-Angled Triangles.
48. In every triangle there are six elements, the
three sides and the three angles. If any three of these
elements, except the three angles, be given, they are, as a
general rule, sufficient to enable the other three to be
determined. The process by means of which the remain-
ing elements are found is called the solution of the
triangle. In the present chapter, we confine ourselves to
the solution of right-angled triangles.
It is obvious that, if two angles of a triangle be known,
the third may be determined from the fact that the three
angles are together equal to two right angles. And, in
a right-angled triangle, if one acute angle be known, the
second may be found in the same way.
We have thus four cases to consider, in which we are
given (1) the hypotenuse and one acute angle; (2) one
side and one acute angle; (3) the hypotenuse and one
side; and (4) the two sides.
49. If ABC be any triangle, the length of the sides
BG, GA, and AB, opposite respectively to the angles J.,
j5, and 0, will be denoted by the letters a, 6, and c.
54
En^Fish Miles
RIGHT-ANGLED TRIANGLES. 55
50. Case I. — The hypotenuse and one acute angle being
given, to find the two sides and the other angle.
Let C be the right angle, and let the hypotenuse c and
the angle A be given. ^
We have 4+^ = ^0",
B = 90°-A.
Also, a/c = sin A and b/c = sin B,
a = c sin A and 6 = c sin i?, a
.-. log a = log c+ log sin J. and log6 = logc + logsin5.
51. Example l.~If c = 3701 and ^ = 41° 13' 24", to find a, b ami B.
B=m°- 41° 13' 24" =48° 46' 36".
log sin 41° 14' =1-8189692,
log sin 41° 13' 24" = 1-8188250 + 5,
log sin 41° 13' =1-8188260,
60'': 24":: -0001442: 8,
8= -0000577,
log sin 41° 13' 24"= 1-8188827.
log a=log 3701 + log sin 41° 13' 24"
= 3-5683191
+ 1-8188827
= 3-3872018.
Now, 3-3872118 = log 2439-0,
3-38720 1 8 = log(2438-9 + S),
3-3871940 = log 2438-.9,
178:78 :: -1 : 8,
8=0-04, nearly,
« = 243^-94.
Similarly, we find log sin 48° 46' 36" =1-8763025, log 6 = 3-4446216,
and therefore 6 = 2783-69.
i?=48°46'36",]
a = 2438 -944, ^
6=2783-69. J
52. Case II. — One side and one acute angle being
given, to find the hypotenuse, the other side and the
other angle.
56 RIGHT-ANGLED TRIANGLES
Let G be the right angle, and let the side a and the
angle B be given.
Then, ^ = 90' -5.
Also, cja = sec B and hja = tan B,
c = a sec jB and 6 = 66 tan 5,
log c = log a H- log sec B and log 6 = log a + log tan B.
53. Example 2.— If a = 374 and B=2V 34', to find c, 6 and A.
^ = 90°-2r34'=68°26'.
Now, c = a sec i5,
log c = log 374 + log sec 21° 34'
= 2-5728716
+ 0 0315215
= 2-6043931.
Now, 2-6043989 = log 402-16,
2-6043931 = log(402-15 + 8),
2-6043881= log 402-15,
108 : 50:: -01 : 8,
8 =-005, nearly,
c = 402-155.
Again, h = a tan 5,
log 6 = log 374 + log tan 2 r 34'
= 2-5728716
+ 1-5968776
= 2-1697492.
Now, 2-1697626 = log 147-83,
2-1697492 = log(147-82 + 8), •
2-1697332 = log 147-82,
294 : 160:: -01 : 8,
8 =-005, nearly,
6=147-825.
A = i
c-
54. Case III. — The hypotenuse and one side being
given, tojind the other side and the tivo acute angles.
I
AND PRACTICAL APPLICATIONS. 57
Let C be the right angle, and the hypotenuse c and the
iide a be given.
' Now, c2 = a2 + &^
62^c2-a2 = (c + aXc-«),
log6 = Hlog(c+«) + log(c-a)}.
Also, H\nB = hlc,
log sin B = log 6 — log c.
55. Example 3. — If c = 569, and a =435, to find b, A, and B.
b = J{c + aXc - «) = Vl004 X 1 34,
log6=Klog 1004 + log 134)
^i/ 3-001 7337 \
^V + 2-1271048/
= 1(5-1288385)
= 2-5644193.
Now, 2-5644293 = log 366-80,
2-5644193 = log(366-79 + 8),
2-5644175 = log 366-79,
118: 18:: -01: 6,
8 = -0015, nearly,
6 = 366-7915.
Again, sin B=blc,
log sin ^ = log 6 - log 569
= 2-5644193
-2-7551123
= 1.8093070.
Now, 1 -80941 89 = log sin 40° 9',
1 -8093070= log sin 40° 8' 8",
1-809269 1 = log sin 40° 8',
1498: 379:: 60": 8,
8=15", nearly,
5=40° 8' 15", and therefore yl = 49° 51' 45",
6 = 366-7915
.4 = 49° 51' 45"
5=40° 8' 15"
58 iii(!iiT-Ai\(Uj':ii rniAMUJCS
66. Case IV. — The. Uvo sidcH co7itainin<j the right
aiujta heiiKj <jlven, to find the hypotenuse and the two
acute angles.
Lot C bo tho right angle, and let the sidoH a and h bo
given.
Then, tan A = a/6,
log tan A = log a — log h.
From this equation A can be fouud and B from the
equation B=iW — A.
Also, cja = cosec A ,
c=acosecil,
log c = log a + log cosec A .
57. Example 4.— If a -371 and 6-504, to find 0, /I, and B.
tan yl -a/6-371/504,
log tan i( - log 371 - log 604
- 25693739
-2-7024305
- 1-8669434.
Now, T'8070937 - log tan 36" 22',
1 •8(J60434 - log tan 30° 2 1' 6",
i-8668291-logtan36°21',
2646: 1143:: 60": 8",
8-26", nearly,
il-36'21'20".
5-53' 38' 34".
Again, 0— aaec/?,
log c - log 37 1 + log Bee 53° 38' 34".
Now, log Hec 53" 39' -0-2271532,
log HOC n:r 38' 34" -0-226981 6 + 8,
log sec 53° 38' -0*2269815,
60": 34":: -0001717: 8,
8 --0000973,
log sec 63' 38' 34" -0-2270788.
AJ^D VRAGTICAL APPLICATIONS, 50
logo- 2*5693730
+0-2270788
- 2-7964627.
N o\y, 2-7064564- log 625-88,
2-7964527 - log(625 -82 + S),
2-7064404-log 625-82,
70 ! 33 :: -01 : 8,
fi-«-005, ueftrly,
0-625-825.
0-625-
il-36'
J5-53"
15-825 ]
j»2r26" Y
r 38' 84" J
Examples VII. a.
(-Yoee.— -Angles shouUl bo fouiul to tho nearest second,
sides to six signilicai^t iignvOvS.)
Solve the following triangles, having given 0^= 00 \ an»l ;
1. 0=172,^1 = 85'' 27'. 7. a=821, 6=1002.
2. 0 = 3041, B = 24M 9'. 8. a = 792, 6 = 562.
X a = 004, B = 41° 17'. 9. a = 249, A = 02*^ 15' 42".
1. 6 =40128, ^ = 64* 8'. 10. o =204137, ^, = 91423.
». o=712,a=408. 11. c =898, 2^=15^4' 8".
(I. 0 =3052, 6 = 796. 12. tt = 401, 6=602.
Examples VII. b.
(See note at the head of Exaiuph^s \\\ \.)
Solve the following triangles, having' K^^ ^'^ ^' = ^^^^ \ ^^^ •
1. o-4018,il-7ri6r. 7. a =518, 6 = 927.
2. 0 = 5702, li = 59** 48'. 8. a = 8012, 6 = 5148.
'X a=48, ^=10' 12'. 9. a=700, B^Sir 14' 16".
k 6-619, 5=25'* 48'. 10. a-69, 6=78.
>. 0 =6941, a=1026. 11. o =621, 6=507.
(). 0 =57102, 6=31040. 12. c =7205, £-31" 14'66\
60 RIGHT-ANGLED TRIANGLES
Practical Applications.
58. Several simple problems in the measurement of
heights and distances may be solved as applications of
the preceding propositions. The principal instruments
required for the purpose, and the method of using them,
are described in Chapter IX. They are the chain, for
measuring distances, and the theodolite, for measuring
angles in a vertical or horizontal plane.
59. Definitions. — The angle of elevation (occasionally
called the elevation or the altitude) of any point above
tKe horizontal plane through the observer's eye is the
inclination to the plane of the straight line joining the
point and the observer's eye. If the point be below this
plane, the angle is then called the angle of depression
(or the depression) of the point.
60. To find the height of a toiver standing on a
horizontal plane, the foot of the tower being accessible.
Let BC represent the tower.
From the foot, C, measure the
length of any line, GA, called a base-
line, in the horizontal plane on
which the tower stands; and at the
other end, A, of this line measure
the angle of elevation, BA C, of the top of the tower.
Now, BCIAG=t8inBAG,
BG= AG tarn BAG,
i.e., the height of the tower above the observer's eye is
equal to the length of the base-line multiplied by the
tangent of the angle of elevation of the top of the tower
at the other end of the base-line.
AND PRACTICAL APPLICATIONS.
61
61. To find the height and distance of a tower stand-
ing on a horizontal plane, the foot of the toiuer being
inaccessible. ^
Let CD represent the tower.
In the horizontal plane on
which the tower stands, measure
the length of any base-line, AB,
directed towards the foot of the
tower, D ; and, at each end of ^
the base-line, measure the angles of elevation, CAD and
GBD, of the top of the tower.
Now, AD = CD cot A and BD = CD cot B,
AB = AD - BD=^CD(cot A- cot B)
— cnf^^^ ^ — ^^^ ^^
= CD.
Vsin A sin BJ
sin B cos A — cos B sin A
sin A sin B
62.
= CD ^^^(^-^)
' sin A sin B
CD = AB sin A sin B/smiB- A).
To find the breadth of a river, or the distance of an
inaccessible object on a horizontal plane.
Let a base-line, AB, be
measured close to one bank
of the river, the ends, A
and B, being marked by
prominent objects, such as
trees. Selecting some ob- ^ J-> B
ject, C, close to the other bank of the river and visible
from both A and B, measure the angles CAB and CBA.
Draw CD per])endicular to AB.
62
RIGHT-ANGLED TRIANGLES
Now, AD = CD coi A and BD = CD cot B,
AB=AD+BD = GD(cot ^ + cot B)
* sin A sin B'
GD=ABBmA^\nBl^m{A+B).
63. Example 1. — If the base-line, AB^ be 127 yards, and the
angles CAB and CBA be 41° 27' and 35° 14', to find the breadth of
the river to the nearest yard.
log CD=\og AB + log sin A + log sin B - log sin(^ + B)
=log 127 + log sin 41° 27' + log sin 35° 14' - log sin 76° 41'
= 2-1038037
+ 1-8208358
+1-7611063-1-9881628
= 1-6857458
-19881628
= 1-6975830
= log 49-84, nearly,
the breadth of the river is about 50 yards.
64. Dip of the Horizon. — The dip of the horizon at
any point above the earth's surface is the angle of depres-
sion, at that point, of a point on
the horizon.
Let the circle in the annexed
figure represent a section of the
earth (supposed spherical) through
the centre (7, and a point 0 above
the earth's surface. Draw OB
perpendicular to OC, OA touching
the circle, and Al) perpendicular
to 00, Then, the angle BOA is
the dip of the horizon at 0.
Let c denote the radius of the earth, and h the height
of 0 above its suiface, so that OG=c-\-h.
AND PRACTICAL APPLICATIONS.
63
Now, angle 50^ = complement of angle AOG
= angle AGO,
and Qo^AGO = ACIOG=cl{c+h).
the dip of the horizon at 0 = cos~^ y-.
65. The following example illustrates a further applica-
l^ion of the methods explained in this chapter : —
Example 2. — The dip of a stratum is 8 degrees to the east.
Find its apparent dip in the direction 0 degrees south of east.
Let A be any point on the surface of the stratum and AB a
vertical line of any length through A.
From B, draw horizontal lines, BC and
BD^ in directions east and 6 degrees
south of east respectively, meeting the
surface of the stratum in the points C
and D.
Then, angle ACB=8, angle CBD^O,
and angle ADB (denoted by a) is the
apparent dip of the stratum in the given
direction. Also, the angle BCD is a
right angle, since CD is a horizontal
line on the surface of the stratum.
Now, BD=BCsece
=AB cot 8 sec 6.
tan a = ABjBD = AB/AB cot S sec 9
= tan 8 cos 6.
^o-f ^
t
Examples VIII. a.
(Note. — In the following examples, unless otherwise
stated, the ground is supposed to be level, and the height
of the observer's eye above the ground is not to be taken
J:
64. RIGHT-ANGLED TRIANGLES
into account. See also the note at the head of Examples
VILA.)
1. The angle of elevation of the top of a vertical cliff at
a point 660 feet from its foot is 34° 16', find the
height of the cliff.
2. Find the length (to the nearest inch) of the shadow
of a vertical flagstaff, 53 feet high, the altitude
of the sun being 22°.
3. At a spot 67 ft. distant from the base of a chimney,
the angle of elevation of its top is 23° 10'; find
the height of the chimney to the nearest inch, the
observer's eye being 5 ft. 6 ins. above the ground.
4. The angle of elevation of the cairn of a mountain,
2314 feet high, from a point on the sea-level
is 11° 14'; find, to within a hundredth of an inch,
the distance between the two points as represented
on a map drawn on a scale of 6 ins. to the mile.
5. A ship sailing due east at the rate of 10 miles an
hour is observed from a lighthouse to bear due
north at a certain moment, and, half an hour later,
to bear N. 32° 20' E. Find the distance between
the ship and the lighthouse in both positions.
6. A statue is placed on the top of a column. At a
point on the ground, 1 24 feet distant from the base
of the column, the angles of elevation of the tops
of the statue and column are 45° 27' and 43° 12',
respectively ; find the heights of the statue and
column to the nearest inch.
7. Find the height of a chimney, when it is found that
walking towards it 100 feet along a straight line
through its base, changes the angle of elevation
from 24° 40' to 46° 10'.
AND PRACTICAL APPLICATIONS. 65
8. From the top of a hill two consecutive milestones
are seen on a horizontal road running directly
Mk from its base. Their angles of depression are
m. found to be 14° 3' and 3° m'. Find the height
W of the hill to the nearest foot.
9. Three consecutive milestones on a road running east
and west are seen from a hill due south of the
first milestone. The second milestone bears N.
24° 30' E. Find the distance of the hill from the
road, and the bearing from the hill of the third
milestone.
10. At two points 1200 yds. apart on a road running
east and west, the bearings of a spire are observed
to be N. 57° E. and N. 43° W. ; find the distance
of the spire from the road.
11. The centre of the base of a tower, surmounted b}?- a
spire, lies directly between two stations 200 feet
apart, and the angles of elevation of the top of the
spire at these two points are 43° 17' and 31° 23'.
Find the height of the top of the spire.
12. A mountain, 4360 feet high, rises from a narrow
peninsula. At two points on the sea-level, on
opposite sides of the mountain, the angles of
elevation of its summit are 10° 50' and 13° 12'.
Find the distance between these points, supposing
the straight line joining them to pass vertically
below the top of the mountain.
13. The dip of a stratum is 38° in the direction K 47° E.
Find its apparent dip in the direction E. 5° S.
14. Find the dip of the horizon at the top of a mountain
15000 feet high, the radius of the earth being
4000 miles.
66 RIQHT-ANQLED TRIANGLES
15. An observatory and distant chimney are 3520 feet
apart. The angle of depression at the observatory
of the top of the chimney was found at a given
moment to be 5° 17' 2", and some years later to be
5° 19' 23". Find, to within a hundredth of an inch,
the amount of subsidence of the ground on which
the chimney is built, supposing the observatory to
be stationary.
16. A base-line, I feet long, is drawn from a point on a
horizontal plane in a direction at right angles to
the line joining that point to the base of a tower
standing on the plane. The angle of elevation of
the top of the tower from the two ends of the base-
line are 30° and 18°. Find the height of the tower.
17. The angle of elevation of a column, as viewed from a
station due north of it, being a, and, as viewed
from a station due east of the former and at a dis-
tance c from it, being /3 ; shew that the height of
,1 1 . c sin a sin /3
the column is •- =-.
{sin(a + /3)sin(a-^)}*
18. At the top of a castle, which stood on a hill near the
sea-shore, the angle of depression of a ship at
anchor was observed to be 5° 10'; and, at the
bottom of the castle, it was observed to be 4° 25'.
If the castle be 60 feet high, find the horizontal
distance of the vessel, and the height of the hill
above the sea-level.
19. A flagstaff, a feet high, stands on a plane inclined
towards the south at an angle of a degrees. If the
altitude of the sun at noon be ^ degrees, find the
length of the shadow then cast by the flagstaff on
the plane.
AND PRACTICAL APPLICATIONS. 67
Find the length of the shadow, if a = 50 feet,
a = 12°, /5=37°.
20. Three consecutive milestones are situated on a road
inclined at an uniform angle to the horizon, the
road running east and west. From a station due
south of the lowest milestone, and in the same
horizontal plane with it, the angles of elevation of
the second and third are found to be a and ^,
respectively. Find the inclination of the road and
its distance from the station.
Shew that tan/3 must lie between tana and
2 tan a.
Examples VIII. b.
(See note at the head of Examples YIII. A.)
1. At a point 300 feet distant from the base of a monu-
ment, the angle of elevation of the top is 18° 42';
find the height of the monument.
2. The shadow cast on the ground by a statue on the
top of a column is 14 "2 2 feet long, the altitude of
the sun being 41° 10' ; find the height of the statue,
and also that of the column if the shadow of the
head be 56*14 feet from the foot of the column.
8. The angle of elevation of the top of a cathedral tower
is 18° 30' at a point on the ground 320 feet from
its base ; find the height of the tower.
4. The distance on a map drawn to the scale of 6 miles
to the inch, of the points representing the summit
of a mountain, 2412 feet high, and a point 134 feet
above the sea-level, is 32*54 inches ; find the angle
of elevation at this point of the mountain summit.
68 RIGHT-ANGLED TRIANGLES
5. A balloon passes vertically over two points, A and B,
12 miles apart. When vertically over A, its angle
y of elevation at J? is 12° 40' ; and, when vertically
over B, its angle of elevation at ^ is 14° 12'. Find
the inclination of its path to the horizon.
6. At the top of a castle, built on the edge of a vertical
cliff, the angle of depression of a ship anchored at
a distance of one mile from the foot of the cliff is
4° 52' ; at the bottom of the castle, it is 4° 2'. Find
the heights of the castle and the cliff.
7. In order to ascertain the height of a tower, a base-
line, 93 feet long, was measured in a direct line
through its base, and the angles of elevation at the
two ends of the line were found to be 33° 20' and
55° 54'; find the height of the tower.
8. From the top of a hill, the angle of depression of two
consecutive milestones on a road running directly
towards the hill are 14° 3' and 3° 56'; find the
height of the hill.
9. Find the distance between two objects lying in a
horizontal straight line from the foot of a tower,
84 feet high, when their angles of depression at
the top of the tower are 57° 30' and 25° 15'.
10. To determine the breadth of a river, a base-line,
516 feet long, is measured close to one bank, and
it is found that the direction of this base-line is
N.W. and S.E. From the two ends of the base-
line, the bearings of a tree on the opposite bank
are observed to be E. 7° S. and N. 10° W. Find
the breadth of the river.
11. ^ and B are two stations, 2 miles apart, A being due
north of B. At A^ the altitude of a cloud is 32°
AND PRACTICAL APPLICATIONS. 69
to the S. ; and, at B, it is 41° to the N. Find the
height of the cloud.
12. On a map, drawn to the scale of 25 inches to a mile,
the distance between a church spire and a road
running N. and S. is represented as 9*5 inches.
Find the actual distance between two points on
the road at which the spire bears N. 14° E. and
N. 35° E.
13. The dip of a stratum is 18° N. Find its apparent
dip in the direction W. 10° N.
14. Find the dip of the horizon at the top of a mountain,
17500 feet high, the radius of the earth being
4000 miles.
15. The summit of a spire is vertically over the centre
of a horizontal square enclosure, whose side is
a feet long ; the height of the spire is h feet above
the level of the square. If the shadow of the spire
just reaches the corner of the square when the
sun has an altitude 6, shew that h,^2 = a tan 0.
Find h, having given a = 1000 feet, 0 = 27° 29' 48''.
16. The angle of elevation of a steeple at a place due
south of it is 45°, and at another place due west
of the former it is 15° ; shew that the height of
the steeple is |(x(3i — 3"*), a being the distance
between the places.
17. A balloon at starting is a miles north of an observer,
and, after travelling due east for a given interval,
(is seen by him in a direction 0 degrees E. of N.
and at an altitude of a degrees. Find the height
of the balloon at this instant.
A building on a square base, A BCD, has two of its
sides, AB and CD, parallel to the bank of a river.
70 RIGHT-ANGLED TRIANGLES
An observer, standing on the river's bank in the
same straight line with DA, finds that the side
AB subtends at his eye an angle of 45°. Having
walked a yards along the bank, he finds that the
side DA subtends an angle sin"^J. Prove that
the length of each side of the building is .aj^^
yards.
19. If the dip of the horizon at the summit of a moun-
tain, 3 miles high, be 2° 13' 27", find the diameter
of the earth.
20. A statue, 10 feet high, stands on the top of a column,
and, at a point in the horizontal plane through
the base of the column, the angles of elevation of
the top of the statue and the top of the column
are 32° 20' and 30°, respectively. Find the height
of the column.
Miscellaneous Examples. — I.
a.
1. Reduce 95741'' to degrees, and find the number of
degrees in the angle of a regulg-r dodecagon.
2. cot^a — cos^a = cot^a cos^a.
8. Solve the equation cosec20 = l + cot0.
4. If cos a = ff and tan ^8 = H, find the value of cos(a — /8)
and sin (a + /3).
5. (cos a + sin a)* + (cos a — sin a)^ = 3 — cos 4a.
6. cos 60° + cos 72° = cos 36°.
7. If the shadow of a column 120 feet high be 164 feet
long, find the altitude of the sun.
/3.
1. Each angle of a regular polygon contains 174°; find
the number of its sides.
AND PRACTICAL APPLICATIONS. 71
2. sin2a(4 cos^a - 1)^ + cos^aC^ cos^a - 3)2 = 1.
3. If cot a = 2 - JZ, then sec a = jQ-\- jj2.
'4. The tangents of 60°, 45°, and 15° are in arithmetical
progression.
K cu irx. I /. X sin(a-/3) + sina + sin(a + /3)_sin a
5. Shew that: (i.) -7—7 ^^f- — . ; — ^^ — ro\~- •
^ ' sin(y-/3) + siny + sin(y + /3) sin y
(ii.) cos 40° + cos 80° = cos 20°.
6. Solve the triangle in which c = 14157, 6 = 10253,
(7=90°.
7. A tower stands on level ground. At two points in
the same straight line through the foot of the
tower and 44 yards apart, the angles of elevation
of the top are 26°14' and 63°47'. Find the height
of the tower.
y-
1. The angles of a triangle are in arithmetical progression,
and the difference between the greatest and least
is equal to half the angle of a regular pentagon ;
find the number of degrees in each angle.
2- { ^(cosec a + cot a) - ^(cosec a - cot a) } ^ = 2 (cosec a - 1 ).
8. Solve the equation 8 cos 0 + 4 sec 0 = 7.
4. Shew that tan 36° = V(5- 2^5).
5. Express cos 7a in terms of cos a.
6. Simplify si" 3a + 2 sin 5a + sin 7a ^„j ^^^^ ^^^^
sin 5a + 2 sm 7a + sm 9a
sin2(22i° + |)-sin2(22i°-|) = ^sina.
7. When the altitude of the sun is 32°40', the cross of the
top of a church-spire casts a shadow 18 feet long
on level ground. Find the length of the cross to
the nearest inch.
72 RIGHT-ANGLED TRIANGLES
S.
^ cos a , cos^ _ cos a , cos ,8
sina + cos^ sin /3 — cos a sin a — cos ^8 sin^+cosa*
2. The sine of a certain angle is K^+^g) ^ : find the
values of its other trigonometrical ratios.
3. If tan a = J and tan /3 = J, find the values of tan(2a + 1^)
and tan(2a — /3).
4. (1 + cot a + cosec a)(l + cot a — cosec a) = cot ^ — tan J
5. Shew that: (i.) sin(GO° + a) - sin(60° - a) = sin a.
,. . . cos 2a — cos 4a cos a — cos 3a sin a
sin 4a — sin 'la sin 3a — sin a cos 2a cos 3a"
6. Solve the triangle in which a = 1025, 6 = 2531, 0 = 90".
7. A lighthouse, 65 feet high, is built on the top of a cliff*.
The angles of depression of a ship from the top
and bottom of the lighthouse are 14°30' and 12°46'.
Find the height of the cliff" and the distance of the
ship from it.
e.
1. If the measures of the angles of a triangle, referred to
1°, 100', and 10000" as units, be in the proportion
of 2, 1, 3, find the angles.
2. cosec^a — cot^a = 3cosec2acot2a + l.
3. Solve the equation cos Q sin 0+cos 0 = cos2^+sin 0.
4. Find the values of cos a and sin a, when cos 2a = -^.
5. 2-%in « - 2- %in 2|ri = 2«sin |.(sin ^^ .
6. Eliminate 0 between tan 20 = a and cot 0 = 6.
7. A cloud just grazing the top of a mountain 2310 feet
high, is seen at an altitude of 30° 15' by a man at
the sea-level. It is driven along by the wind at
AND PRACTICAL APPLICATIONS. 73
the same height and directly away from him, and,
twenty minutes later, he finds its altitude to be
12°S0'. Find the velocity of the wind in feet per
second.
1. If tan a = ^ \ find the other trigonometrical
ratios of a.
2. Eliminate 0 between
tan 0 + cot 0 = a and c cos 0 = 6 — c sin 6.
3. cotra — cot(r+l)a =
sm a
sin7"asin(r+l)a
l-6tan2| + tan*|
4. cos 2a =
(l + K)^
5. Shew that: (i^ ^^^^^4^^±^^^ = 4cos2acos4a.
tan oa — tan da
(ii.) tan 70° + tan 20° = 2 sec 50°.
6. Solve the triangle in which c = 1000, A = 19°36', C= 90°.
7. A man, observing at noon a cloud due south of him,
finds its angle of elevation to be 33°10', and that
of the sun 54°20. At the same time he notes the
position of its shadow, and afterwards ascertains
its distance to be 1020 feet. Find the height of
the cloud.
V'
1. If A be the right angle of an isosceles right-angled
triangle ABC, and if BL, BM be drawn to make
equal angles with BG and to meet ^(7 in X, M; then
AL.GM+AM.GL = AB.LM.
74 RIGHT-ANGLED TRIANGLES
2. (3 - taii2a)cot a = (cot^a - 3)tan 3a.
3. Find the value of cos 3a when cosa = f, and of sin 3a
when sin a = J.
4. sin(a + /3)cos/3-sin(a + y)cosy = sin(|8-y)cos(a + )Q+y).
5. Solve the equatioDS : (i.) cos 90+ cos 30 = cos 60.
(ii.) tan 20 = 3 tan a
6. Solve the triangle in which a = 7l, ^ = 9°54', 0=90°.
7. The dip of a stratum is 24°30' towards the N.E. ; find
its apparent dip in the direction E. 1 5° S.
0.
(Examples on Chapter II.)
- - sin^a „ /. tan^aX
2. cot% + cot2a = cosec*a — cosec^a. Also verify this for-
mula when a = 30°.
3. If cos a= j^ and cos/5 = ff, find the value of
tan a sin ^ — sin a tan ^.
4. (sin 30° + 2 cos 45° - 1)^ = 3 sin260° - cosec 45°.
5. Solve the equation 5(1 — sin 0) = 008^0(5 — 2 sin 0).
6. Eliminate 0 between
sec^0 + cosec^0 = a, and tan 0 = 6 sec^0.
7. Find the least value of cos^O+sec^O.
K.
(Examples on Chapter III.)
1. If sin a = -5^ and sing= ^ ^,, find the value of
tan(a+/3).
2. ia,n(a + r/3)-tan(a + r^/3)= — , , —^'1 / _l /3^•
^ '^ '^ cos(a+r-l|8)cos(a + r/3)
3. Express as a single term 2 cos a cos 2a + sin a sin 2a.
AND PRACTICAL APPLICATIONS. 75
4. Shew that :
(i.) sin(0 + ^) - sin 0 = cos S sin s(l - tan 6 tan |\
(ii.) tan 50° + tan 40° = 2 sec 10°.
5. Solve the equation
cos 20+ cos 40 -cos 80 -cos 100 = 2^2 cos 0 sin 30.
6. If cos 80 + cos 40 = a and sin 80 + sin 40 = 6, then
2cos20 = V(»H6').
7. Two parallel chords of a circle, lying on the same side
of the centre, subtend respectively 72° and 144° at
the centre. Shew that the distance between the
chords is half the radius of the circle.
X.
(Examples on Chapter III.)
1. tan 5a — tan 3a — tan 2a = tan ha tan 8a tan 2a.
2. Find the value of tan a, when tan 2a = x/l5.
3. Simplify - cos2a + sin2«
2 cos a + sm a — 2(cos'^a + sm^a)
4. Shew that :
/. X sin a + 2 sin 3a + sin 5a_4sin a — 3coseca
cosa — 2 cos3a + cos5a 4cosa — 3seca"
(ii.) l+cosl8° + cos36° + cos54° = 4cos9°cosl8°cos27°.
5. Solve the equation tan(0 + 3O') = (7 + 4v^3)tan(0-3O°).
6. Eliminate 0 between
acos0 + 6sin0 = cand2a6cos20-(a2-62)sm20 = d2
7. Points A, B, G, D are taken on the circumference of a
circle, so that the arcs AB, BG, and GD subtend
respectively at the centre angles of 108°, 60°, and
36°. Shew that
AB=BG+GR
PART II.
REAL ALGEBRAICAL QUANTITY.
CHAPTER VI.
CIECULAR MEASUEE OF ANGLES.
66. For practical purposes, as we have seen (art. 1), it
is essential that the unit of angular measurement should
be constant, easily obtained and of a convenient magni-
tude for measuring the angles most frequently in use.
For theoretical purposes, it is essential that the unit
should be constant, and that it should be so chosen that
the expressions and formulae in which it is employed
should by means of it be reduced to the simplest attain-
able form.
67. Definitions. — The length of the circumference of a
circle is the limit of the length of the perimeter of a
regular inscribed polygon when the number of sides in
the polygon is indefinitely increased.
The length of the arc of a circle is the limit of the
length of a broken-line which consists of a series of
CIRCULAR MEASURE OF ANGLES. 77
consecutive equal chords of the arc, when the number
of these chords is indefinitely increased.
Two assumptions are made in these definitions, namely :
(1) That the length of the perimeter of the inscribed polygon
(or broken-line) tends to a limit ;
(2) That this limit is the same for all inscribed polygons (or
broken-lines) when the number of sides is indefinitely increased.
For example, we may first inscribe a square in the circle, then, by
bisecting the arcs, we get regular figures of 8, 16, 32, ...sides ; or
beginning with an equilateral triangle, we may suppose regular
figures of 6, 12, 24, ... sides, or of 9, 27, 81, ... sides, to be inscribed,
or we may proceed by any other law. The assumption is that the
perimeter of the polygon tends to the same limit in all cases.
(For proofs of these j)ropositions, see Kouch^ et De Comberousse,
Geometrie Elementaire, art. 290.)
68. If two circular arcs stand upon the same straight
line, the length of the exterior arc is greater than that of
the interior.
Let AG DEB and AFOHB be two convex broken-lines
A "B
standing upon the same straight line AB. Produce AF,
FG, GH to meet the outer line in a, h, c.
Then, AG-\-Ga> AF+Fa,
Fa+aD-\-Db>FG+Gb,
Gh + hE+EoGH+Hc,
and Hc+cB>HB.
:. by addition, the length of the broken-line AGDEB is
greater than the length of the broken-line AFGHB.
78 CIRCULAR MEASURE OF ANGLES.
Now, suppose AG, CD, BE, EB and AF, FG, GH, HB
to be series of consecutive equal chords of two circular
arcs standing upon the straight line AB; then, the length
of the exterior broken-line is greater than the length of
the interior broken-line. This is true whatever be the
number of chords in the two arcs, and therefore when
the number in each is infinite, in which case, the lengths
of the broken-lines become the lengths of the correspond-
ing arcs.
Hence, the length of the exterior circular arc is greater
than the length of the interior.
Gov. — In the same way, if BAG be a circular arc
standing on a straight line BG
and if jBT, GT be the tangents
at B and (7, it may be shewn
that the length of the sum of
the tangents BT, TG is greater
than that of the arc BAG^ and the length of the arc BAG
greater than that of the chord BG.
69. The circumference of a circle varies as its radius.
Let 0, 0' be the centres of two circles of different radii,
and let regular polygons ABGD ... , A'BG'B' . . . , be in-
scribed in them, both having the same number {n) of
sides. Join OA, OB, ..., 0'A\ 0'B\ ....
CIRCULAR MEASURE OF ANGLES. 79
Then, in the triangles OAB, O'A'B', the angles AOB,
A'O'E are equal, each being the same fraction of four
right angles.
Again, the sum of the angles OAB, OBA is equal to
the sum of the angles O'A'B', O'B'A' (Eucl. I. 32) ; and
the angles OAB, OBA are equal to one another, and also
the angles O'A'B', O'B'A' (Eucl. L 5);
.-. angle 0^5=angle O'A'B'd^ndi angle 05^= angle 0'FA\
:. the triangles OAB, O'A'F are similar (Eucl. VI. 4).
AB:A'E=OA:0'A\
n.AB'.n.A'E=OA'.0'A\
.'. perim. of polygon A BCD. . . : perim. of polygon A'B'G'D'
= radius OA : radius O'A'.
This is true whatever be th^ number of sides in the
two polygons ; and it is therefore true when the number
of sides in both is infinitely great, in which case the
length of the perimeter of either polygon is the length of
the circumference of the circle in which it is inscribed,
circumf. of circle ABG : circumf of circle A'B'G'
= radius OA : radius 0'A\
i.e. the circumference of a circle varies as its radius.
70. It follows, from this proposition, that the ratio of
the circumference of a circle to its diameter is constant.
This ratio is denoted by the letter tt.
Hence, if r be the length of the radius of a circle, the
length of the circumference is 27rr.
It has been shewn that the number tt is incommensur-
able with unity.* Various methods have been employed
* For the proof of this proposition, see Chrystal's Algebra, chap.
80 CIRCULAR MEASURE OF ANGLES.
for obtaining its approximate value, several of them
being given in later chapters. By one of these methods
its value has been found to more than 700 places of
decimals. To the first 20 places it is
3-14159265358979323846.
Convenient approximate values of tt are -^ and W\.
In the first of these, the error is about ^-nnr* ^.nd, in the
second, less than TirgTroooo? ^^ ^^ ^^'^^ value.
71. Definition. — The unit of circular measure is the
angle at the centre of a circle which stands on an arc
equal in length to the radius.
This unit is called a radian.
72. The radian is a constant angle.
Let 0 be the centre of a circle of any radius (see figure
of next article), and let ^-S be an arc equal in length to
the radius. Join OA, OB.
Then, angle A OB : 4 right angles
:: arc AB : circumference of circle (Eucl. YI 33).
::r : lirr
::l:2x,
angle A OB x 27r = 4 right angles.
Hence, the angle AOB is constant, whatever be the
radius of the circle.
Cor. 1. — Since the angle AOB is one radian, it follows
from the last equation that
4 right angles = 27r radians.
2 „ =7r „
1 ,, =7r/2 „
Cor. 2— One radian = 18073-14159265... =57° 17' 45'',
nearly.
CIRCULAR MEASURE OF ANGLES.
81
73. The number of radians in any angle at the centre
of a circle is equal to the length of the arc on which the
angle stands divided by the radius of the circle.
Let AOC be the angle, 0 being the centre of a circle
of any radius ; and let AB be an arc equal in length to
the radius, so that the angle AOB is one radian.
Then, angle AOC : angle AOB
f. =arc ^0 : arc AB (Eucl. VI. 33)
= arc AG : radius.
.-. angle AOC
arc AG , . ^ „
= — T-. — X angle A OB.
radius °
.•. the number of radians in the
angle AOG= arc A (7/radius.
74. If 0 be the number of radians in an acute angle,
0 is greater than sin 0, but less than tan 0.
Let AOB be the given angle, AB the arc of a circle
with centre 0 and any radius.
Draw BN perpendicular to OA
and produce it to G so that NO is
equal to BN. Then OG is equal
to OB (Eucl. I. 4), and therefore q.
the circle of which AB is an arc
passes through G. Also, CT is
equal to BT, and GT touches the
circle (Eucl. I. 4, IIL 16).
Now, sum of tangents BT, GT > slyc BAG > chord BG
(art. 68. Cor.).
BT> SiYcAB >B]Sr.
BT/OB > arc AB/OA > BN/OB.
tan0> e >smO,
that is, 0 is intermediate between sin 0 and tan 0.
F
82 CIRCULAR MEASURE OF ANGLES.
75. If 6 he the number of radians in an acute angle,
tJie limiting values of sin 0/0 and t&n 6/0, when 0 is
indefinitely diminished, are each unity.
By the last article, we have
tan 0 > 0 > sin 0,
tan O/sin 0 > O/sin 0>1,
sec 0 > O/sin 0>1.
Now, when 0 is indefinitely diminished, the limit of
sec 0 is unity.
.*. the limit of O/sin 0, and therefore also of sin 0/0,
when 0 is indefinitely diminished, is unity.
. . tan 0 sin 0 1
Again, -^^- = — ^r— . -,
^ 0 0 cosO
and, when 0 is indefinitely diminished, the limits of both
sin 0/0 and 1/cos 0 are unity,
the limit of tan 0/0 is unity.
Hence, sin 0, 0 and tan 0 vanish in a ratio of equality.
76. If 0 he the number of radians in an acute angle,
0^ . .
cos 0 is greater than 1 — q-, ctnd sinO is greater than
'-T
We have
cose=l-2sin2|,
and
. 0^0
«^^2"^2'
hence,
cos 0> 1-2.^,
i.e.
>'-f-
Again,
0 6
sin 0 = 2 cos ^ ^^^9'
and
.0,0 0
smg^tan^cosg,
CIRCULAR MEASURE OF ANGLES. 83
hence, sin 6 = 2 tan ^ cos^^
%.e.
= 2tau|(l-sin^|)
>-f{'-(Dl.
/^ 6'
>^-4-
(art. 74)
Change of Units.
77. In this section, and in the corresponding examples,
reference will be made to another system of angular
measurement no longer in use, the centesimal system,
in which the right angle is the fundamental unit'. A
right angle is divided into 100 equal parts called grades,
a grade into 100 equal parts called minutes, and a minute
into 100 equal parts called seconds. An angle of 89
grades, 71 minutes, 47 seconds, is written 89^ 71' 47".
It will be noticed that any angle whose centesimal
measure is known can be at once expressed as a decimal
of a right angle. Thus, 89^ 71' 47'' is -897147 of a right
angle ; and 7^ 45' 3" is '074503 of a right angle.
So, also, -157423 of a right angle is equal to 15^ 74' 23",
and '000503 of a right angle is equal to 5' 3".
78. If D degrees, G grades and 0 radians be the
sexagesimal, centesimal and circular measures of the
same angle, then D/180 = G/200 = O/tt.
1 degree = 1/180 of two right angles,
.0 degrees = D/180 „
Also, 1 grade = 1/200
G^ grades = (?/200 -„
84 CIRCULAR MEASURE OF ANGLES.
And, 1 radian = I/tt of two right angles,
0 radians = O/tt „ „
But the given angle is the same fraction of two right
angles in whatever measure it is expressed,
D/18O = G^/2OO = 0/7r.
79. Example 1. — To find the centesimal measure of 23° 18' 36".
23° 18' 36" = 23° 18' -6
=23°-31
= -259 of a right angle
= 25^90\
Example 2.— To find the sexagesimal measure of 13^ 38^ 12^
•133812
13^ 38^ 12^'= -133812 of a right angle 90
= 12° -04308 12-04308
= 12°2'-5848 ^^
= 12°2'35"-088. ^^^1o
35-088
Example 3.— To find the circular measure of 12° 16'.
Let 6 radians be the circular measure of 12° 16', i.e., of 736'.
Now, ^'—TEK — ^ of two right angles
60
36
60
18-6
90
23-31
•259
12° 16'=
180x60 "" ' ~°"~'
736
180 X 60 "
and ^ radians =^/7r „ „
n_ 7367r _467r
180x60 676"
Example 4. — Find the number of seconds in the angle sub-
tended at the centre of a circle, whose radius is one mile, by an
arc 5^ inches long (tt being taken equal to 3^).
The number of radians in the angle — — 1^=
63360 11520
Let 8 be the number of seconds in the same angle,
*^^''' 180x60x6o"Tl520"^'''
/ ^_180x60x60_1575
1152077 88
- =1711 seconds.
CIRCULAR MEASURE OF ANGLES. 85
Viva Voce Examples.
Find the number of degrees in the angles whose
measures in radians are :
1. 7r/4.
9. tt/IO.
17. 47r/3.
2. x/3.
10. 7r/12.
18. 57r/12.
3. S7r/2.
11. 7r/15.
19. 77r/6.
4. 87r/4.
12. 7r/18.
20. Il7r/18.
5. 7r/8.
13. 7r/36.
21. 47r/9.
6. 7r/6.
14. 57r/6.
22. 27r/5.
7. 37r/8.
15. 57r/3.
23. 137r/9.
8. 57r/4.
16. 27r/3.
24. 77r/12.
Express
the following: ansfles in
radians :
25. r.
30. 270°.
35. 120°.
26. 30^.
31. 225°.
36. 11° 15'.
27. 15°.
82. 22° 30'.
37. 7° 30'.
28. 60°.
33. 330°.
38. 0° 30'.
29. 135°.
34. 210°.
39. 67° 30'.
Express
the
following angles
as decimals of
angle :
40. 39fi'14^24^ 43. 91^8^3". 46. 4^ 2^ r\
41. 485^2^31". 44. 4^' 15' 87". 47. 3' 2'\
42. 19^ 34' 5". 45. 15^ 3' 2". 48. 17".
Express the following decimals of a right angle in
centesimal measure :
49. -375984. 52. -051403. 55. -00003.
50. -58441. 53. 70001. 56. -91005.
51. -793602. 54. -006017. 57. '385.
86 CIRCULAR MEASURE OF ANGLES.
Examples IX. a.
1. A railway train is travelling on a circular arc of half
a mile radius at the rate of 20 miles an hour;
through what angle does it turn in 10 seconds ?
2. Find the length of an arc which subtends an angle
of 50° at the centre of a circle of radius 8 feet
(7r = 31416).
3. The length of a degree on the earth's surface in the
neighbourhood of the equator is G9'07 miles ; find
the radius of the earth, to the nearest mile.
4. On a circle, 10 feet in radius, it was found that an
angle of 22° 30' at the centre was subtended by an
arc 3 feet 11 J inches in length ; hence find the
value of TT to three places of decimals.
5. Prove geometrically that a radian is less than 60°.
6. Find approximately the distance of a tower whose
height is 51 feet, and which subtends at the eye
an angle of h^\' (tt = 3|).
7. By considering an angle of 15°, shew that tt is greater
than 310 and less than 322 (art. 74).
8. Find the limit of ^inn^jn, when n is indefinitely
diminished.
9. Find the limit of -^ir^tan-, when n is indefinitely
increased.
10. Find the sine of 1° to 5 places of decimals, by means
of art. 76 (7r = 314159).
11. If TYh and fjL be the numbers of sexagesimal and
centesimal minutes in the same angle, find the
relation between m and />t.
12. Find the sexagesimal measures of the following angles :
(1) 12^ 47^ 93". (2) 0^ 1^ 2". (3) 87^ 0^ 13".
CIRCULAR MEASURE OF ANGLES. 87
13. Find the centesimal measures of the following angles:
(1) 29°14'15"'. (2) 75° 47' 22". (8) 43° 19' 20''.
14. Find the number of radians in the following angles :
(1) 35° 1' 12". (2) 47^ 0^ 30".
Examples IX. b.
1. Find the circular measure of the angle subtended at
the centre of a circle, whose radius is one mile, by
an arc 22 inches long.
2. If the radius of a circle be 4000 miles, find the
length of an arc which subtends at the centre an
angle of ^ radians.
3. A rail way- train, travelling due north, begins to move
on a line in the form of a circular arc. After
travelling 5 J miles, its direction of motion is N.W.
Find the radius of the circle (7r = 3^).
4. If an arc of 6*283 inches subtends an angle of 30° at
the centre of a circle one foot in radius, find the
value of TT to two places of decimals.
5. Prove that tt is greater than 3 and less than 2^3,
by considering the lengths of the perimeters of
regular hexagons described in and about a given
circle.
6. Find approximately the diameter of the sun, which
subtends at the earth an angle of 32' 1", its dis-
tance from the earth being 91 million miles
7. By considering an angle of 18°, shew that tt is greater
than 309 (art. 74).
8. Find the limit of tan7i°/7i, when n is indefinitely
diminished.
88 CIRCULAR MEASURE OF ANCLES.
9. Find the limit of Jnr^in— , when n is indefinitely
increased.
10. Find the sine of 3° to 3 places of decimals, by means
of art. 76 (7r = 3-14159).
11. If s and (T be the numbers of sexagesimal and
centesimal seconds in the same angle, find the
relation between s and o-.
12. Find the sexagesimal measure of the following angles:
(1) 53^ 81^ 7". (2) 5^ 12^ 9". (3) 0^ 0^ 3".
13. Find the centesimal measure of the following angles:
(1) 5° 3' 14". (2) Sr 13' 17". (3) 0° 0' 5".
14. Find the number of radians in the following angles :
(1) 87° 37' 15". (2) 1^ 25' 4".
Examples X.
1. What must be the unit angle, if the sum of the
measures of a degree and grade be 3 ?
2. What unit is such that the number of units in a
radian is equal to the circular measure of a
grade ?
3. Divide the angle 77° into two parts, so that the
number of sexagesimal minutes in one part may
equal the number of centesimal minutes in the
other part.
4. The number of degrees in an angle is n times the
number of minutes in its complement; find the
number of radians in the angle.
5. The angles of a triangle are in arithmetical pro-
gression, and the greatest contains as many
degrees as the smallest contains grades ; find the
angles in degrees.
CIRCULAR MEASURE OF ANGLES. 89
6. Divide a right angle into two parts so that the
number of grades in their difference may be
equal to the number of degrees in the whole
angle.
7. The difference between two angles which contain the
same number of degrees and grades, respectively,
is -^ radians ; find the angles.
8. One angle of a triangle is 45°, and another is IJ
radians. Find the third, both in degrees and
radians (7r = 8f).
9. The angles of a quadrilateral are in arithmetical pro-
gression, and the difference between the greatest
and least is a right angle. Find the number of
degrees, and also the number of radians, in each
angle.
10. If the circumference of a circle be divided into
five parts in arithmetical progression, the greatest
part being six times the least, express in
radians the angle which each subtends at the
centre.
11. The angles of a triangle are in arithmetical pro-
gression, and the ratio of the number of radians
in the least to the number of degrees in the mean
is 1 : 120. Find what multiple of a right angle
is the greatest angle.
12. The perimeter of a certain sector of a circle is equal
to the length of the arc of a semicircle having the
same radius. Express the angle of the sector in
degrees, etc. (tt = 8^).
13. Prove geometrically that cos 0>1-^.
90 CIRCULAR MEASURE OF ANGLES.
14. The angles of a triangle, when referred, in ascending
order of magnitude, to three units in geometrical
progression, are represented by numbers in arith-
metical progression. The mean angle is equal to
the sum of the first two units or to half the sura
of the last two, and the greatest angle is four
times the mean unit. Determine the angles.
CHAPTER VII.
GENEKAL DEFINITIONS OF THE CTECULAR
FUNCTIONS. FORMULA INVOLVING ONE VARIABLE
ANGLE.
§ L Definitions.
80. Negative Lines. — A straight line may be generated
by the motion of a point. The point may move in either
of two opposite directions or senses.
The sense of a line is denoted by the order of the
letters. Thus, " the line AB " means " the line generated
by a point moving from A to B!'
If one of the two senses in which a line is generated
be regarded as positive, the opposite sense is said to be
negative. Thus AB and BA are equal in magnitude but
opposite in sense, and, if AB be regarded as positive, BA
is equal to AB in magnitude and is negative. Hence,
AB+BA=0, or AB=-BA.
A C B B A O
If A, B, C be points in any order in the same line, then
AB = AC-\-GB in all cases.
81. Negative Angles.— An angle may be generated by
the rotation of a line about a fixed point. The line may
rotate in either of two opposite dii'ections or senses.
91
92 DEFINITIONS OF THE
The sense of an angle is denoted by the order of the
letters. Thus, "the angle AOB" means "the angle gen-
erated by a line rotating about 0 from OA to OB."
If one of the two senses in which an angle is generated
be regarded as positive, the opposite sense is said to be
negative. Thus, the angles AOB and BOA are equal in
magnitude but opposite in sense, and, if AOB be regarded
as positive, BOA is equal to AOB in magnitude and is
negative. Hence, lAOB-^-lBOA = 0, or lAOB= -lBOA.
If OA, OB, OC he lines drawn from 0 in any order,
then lAOB==lAOG-\-lCOB in all cases.
82. The positive sense of a line is arbitrary, and must
therefore be defined explicitly or implicitly in each case
considered.
The positive sense of an angle is likewise arbitrary,
but, for the purposes of Plane Trigonometry, it is sufficient
in all cases to fix as the negative sense that in which the
hands of a watch placed face upwards on the plane rotate.
Hence, in all cases of rotation, positive and counter-
clockwise are equivalent terms, and so also negative and
clockwise are equivalent.
83. Def. — The foot of the perpendicular from a point
on a line is called the 'projection of the point on the line.
If the point be on the line, the point and its projection
coincide.
CIRCULAR FUiVCTIOJVS.
93
Def. — The intercept between the projections of the
ends of a line on another line is called the projection of
the first line on the second.
The projection of a line is to be considered with regard
to magnitude and sense.
The sum of the projections of the parts of a continuous
B
b d
broken-line is equal to the projection of the line joining
its extremities. Thus,
ah-\-hc-\-cd = ad.
Hence, the sum of the projections of the sides of a
closed polygon is zero.
84. Defs. — Let a line rotate about 0 from OX through
any positive or negative angle a to the position OA ; let
X'
if>
o
N
M
X'
N
/
T
M
X
OF be a line making an angle |^ in the positive sense
with 0X\ and letO^, OX, OF be the positive senses of
94 FUNDAMENTAL PROPERTIES OF THE
the lines OA, OX, OY. Let a length OP, of any mag-
nitude and of either sense, be measured along OA ; and
let OM, ON be the projections of OP on OX, OY,
respectively.
The ratio OM : OP is called the coaiTie of the angle a,
ON : OP the sine of a, OiV^ : OM the tow^eTi^ of a, OP : Oif
the secant of a, OP : ON the cosecant of a, and Oilf : OJV
the cotangent of a.
These ratios are called the Circular Functions of the
angle a.
Two other ratios are occasionally used, and are defined as follows :
If the length OP be equal in magnitude to OX, and positive in
sense, and if OY=OX, the ratio MX : OP is called the versine of a,
and NY : OP the coverdne of a.
The abbreviations of these ratios are vers a and covers a.
§ 2. Fundamental Properties of the Circular
Functions.
85. Each circular function of a given angle has one
value only.
Consider the cosine of a given angle XOA or a, and
suppose that OP, Op are lengths of any magnitude and
either sense measured along OA. We shall shew that
OM : 0P= Om-.Op, where OM and Om are the projections
of OP and Op on OX.
The angles POM, pOm are equal, and the right angles
PMOy pmO are equal, therefore the triangles POM, pOm,
CIRCULAR FUNCTIONS.
95
X"
m M X
F'
are similar (Eucl. YI. 4), and therefore, regarding the
magnitude only of the lines,
0M:0P==0m:0p.
Again, if P and ^ are on
the same side of 0, OP and
Ojp are of the same sense, and
so also are OM and Om ;
while, if P and p are on op-
posite sides of 0, OP and Op
are of opposite sense, and so
also are OM and Om. Hence,
regarding the sense as well as the magnitude of the lines,
0M:0P = 0m:0p,
i.e., the cosine of a has one definite value independent of
the magnitude or sense of the length OP.
In like manner, it may be shewn that sin a, tan a, sec a,
etc., are one- valued functions of a.
86. Signs of the Circular Functions. — If an angle lies
between 0 and ^, it is said to be in the first right angle ;
if between J and tt, in the second right angle ; and so on.
TT
If an angle lies between 0 and — ^, it is said to be in the
first negative right angle ; if between — ^ and — tt, in the
second negative right angle ; and so on.
Referring to the diagrams of art. 84, and supposing
OP to be measured in the positive sense in all cases, we
see that :
When a is in the first right angle, OM and ON are
positive, and therefore all the ratios are positive ;
When a is in the second right angle, OM is negative
96 FUNDAMENTAL PROPERTIES OF THE
and ON positive, and therefore the sine and cosecant are
positive, and the other ratios negative ;
When a is in the third right angle, OM and ON are
both negative, and therefore the tangent and cotangent
are positive, and the other ratios negative ;
When a is in the fourth right angle, OM is positive
and ON negative, and therefore the cosine and secant are
positive, and the other ratios negative.
Thus, the succession of signs is as follows :
cosine and secant, + — — +
sine and cosecant, + + — —
tangent and cotangent, + — + —
87. Values of the Functions of 0, ^, tt, etc.— Let the
angle a (see fig. of art. 84) have the series of values
0» 9> '^f IT' ^"^^ •••' ^^®^ ^^ ^^^ ^^® series of values
OP, 0, -OP, 0, OP, ..., and ON the series 0, OP, 0,
— OP, 0, .... Hence, we see that
cosO = l, cosj=0, cos7r=-l, cos~=0, cos27r=l,...
sinO=0, sin- = l, sm7r=0, sin— =-1, sin27r=0,...
tan 0=0, tan ^= GO, tan7r=0, tan— =00, tan27r=0, ...
secO = l, sec^=oo, sec7r=-l, sec — = qo, sec27r=l, ...
2t 2
cosecO = oo, cosec^=l, cosec7r=Qo, cosec— =-1, cosec27r=oo, ...
Z 2
cotO=oo, cot^=0, cot7r=Q0, cot-J = 0, cot27r=oo, ...
Example. — Trace the changes in the sign and magnitude of
sin 0, as 0 increases from 0 to 27r :
As d increases from 0 to ^, sin Q is positive, and increases from
Otol:
CIRCULAR FUNCTIONS. 97
As ^ increases from - to tt, sin 9 is positive, and decreases
from 1 to 0 ;
As 0 increases from tt to -^, sin 6 is negative, and increases
numerically from 0 to - 1 ;
As 9 increases from ~ to Stt, sin 9 is negative, and decreases
numerically from - 1 to 0.
88. Periods of the Circular Functions.— If an angle
d pass in order through a series of values from 0 to 27r,
the cosine of 0 passes through a series of values ranging
from +1 to — 1, and returning from — 1 to + 1 ; if 0
pass in order through a second series of values from 27r
to 47r, the values of cos Q recur in the same order as in
the first series ; and this series of values of the cosine is
continually repeated after each complete revolution of
the generating line of the angle 0. Thus, if a, a + 'lnir
be values of 6 differing by an}' multiple of 27r, then
cos a = cos (a + Stitt). From this property of the recur-
rence of its values, the cosine is called a periodic function
of the angle, and the increment Stt of the angle after
which the values of the cosine recur is called the period
of the cosine.
In like manner, sin 0, sec 6 and cosec 0 are periodic
functions of 0 of period 2x.
In the case of the tangent or cotangent, the values
recur after an increment tt of the angle, hence tan 6 and
cot 6 are periodic functions of 0 of period tt.
In any given period, a given value of cos 6, sin 0,
sec 0 or cosec 0 corresponds to two values of 0, while
a given value of tan 0 or cot 6 occurs once only in each
period.
98
FUNDAMENTAL PROPERTIES OF THE
89. Continuity of the Circular Functions.— Def. If a
circular function of an angle be such that, as the angle
increases from one given value to another, an infinitely
small change in tlie function corresponds to an infinitely
small change in the angle, then the function is said to be
continuous between those two given values of the angle.
For example, if 6 and O' be two values of an angle,
and if we can shew that cos 0 r^ cos $' is infinitely small
when 0 '-' 6' is infinitely small, then cos 0 will be a
continuous function of 6.
If 0 and 6' be in the same right angle, we need only
consider the case in which 0 and S' are in the first right
angle ; for, if 0 and 0' be in any other right angle, the
only difference is in the sign of the function and the
order in which it passes through its diflferent numerical
values. If 6 and 6' be in consecutive right angles, as
might be tlie case when 0 is less than, but very nearly
equal to, a multiple of Jtt, it will be seen that the same
reasoning holds true.*
Let the line OP rotate through a right angle from the
position OA, in the positive sense,
to the position OB ; and, from any
intermediate position OP, through
an infinitely small angle, in the
same sense, to the position OP'.
Let 0, 0' denote the angles AOP,
AOP\ Draw PM, P'M perpen-
dicular to OA, and PL perpendic-
ular to P'M'.
* In this case, however, in considering the continuity of the tangent
and cotangent, the formula for sin {&' - 6) is supposed true when 0' is
greater than 5" ; this is proved in the next Chapter (art. 1 10).
CIRCULAR FUNCTIONS, 99
Cosine and Sine. — We have, from the figure,
OM^OM' MM LP
cos 0 — COS 0' =
OP ~OP~OP
PP' SiYcPP' . ^a' a
<OP^-^P'''-''^^-^'
Since 6' — 0 is infinitely small, it follows that cos 0
— cos & is infinitely small ; and this is true for all values
of the angle 0 between 0 = 0 and 0 = Jtt.
Hence, the cosine is continuous for all values of the
angle 6.
Similarly, the sine is continuous for all values of the
angle Q.
Secant and Cosecant — Again,
a' a nv( ^ 1 ^ OP. MM
sec 0 — sec 0 = 0P\ —yt, —
UM OMJ ~ OM' OM
OP . arc PP' . OP^ arcPP'
'^ OM'.OM ''^'^'^OM .OM' OP
OP^
Since O' — O is infinitely small, sec 0' — sec 0 is so also,
provided OM' . OM is not infinitely small. If, however,
0 be very nearly equal to Itt, we can make OM' . OM
comparable in magnitude with OP^(0' — 6), or less than
this, if we please, by making the difference between 6
and Jtt small enough. In this case, it does not follow that
sec 0' — sec 6 is infinitely small. We know, indeed, that
it is not, for, as 0 passes from a value a little less than
Jx up to Itt, sec 0 changes from a quantity that is finite
to one that is infinitely great.
Hence, the secant is continuous for all values of the
angle 0, except those which differ by an infinitely small
100 FUNDAMENTAL PROPERTIES OF THE
quantity, either in defect or excess, from an odd number
of right angles.
Similarly, the cosecant is continuous for all values of
the angle ^, except those which differ by an infinitely
small quantity, either in excess or defect, from zero or
any even number of right angles.
Tangent and Cotangent — Lastly,
, ^ , ^ sin 0" sin 0
tan Q —tan Q = yr, ^
cos Q cos Q
_sin ^cos 0 — cos Q' sin 0_sin {B' — 0)
~ cos & COS 0 ~ cos 6' cos 0
.-. tan0'-tan0< -^ A^-0).
cos V COS d
Since O' — O is infinitely small, tan 0' — tan ^ is also
infinitely small, provided cos & cos 0 is not infinitely
small, i.e. provided 0 is not very nearly equal to Jtt.
Hence, as in the case of the secant, the tangent is
continuous for all values of the angle 0, except those
which differ by an infinitely small quantity, either in
defect or excess, from an odd number of right angles.
Similarly, the cotangent is continuous for all values of
the angle 0, except those which difter by an infinitely
small quantity, either in excess or defect, from zero or
any even number of right angles.
90. Fundamental Formulae.— The following formulae
may be obtained immediately from the definitions of
art. 84, as in arts. 8, 9 : ^ ::
f cos 0 sec 0 = 1, sin Q cosec 0 = 1, tan 0 cot 0 = 1 ;
. ^ sin 0 , r. cos d
^ tan0 = ^, cotO = ^— ^. r
cos Q sm 6
We have, also, by Eucl. I. 34 and 47,
OM''+ON^=OF^;
CIRCULAR FUNCTIONS. 101
whence, dividing by OP'^, OM'^, and ON"^, we get, iu
succession, (^ cos^^4-sin20=l,
? l + tan''^0 = sec2e,
'> cot20+l=cosec2a "
Of these eight formulae, five only are independent ; the
fifth, seventh, and eighth may easily be obtained from
the other five.
"We have, also, from the definitions of Art. 84,
n MX OX-OM , OM . n
-"''^^^ OP^—OF- = ^- 0P = ^-'''^^
-, n NY OF- ON , ON , . n
and covers ^=-^-=—^^^ =l-_p = l-sm^.
Viva Voce Examples.
What are the signs of:
1. sin 120°.
10.
11.
12.
13.
cosjf.
15.
cot^.
4
. Sir
sm -p.
4
16.
27r
sec-—.
a
tan^=-.
17.
cos-^.
. ox
sm~.
18.
tan-^.
4
2. cos 150°.
3. tan 210°.
4. cot 300°.
5. sec 260°.
6. cosec 250°.
7. cos 320°.
8. tan 140°.
9. sin 245°. .. «^„^« ^"^
14. cosec -J-.
4
19. Trace the changes in sign and magnitude of cos 6 as
0 increases from 0 to 2x.
20. Trace the changes in sign and magnitude of tan 0 as
0 increases from ~ tt to + tt.
102
FUi\DAMENTAL VROPELiTIES OF THE
e
§ 3. Reduction of Functions of
91. Even and Odd Functions.— If, in any integral
rational expression involving even powers only of a
x^ x^
variable x, such as I — ^-f ^, we change x into —x, the
expression is unaltered in magnitude and sign; if the
OC/ X
expression involves odd powers only, such as a;- rr + j^y
its magnitude is unaltered but its sign reversed by the
reversal of the sign of the variable. Hence, by a simple
extension of the meaning of the words even and odd, we
have the following definition :
Def. — l{f{ — x) be equal to f(x) for all values of x, then
f(x) is said to be an even function of x; and if /( — a?) be
equal to -f(x) for all values of x, then f{x) is said to be an
odd function of x.
92. To prove that cos 0 and sec 0 are even functions of
0, and that sin 0, cosec 0, tarn 0, and cot 0 are odd func-
tions of 0.
Let a line rotate about 0 from the position OX, through
an angle 6 to the position OF ;
also, let a line rotate from OX
through an angle equal to 0
in magnitude and of contrary
X sense, to the position Op ; then
OP and Op have the same pro-
jection OM on XX\ while their
projections on YT, namely, ON"
and On, are in all cases equal
r
/
0^
ir^^^^
/
o
P
\^_
[_^^
y
in magnitude and of contrary sense.
Hence, cos 0 =-rrT, = -it- = ^^K ~ ^)'
OJr Up
. . /, ON —On . / m
and sm0=^^= ^^— -=-sin(-0),
OP Oj9
Similarly, sec 0 = sec( — 0), cosec 6= — cosec( — 0),
tan0=-tan(-0), cot 0= -cot(-O).
93. To _proi;e that cos\Q-{-'^] = —sin 0.
Let a line rotate about 0, from the position OX, through
Y
an angle 0, to the position OP ; let it further rotate from
the position OP, through an angle ^ in the positive sense,
to the position OQ ; then angle X0P = 6 and angle
Leti M be the projection of P on OF, and N that of Q
on XX\
Then, in the triangles POM, QON, the angles POM and
QON are equal (since POQ and ifOiV are right angles),
the right angles PMO and QNO are equal, and OP = OQ;
therefore, the triangles are geometrically equal in every
respect, and, therefore, OM=ON in magnitude.
104
FUNDAMENTAL PROPERTIES OF THE
Again, since OQ is always a right angle in the positive
sense in advance of OP, therefore, when P is on the same
side of XX' as F, Q is on the opposite side of YY' to X ;
and, therefore, when OM is positive, ON is negative. In
like manner, when OM is negative, ON is positive.
Hence, in magnitude and sense, ON— — OM,
■ m^_qM
OQ OP'
{e+|)=-sina
cos
94. To prove thit dniO + 1-) = (^os 0.
Let a line rotate about 0, from the position OX,
Y' Y'
through an angle Q, to the position OP ; let it further
rotate from the position OP, through an angle ^ in the
positive sense, to the position OQ, ; then, angle XOP = 6
and angle XOQ = 0-\--^-
Let M be the projection of P on XX\ N that of Q
on YT.
Then, in the triangles POM, QON, the angles P031,
QON are equal, since POQ and AWN are right angles,
CIRCULAR FUNCTIONS. 105
the right angles PMO, QNO are equal, and OP = OQ;
therefore the triangles are geometrically equal in e very-
respect, and therefore Oili = OiV in magnitude.
Again, since OQ is always a right angle in the positive
sense in advance of OP, therefore, when P is on the
same side of YY' as X, Q is on the same side of XX'
as F; and therefore, when OM is positive, OiV is also
positive. In like manner, when OM is negative, ON is
also negative. Hence, in magnitude and sense, 0N= OM.
ON_OM
OQ~OP'
sin( 0+« ) = cos0.
95. Since, cosf 0 + |^J= —sin Q and sinf 0 + |^j = cos 0,
( 0 + ^ ) = 7 r = : — ^ = - cosec 0,
V -1) ^^^n^ -sm0
sec
cos
A , fn^A '^'"V^"^27 cose ,.
COS
So, also, cosecf 0 + ^j =sec 0,
and cot f 0 + 1^ j = — tan 0.
96. By arts. 93-95, we can express any circular
function of an angle in terms of a function of an angle
less or greater than the original angle by a right angle,
and, therefore, by repeated operations, we may diminish
or increase the angle by any multiple of a right angle ;
106 FCyDAMENTAL PROPERTIES OF THE
by art. 92 we may change the sign of the angle ; hence,
we can express the circular functions of an angle n^ ±6,
where n is an integer, in terms of a function of 0.
Thus, we have
sinf ^ — 0 j = cos( — 0) = cos 0,
sm{7r — 0) = cos(~ — 0]=: — sin( — 0) = sin 0,
and so on.
97. From the above propositious, we see that, when
n is even, any circular function of (n^±0] is equal in
magnitude to the same function of 0; and that, when
71 is odd, the sine, tangent and secant of the one angle
are respectively equal in magnitude to the cosine,
cotangent and cosecant of the other.
The relation between the signs of the functions of the
two angles can be readily seen by inspection of the
figure in which ^ is a positive acute angle.
Example 1.— Simplify cos(180° - a).
Here, the number of right angles is even^ therefore we retain the
cosine; in the simplest case i80°-a is in the second right angle,
and therefore cos(180° - a) is negative ;
cos(180° - a)= - cos a.
Example 2. —Simplify sin(270° - a).
Here, the number of right angles is odd. and the sine is negative
in the third right angle,
sin(270° - a) = - cos a.
Example 3.— Simplify tan 210".
Rejecting two right angles, and observing that the tangent is
positive in the third right angle, we get
tan2I0° = tan30°=-L,
v3
CIRCULAR FUNCTIONS. I07
Viva Voce Examples.
Simplify :
1. sin (180° - a). 14. cos (270° + a).
2. tan (180° -a). 15. sin (270° + a),
a sec (180° -a). _ . /x^
4. cot (180° -a). ^^' ^^H2+«
5. cosec (180° — a)
6. sin(7r + a).
7. tan(27r-a).
8. tan(7r + a).
9. sec (27r-a). ^^ . /Stt
1"
17. tan(|^ + a).
18. cot(^-a).
in «^. / I \ 19. sm (-^ + a
JO. cos (-TT + a). \ 2
11. sin(90° + a). 4'ur^o. _. /Stt \
12. cos(90° + a). -^. 20. sec^-^-aj.
13. tan (270° -a).
Find the values of:
21. cos 150°. :V
22. sin 150°. '^^- ^^^ T'
23. tan 120°. . 97r
24. cot 135°. '^^- ^'" To*
25. sec 135°. . IItt
26. sin 210°. '^^- ""^^ 10 ■
27. cos 300°. .,^ 77r
28. cot 330°. 37. cos ^.
29. sec 240°. qq . ^tt
30. cosec 225°. ^^' *^^" 4 *
31. tan ^. 39. cot ^.
4 o
32. sin ?J. 40. sec ^^
33. cos ,
o
3 3
57r
108
FUNDAMENTAL PROFEHTIES OF THE
98. All the formuloe connecting the functions of
71+0 with those of 6 may be proved directly from a
diagram for all values of 0 by the method of arts. 92-94.
Example 1. — To prove that sin(180°- a) = sina.
Let a line rotate about 0 from the position OX, through an
Y angle a, to the position OP ; also,
let a line rotate from OX through
180° in the positive sense to the
position OX', and then let it further
rotate from OX', through an angle
X'\ ZP^ 1^' <^f magnitude a and of contrary
sense to a, to the position OP'
so that angle XOP = a and angle
XOP' = 180" -a] then OP and OP'
have the same projection ON on
y YT,
sin(180° - a) = sin a.
Example 2. — To prove that cos(a+18()°;= -cos a.
Let a line rotate about 0 from the position OX, through an
angle a, to the position OP ; also, let it further rotate from the
0 \
\^
^\/
P'\
A^ >
position OP, through an angle of 180° in the positive sense, to the
position OQ, so that angle XOP=a and angle XOQ = a+ 180°.
Let M be the projection of P on XX'.
CIRCULAR FUNCTIONS.
109
Regarding P as a point in OQ, we have
cos(a+180°)=y-^, where OP is negative,
and, regarding P as a point in OP, we have
cos a=77pj where OP is positive,
cos(a + 1 80°) = - cos a.
Example 3. — To prove that cot(270°-a)=tan a.
Let a line rotate about 0 from the position OX, through an
angle a, to the position OP; also, y
let a line rotate from OX, through
an angle 270° in the positive sense,
to the position OY', and then let it
further rotate, through an angle
equal in magnitude to a, but of x'\ ^ ^ ^,X
contrary sense, to the position Op.
Let M, m be the projections of
P, p on XX', N, n the projections
on YY', then 0?n = OiV'and On=OM
in magnitude.
Also, since angle Y'Op= -angle XOP, we see that p crosses the
line YY' when P crosses XX' ; hence Om= - ON.
Similarly, On=-OM,
.^f/o^A° N Om -ON ON ,
cot(270 -«)-^-ZCT=^^=tana.
X
"'^
"^^
P
M
^
\
\
/
0 '« 1
\
/
A^^^^X
/
P
^-~-
§ 4. Inverse Functions.
99. Defs. — If cos 0 = a, the angle 6 is called an inverse-
cosine of a. Thus, if cos0 = J, any one of the angles
TT Stt Vtt tt' Stt 77r
3' 3 ' 3
3'
. . ., is an inverse-cosine of J.
110 J'UNDAMENTAL PROPERTIES OF THE
The symbol Cos"^a is used as an abbreviation of the
words " inverse-cosine of «," or " any angle whose cosine
is a," the capital letter being used here, and in other
similar cases, to indicate that the symbol is many-
valued.
Similar definitions may be given of the inverse-sine,
inverse-tangent, etc., and a similar notation will be em-
ployed ; thus, Tan-^1 denotes any one of the group of
, TT Stt Qtt Stt Ttt IItt
angles -r, -j-, -j-, .... — r-, — j-, r-, ....
o 4'4'4' ' 4' 4 4'
It is convenient to define some one of the group of
angles denoted by an inverse circular function as the
piHncipal value of that inverse function. We select as
the principal value the numerically least angle, taking
the positive one when there are two of equal numerical
value. Tt is evident from the definitions of art. 84, or by
considering the curves of the functions (see art. 107), that
the principal values of Cos"^a, and Sec"% lie between 0
and TT, while those of Sin " ^a, Tan ~ ^a, Cosec " ^a, and Cot " ^a
lie between — ^ and -^.
The symbol cos-^a is used as an abbreviation of the
words " principal value of the angle whose cosine is a,"
the small initial letter being used to indicate that the
symbol is one-valued. Similarly, the principal values of
Sin~%, Tan"^a, etc., are represented by sin~%, tan "%, etc.
Thus, tan-il=^, cos-i(- J) = ^|^, sin-i(-i)= -|.
co8ec"^^2 = Y, tan-^(-f-x)= , tan-^( — oo ) = — ^,
TT
cot-K + 0) = |, cot-V-0)=-|.
CIRCULAR FUNCTIONS.
Ill
100. To find an expreasion for all angles which have a
given cosine.
Let a be the given cosine.
Describe a circle with centre 0, and radius OX equal to
unity. On OX take a length
OM equal in magnitude and
sense to the given quantity a.
Through M draw the chord
PP' at right angles to OX, and
join OP, OF.
Then all angles bounded by
OX and OP, and all angles
bounded by OX and OP', and
no other angles, have their cosines equal to a.
If a be one of the angles bounded by OX and OP, then
since any other of the angles bounded by these lines
differs from a by a multiple of four right angles, we see
that all the angles bounded by OX and OP are included
in the formula 2ti7r + a, (1)
where n is zero or any integer positive or negative.
Again, since the angles XOP, XOP' are geometrically
equal and of conti*ary sense, it follows that one of the
angles bounded by OX and OP' is —a, and consequently
all the angles bounded by OX and OP' are included in
the formula 2nir — a, (2)
where 71 is zero or any integer positive or negative.
The formulse (1) and (2) are together equivalent to the
single formula ^nir ±a (o)
Got. 1. — If a — cos"^a, the theorem becomes
Cos " % = 2')i7r ± cos " ^a.
Cor. 2. — If a=\,P and P' coincide with X, and we get
Cos-n = 27i7r.
112
FUNDAMENTAL PROPERTIES OF THE
2n7r±-^ or Cos-i0 = m'7r+2.
If a= — 1, P and P' coincide with X\ the other ex-
tremity of the diameter through X, and we get
Cos-i(-l)=(27i + l)7r.
If a = 0, P and P' coincide with Fand Y\ and we get
Cos -10:
Cor. 3. — In like manner, if a be one of the angles which
have a given secant a, we can prove, by taking 0M= 1/a,
that all the angles with the given secant are included in
the formula Snx ± a.
101. Example.— Solve the equation
sec ^- 2 cos ^=1.
Multiply both sides of the equation by cos ^,
then l-2cos2^=cos^.
2cos2^+cos^-l=0,
(2cos6>-lXcos6'+l)=0,
cos 6=^ or -1,
0 =27117 ±l7r or (2n+l)7r.
102. To find an expression for all angles which have
a given sine.
Let a be the given sine.
Describe a circle with centre 0, and radius OX equal
to unity. Draw YY' the dia-
meter at jight angles to XX\
On 0 F take a length OJN' equal
in magnitude and sense to the
X given quantity a. Through N,
draw the chord PQ parallel to
OX, and join OP, OQ.
Then, all angles bounded by
OX and OP, and all angles
bounded by OX and OQ, and no other angles, have their
sine equal to a.
CIRCULAR FUNCTIONS. 113
If a be one of the angles bounded by OX and OP, then,
since any other of the angles bounded by these lines
differs from a by a multiple of four right angles, we see
that all the angles bounded by OX and OP are included
in the formula 2m7r + a, (1)
when m is zero or any integer positive or negative.
Again, since the angles XOP, X'OQ are geometrically
equal and of contrary sense, it follows that one of the
angles bounded by OX and OQ is tt — a, and, conse-
quently, all the angles bounded by OX and OQ are
included in the formula
^mTT+TF-a (2)
where m is zero or any integer positive or negative.
The formulae (1) and (2) are together equivalent to the
sentence ; " Take any multiple of tt and add or subtract
the angle a according as the multiple of tt is even or odd";
or, in symbols, 7i7r+( — l)^a (3)
Cor. 1. — If a = sin"%, the theorem becomes
Sin ~ ^a = riTT + ( -- 1 )"sin - ^a.
Gov. 2. — If a = l, P and Q coincide with F, and we get
Sin"il = 27i7r+|.
If a= — 1, P and Q coincide with Y', and we get
Sin-i(-l) = 27i7r-^.
If a = 0, P and Q coincide with X and X\ and we get
Sin-i0 = 7i7r.
Cor. 3. — In like manner, if a be one of the angles which
have a given cosecant ct, we can prove, by taking OH
equal to I/a, that all the angles having the given cosecant
are included in the formula
H
114
FUNDAMENTAL P/iOPE/tT/FS OF THE
103. Example.— Solve the equation
cosec ^ — 4 sill $=%
Multiply both sides of the equation by sin 6,
theu, 1-4 sin2^= 2 sin d,
4sm2^+2siii(9-l=0, "^l**^*
^ .w ^,.
6^
-2±v^+16
8
±J5-1
104. To find an expi^ession for all angles which have
a given tangent
Let a be the given tangent.
Describe a circle with centre 0, and radius OX equal to
unity. Through X draw TT
at right angles to OX, and let
XTy the direction in which the
intersection of TT and a line
rotating about 0 in the posi-
tive sense would move, be the
positive sense of TT. On XT
take a length XA equal in
magnitude and sense to the
given quantity a. Through A draw the diameter PP'.
Then, all the angles bounded by OX and OP, and all
the angles bounded by OX and 0P\ and no other angles,
have their tangent equal to a.
If a be one of the angles bounded by OX and OP, then,
since any other of the angles bounded by these lines
diflfers from a by a multiple of four right angles, we see
that all the angles bounded by OX and OP are included
in the formula 2m7r + a, (1)
where m is zero or any integer positive or negative,
CIRCULAR FUNCTIONS. 115
Again, since the angles XOP, X'OP' are geometrically
equal and of the same sense, it follows that one of the
angles bounded by OX and OP' is ir + a, and, conse-
quently, all the angles bounded by OX and OP' are
included in the formula
2m7r + 7r + a, (2)
where m is zero or any integer positive or negative.
The formulse (1) and (2) are equivalent to the single
formula nir + a (3)
Gov. 1. — If a = tan-^a, the theorem becomes
Tan - hi = nir-\- tan " ^a.
Cor. 2. — If a =00,
IT
Ifa = 0, Tan-i0 = 7i7r.
Cor. 3. — In like manner, if a be one of the angles which
have a given cotangent ot, we can prove, by taking XA
equal to 1/a, that all the angles with the given cotangent
are included in the formula
nir + a.
105. Example 1. — Solve the equation
V3 tair(9+ 1 =(1 +v/3) tan ^.
We have v^3tan2(9- (1 +V3)tan ^+1=0,
(V3tan(9-lXtan(9-l)=0,
tan 6= — or 1,
\'3
0=9177 + '^ or mr + '^.
6 4
Example 2. — Write down the four smallest angles which satisfy
the equation 3 cot-^ -1=0,
Since cot^=±-)-,
V3
the four smallest values of 9 are ^, -"^ ^, - ^^.
3 3 3 3
116 FUNDAMENTAL PROPERTIES OF THE
106. In translating formulae expressed in terms of the
direct circular functions into the notation of the inverse
functions, attention should be paid to the many-valued
nature of the inverse functions.
The following relations are always true :
sec~iic = cos"i-, cosec"^ir = sin-^-, cot"^ir = tan"^-,
X XX
cos"^a;+sin"'^a;= »-, sec"^i:c + cosec'^aj = — ,
but , tan~^a;+cot"%=,. or — ^,
according as x is positive or negative.
Again, from the formula cos20 + sin'^O = l, we deduce, by
putting sinO = aj, and taking the positive value of v 1 -^^
sin-ia; = cos"^>v/l— a;^ when x is positive,
and sin"^a;= —q,o^~'^s/\—x^, when x is negative.
Similarly, if cos Q = x, we have
cos"^a; = sin~^/v/l— ic^ when x is positive,
and cos"^ic = 7r — sin"i>v/l— a;^, when x is negative.
Vivi Voce ExAii4>LES.
State, in degrees, the value of :
1. cos-U. 8. tan-X-V3). 17 cofiC— ^-
2. sin-ij. 9. sec-ix. ' V ^3
3. tan-i(-l). 10. sin-(-l). ' ^3 ^^^.,1
^ 1 11. cot-loo. ^2
• ''' JS- 12. cosec-(-l). 10, eos-<-4-\
5. sec-i(-l). 13. sec-n. ' \ J'^
1 4 14 tan-V-x). 20. sin" i(- J)
6. cosec-i /- T -,- . n oi ^ 1/ .
V^ — 1 lo. sin-U. 21. tan-X + 00
7. cos-Y-l). 16. cosec-V^- 22. cot-i( + 0>
CIRCULAR FUNCTIONS. 117
23. sec-i2. 26. tan-^l. 29. cofX-O).
Zo. sin ^-2 ;. 28. cos-i(-4).
Give, in circular measure, the values of:
32. Sin-H. 38. Sin-i(-i). ^''^ ^"^ V JiJ-
33. Tan->(-l). 39. Cos-U. 43. Sin-'O.
34. Sec-H-1). 40. Cosec->^i^.**- C°t-'^-
35. Cosec-12. sj^-^io. Tan "1(^3 -2).
36. Cot-n. 4i_ Sec-'(-A).
§ 5. Curves of the Circular Functions.
107. In the follov^ng figures, let OX and OY be
any two lines at right angles to one another. Let
a line of any length measured from 0 along OX in
the positive sense be chosen to represent one radian ;
then, if 0 be a positive number, an angle containing
0 radians will be represented by a line OM measured
from 0 along OX in the positive sense and 6 units in
length; and an angle containing —Q radians will be
represented by a line of the same length measured from
0 along OX in the negative -sense. From the end M
of the line OM, draw MP at right angles to OX, to
represent, both in magnitude and sense, any circular
118
FUNDAMENTAL PliOPEliTlES OF THE
function of Q. The line chosen to represent a circular
function whose value is unity may be of any length, but,
in the diagrams here given, it is equal to the length of
the line representing one radian. Then, as M passes
alono^ the line OX, from an infinite distance in the
Curves of the Cosine and Secant.
Cosine
Secant
negative sense, through 0, to an infinite distance in
the positive sense, the point P traces out a curve, which
is the curve of the particular circular function considered.
By aid of the results of arts. 86-88 and 93-97, and
the numerical values of the ratios of the angles 0,
12' 6'
CIRCULAR FIWCTIONS.
119
T, Z, iT. T» given in arts. 15, 16, 18, 19 and 28 (and,
4 o 12 2
if greater accuracy is required, of other intermediate
angles whose ratios are given in Mathematical Tables),
we may determine any number of points on the curves
of the circular functions, by means of which points the
curve may be drawn.
Curves of the Sine and Cosecant.
Sine
Cosecant
In the first of the figures, the continuous line belongs
to the cosine and the dotted line to the secant; in the
second figure, to the sine and cosecant, respectively ; and,
in the third, to the tangent and cotangent, respectively.
In each case two complete periods are given.
120
FUNDAMENTAL PliOrERTlES OF THE
It is convenient to define some part of each curve
representing every value of the function without re-
petition as the principal part of the curve. The part
from 0 to TT will be regarded as the principal part for
the cosine and secant (or even functions) ; that from
— ^ to ^ as the principal part for the remaining (or odd)
functions. In the figures, the principal parts of the
curves are indicated by broader lines.
108. It will be seen,
by inspection of the
figures, that a straight
line drawn through a
given point in OX and
parallel to OF cuts each
of the curves in one,
and only one, point; and
this represents graphi-
cally the fact, assumed
in drawing the curves,
that, corresponding to
a given value of the
angle, each of the cir-
cular functions has one,
and only one, value, or,
as in art. 85, the circular
functions are one- valued
functions of the angle.
Again, if a straight
■I
Y
\ /
\ /
J
\ J
/ \
s^
.-.-.- -<
0 '
\
\ /
A
/ \
/ \
/ \
/ \
/ »
/ \
1
1
1
j
1
X
Curves of the Tangent and Cotangent
Tangent
Cotangent
line be drawn through a given point on one of the curves
and parallel to OX, it will cut the curve in an infinite
number of points; and this represents graphically the
CIRCULAR FUNCTIONS. 121
fact that, corresponding to a given value of the function,
there are an infinite number of values of the angle.
The curves also illustrate the following facts : (1) the
continuity of the cosine and sine for all values of the
angle, the discontinuity of the tangent and secant in the
immediate neighbourhood of the angles ^, -^-, etc., and
of the cotangent and cosecant in the immediate neigh-
bourhood of the angles 0, tt, 2x, etc., and the continuity
of the last four functions for all other angles (art. 89) :
(2) the fact that the cosine and sine of an angle lie
between +1 and —1, while the secant and cosecant
have all values except those between +1 and —1, and
the tangent and cotangent have any values whatever
(art. 21).
If the unit angle and the unit circular function be
represented on the same scale, the graphical representa-
tion of the facts that the limits, when 0 is zero, of
sin QjQ and tan QjQ are both unity (art. 75), is that the
sine-curve and the tangent-curve cut OX at 0 at an angle
Itt ; for, if P be a point on either of these curves and PM
be perpendicular to OX, these limits shew that the triangle
PMO is ultimately isosceles when P coincides with 0.
Again, we may employ the curves to discuss the
number of solutions of such an equation as cos 0 = 0, or
tan Q = kO, where k is a constant.
Using the cosine-curve, take any line OM in OX, and
draw MP parallel to OF and equal to OM, then we infer,
from the figure, that the indefinite line OP cuts the
curve in one point only; thus there is one solution of
the equation cos 0 = 0, and that between 0 and J.
122 FCiVDAMtWTAL PlwrEUTlES OF Till-:
If we consider the tangent-curve, and make MP = k.OM,
we may similarly infer that the equation tanO — kO has
an infinite number of roots.
Examples XL a.
1. cosl05' + sinl05° = cos45°.
2. cos 30° + cos 60' -f cos 210° + cos 270° = I
^ T?- A ^.u 1 ^ sin 495° -f cos 390°
3. Find the value of ^a-o .— •— on/Vo-
cos49o -f sin390
Solve the equations (4-8) :
4. 4cos20-|-2cos0=l.
5. sec30-2tan2O = 2.
6. sec2e(tan0-l) = 3tane-l.
7. cos20+2sin22e = l.
8. 7sin2(x + e) + 3sin2(^| + e) = 4. "
9. Find all the angles between 0 and tt which satisfy
the equation 3 tan^O + 8 cos^O = 7.
10. Find all the angles between 0 and tt which satisfy
the equation 3(tan220+cot220) = lO.
11. Find the general value of 0 which satisfies simul-
taneously the equations
sin0= —"^ and cos0 = J.
12. Find the general value of 0 which satisfies simul-
taneously the equations
tan 0= —I and cos 0 = — ^.
13. Shew that the same series of angles are given by the
formula ('>^-i)7^-h(-l)4 and (2?i -f i)7r ± |^.
14. Shew that the same series of angles are given by the
formulc^ (27i-l)|-f(-l)"|and 2ti7r±|.
^B CIlWULAli FUNCTIONS. ]23
^H Examples XL b.
^L cos 165° + sin J (35° = cos 135°.
2. Find the value of sin 30° + sin 60' + sin 210° + sin 300°.
o -r.- J ,1 1 p cos 435° + sin 315°
3. Find the value of -^-tw^-, i^^v-o.
sm435 +cos31d
Solve the equations (4-8) :
4. sec2O + 3cosec20 = 8.
5. V*^(cosec2^-2) + 2cot0 = O.
6. cot^0(cosec 0 — I) = 1 + cosec Q.
7. 3(l+cos30) = 2sin23a
8. 6cos2(7r + 0) + 5sin(j + 0) + l = O.
9. Find all the angles between 0° and 500° which satisfy
the equation sin20 = j.
10.- Find all the angles between 0 and tt which satisfy the
equation sec*0 - 6 sec^O + 8 = 0.
11. Find the general value of 0 which satisfies simul-
taneously the equations
tan Q = ^3 and sec 0 = — 2.
12. Find the general value of 6 which satisfies simul-
taneously the equations
cos 0= ——7^ and cotO= —1.
13. Shew that the same series of angles are given by the
TT T , TT
formulfe (27i±l;;7- and '}27r
7".
4 '-'-"-4
14. Shew that the same series of angles are given by the
formulae (2n + D7r± a and Oi-i)x+(-l)"(|^- a).
CHAPTER VIII.
CIRCULAR FUNCTIONS OF TWO OR MORE VARIABLE
ANGLES.
109. To prove that cos(a + jS) = cos a cos ^8 — sin a sin ^.
Let OF be a line making an angle ^ in the positive
sense with OX. Let a line rotate about 0 from the
position OXy through an angle a, to the position OP; let
it further rotate from the position OP, through an angle
13, to the position OQ; then, the angle XOQ = a + ^.
124
TWO OR MORE VARIABLE ANGLES.
125
Let Op be a line making an angle -^ in the positive
sense with OP ; let if, N be the projections of Q on OP,
Op respectively, and H, K, L those of Q, M, N on OX.
^ P
Since MQ is equal to, parallel to, and of the same sense
as, ON,
KH= OL in magnitude and sense.
We have, in all cases,
OH = OK+KH - '*" - -'^■
= OK+OL
OK .,,. , OL ^,.
= OM-^^-^ON-^^
= cosa. Oif+cosU + IJ . ON.
But, cosU + Z)= — sin a,
OH = cos a . OM-sin a . ON,
OH OM . ON
cos(a -h ^) = cos a cos ^ — sin a sin /5.
126
CmCULA R FUXCTIONS OF
Cor.— Changing /3 into — /3, we get
cos(a — /3) = cos a cos( — y8) — sin a sin ( — /8) ;
but cos( — ^) = coS|8 and sin( — /3)= --sin/3,
cos(a — /3) = cos a cos j8 + si n a sin /3.
110. To prove that sin(a + /3) = sin a cos /8 + cos a sin 6.
Let OF be a line making an angle ^ in the positive
sense with OX. Let a line rotate about 0 from the
position OX, through an angle a, to the position OP ; let
it further rotate from the position OP, through an angle
^, to the position OQ; then, the angle XOQ = a + ^.
Let Op be a line making an angle ^ in the positive
sense with 0P\ let if, iV be the projections of Q on OP,
0^9 respectively, and H, K, L those of Q, M, iV on OY.
Since i/Q is equal to, parallel to, and of the same sense
as, ON,
KH=OL m magnitude and sense.
TWO OR MORE VARIABLE A NG LBS.
We have, in all cases,
OH=OK+KH
= OK+OL
127
sina.0if4-sm(a + |).0i\^.
But,
sin(a+.^ ) = eosa,
0^=sin a . Oi/+eos a . ON,
OH . OM ^ ON
-^ = sma.-^.fcosa.^,
sin(a + ^) = sin a cos ^ + cos a sin ^.
Cor. — Changing /3 into — /3, we get
sin (a - /3) = sin a cos( -^) + cos a sin( - /5)
— sin a cos /5 — cos « sin /5.
128
CIRCULAR FUNCTIONS OF
111. We may deduce either of the formulae for cos(a + /3)
and sin(a + /S) from the other.
Thus, in the formula for cos(a+/3), change a into
l'^ a, then
but
and
cos(^^ + a + /3J = cos(^^ + ajcos^-sin(^|' + ajsin^;
cos(| + a + /3) = - sin(a + ^),
cosg + a)
sing + a)
— sm a,
cos a ;
hence, we have
sin(a + /8) = sin a cos /3 + cos a sin ^.
The formulae for cos(a + )8), sin(a + |8), cos(a — ;8) and
sin(a — j8) are called the Addition Formulce for the
Circular Functions.
112. Since the Addition Formulce have been proved
for all angles, their consequences, as given in Chapter IIL,
are also, as before remarked (art. 24), universally true.
Thus we have, for all angles :
2 cos a cos ^ = cos(a — j8) + cos(a + ^y
2 sin a sin /3 = cos(a — ^5) — cos(a + P)
2 sin a cos ^ = sin(a + P) + sin(a — /5)
2 cos a sin |8 = sin(a + /3) — sin(a — 18)
, cos a + cos ^ = 2 cosh{a + /3)cos J(a — jSY
cos a — cos /3 = 2 sin J(a + /3)sin J(/3 — a)
sin a + sin/3 = 2sinJ(a + /S)cosJ(a — /3)
sin a — sin /3 = 2 cos J(a + /5)sin |(a — 13} ^
J. / j./o\ tana±tan/3
tan(a±^)-__- -r— ^.
^ '^ l + tanatan^
TWO OR MORE VARIABLE ANGLES. 129
fcos 2a = cos^a — sin^a = 2 cos^a --1 = 1 — 2 sin^a,
'2 cos^a = 1 + cos 2a, 2 sin^a = 1 — cos 2a,
sin 2a = 2 cos a sin a,
, ^^ n 2 tan a
tan2a = :j — t — r"'
1 — tan^a
T
i COS 3a = 4 cos^a — 3 COS a,
L sin 3a = 3 sin a — 4 sin^a,
3 tan a — tan^a
tan 3a
1 -Stanza
113. To find cos a and sin a in terms of cos 2a.
We have seen (art. 80) that
2 cos^a = 1 + cos 2a and 2 sin^a = 1 — cos 2a,
/H-cos2a -, . . /l
;a/ — —— and sina=±A/ —
^ . . - - _ _ , — cos2a
cos a ~'"
2
Hence, we have two values of cos a, and also two of
sin a, in terms of cos 2a; the two values of each pair
being equal in magnitude but of opposite sign.
That there should be two values of each may be shewn
as follows : —
(1) Algebraically. — If 2a be an angle which has a
given cosine, then all the angles which have this cosine
are included in the formula 2?i'7r±2a.
Hence, in finding cos a in terms of cos 2a, we are find-
ing the cosines of all angles included in the formula
J(2'7i7r±2a) or 7i7r±a.
Now, cos('}i7r±a) = cos a, if n be even, and —cos a, if
n be odd.
Hence, there are two values of cos a, and, similarly,
two of sin a, in terms of cos 2a, equal in magnitude and
of opposite sign.
130
CIRCULAR FUNCTIONS OF
(2) Geometrically. — Take a circle of radius OX equal
to unity, and along OX make
ON equal to cos 2a, and, through
N, draw the chord PP' per-
pendicular to OX.
Then, by bisecting the group
of angles bounded by OX and
OP, we obtain two positions
of the second bounding line of
a, namely, OQ^ and OQ^ in the figure.
Also, by bisecting the group of angles bounded by OX
and OP', we obtain two more positions of the second
bounding line of a, namely, OQg ^^^ OQ^ in the figure.
It may be shewn that Q^Q^ and Q^Q^ are diameters.
Hence, we obtain two values of cos a, and two of sin a,
in terms of cos 2a, the values of each being equal in
magnitude and of opposite sign.
114. To find cos a and sin a in terms of sin 2a.
We know (arts. 10 and 31) that
cos2a + sin2a = l,
2 cos a sin a = sin 2a,
(cos a + sin a)^ = 1 + sin 2a,
(cos a — sin a)^ = 1 — sin 2a,
cosa + sina= ±>v/l+sin2a, (1)
±>s/r-sin2a, ...(2)
and
and
and
cos a — sin a
2 cos a = ± V 1 + sin 2a ± /v/l — sin 2a,
and 2sina= ±x/l+sin 2a + x/l — sin 2a.
Hence, we have four values of cos a, and also four of
sin a, in terms of sin 2a ; the values occurring in pairs of
equal magnitude and of opposite sign.
TWO OR MORE VARIABLE ANGLES.
131
That there should be four values of each, may be shewn
as follows :
(1) Algebraically. — If 2a be an angle which has a
given sine, then all the angles which have this sine are
included in the formula 7i7r+( — l)"2a.
Hence, in finding cos a in terms of sin 2a, we are finding
the cosines of all angles included in the formula
n-TT
J{7i7r + (-ir2a} or :^+(_l)-a.
Now.
eos|-y+i
( — l)*^a[ = cosa, if '7i = 4m,
= sina, if 7i = 4m + l,
= —cos a, if ?i = 4m + 2,
= —sin a, if '7i = 4m + 3,
where m is zero or an integer.
Hence, there are four values of cos a in terms of sin 2a,
and similarly, four of sin a, the values in each case occur-
ring in pairs of equal magnitude and of opposite sign.
(2) Geometrically. — Take a circle of radius OX equal
to unity, and let OF be a radius
making an angle -x in the posi-
tive sense with OX. Along 0 Y
make ON equal to sin 2a, and,
through Ny draw the chord PP'
parallel to OX.
Then, by bisecting the group
of angles bounded by OX and
OP, we obtain two positions of
the second bounding line of a, namely, OQ^ and OQ^ in
the figure.
Also, by bisecting the group of angles bounded by OX
132 CIRCULAR FUNCTIONS OF
and OP', we obtain two more positions of the second
bounding line of a, namely, OQ^ and OQ^ in the figure.
It may be shewn that Q^Q^ and Q2O4 ^^^ diameters.
Hence, we obtain four values of cos a, and four of sin a, in
terms of sin 2a, the values in each case occurring in pairs
of equal magnitude and of opposite sign.
115. The proper signs to be taken before the radicals
in equations (1) and (2) of the preceding article may be
determined as follows :
If a lie between 2?i7r — t and 2n7r-{-T, cos a is always
positive and of greater magnitude than sin a ;
.'. the + sign must be taken in both equations.
If a lie between 2w7r+T and 2n'7r-{--T^, sin a is alwa,ys
positive and of greater magnitude than cos a ;
.*. the + sign must be taken in equation (1), and the
— sign in equation (2).
If a lie between 2n7r+-T- and 2ti7rH--j-, cos a is always
negative and of greater magnitude than sin a ;
.*. the — sign must be taken in both equations.
If a lie between 2'7i7r+-T- and 27i7r+ x, ^^^ « ^^ always
negative and of greater magnitude than cos a ;
/. the — sign must be taken in equation (1), and the
+ sign in equation (2).
Example. — Given sin 210°= -^, find cos 105° and sin 105°.
The sine of 105° is positive and of greater magnitude than cos 105°,
cos 105° + sin 105°= +-^,
"^^ cos 105° - sin 105° = - ^f ,
cos 105°=!^^ and sin 105° = 1±^.
2>J2 2 v/2
TWO OR MORE VARIABLE ANGLES. 133
YiVA Voce Examples.
State the signs of cos a + sin a and cos a — sin a when a is :
1. 98°. 5. 235°. 27r ., Utt
2. 174°. 6. 300°. 3* ^^- 6 *
3. 87°. 7. 14°. 37r 12. 4x.
4. -12°. 8. 325°. 2'
116. Example 1.— If J +^+C=7r, then
ABC
sin ^ + sin ^ + sin C= 4 cos — cos — cos — .
2 2 2
sin B + sin C= 2 sin ^±^ cos ^^=^ 2 cos | cos ^^H^,
• 2 2 2 2'
sin^ = 2cos-siu- = 2cos-cos^+^;
2 2 2 2 '
.-. sin^ + sin^+sin(7=2cos^fcos:?±^+cos:^Il^')
2\ 2 2 /
. ^ 5 C
= 4 cos — cos - cos —
2 2 2
. '- -^
Example 2.— If ^+J5 + (7=7r, then
cos^ J + cos^^ + cos^(7+ 2 cos A cos J5 cos C= 1.
cos25 + cos^C- 1 = cos^^ - sin^C
= cos(5+ C)cos(5- C)= -cos.4cos(^ - C\
. •. cos^^ + cos^i? + cos^C - 1 = - cos ^ { cos(5 + C) + cos(5 - C)}
= — 2 cos A cos B cos (7,
.*. cos^^ +cos25+cos2C+2 cos A cos 5 cos (7=1.
Example 3. — Shew that
cos-i|| + 2tan-ii=sin-^f.
Let a=sin-^f and yS = tan-^i
Then tan 2/3 = ^'\ = _5_
COS 2^ = if and sin 2^ = /^,
cos(a-2^)=|.l| + |.-3-%=|f.
sin-i I - 2 tan-i -J- = cos-i f f .
M^
*t,j]
134 CIRCULAR FUNCTIONS OF
Example 4.— Shew that
l-xy
where 9i=l, 0 or - 1, according as tan-^a? + tan-^y>|^, lies between
-J and ^, or< -^, t.e. as iry > 1 (^ and y being positiveX
2 2 2
^ < 1, or ^ > 1 (:r and y being negative).
We have tan(tan-^^ + tan" H/) = -^^ ;
hence, tau~^,a;+tan~^y is one of the group of angles given by
Tan-i^±^-, or mr+ta.n-'-p^ .
\-xy \-xy
Now, each of the angles tan~^^, tan~^v and tan~^^ -^ lies
l-xy
between — - and -; therefore,
2 2' '
if tan~^;r+tan~^y > J, we must have w = l,
2
if tan~-'^+tan~^y < -^, we must have w= — 1,
but if - 1^ < tan-^^ + tan-^y < ^, then n=0.
The three cases may be more readily distinguished by consider-
ing the values of the product xy :
If CO and y be positive and xy>lf we have
tan~^^+tan-^y > tan-^-p+tan"^-, i.e. > % ;
X 2
and, therefore, tan"^^ + tan-^y = tt + tan"^:^ — ^.
l-xy
Similarly, if x and y be negative and xy>l, we have
tan-^;r + tan" V < tan-^^ + tan"^ , i.e. <~ ;
X 2
and, therefore, tan-^a;+tan-V= -Tr + tau"^,^ — ^.
l-xy
In all other cases,
tan-^a: + tan- V = tan-^^:t2^.
1-Xlf
TWO OR MORE VARIABLE ANGLES. 135
It follows that for all positive values of x and y,
tan- ^1' - tan- V = taD-\'^~^.
Example 5. — To prove that
(1) tan-i^ + tan-i^=J (Euler's formula).
(2) 4 tan-^i + tan-^ g J-g = | (Machin's formula).
(3) tan-^YT9 = tan- VV - tan"^ ^V (Rutherford's formula).
(1) Let a=tan-H, ^ = tan-i^,
^, . , , o\ tan a + tan/? \-\r\ f .
then tan(a + jS) = - — r -^ = —-y^ = -l = i-
^ '^ 1 - tan a tan ^ 1 - i • f f
tan-i^ + tan-i^=|'.
(2) Let a=tan-^i, /?=tan-Vi-9:
, , X o 2 tan a
then tan 2a = . — - — 2
1-tan^a 1-oV
o 5
fan Ar,— ^-12 —120
tan 4a-- ^^ -TX95
tanaa-m- TT? " ¥¥9- _ 2 8 6 8 0-1 1 9 _ 2 8 5 6 1
tan(^4a P;-^ ^-^^-^ ,— 2844 1 + 1 2028561
4 tan-^l^ - tan
(3) Let a=tan-V^, /? = tan-V9J
2T9^
fTiPn foT^/^^_/?N— 70 99 _ 99-70 _ 29 _ 1
tnen tan(^a /ij - — — - ggg^q.^ - btft - TT^-S"-
^ + T (T • ■9 "9
tan-^a'W = tan-^y\^ - tan" V¥'
Hence, |= 4 tan-^^ - tan-^^ + tan" V9 •
Example 6. — Solve the equation
sin2^ + sin22 9 + sin23 9 + sin24 (9=2.
Multiplying both sides of the equation by 2, it becomes
- cos 2^ + 1 - cos 4^+ 1 - cos 6^ + 1 - cos 8(9=4,
cos 2^+ cos 4^ + cos 6^ + cos 8^=0.
136
CIRCULAR FUNCTIONS OF
Now,
and
cos 2^ + COS 8^ = 2 cos bd cos 3^,
cos 4^+ cos 6^ = 2 cos bd cos ^,
2 cos 5^cos 3^+ cos ^)=0,
4 cos bS cos 26 cos 9=0.
66:
4-1 or
'=W7r + ^ or d=mr+
2'
(9=^^!■+•
f^or.|-|.|or..+|.
It is obvious that all the angles given by the formula
-+|
are included in the formula
5 10
This is also shewn by the accom-
panying figure, in which XOX' and
F^OPg are lines at right angles ;
Pj, Pgj Aj ••• Ao ^^® *^® points in
which the second bounding lines of
the angles given by the formula
5 10
meet the circumference of a circle with 0 as centre. Qi, Q2, $3, Qi
■ are the corresponding points for the angles given by the formula
""2^4'
and P3, Pg tliose for the angles given by the third formula
nir + l.
Thus, 6'=7i| + ^'^ or
given equation.
i- + - is the complete solution of the
Example 7. — Solve the equation
v/3cos^+sin ^=1.
Dividing both sides by 2, the equation becomes
^cos(9+^sin(9=i
.(1)
4b&;%
TWO OR MORE VARIABLE ANGLES. 137
cos J COS ^ + sin J sin 6 = cos J,
6 6 3
cosf ^-^J = cos
e = 2n7r + '^ or 2?i7r-| (2)
Equation (1) might, however, have been written
sin ^ cos $ + cos ^ sin ^ = sin ~,
3 3 6
e+|=«x+(-l)».|,
0=«x + (-l)''.|-|. (3)
It may easily be shewn that the formulae (2) and (3) give the
same series of angles.
For, if n be even and equal to 2m, the formula (3) becomes
2m7r+|-|or2m7r-|,
the second of the formulae (2).
If n be odd and equal to 2m + 1, the formula (3) becomes
(2m+l)7r-|-|or2m^ + |,
the first of the formulae (3).
The equation here solved is a particular case of the equation
whose method of solution is given in the next example.
Example 8.— Solve the equation a cos ^+6 sin 0=c.
Suppose c to be positive. Dividing both sides by sJd^+W, the
equation becomes
« cos ^ + —1= sin ^=. ""
Let
a
and sin a = , then tan a = -
b
138
CIRCULAR FUNCTIONS OF
cos a COS ^ + sin a sin 6^=
cos(^-a)
Voi^+P
^-a = 2n7r±cos~^-
0=2w7r±cos-^-
+ a.
Va2 4-62
To determine a, we first obtain from the tables the angle a! whose
tangent is equal to the numerical value of hja. "We then have
a=a', if a is positive and h positive ;
a= — a', if a is positive and h negative ;
a=7r-a', if a is negative and h positive ;
a = — TT 4- a', if a is negative and h negative.
The following geometrical construction illustrates the solution of
this equation : —
From any line OX cut oflf a part OA equal to a in magnitude and
sense ; from A draw AB dit right
angles to OA and equal to h in
magnitude and sense ; so that
the angle XOB=a. With 0 as
centre and OB as radius, describe
a circle. From OB cut off ON
equal to c ; and through N draw
(if possible) the chord PP' per-
pendicular to OB.
Then,
cos(^-a).
Hence, all the angles bounded by OX and OP, and all the angles
bounded by OX and OP, satisfy the given equation.
If ON<OB, i.e. if c<\fa^+^, there are two series of values
of 6, real and different ; if c=^/a^-\-b^, there are two series of
values, real and equal ; and if c> JcfiTW\ the line PP does not
cut the circle in real points, and therefore there are two series of
values, but imaginary and different.
This is also evident from the solution obtained above, since the
cosine of an angle is never greater than unity.
TWO OR MORE VARIABLE ANGLES. 139
Example 9. — Trace the changes in sign and magnitude of
cos 0+sin 0, as 6 increases from 0 to 27r.
cos ^ + sin 0=j2(Ji—cos ^+-i. sin d
= V2[cos ^ cos ^+ sin J sin e\
=v^2cos(^-|).
As 0 increases from 0 to — , cos^+sin^ is positive and in-
creases from 1 to ^^2 ;
As 6 increases from j to -— , cos^+sin^ is positive and de-
creases from ;^2 to 0 ;
As 6 increases from -- to -—, cos^ + sin^ is negative and
decreases from 0 to —^2;
As 0 increases from -- to ■—, cos^ + sin^ is negative and
increases from - ^2 to 0 ;
As 6 increases from - to 27r, cos^ + sin^ is positive and
increases from 0 to 1.
The changes in the value of cos ^+sin ^ may be iUustrated by
a curve as in the case of the cosine and other trigonometrical
ratios.
Examples XII. a.
1. cosa + cosfa+~j + cosfa + -^j = 0.
2. cosec a + cosecf a + — ] + cosecf a + ^^ j = 3 cosec 3a.
ty ,a, ,a + 7r, .q + Stt „ ,
3. cot ^ + cot — H — h cot — ^ — = 3 cot a.
COS
COS
140 CIRCULAR FUNCTIONS OF
4. cos2a = 2sinfa + T)''^i"fa+^j.
i3a = 22sinfa+^Jsinfa+^jsinfa+^j.
5. 4 sin a sin ^ sin y = sin (^ + y — a) + sin(y + a — /8)
+ sin(a + )8-y)-sin(a + ^+y).
6. Express 4(cos a cos ^ cos y cos ^ + sin a sin /3 sin y sin ^)
as the sum of four cosines.
7. sin(/34-y)+sin(y+a)+sin(a+/3)
. . a . B . y a + B-\-y
= 4 sm ^ sm ^ sm ^ cos ^ — '-
, A « /5 y • a + i8+y
+ 4 cos ^ cos ^ cos ^ sm ^^ — ^.
^ sin(0 — |8)sin(<^ — y) — sin(0 -- p)sin{0 — y)
sin(^-y)
_ sin(^ — y)sin(0 — a) — sin(0 — y) sin(0 — a)
~" sin(y — a)
_ sin(0 — a)sin(0 — ^) — sin(0 — a)sin(^ — ^)
~ sin(a — )8)
g sin(^-y) ■ 8in(y-a) ^ sin(a-/3)^Q^
cos |8 COS y cos y cos a cos a cos /3
10. cos2(/3-y) + cos2(y-a)+cos2(a -/3)
= 1+2 cos(/3 — y)cos(y — a)cos(a — P).
11. cos /3 cos y sin(/3 - y) + cos y cos a sin(y — a)
+ cosacos/3sin(a-/3) + sin(^-y)sin(y-a)sin(a-/5) = 0.
12. sm -y- + sm — — sm y = 4 sm =: sm =- sm -=-.
If ^+5+a=7r, prove that (13-21):
13. sin 2J. + sin 25+ sin 2(7= 4 sin ^ sin 5 sin (7.
TWO OR MORE VARIABLE ANGLES. 141
ABC
14. cos ^ + cos 5 + cos (7= 1 + 4 sin — sin ^ sin ^.
15. cos2^ + cos^^ + cos^^ = 2( 1+sin ^- sin ^ sin — ).
16. tan A + tan 5+ tan 0= tan A tan 5 tan G.
17. cot5cot(7+cot(7cot J.+cot J.cot5 = l.
18. sin 6J.+sin65 + sin 6(7=4 sin SJ: sin 85 sin 3(7.
19. (sin ^ +sin 5 + sin (7)(sin 5+sin G— sin J.)
X (sin (7+ sin ^ — sin 5) (sin A-i-sinB- sin C)
= 4sin2^sin2J5sin2a
sin 2^ , sin 25 . sin 2(7
20
l+COS^^ ■ l + C0S2i^ ■ l-\-G0S2U
sin 9.4 -Lain 9 7? -Lain OH
' l+cos2^ + cos2jB+cos2a'
21. The three expressions sinM + cos A sin B sin (7,
sin^^ + cos 5 sin G sin J. and sin2(7+ cos G sin ^ sin B
are equal.
09 T£ >i_L R4_r' —'^ sin J. + cos 5 — sin G _ 1 + tan|5
~2' sin^ + cos(7-sin5~l + tani(7'
23. If the sum of four angles be ir, the sum of the pro-
ducts of their sines taken two and two together is
equal to the sum of the products of their cosines
taken two and two together.
Prove geometrically the formulse (24-26) :
OA c^ l — tan^a
24. cos2a = .i-— — 2 .
l+tan^a
25. tana+tan^ = ii5Mj^.
cos a cos p
26. tan^ = g^"°+"'"^.
2 cos a + cos p
27. Prove the formula for cos(a + ^), when
a>Jand a + /3<7r.
142 CIRCULAR FUNCTIONS OF
28. Prove the formula for sin(a+/8), when
a > TT and < — , and a + i8 > -^ and < 27r.
29. cos-iT^+sin-iy«^ = cos-iAV
30. cot-i2+cosec-VlO = ^-
31. 8in-ii+sin-i3^+sin-i-l^ = |.
32. 2tan-H + tan-i|=|'.
33. tan-iJ + tan-4+tan-H+tan-i^S=j.
nA -m- 1 j.1. J. i. i?i. 1 ajcosa , ,«;— sina
34. Find the tangent of tan-\- —--. tan"^ .
^ 1 — aj sin a cos a
Solve the equations (35-46) :
35. cos50+cos30+cos0 = O.
30. 4 sine sin 30 = 1.
37. sin2/O+sin2s0 = cos(r~s)a
38. sin3a-sine = 0.
'39. 4sine = sec2a
40. tan 0 + tan 20 = tan 3a
41. sin50=16sin5a
42. sin2r0-sin(r-l)0 = sin2a
43. tan(^ + 0) = 3tan(|-0).
44. cos 0 — sin 0 = — ^.
45. cos(a + 0) = sin(a + 0) + V2cos/3.
46. 3cos0-|-sin0 = 2.
Trace the changes in sign and magnitude, as 0 increases
from 0 to 27r, of (47-49) :
47. ^3 cos 0 + sin a
TWO OR MORE VARIABLE ANGLES. 143
sin 6 — ^3 cos 0
^Ssir
cos 20
4g .
;^3 sin 6 + cos 6
49. .-
cos 6
50. Find the values of cos 9°, sin52J°, sin 97J°, and
cos 195°.
51. Find the limits between which 2a must lie, when
2sina= — x/l + sin 2a + s/l — sin 2a.
52. Prove that tan a, when expressed in terms of tan 2a,
has two values.
53. Prove that cos a, when expressed in terms of cos 3a,
has three values ; and that sin a, when expressed
in terms of cos 3a, has six values.
54. Prove that tan a, when expressed in terms of sin 4a,
has four values.
55. Eliminate 0 between
x = 2a sin 0 sin 20 — a cos 0,
2/ = 26 sin 0 cos 20 + 6 sinO.
56. If, in a triangle ABC, cos J. = sin 5 sin C, then the
triangle is right-angled.
57. sm\0-{-a)+sm\0 + l3)-2cos(a-l3)sm{0 + a)sm{0 + /3)
is independent of 0.
58. If (l + cosO)(l4-cos0) = sin0sin 0, then 0 or 0 or
0 + ^ = (2^ + l)7r.
59. Find sec(a + /3) in terms of sec a and sec/3, and prove
that sec 105° = - ^2(1 + V^)-
60. cos 12° + cos 60° + cos 84° = cos 24° + cos 48°.
61. tan 70° = tan 20° + 2 tan 40° + 4 tan 10°.
62. sin2l0° + cos220°-sinl0°cos20°
= sinnO° + cos240° + sin2l0°cos 40° = f .
4||^fc.
144 CIRCULAR FUNCTIONS OF
Examples XII. b.
1. sma + sin(a + ^3^) + sin(a + ^) = 0.
2. COs2a + cos2(a + ^) + Co.s2(a + y) = f.
3. sm2a = 2sinasiD(a + ^),
sin3a = 22sin a sin(a + |)sin(a + ^\
sin4a = 23sin a siD(a + ^)sin(a + |)sin(a + ^).
4. cos W — cos 2a = 2(cos Q — cos a)(cos 0 — cos o + tt),
cos 30— cos 3a
= 22(cos 6 - cos a)rcos 0 - cos a + -^)
X fcosO— cosa4- >r),
cos 40 — cos 4a
= 23(cos 6 - cos a)f cos 0 - cos a + 5)(cos 0 - cosa + 7r)
X fcosO — cosa + -^j.
5. tan(3O-gtan(o+-^^) = tan(0+^)tan(e-^).
6. 4 cos(/5 + y - a)cos(y + a - /5)cos(a + /3 - y)
= cos(a + /5 + y) + cos(^ + y _ 3a) + cos(y + a - 3^)
+ cos(a + ^-3y).
7. sin /3 sin y sin(^ — y) + sin y sin a sin(y — a)
+ sinasin/3sin(a-/3) + sin(/3-y)sin(y-a)sin(a-^)=0.
8. sin(^ + 2y) + sin(y+2a)+sin(a + 2/5)
+ sin(2/3 + y) + sin(2y + a) + siD(2a + /S)
= 2sin(a + i8+y){4cosi(/3-y)cosi(y-a)cosKa-/3)-l}.
TWO OR MORE VARIABLE ANGLES. 145
9. sin2(/5-y) + sm2(y-a) + sin2(a-/3)
+ 4 sin(/3 — y)sin(y — a)sin(a — /3) = 0.
10. COS -y- + COS -=r + COS -=- + 4 COS -j^ COS -1^ COS -^- + 1=0.
If ^+5 + (7=7r, prove that (11-20) :
11. COS 2J. +COS 25-f-cos 2(7+4 cos A cos 5 cos C+ 1 =0.
12. sin-^ + sin -^ + sin ^ = 1 + 4 cos — 7 — cos — y- cos — j—.
13. sm2^- + sm^ + sin^-^ + 2 sm ^ sin -^ sin 9 = 1.
1 4. cot ^ + cot ^ 4- cot ^ = cot ^ cot -^ cot j^.
15. sin 4J. + sin 4j5 + sin 4(7+4 sin 2^ sin 25 sin 2(7= 0.
16. sin22^ + sin225+sin22a= 2(1 - cos 2A cos 25 cos 2(7).
17. sin*^+sin45+sin^(7
= f + 2 cos ^ cos 5 cos (7+ Jcos 2 J. cos 25 cos 2(7.
-„ . 2 _1— cosu4+cos5+cos(7
(7 1 — cos (7+ cos J. + cos 5
tan 2.
19. cos ^+ cos 5+ cos (7
. A B-C , . B 0-A , . 0 A-B
= sin ^ cos — ^ ("Si'^ o" cos — ^ hsin -^ cos — ^ — .
20. cos 2^ (cot B - cot (7) + cos 25(cot (7- cot .4)
+ cos 2(7(cot ^ - cot B) = 0.
21. If^+5+(7+i) = 27r,
. , J, , ^ , ;n . ^+C C+A A+B
cos^ + cos5 + cos G'+ cosZ) = 4cos— ^— cos— ^^— cos— ;i— .
^ Zi Z
22. If ^+5+ (7= J, then
tan 5 tan (7+ tan (7 tan A + tan j! tan 5 = 1.
146 CIRCULAR FUNCTIONS OF
Hence, shew that
2(tan2^ + tan25 + tan^O) - 2
= (tan5-tanC)2 + (tan(7-tanJ.)2 + (tan^-tan5)2,
and that the expression tanM + tan^JJ+tan^O is
never less than unity.
23. If the sum of four angles be two right angles, the
sum of their tangents is equal to the sum of the
products of the tangents taken three and three.
Prove geometrically the formulae (24-26) :
24. tan^° l^f^ii'.
2 1 + cos a
g„ tana + tan/3_sin(a-|-i3)
tan a — tan |8 ~ sin(a — ^)'
26. cot^±^ = "'°°-"'°^.
2 cos p — cos a
27. Prove the formula for cos(a-f/3), when a<~, and
a + /3>7r and<^.
28. Prove the formula for sin(a-f-/3), when a>'7r, and
29. sin-i| + sin-W = sin-i|-^,
30. 4(cot-i3 + cosec-V5) = 7r.
31. tan-4 + cot-4 + sin-i'^^ = 7r.
32. tan-iT2__|_2tan-i| = tan-i|.
38. tan-4 + tan-if + tan-H + tan-4 = |^.
34. Tan-i 7^^^"" +Tan-^^HL^ = r^7r + a, n being
l+w-cosa m + cosa
any integer.
TWO OR MORE VARIABLE ANGLES. 147
Solve the equations (35-46) :
35. sin 0+sm3O+sino0 + sin 70 = 0.
36. cos 0 cos 30 = cos 20 cos 6a
37. cosr0+cos(r-2)0 = cosa
38. sin 40 -sin 0 = 0.
39. sin 0 = cos 40.
40. tan0 + tan30 = 2tan20.
41. 27sin80 = cos80-4cos40 + 3.
42. sin 30-2 sin30 = f.
^^' ^^^(252'^^^) = 'K272''^'4
44. cos 0 + sin 0 = c.
45. 1 + cos 20 -sin 20 = 0.
46. 2sin0-cos0 = J.
Trace the changes in sign and magnitude, as 0 increases
from 0 to 27r, of (47-49) :
47. cos 0 — sin 0.
48. tan0-2cosec2a
V3-|-tan0
^' V3-tan0*
50. Find the values of sin7J°, cos22J°, cos 127 J°, and
sin 1874°.
51. Find the limits between which 2a must lie when
2 cos a = — v/l + sin 2a — aJi — sin 2a.
52. Prove that sin 2a, when expressed in terms of sin a,
has two values of equal magnitude and opposite
sign ; and that cos 2a, when expressed in terms of
cos a, has only one value.
58. Prove that sin a, when expressed in terms of sin 3a
has three values ; and that cos a, when expressed
in terms of sin 3 a, has six values.
148 CIRCULAR FUNCTIONS OF
54. Prove that sin a, when expressed in terms of tan 2a,
has four values.
55. Eliminate 0 between
X = 2a sin W cos 0 — a sin 20,
y = 2h cos SO cos 0 — 6 cos 20.
56. Evaluate tanf|^+ a jtanf—+ a V
57. The tangents of two of the angles of a triangle are 2
and 3, find the third angle.
58. Prove that
cos(|8+y-a)+cos(y+a-/3)+cos(a+^-y)-4cosacos^COSy
vanishes when a + fi + y is an odd multiple of a
right angle.
59. cos20° + cosl00' + cosl40° = 0.
60. cos 12° 4- cos 108° + cos 1 32° = 0,
cos 108°cos 132°+cos 132°cos 12° + cos 12°cos 108°=-|,
cos 12°cos 108°cos 132° = 1±^.
lb
61. (2cos^+10cos|^y+(4sin^)'=7].
62. If sin 3a = 71 sin a be true for any values of a besides
0 or a multiple of ^, then n must be less than 3
and not less than - 1.
117. Example 1. — If cos2a+cos^/3+cos^y + 2cosacos/?cos'y = l
find the relations which must exist between a, /? and y.
cos'^a+cos^jS + cos^y + 2 cos a cos /5 cos y - 1
= (cos a + cos /? cos y)'^ - cos^/? cos^y + cos^^ + cos^y - 1
=(cos a+cos /? cos y)^ - (1 - coa^/3){l - cos^y)
= (cos a + cos )8 cos y )2 - sin^/? sin^y
= (cos a + co=' /? cos y - sin /? sin y)(cos a + cos ft cos y + sin ^ sin y)
TWO OR MORE VARIABLE ANGLES. 149
= {cos a + cos(/^4-y)}{cos a + cos(^ — y)}
a-\-B + y B + y-a y + a-^ a + /5-y ^
2 A 2 Z
.'. either a + /3 + y, /3 + y-a, y + a- /3, or a + f3-y must be an odd
multiple of tt.
Example 2.— If v = taii-i^^^i±^i:i + tan-^-^,,
express x in. terms of y in its simplest form.
Suppose X numerically less than unity; let ^ = tan~^^, then
^=tan 0.
vT+^-l__sec ^-l„l-cos6^^^^^ $
X
tan e sin Q 2'
•1 'ix 2 tan Q x. r.n
and = - — — -— = tan 2^ ;
6'=tan-\r=^.
5
If X be positive and > 1, then it may be shewn that
.r = tan-|(y + 7r);
if X be negative and numerically > 1, that
^=tan|(2/-7r).
If X have any other value, then
^=tan-^.
5
Example 3. — Eliminate Q between the equations
(a + 6)(^ +^) = cos ^( 1 + 2 sin26'),
(a -h){x-y)= sin ^(1+2 cos26').
We have (a + 6) (.^ + ?/) = cos ^ + sin Q sin 2 ^,
{a-})){x-y)—%va. ^ + cos ^sin 2^ ;
2(a^ + 6y)= (cos ^ + sin ^)(1 + sin 2^)
= (cos^ + sin6')3,
and %ciy + 6ji;) = (cos Q - sin ^)3 ;
(a^ + 6?/)^ + (a^/ + 6^)3 = 2-i-2^ = 2*
150 CIRCULAR FUNCTIONS OF
Examples XIII.
1. Prove that
tan a -f tan 2a + tan 3a + tan a tan 3a tan 4a =
cos 2a cos 4a'
and verify this formula when a = t ^^^ when a = ^.
2. sin(^ + y — a)sin(^ — y)cos(^ — y)
+ sin(y 4- a — /8)sin(y — a)cos(y — a)
+ sin(a + /3 — y)sin(a — /3)cos(a — /3) = 0.
3. sin K/S - y)sin f (/? + y) + sin Ky - a)sin f (y + a)
+ sinJ(a-/3)sinf(a+)8)
= 4 sin J(/3 - y)sin J(y - a)sin J(a - /3)sin(a + /3 + y).
4. COs(^ + y)cos(y+a)cos(a + i5)
= cos a cos /3 cos y cos(a + jS + y)
+ sinasin^sinysin(a + /3+y).
5. cos2acos2(^ + y) + C0S 2/5cos2(y + «) + cos 2ycos2(a+y8)
= cos2acos2^cos2y + 2cos(^+y)cos(y+a)cos(a+^).
6. cos(a + /3)cos(a — |8)cos(y + ^)cos(y — ^)
— sin(a + /3)sin(a — /3)sin(y + (5)sin(y — (5)
= l-Jsin2(^+y)-Jsin208-y)-Jsin2(a + (5)-Jsin2(a-^).
7. cos(a + 18 + y)cos(^ + y — a)cos(y + a — /8)cos(a + )5 — y)
+ sin(a + /3 + y)sin(/3 + y - a)sin(y + a - /3)sin(a + /3 - y)
= cos 2 a cos 2/3 cos 2y.
9. {sec a + cosec a(l + sec a)}(l — tan2Ja)(l— tan^Ja)
= (sec |a + cosec Ja)sec2Ja.
10. If^+5+C^=7r, then
sinM sin 2^ + sin^^ sin 2^+sin2(7sin 2G
= 2 sin J. sin B sin (7+ sin 2^1 sin 2B sin 2(7.
11. If ^+5+ (7= TT, then ^ 5 (7
tan^+tan^+tan^ = 4. ^^^^^,:,^B+.inG
TWO OR MORE VARIABLE ANGLES. 151
12. Prove, geometrically, that
sin(a + fi)sin{a — /3) = sin^a — sin^/^.
13. If the sum of the sines of three angles is equal to the
sine of their sum, the sum of two of the angles
must be a multiple of four right angles.
Solve the equations (14-21), 6, 0 and x being the
unknown quantities :
14. cosec 4a — cosec 40 = cot 4a — cot 40.
15. 0 + 9^ = 240° and vers 0 = 4 vers ^.
16. tan20 = 8cos20-cota
17. sin0-cos0-4sin0cos20 = O.
18. tan0 + tan(^^ + 0) = 2.
- Q sin a cos(/3 + 0) _ tan ^
sin/3 cos(a-|-0) tana'
20. sin-i:r^, + tan-i ^^
TT
1+x^ ' 1-x^ 2
21. tsin-'^x + t3i.n-\l-x) = 2t&n-'^Ajx-xi
22. Find all the values of u and 0, for which
_ tan 0
'^~l-ta.n20
changes sign, as 6 passes from a small negative
quantity, through zero, to four right angles.
23. Trace the changes in sign and magnitude of
sin 0 + cos 0
sin 0 — cos 0
as 0 changes from 0 to 27r.
Eliminate 0 between the equations (24-26) :
24. cos20-hsin20 = cos0 + sin0 = (X.
OK • _ ^^^ " _ 1
'^^' '''' "" - 73^inY0 ~ 2-hV3cos20'
26. asin 04-6 cos 0 = c and acosec0H-6sec0 = d
152 CIRCULAR FUNCTIONS OF
27. Express as a single term
1 I - 1 .
^2 cot |a — cosec Ja ^2 cot \a + cosec J a*
28. If sm(a + )8)cosy = sin(a + y)cos/3, then, either ^ — y
is a multiple of tt, or a is an odd multiple of ^.
29. Prove that 16 sin50 = sin 5^-5 sin 30 + 10 sin 0, and
deduce the value of 32 sin^0 in terms of cosines of
multiples of 0.
30. If atana + 6tani8 = (a + 6)tan^^^, then ?=^^.
'^ ^ ^ 2 b cos/3
31. If !!£i4^=!i5g±^, then either a and ^. or 6
sm(a + 0) sm(/3 + </)) '^'
and (p, differ by a multiple of tt.
32. If sec(0 — a), sec 0, and sec(0 + a) be in arithmetical
progression, then cos (p = ^1 . cos J a.
oo Tr- • . cos^O— sin0 J ^ sin^^ — COS0 .,
33. If sm 0 = r jr-. — 7^ and cos <h = .; ^ . ^, then
^ 1— cos^sin^ ^ 1— cosOsm^
. /^ cos^(^ — sin 0 , ^ s\v?(h — cos 0
sin Q = - — -^ r-^- and cos 6 = z: ^ , . ^ .
1 — cos (p sin 0 1 — cos <p sm 0
34. If cos a = f and /3 — a = -r, find tan^ and tan(a + /3).
2 2 2
35. If -„ cos 0 = ^ cos 0+^2 cos a,
a^ a^ 0^ ^
and ^ - y - ^
sin(0+0i) sin(0-0i) sin 20'
, , sin 0 W
then ^_^ =
sin 0^ a^
36. If tan a; = cos a tan y,
tan^" sin 2^/
• then tan(2/ — a;)=■
l+tan2^cos22/
TWO on MORE VARIABLE ANGLES. 153
cos U—e
87. If cosF=
1 — ecos V
U
then tan2 = yj^tan-2-
88. If cot 0 = 71 cot(a-O),
39. If sin a and sin /3 be two values of sin 0 satisfying
the equation a cos 20+ 6 sin 20 = c,
then cos^a — sin^-S = -^^-^ „.
40. cos8(a) + a)sin(/3 — y) + cos8(ir + ^)sin(y — a)
+ cos Z{x-\- y)sin (a — /3)
= 4 cos(3cc + a + )8 + y)sin(/3 — y)sin(y — a)siu (a — /3).
41. (cos a + cos |Q + cos y)
x{cos2a+cos2/3-fcos2y-cos(|8+y)-cos(y+a)-cos(a+/3)}
— (sin a + sin /5 + sin y)
X {sin2a+sin2|8+sin2y-sin(/3+y)-sin(y+a)-sin(a+j8)}
= cos 8a + cos 8y8 + cos 8y — 3 cos(a + /3 + y).
42. cos A cos B cos G cos D + sin A sin 5 sin C sin i)
= cos a cos /3 cos y cos (5+ sin a sin /5 sin y sin ^,
where 2a=5 +(7 +i)-^,
2p = G^D+A-B,
2y = D+A+B-G,
^6=A-\-B+G -D.
48. If a, i8, y be all unequal, and if no two of them
differ by a multiple of tt,
{tan(/3 - y) + tan(y — a) + tan(a — /3)}
X {cot(^-y) + cot(y-a) + COt(a-/5)}
= 1 — sec(|8 — y)sec(y — a)sec(a — P).
154 CIRCULAR FUNCTIONS OF
,. (sec g sec /3 + tan « tan /3)^ — (tan g sec /3 + sec a tan /3)^
• 2(1 + tan^g tan2/3) - sec^g sec^^
_ sec 2g sec 2/3
~ sec^gsec^jS *
.p, 3sin3g 1
sing 'cos3g — cos3^
= \ + 1 + \
cosg-cos/3^ , /tt ^\ /27r ^V
'^ cosg + cos( K- — p) cosg — cos! -^ — /5)
46. (a sin ^ + 6 cos 0)(a sin 1/^ + 6 cos i/r)sin(0 — yp)
+ (a sin \/r + 6 cos V^)(a sin 0 + 6 cos 0)sin(i/r — 0)
4- (a sin 0 + 6 cos 0){a sin ^ + 6 cos 0)sin(0 — 0)
+ 4(a2 + 62)sin(9!> - V^)sin(^ - 0)sin(0 - 0) = 0.
47. 2 (cos y8 cosy -cos g)(cosy COS g-cos/3)(cos g cos^-cosy)
+ sin^g sin^/? sin^y — sin2g(cos ^ cos y — cos g)^
— sin2/3(cos y cos g — cos /3)2 + sin2y(cos g cos /5 — cos y)^
= (1 — COS^g — COS^^ — COS^y + 2 COS g COS /3 COS yf.
48. If^+5 + C' = 7r, then
sin(^-a) mi\{G-A) ^m{A-B)
sin -4 sin B sin (7
4sin(^-(7)sin((7-^)sin(^~^)^
■*■ sin 2^ + sin 2j5+sin 2(7
49. If ^+5 + (7= X, then
(2/4-2; COS ^ )(0 4-a5 cos 5)(fl?+2/ cos (7)
+ (2/ cos A-\-z){z cos -B + a;)(a; cos (74- 2/)
vanishes, if x sin J. 4- ?/ sin 5 4-0 sin (7= 0.
50. If g, ^, y, ^ be the angles of a quadrilateral, then
tan a tan ;8 tan y tan ^^tan « + tan ^+tan y+tan 6
cot g4-cot p4-cot y4-cot S
51. If ^4-54- (7= TT, then
sin A cos(J. -5)cos(^ - (7)4- sin B cos(B - C)cos{B - A)
H-sin (7cos((7-^)cos((7-5)
= 3 sin J. sin B sin (7+ sin 2 J. sin 2B sin 2(7.
TWO OR MORE VARIABLE ANGLES. 155
52. If J. +^+(7= TT, then
sin^^sin^a+ sin^asinM + sin^^ sin^^ - sinM sin^^sip^g
cos^ J. cos^^ cos^C
= (tan B tan (7+ tan G tan ^ + tan A tan -B)^.
53. If ^+5+(7=7r, and n be any integer, then
tan nA +tan 7i5 + tan nG= tan -Ji^ . tan nB . tan ^iC.
54. If J.+5+a=x, then
cos 2^ (tan 5 — tan G) + cos 25(tan G— tan ^)
+ cos 2(7(tan J. — tan B)
2 sin(^- (7)sin(a-^)sin(^ -^)
~ cos A cos J5 cos (7
•cos a;
cos a;*
trtr To ^ l/x «x ^\n COSa + (
55. COS 2 tan-^tan rrtan ^ =t-;
L \ 2 2/ J 1 + cosa
56. tanr2 tan-^tan % tan(j-f )|1 =_ii5^£2i^.
L I 2 \4 2/JJ sm/3 + cosa
57. If 2^/ = aj + sin ~ \a sin ic), then
tan(a;-2/)=Y^'tan2/.
58. If It = cot ~ ^x/cos a — tan ~ ^Vcos a, then
sin u = tan^-.
59. If 2/ = tan"^ . , find the value of a; in
vl + aJ +v 1— a?
terms of 2/.
60. If sin(0+</)-\/r) = sin(a + /3), cos(V^ + ^-9!>)=cos(y + a),
and tan(0 + i/r — 0) = tan(y8 + y), find the general
values of 6, 0, yjr in terms of a, /5, y.
61. Find the general value of 20 from the equation
tan0+tan(j + 0) = 2.
156 CIRCULAR FUNCTIONS OF
Solve the following equations (62-75), 0, cp and x being
the unknown quantities :
6 2. cos^^ 4- cos 0 = 1 = sin^O + sin (p.
63. p sin^0 — q sin*0 =p,
p cos^O — q cos*0 = q.
64. ^3 cos 20 -^2 cos a = ^2 sine -sin 2a
^„ ^ 2cos(0 + a)cos(O — a)
•cos(0 + a) + cos(0 — a)
66. cos(20 4- 3a)cos(2e - 3a) - 2 cos a cos 3a cos 20 + cos^a = 0.
67. 1 — cos 20 = 2(cos acos 0— cos 2a).
68. (1 + sin 0)(1 - 2 sin Of = (1 - cos a)(l + 2 cos of.
69. sec 40 -sec 20 = 2.
70. a tan 0+6 cot 0 = c, by the aid of trigonometrical
tables.
K^ mtan(a--0)_J cos 0 \^
n tan 0 ~ \cos(a — 0)/
72. COS0 + COS ^ + cos a = sin 0 + sin 0 + sina, 0 + 0 = 2a.
73. cos30 - cos 0 sin 0 - sin30 = I.
74. cos(20 + 0) = sin(0-20),
cos(0 + 20) = sin(20-0).
75. sm ^ — hsin ^ — = ^.
76. If sin? = =^^ — ^, trace the changes in 0 as 0 in-
2 1 + COS0 ° ^
creases from 0 to ^- ; and find cos 0, sin 0 and
tan <p in their simplest forms as functions of 0.
Trace the changes in sign and magnitude, as 0 increases
from 0 to 27r, of (77-82) :
^y sin 30
'' cos 20*
78. sec 0-3 + 2 cos 0.
TWO OR MORE VARIABLE ANGLES. 157
79. cos(7r sin 0).
80. sm(7r cos 20).
81. cos(7r sin 26).
82. cos(j7r cos 6) — sin( Jtt cos 6).
Eliminate 0, or 6 and (^, from the equations (83-88) :
83. tanO + tan^ = a, cot0 + cot0 = 6, 0 — (p = c.
84. sin0 + sin^ = a, cos 0 + cos 0 = 6, (1 — c^)tan(O+0) = 2c.
85. c = acos 0 + 5 sin 0 = a cos{0 — a) + h sin{0 — a).
86. - = cos0 + cos20, | = sin0 + sin2a
87. a sec(0 -a) = h sec(0 - jS), a'sec(0 - a) = h'sec{e - /^O-
88. tan 0 + tan ^ = a, tan 6 tan ^(cosec 20 + cosec 2^) = 6,
cos(0 + 0) = c cos(0 - 0) ;
and, if CL = —m and 6 = 2(^^2 — 1), find the least
positive values of 6 and 0.
89. If !ir^=!;z&=c,
cos t7 sm 0
and =——^^=:c,
cos 0 sin 6
where c and c' are both positive or both negative
quantities, then (a — ay + {b — h')^ cannot be greater
than (c + c')2.
90. If a and /5 be two different values of 0 satisfying the
, . cos 0 . sin 0 1
equation — = — = -,
a 0 c
then a cos — ~- = h sm '^ = c cos — q^-
A /!i A
91. If a and ^8 be two different values of 0 satisfying the
,• cos 0 . sin 0 1
equation -+
then ^-M^-
158 CIRCULAR FUNCTIONS OF
92. Prove that
2 cos(/3 - y)co8(0 + ^)cos(0 + y)
+ 2 cos(y — a)cos(0 + y)cos(0 + a)
+ 2 cos(a - /3)cos(0 + a)cos(0 + /3)
-cos2(e+a)-cos2(O+)5)-co.s2(0 + y)-l
is independent of 0, and exhibit its value as the
product of cosines.
gg j£. sin(0 + g) _ 1 + e cos(0 + ffl
sin a lH-ecos/3 '
/I
find tan - in terms ot a, /8, and e.
94. If a; = acos0 — rcosf^+^j,
2/ = 6 sin 0 + r sin(^| + 0 Y
^ , /cos Q , sin 0\
2cos<^ = r(— +-^-)
^ . ^ fcoaO sin6\
(a + h^Sf-^ib + ajSy-^-
95. If tan0tan0 = J(|^),
then (a - 6 cos 20) (a - 6 cos 20) = a^ - h^
96. If ^3-tan0 = -l--tan0,
x/3
then tan(0-|) = 3tan(0-|).
97. If tan ^0 = tan^ J0, and tan ^ = 2 tan a,
then 0+0 = 2'M7r + 2a.
98. Extract the square root of
(2 + 2 sin 0 + cos 0)2+ (2 + sin 0 + 2 cos 0)2.
TWO OR MORE VARIABLE ANGLES. 159
99. If Q and 0 be acute angles, such that
3sin20 + 2sm20 = l,
Ssin 20-2 sin 20 = 0,
prove that 0 + 20 = J;
and find sin Q and sin 0.
100. If ^2cos0 = cos0 + cos30,
and ^/2 sin 0 = sin 0 — sin^0,
then , ±sin(0 — O) = cos20 = J.
101. If^cos0 + 5sin0-(7vanishfora = a,0 = ^,0 = a+/3,
respectively, then A — G.
102. If cos0 + cos0 = a and sin0+sin0 = 6,
find the value of cos(0 — 0) and cos(O + 0) in terras
of a and h.
103. If 0^ aDd ^2 be two roots of the equation
Aco^e+B^me+G = Q,
such that the cosines of 0^ and Og ^^^ ^^^ equal,
then cos(0, + 0,) = ^!=J'.
104. 16 cos t;— cos -zr^ cos -=^ cos -=^ = 1.
15 15 15 15
105. If ; S = l, prove that both the numerator
sec a + sec /5 ^
and denominator of this fraction must vanish un-
less
a = 27i7r + |,
or .
^ = 2^7r + |,
or
a + ^ = (2n + l)7r.
160 CIRCULAR FUNCTIONS OF
106. If a, P, y, S be all different, then
cot(a-/3)cot(a-y)cot(a-6) + cot(/3-y)cot(/3-^)cot(^-a)
+ cot (y-^)cot(y-a)cot(y-^) + Cot(^-a)cot(^-^)cot(^-y) = 0.
107. If a, P, y, 6 be all different, then
Cot(a-y)cot(a-^) + cot(a-^)cot(a-/3) + cot(a-j8)cot(a--y)
+ cot(^-^)cot(^-a) + cot (^-a)cot(/3-y) + cot(;8-y)cot(/3-^)
+ COt(y-a)cot(y-^) + cot(y-^)cot(y-^) +COt(y-(5)cot(y-a)
+ cot(^-^)cot((5-y)+cot(^-y)cot((5-a)+cot((5-a)cot(5-^)=-4.
108. If a, A y, S be all different, then
cos 2a , cos 2^
. a-B . a-y . a-^ T~^-y . /3-5 . S-a
sin — ^ '^ sin —^ sm — ^ sm s^ sm ^— «— sm -'— ^ —
cos 2y , cos 2^
. y — ^ . y — a . y — & - S—a . S — B . ^ — y
sin ^^-^ sm ^^-s— sm ^ '^ sm —^ sin — o^ sin — ^
109. If (a+6)tan(e-0) = (a-6)tan(e+0),
and a cos 20 + 6 cos 2^ = c,
then fe2 _ ^2 _ 2c<x cos 2^ + a\
110. If ajcos20+2/sin20 = a(cos0X,
and ccsin 20 — 2/cos2O = a(cos0)*'-isin0,
l-C+S)'
111. Eliminate 0 from the equations
, , sin a cos 0 — sin i8 sin 0
tan0 = ^ . ^,
cos a cos t^ — cos p sm t7
, , sin a sin 0 — sin i8 cos 0
tanY^ = r— 5 % -.
cos a sm t7 — cos p cos t7
TWO OR MORE VARIABLE ANGLES. 161
112. Eliminate 0 from the equations
cos^e 3sin2e sin^e_ 4
a^ ac c^ ac
cos6.8m6_ 2
c a J'oi'c
113. If cosOcos0 = sin(a — /3)sin(a + j8),
and sin(0 — 0)sin(04-0) = 4cosacos/3,
find cos 0 and cos cj).
114. If a; + 2/+2; = 2sin0, a;2+2/H2;2 = cos20,
2(a;3 + 2/^+;s^) = 3 sin 30 and 2iC2/^ = 3 sin 0,
then f) = TiTT or TiTT ± J.
o
115. Find x from the equation
3 tan-i(aj + l) = 2 tan-X!:c-l)+tan-i
2-a5
116. Trace the chancres in siffn of — j^- . ' J., as 0
^ ^ cos(7r sm 0)
increases from 0 to tt.
117. If cos i/r = cos 0(tan a sin ^ — cos 0)
= sin 0(sin 0 —cot a cos 0),
prove that either yp- is an odd multiple of -^^ or else
i/r=(2'?i+l)7r±(0+^), where -n is an integer.
118. Prove that the expression
a cos^O + 26 cos 0 sin 0 + c sin^O
may be written in the form
a + c
where m=— ^ , 7i = o, tan0 = — .
2 ^ n
Hence, prove that the greatest and least values
of the given expression are the roots of the equation
{x — a)(x — c) = h^.
162 CIRCULAR FUNCTIONS OF
119. If a' = acos20+2^cos^sia04-?>sm2a,
W ={h — a) cos 0 sin ^ + A, cos 20,
})=a sin^^ — 2/i cos 0 sin 6+h coa^O,
then
a'cos\<p -6) + 2h'cos{<p - 0)sin(9!) - 0) + 6'sin2(0 - 6)
= (X cos^^ + 2h cos 0 sin ^ + 6 sin^^.
120. If y^-{-2yz cos e + z^ = p\
z^ + 2zxcos<l) + x^ = q\
x^+2xy cos \jr + y^ = r\
and 0 + ^ + V^ = 7r,
then 4(2/0 sin O + zxsiuip + xysm \]rf
= 2 (^V + ry +^2^2) _p4 _ ^4 _ ^_
121. If a sin a = 6 sin ft a sin |8 = 6 sin 0, a — /3 = 0 — (l),
and a, 6 be not numericall}^ equal, then a = /5 + '>?x,
0 = (p+n'7r, where ti is an integer.
122. If x = y cos Z+z cos Y,
2/ = 0 cos X+ a? cos Z,
and if jr+ Y+Z be an odd multiple of tt, prove
that 0 = oj cos F+ 2/ cos X.
Hence, prove that
C0SZ = ^^^— ^TT .
2yz
^go Tf sin ra _ sin (9* + 1 )a _ sin(r + 2)a
I ~ 7n ~ n '
, cos ?-a _ cos(r + l)a _ cos(r + 2)a
^., 2m2 — ^(Z+'yi)~ 7n{n — l) ~n{l+n) — 2m^'
l24i. If tan (cot 0) = cot (tan 6), shew that the real values
1 of 0 are given b}^
sin 20 = 7^5 TT^'
72, being any integer, positive or negative, except
TWO OR MORE VARIABLE ANGLES. 163
1 2o. cos*^ + cos^-^- + cos*-^ + cos^^ = jg.
126. sec*^+sec4^ +sec*-|" + sec'i^ = 1120.
10H. TT 27r 3x 47r Stt Btt Ttt /IV
127. cos ZTF cos vv cos 7^ cos :rr cos TV cos ^r^ cos T^ = 7: ) '
15 lo 15 15 15 15 lo \2/
1 28. Solve the equations
cos (0-\-a) = sin 0 sin ^8,
cos(^ + /3) = sin 0 sin a,
and shew that, if 0^, ^g ^c two values of 0 not
differing by a multiple of tt,
, /^ , J \ sin 2^
tan (0, + 02) =sIn2-^_cos2^s^^-
129. If -^ — h 7- 7) have its least positive value, prove
that 6 is greater than ^3 — 1.
130. If A, B, G be the angles of a triangle, and x, y, z
any real quantities satisfying the equation
2/ sin (7—0 sin J5 _ 0 sin J. — ic sin C
aj— 2/ cos 0—0 cos ^ 2/~^cos^— ajcos C
then
y
sin A sin B sin C
131. If (sin^a — sin2^)(sin2a — sin^y) = sin2^sin2y cos*a,
then tan^a = tan^/? + tan^y.
132. If ^+5+0=7r,
and sin^ft) = sin (J. — w) sin {B — o)) sin ((7— o)),
then cot ft) = cot^+cot.B+cot 0,
and cosec^a) = cosecM+cosec^-B + cosec^C.
133. If C0s(^-y) + C0s(y-a) + C0s(a-/5)=-f,
then cos(a + 0) + cos(/3 + 0)+cos(y + 0)
and sin (a + 0) + sin (^8 + 0) + sin (y + 0)
vanish whatever be the value of 0.
164 CIRCULAR FUNCTIONS, ETC.
134. If cos(/3-y) + cos(y--a) + cos(a-)8) = -f,
prove that cos na + cos ^1/3 + cos ny is equal to zero,
unless 71 is a multiple of 8, and that, if ti be a
multiple of 3, it is equal to 3cos j7i(a+/3-f y).
135. If cos2a(2/^cos2y8 + 2;2cos2y — aj^cos^a)
= cos2|8(2;2cos2y + aj^cos^a — y^cos^fi)
= cos2y(a32cos2a + y^co^^^ — z^co^^y),
and if cos^a + cos^/S + cos'^y = 1 ,
then ±^ — = ± . o= ^— — •
sm a sin p sm y
136. If coso+cos/8+cosy+cosacos/3cosy = 0,
then
cosec^a + cosec^/3 + cosec^y ± 2 cosec a cosec /3 cosec y = 1 .
137. If ^+5+a=27r,
and if
2/^+2;2_22/2;cos-4=2;^+a;2 — 22^fl3cos5=a;2+2/^- 2£C2/cos C,
prove that each of these quantities is equal to
2 .
'^z sin J. +0£c sin B+xy sin 0).
V3'
CHAPTER IX.
RELATIONS BETWEEN THE ELEMENTS OF A
TRIANGLE, SOLUTION OF TRIANGLES
AND PRACTICAL APPLICATIONS.
§ 1. Relations between the Elements of a Triangle,
118. If ABC be any triangle, we denote, as in Chapter
v., the lengths of the sides opposite the angles A, B, and
G by a, h, and c respectively.
119. In any triangle, a = b cos (7+c cos B, etc.
Let ABC be the triangle, the angle C being either
C D B C
acute, obtuse or right. Draw AT) perpendicular to BG,
produced if necessary.
Then 5(7=^5 cos 5+^0 cos (7, if G be acute or right,
or AB cos B — AG cos(7r — G), if G be obtuse or right,
BG = AB GOB B + AG cos G, in every case,
i.e. a = 6 cos (7+ c cos B.
165
166 RELATIONS BETWEEN THE
Similarly,
6 = c cos ^ + a cos (7, and c = a cos 5+ 6 cos A.
Cor. — If cZ, e, /denote the lengths of the altitudes AD^
BE, OF respectively, it may be shewn that
2cZ = 6sin(7+csin5, 2e = c sin J. + a sin (7,
and 2/= a sin 5 4- 6 sin J..
120. The sides of a triangle are proportional to the
sines of the opposite angles.
Using the figure and construction of the last article, we
have AD = ABsmB.
Also AD = AG sin G, if G be acute or right,
or AGsm(7r — G), if G be obtuse or right,
AD = AG ain G, in every case.
ABBmB = AGsmG,
6 : c = sin5:sinC.
Similarly, a:b = 8mA : sin B,
a _ b _ c
sin A sin B sin G
It will be seen, in the next chapter, that each of these fractions is
equal to the diameter of the circumcircle of the given triangle.
121. To find the cosines of the angles of a triangle in
terms of the sides.
(See figure and construction of art. 119.)
By Euclid II. 13 and 12, we have
A^ = AG^+BG^-2BG.GD, if G be acute or right,
or AG^+BG^+2BG. GD, if G be obtuse or right.
Now, GD = AG cos G, if G be acute or right,
or AG cos(Tr — G), if G be obtuse or right,
A^ = AG-^+ BG^ - 2BG .AG cos G, in every case,
i.e. c^ = a^ + h^- 2ab cos G..
ELEMENTS OF A TRIANGLE. 167
Similarly,
a2 = 62_j_c2-26ccos^ and h'^ = c^+a^-2caco^B,
and cos C=
2ab
It will be noticed that each of these expressions is a proper
fraction, if (taking the first) b^+c^-a'^<2bc, i.e. if })^ + G^-2hc<a\
i.e. if (6-c)2<a2 or b-c<a or b<c+a, which is the case (Eucl.
I. 20).
Also, that cosil is positive or negative, according as a^< or
> b^+c^, i.e. as A is acute or obtuse.
122. To find the cosines of the semi-angles of a
triangle in terms of the sides.
2 cos^-^ = 1 + cos A
_ 62 + 02-02
~ ^ "^ 26c
_2bc+^+f-a^
~ 26c
(6 + c)2-a2
26c
_(6 + c+a)(6 + c-a)
~~ 26c
Let a + 64-c, or the perimeter of the triangle, be de-
noted by 2s. Then, 6 + c — a = 2s - 2^^,
^A 2s.2(s-a)
C0S^-^= -~r -,
2 46c
A ls(s - a
o- -1 1 ^ Ms -6) , G ks-c)
Similarly, cos ^ = ^^ ^^ and cos^ = ^^ ^^ >
168 RELATIONS BETWEEN THE
Each of the expressions under the radical sign is positive,
for «-a, 8-h and s-c are all positive (Eucl. I. 20). Again, each
is a proper fraction, for (taking the first), s(5-a)<6c, if
(6+c)2-a2<46c, i.e. if h^-'ibc+c'^Ka?^ i.e. \ih-c<a or 6<c+a,
which is the case.
Hence, these expressions give possible values for the required
cosines.
123. To find the sines of the semi-angles of a triangle
in teiins of the sides.
2 sin^-^ = 1 — cos ^
26c
_26c-b^-cHa^
26c
^a^-{h-cf
'2bc
_(a—b+c)(a+h — c)
26c
^^_2(8-6).2(8-c)
^'"^ 2" 463 '
=v
^^^A=J(s-b)is-c)
2 \ 6c
cj- .1 1 • ^" l{s-c)is-a) , . G ks-a){s-h)
Similarly, sm ^ = ^ ^^ and sm ^ = y ^ -^^ — -'.
Also, as in the last article, it may be shewn that these expres-
sions give possible values for the required sines.
124. To find the tangents of the semi-angles of a
triangle in terms of the sides.
^ A . A A
tan ^ = sin -^ h-cos -^
ELEMENTS OF A TRIANGLE. 169
'i
_ l(s-b)(s-c)
s(s — a)
Similarly, tan|=/-l^^|^
and u.l=J^r^B.
2 \ s{s — c)
Each of the fractions under the radical sign is positive, for every
factor is positive ; and, therefore, these expressions give possible
values for the required tangents,
125. To find the sines of the angles of a triangle in
terms of the sides.
Sin A = 2 cos y sin -^
_^ls{s-a) Us-hXs-c)
he V he
o
= 5^Vs(s-a)(s-6)(s-c).
The expression s/s{s — a){s — h){s — e) is usually denoted
hy&
Hence, sin A = 2Slbc, and similarly, sin B = 28/ ca and
sin(7=2>Sf/a6.
As before, the expression under the radical sign is positive.
Also, 2Slbc is a proper fraction,
if As{s-a){s-bXs-c)<bh\
if {a-\-b + c){b + c-a){c + a-bXa + b-c)<Abh\
if 26V + 2c2a2 + 2a^b^ - a^ - 6* - c* < 4b^c\ .
if aH6Hc4 + 26V-2c2a2-2a262^0,
if (a2 - 62 - c2)2 > 0, which is the case.
Hence the expressions obtained above give possible values for
the required sines.
170
RELATIONS BETWEEN THE
126. To 'prove that, in any triangle,
B-G h-c ^A ,
taii^- = j^^cot2,etc.
(1.) Algebraical proof .
6-c^sin^-sin G^ 2 cos i{B+G) sin J(^-C)
h+c sin^-hsin (7~ 2 sin 1{B+G) cos \{B-G)
,B-\-C^ B-G
= cot — ^ — tan — ■= — .
B-G h-c, B+G
tan , =r-— tan ^
2 6+c 2
6-0 ,^
6+c 2
Similarly, tan
G-A c-a ,B
-2- = c-+^^^^2-
ind
tau — 7i — = — r-r cot 7j-
2 a + 6 2
(2.) Geometrical proof. — Let ABG be the triangle.
^ From ^(7 and GA pro-
duced, cut off J.Z) and
AE equal to ^j5. Join
BD and 5^, and draw
D-Fperpendicular to BD,
meeting BG in F. Then,
BBE is a right angle,
and, consequently, BF
C F
and BE are parallel.
Now, angle ^5i) = angle^D5=p^£^:
B+G
and angle DBG= angle ADB — angle A GB
B+G ^_5-C
—2 ^-"2-
Again,
ELEMENTS OF A TRIANGLE. 171
h-c_CD_DF_DF BE
h+c~GE~BE~DB'DB
^ B-G , B+G
= tan — ^ — Htan — ^— •
B-G h-c ,A
tan — rr— = 7—— cot -^.
2 6+c 2
127. Example 1. — From the formulae a = bcoaC+ccosB, etc.,
deduce the expressions for the cosine of an angle of a triangle in
terms of its sides.
We have a = 6 cos C+ c cos B,
b = c cos A-\-a cos C,
c=acosB+bcoaA.
Multiply both sides of these equations by «, 6, c respectively, and
subtract the first from the sum of the second and third : then,
b^-\-c^-a^ = be cos A+ab cos C + ca cos B+ be cos A-ab cos C-ca cos B
= 2bc COS A,
cos -4 = —} .
26c
Example 2. — Having given that the sides of a triangle are pro-
portional to the sines of the opposite angles, deduce the expression
for the cosine of an angle in terms of its sides.
We have a/sin A = 6/sin B = c/sin C.
Let each of these fractions equal d, so that
a=d8in A, b=^dsmB 3ind c=d sin C.
Then, 62+c2 - 26c cos A=d\smW + 8Ui^C- 2 sin B sin Ccos ^)
= d\sm B(sin 5 — sin C cos A)-\- sin (7(sin C— sin 5 cos A )]
= c?2[sin B{sm( 6^ + ^ ) - sin Ccos ^ } + sin r{sin( J + ^) - sin 5 cos ^ }]
=c?2[sin B cos Csin A +sin (7 sin A cos B]
= dhin A sin{B+ C) = dhin^A = a^.
Example 3. — If ^^' be a median of the triangle ABC, then
2cotJ^'^ = cotC-cot^.
Let 6 denote the angle AA'B;
then,
by art. 120,
,«.T '(■>*<-] ■■
wf:M
BA'
sin A'AB
_sin{d+B)
AA'
sin A' B A
sin B '
and ^^'-.
_sinA'AG_
_sin{e-C)
sin C
A A'
sin A'GA
A' AC -+C .: fi! i^t ^ ^ ' (L
172 RELATIONS BETWEEN THE
But BA' = CA\
s,\nCs,m{d-\-B)=BmB«m{d-C),
sin (7(8in 9co^B-\- cos 6 sin B) = sin 5(sin OcoaC- cos ^ sin C),
cos ^(sin 5 sin C+sin B sin C)=sin ^{sin 5 cos C- cos ^ sin C),
o«.^^■/^ sin jB cos C- cos 5 sin (7
2cot6/= : — - . -, ,
sm BamC
2cotAA'B=cotC-cotB.
Example 4. — To find the relation between the lengths of the six
lines joining any four points in a plane.
Outline of proof : Let ABO be a triangle, a, b, c the lengths of
its sides, and let P be any point, either within or without the tri-
angle : join PA, PB, PC and let a, y8, y be the lengths of these
lines. Let the angles between each pair of the last three lines be
denoted by 6, <f>, yjr ; one of these angles being greater than two
right angles when P is without the triangle ABC. Then,
e + cfi + ylr = 27r,
and, consequently,
cos'-^ + cos^^ + cos^yf/ - 2 cos $ cos <^ cos i/r = 1 .
Express each of these cosines in terms of the given lines by art.
121 ; multiply both sides of the resulting equation by 4a^ /3^y^ ; and,
after reducing, the required relation will be obtained, namely,
a2a2(a2 + ^2 _ J2 _ ^2 _ ^2 _ ^2) + 52^(2,2 4. ^2 _ ^2 _ ^2 _ ^2 _ ^^2)
+ C2y 2(c2 + y 2 _ ^2 _ 52 _ ^2 _ ^) + «2^2y2 + ^2 2^2 + c2a2/?2 + a262c.2 = Q.
Examples XIV. a.
1. Find the cosines of the angles of a triangle whose
sides are 10, 13 and 15.
2. The sides of a triangle are x, y and /^(x^ + oi^y + y^),
find the greatest angle.
3. If a = ^5, 6 = 2, c = JS, then 8 cos ^ cos 0= 3 cos B.
4. If ^=45° and 5=60°, then 2c = a(l + V'^).
5. If ^ = 30°, 5 = 45" and a = 8j2, find h and c.
6. Find the cosines of the semi-angles of a triangle
whose sides are 1, 4 and 4.
ELEMENTS OF A TRIANGLE. 173
7. Find the sines of the semi-angles of a triangle whose
sides are 35, 15 and 34.
8. Find the tangents of the semi-angles of a triangle
whose sides are 25, 52 and 63.
9. Find the sines of the angles of a triangle whose sides
are 193, 194 and 195.
10. Find the tangents of the angles of a triangle whose
sides are 10, 35 and 39.
11. If 6 = 5, c = 3 and J. = 120°, find tan J(5-(7).
12. If a = 2, 6 = ^3 and C=30°, find A, B and c,
13. (6 + c)cos J.-l-(c + a)cos5+(a + fe)cosC=aH-6-fc.
14. asinJ[-f-6sinJ5+csin C=2(cZcosJ.+ecosJ54-/cosC).
15. sin {A-B)'. sin 0= a^-h^: c\
16. cosi(^-5):cosi(^-f5) = a+6:c.
17. s^ = >S^ cot -^ cot ^ cot ^.
18. If D be any angle,
a sin (D - 5) -I- 6 sin (D + ^) = c sin Z>.
19. c = 6cos^±^(a2-62sinM).
90 cos 2 J. cos2^_l 1
21. a(sin2|-fsin2g + 6(sin2^+sin2^)
+ c(sin2| -l-sin^l) ^ J(a-f 6-Fc).
Examples XIV. b.
1. The sides of a triangle are 2, ^2 and ^^3 — 1 ; find
its angles.
2. The sides of a triangle are x^-\'Xy-^y^, 2xy + y^ and
x^ — y'^] find the greatest angle.
174 RELATIONS BETWEEN THE
3. The sides of a triangle are 3, 4 and -s/38; shew,
without using tables, that the largest angle is
greater than 120°.
4. If 5 = 15° and (7=30°, shew that a^cj%
5. If^ = 15°, J5=105°and c = ^6, find a and 6.
6. Find the cosines of the semi-angles of a triangle
whose sides are 125, 154 and 169.
7. Find the sines of the semi-angles of a triangle whose
sides are 11, 25 and 30.
8. Find the tangents of the semi-angles of a triangle
whose sides are 25, 51 and 52.
9. Find the sines of the angles of a triangle whose sides
are 125, 123 and 62.
10. Find the tangents of the angles of a triangle whose
sides are 21, 89 and 100.
n. If a = 15, 6 = 8 and 0=90°, find tanJ(^-5).
12. If a=90° and a : 6 = ^3 + 1 : V3-l,find^,5, and c.
13. (6-l-c) sin ^ 4- (c-t-a) sin B-\-{a-\-h) sin 0= 2(c?+6+/).
14. a sin {B- G) + b sin {C-A) + c sin {A -B) = 0.
15. smi(A-B):smi{A + B) = a-b:c.
16. tan(4+5) = ^tan4.
\z / c — o 2
17. 8S= abc cos -^ cos -^ cos » .
Zi z z
18. bccosA-\-cacoaB+ahcosG=i(a^+¥+c^).
19. c2 = (a-6)2cos2^+(a+6)2sin2?.
20. a^cos 2B -f b^cos 2A = a^+b^- 4<ab sin A sin B.
^ /sin A +sin 5-|-sinCY_a cos J. + 6 cos 5H-c cos (7
' \ a + b+c / ~ 2abc *
I
ELEMENTS OF A TRIANGLE. 175
Examples XV.
1. I{ ABGD be a quadrilateral, then
ABgosA-BGgos(A-\-B) + CDcosD = AD,
eindABsmA-BGsm(A + B)-CDsmD = i).
2. ^2 : fe2 : c2 : : cot 5 + cot 0 : cot 0+ cot ^ : cot 4 + cot 5.
3. he sinM + ca sin^^ + ah sm^G
= {a+h + c){a cos B cos G+b cos GcoaA+c cos J. cos B).
4. a2+62+c2<2(6c + ca + a6).
5. The sides of a triangle are a, h, c, and the angle G is
120°. Shew that, if we form the triangles whose
sides are a, a + h, c and a + b, h, c, the angle
opposite c will in each case be 60°.
G, If a, h, c be in arithmetical progression, then
Stan ^tan^ = l.
7. If the sides of a triangle be in arithmetical progres-
sion, and if a be the least side, then
, 4C-86
8. If G=2B, then c^ = h(a + h).
9. If ^ = SB, then sin B = |a/(^^)-
10. If the sides of a triangle be in arithmetical progres-
sion, the tangents of its semi-angles are in har-
monical progression.
11. Shew, trigonometrically, that, if any angle of a tri-
angle be bisected, the segments of the base formed
by the bisecting line, are in the ratio of the sides
of the triangle.
1 2. ahc{l — 2 cos J. cos B cos G)
= a^cos B cos G-\- 6^cos G cos A + c^cos A cos B.
176 RE LA TIONS BET WEEN THE
13. a^sin {B-C)^-hhm {C-A)^chm {A -B) = 0.
14. -^-T-cos^+ 1 , -cosjgH ri-cosC'=a* + 6* + c3.
6*c* c^a^ a^b^
15. If a = (c-a)(a-6), /3 = (a-6)(6-c), y = (6-c)(c-a),
then
aV + 62/32+cV-22)C/8ycos^ - 2cayacos5-2a6a^cosO
= {aa + h^ + cyf.
acos^^ + ^cos^-x + c COSM-
IC. Express ^V-^ -77 r - "tt- i^ terms of the sides.
^ cos^ + cos j5+cosC7
17. In a triangle ^5(7, ^D is drawn to meet BG, or BC
produced, in D, so that AD is equal to J-O; if the
sum of AB and AG is n times 5C, then their
difference is -th of BD.
n
18. If a cos J. = 6 cos jB, the triangle is either isosceles or
right-angled.
19. In any quadrilateral figure, the square on one side is
less than the sum of the squares on the other sides
by twice the sum of the products of these sides
taken two and two together and multiplied by
the cosine of the angle included between them.
20. If (a^ + b^) cos 2 A = b^-a\ the triangle is either right-
angled or two of its angles differ by a right angle.
21. P is any point in the base AB of a triangle ABG,
such that AP : PB = m : n, and the angle GPB is
6 \ shew that
(m+7i) cot 0 = 71 cot A—m cot B.
22. On the base BG of a triangle ABG, two points, Q, R,
are taken, so that BQ = QR = RG; then
sin BAR . sin C^Q= 4 sin BAQ . sin GAR.
ELEMENTS OF A TRIANGLE. 177
23. If a = 26, and A — SB, find the angles of the triangle
and the ratios otAB to BC and AG.
24. a^cos 2{B-C) = h'^cos 2B+c^cos 2C+2hccos{B-G).
25. Express the bisector of an angle of a triangle in
terms of the sides, and shew that the greatest
bisector is that which bisects the smallest angle ;
also, if a, /3, y denote the lengths of the bisectors,
that -cos-r- + ^cos-^+-cos^=-+r + -.
a zpzy z a 0 c
26. If the squares of the sides of a triangle be in arith-
metical progression, the tangents of the angles are
in harmonical progression.
27. The measures of the lengths of sides of a triangle
are three consecutive integers, and the largest
angle is double of the least ; find the sides.
28. If a triangle be such that it is possible to draw a
straight line AD meeting BG in D, so that the
angle BAD is one-third of the angle BAG, and
also BD is one-third of BG, then
a262 = (62_c2)(62 + 8c2).
29. A triangle is turned round each of its three sides
successively till the vertex comes into the same
plane; if the vertices be joined forming a new
triangle whose sides are a', h\ d , then
^=l + 8cos^sin5sina
A . ,B . .G
cot
on
• cot^-fcotj5-hcot(7 a^-{-¥-\-c^'
31. a^cos^-|-6^cos5-f c^cos G=abc(l + 4!CoaAcoaBcosG).
32. If the sides of a triangle be in geometrical progression,
and if the altitudes be taken as the sides of a new
COt^+COt-^+COt^ („ + j^.,)2
178 RELATIONS BETWEEN THE
triangle, then the angles of this triangle will be
equal to those of the original triangle.
33. If the sides BC, GA, AB be divided in G, H, K, so
tha,tBG:GG=GH:HA=AK:KB; then
AG^+BR' + GK'
will be least when G, H, K are the mid-points of
the sides.
34. If p, q be the perpendiculars from Ay B on any arbitrary
line drawn through (7, then
a?-'p^ + ll^q^ — 2abpq cos G = a^bhin^G.
35. A triangle ABG, whose angles are given, is to be
drawn with its angular points on three given
parallel straight lines ; if the middle line, passing
through G, be at distances a, /3 from the other lines
passing through A, B, then the sides of the triangle
are determined by the equations
a _ h _ c
sin^~sini^ sinO
s/a^am^A + 2a^ sin A sin B cos G+ /Shin^B
~ sin A sin B sin G
36. If A and B be the greatest and least angles of a
triangle, the sides of which are in arithmetical
progression, then
4(1 — cos ^)(1 — cos B) = cos ^ + cos B.
37. If the squares of the sides of a triangle be in arith-
metical progression, then
sin3^^/a^-cV
sin jB ~ \ ac J '
38. The altitudes of an acute-angled triangle meet in 0,
and OA, OB, OG are taken for the sides of a new
triangle. Find the condition that this should be
ELEMENTS OF A TRIANGLE. 179
possible ; and, if it be, and the angles of the new
triangle be a, /?, y, then
_ , cosa , cos/3 , cosy . . „ n
l-\ -r-\ ^H ^=*secj4sec5sec(7.
cos ^ cos B cosG ^
39. If P be any point in an equilateral triangle ABC,
PB^4-PG^ — PA^
then co^iBPC- 60°) = 2PB PG
40. The sides of a given triangle are a, h, c, and the
angles A, B, G; if a point be taken within an
equilateral triangle, so that its distances from the
angles of the triangle are proportional to a, h, c;
then the angles between these distances will be
l+J, 1+5 and 1+0.
41. If X, y, z be the distances of any point in the plane
of an equilateral triangle, whose side is a, from
the angular points, then
2/%2 + 02^2 + rj^^yl _ ^4 _ ^4 _ ^.4 _^ ^2(^2 J^ylj^ ^2) _ ^4 ^ Q.
§ 2. Solution of Triangles.
128. In every triangle there are six elements, namely,
the three sides and the three angles ; any three of which
may be independent of one another, except the three
angles. The relations that have been proved, in the
preceding section, between the sides and angles of a
triangle enable us, as a rule, to determine the remaining
elements, if three independent elements be given.
180 SOLUTION OF TRIANGLES.
There are four cases to be considered, those in which
we are given : (1) one side and any two angles, (2) two
sides and the included angle, (3) two sides and the angle
opposite one of them, and (4) the three sides.
In the examples at the end of the last section, some
examples of the solution of triangles have already been
given, not however involving the use of logarithmic
tables. In the articles and examples which follow, we
shall consider more completely the method of solving
triangles with the aid of such tables.
129. Case I. — One side and any two angles being
given, to solve the triangle.
Let the side a and the angles B and G be given ; to
find 6, c and A.
By Eucl. T. 32, A^im^-B-C.
Also, &/a = sin5/sin^, (art. 120)
log 6 = log a + log sin J5 — log sin ^ .
Similarly, log c = log a + log sin (7 — log sin ^ .
130. Example 1.— Given a = 357, Br=AT 19' and C=58° 23'; to
find 6, c and A.
J = 180° -47° 19' -58° 23' = 74° 18'.
log 6 = log a + log sin B - log sin A
= log 357 + log sin 47° 19' -log sin 74° 18'
= 2-5526682
+ 1-8663534 - 1-9834872
= 2-4190216
-i -9834872
= * 2-4355344.
2-4355418 = log 272-61,
2-4355344 = log(272-60 + S),
2-4355259 = log 272-60,
159 : 85 :: -01 : 8.
SOLUTION OF TRIANGLES. 181
159) -85 (-005 .
795
55
6 = 272-605.
Similarly, log c = 2-4994036,
c = 315-794.
A=7i° 18'
b:
c-
= 74° 18' ^
= 272-6051.
= 31 5-794 j
131. Case II. — Two sides and the included angle being
given, to solve the triangle.
Let the sides 6, c and the angle A be given; to find
a, B and G. ' ' . ^
B—Gh — c,A , , ■,or>\"
tan — ^ — = J—- cot ^, (art. 126)
B—G A
log tan -y- = log(6 - c) + log cot ^ - log(6 + c).
From this equation, we can find ^(B—G), and we know
that 1(5+0) = 90°- J^. Hence, B and G can be found.
Again, as in Case I.,
log a = log h + log sin ^ — log sin B.
132. We may, if we please, determine the side a, with
the aid of a subsidiary angle, from the formula
a^ = h^ + c^-2hc cos A,
without first finding the angle B or G. Thus
a^ = h^-2bc + c^-\-2bc(l-cosA) ■
=:{b^c)^ + 4>bcsm^~
=(^-Ki+A^^^4}
If, now, we put tsin^O = -pi r^sin^— ,
^ (6 '- c)2 2
the last equation becomes a = (6 -- c)sec 9.
182 SOLUTION OF TRIANGLES.
133. Example 2.— Given 6 = 541, c = 2b% ^=48° 26'; to find
B, C and a.
B — C A
log tan = log(6 -c)~ log(6 + c) + log cot —
2 Z
= log 282 - log 800 + log cot 24° 1 3'
= 2-4502491
+ 0-3470119 - 2-9030900
= 2-7972610
-2-9030900
= T-8941710.
1-8943715 = log tan 38° 6',
1-8941710= log tan 38° 5' S",
1-8941114= log tan 38° 5',
2601 : 596:: 60: S.
596
60
2601)35760(13-7
2601
9750
7803
19470
18207
|(5-C) = 38°5'14".
Now,
Ki?+O=65°47'0",
i=103°52' 14",
^=27° 41' 46".
Again, as in art. 129,
log a = log b + log sin A — log sin B
= log 541 + log sin 48° 26' - log sin
103°
52' 14'
= log 541 + log sin 48° 26' - log sin
76°:
r46"
= 2-6200582,
a=416-925.
^ = 103° 52' 14""
C=27°41'46" ■•
a =416-925
I
SOLUTION OF TRIANGLES. 183
134. Case III. — Two sides and the angle opposite one
of them being given, to solve the triangle.
Let the sides a, h and the angle A be given ; to find
c, B and C.
sin j5/sin A = hja,
sin jB = 6sin^/a;
log sin 5 = log 6 + log sin J. — log a.
This equation does not, however, always lead to a
definite result. For, since the angle is to be determined
from its sine, it is possible that there may in certain
cases be two angles satisfying the equation, both less
than two right angles and one the supplement of the
other. The following cases may occur : —
(1) a>h. Then A is greater than B, and, therefore,
the less only of the two values obtained for B is
admissible, for there cannot be two obtuse angles in a
triangle.
(2) a = 6. Then A is equal to B, and, again, only the
less of the two values obtained for B is admissible.
(3) a<h. There are here three sub-cases, according as
a is greater than, equal to, or less than 6 sin ^.
If a>6sin^, the expression for sin 5 is a proper
fraction, and both values obtained for B are admissible,
since B is greater than A. Thus, there are two triangles
which satisfy the given conditions, the value of B in one
being the supplement of that in the other. This is
known as the ambiguous case in the Solution of Triangles.
If a = b sin A, then sin 5 = 1, and B = 90°. Since the
supplement of 90° is also 90°, there is only one triangle,
or, rather, there are two coincident triangles, satisfying
the given conditions, the angle B in each case being a
right angle. There is thus no ambiguity in the solution.
184
SOLUTION OF TRIANGLES.
If a < 6 sin A, the expression for sin 5 is greater than
unity, and there is no real value of B. Thus, there is no
triangle satisfying the given conditions.
We may sum up these three sub-cases (when a<h)
as follows: There are in each sub-case two triangles,
and they are real and different, real and coincident, or
imaginary and different, according as a is greater than,
equal to, or less than, 6 sin J..
Having found one or both values of B, the correspond-
ing value or values of G may be obtained ; and, lastly,
the value or values of c, as before, from the equation
log c = log a -f log sin G— log sin A .
135. The third case in the Solution of Triangles may
be illustrated as follows :
Let AG=b', and angle GAB = A. With centre C and
radius equal to a, describe a circle cutting BA, produced,
if necessary, in B and B\
SOLUTION OF TRIANGLES. 185
(1) If a > 6, then B and B' are on opposite sides of A,
and there is only one triangle having the given angle A,
as well as the given sides a and 6, namely, the triangle
GAB.
(2) If a = h, then B' coincides with A, and there is
only one triangle.
(3) If a<h, then the circle will cut AB in two real
and different points, B and B\ on the same side of A, if
CB be greater than the perpendicular distance of G from
AB, i.e. if a is greater than 6 sin J. ; and there will then
be two triangles, GAB and GAB\ the angles GBA and
GB'A being supplementary. If a is equal to 6sin^,
the circle will cut AB in two real and coincident points,
i.e. it will touch it at 5 ; and the two triangles will be
coincident and right-angled at B. Lastly, if a is less
than &sin^, the circle will cut AB in two imaginary
points, and there will be no real triangle having the
given elements.
136. We may, again, investigate the third case in the
Solution of Triangles by considering the values of the
side c, instead of the angle B, in terms of the given
elements a, h and A.
For a- = ¥ + 6^ — 2hcQ0^A,
and, from this equation, which is a quadratic in c, two
values of c can be found, namely,
6 cos ^ ± VC^'cosM - 62 + a2),
or hQO^A±J{ci}-h^^m''A).
We have, as before, the following cases :
{I) a>h. Then, ^(a^-^^inM) is greater than
;^(5'^ — 6%inM), or 6 cos J.. Hence, there is only one
positive value of c, and therefore only one triangle.
186 SOLUTION OB' TRIANGLES.
(2) a = h. Then, ^(a^ — h^An^A) is equal to 6cos^.
Hence, the values of c are 26 cos A and 0 : i.e. there is
only one triangle.
(3) a<h. Then, J{a?-h'^sm^A) is less than 6cos^.
Hence, there are two values of c, which are real and
different, real and equal, or imaginary and different,
according as a is greater than, equal to, or less than,
h sin A ; and, consequently, two triangles which are real
and different, real and equal, or imaginary and different,
according as a is greater than, equal to, or less than,
h sin A.
137. Example 3.— Given a = 273, 6 = 392 and ^ = 37n4'; to find
ByC^ and c.
Here, a being less than 6, it is possible that there may be two
values of B.
log sin 5= log sin ^ + log 6 - log a
=log sin 37°14' + log 392 - log 273
= T' 7818002
4-2-5932861 -2-4361626
= 2-3750863
- 2-4361626
= 1-9389237.
1-9389796 = log sin 60° 20',
1-9389237 = log sin 60° 19' 6",
T-9389076 = logsin60°19'.
720 : 161 : : 60 : S,
8=161x^^=13",
^=60° 19' 13" or 119° 40' 47",
and C=82° 26' 47" or 23° 5' 13".
Corresponding to these pairs of values, we find, as in Case I.,
c= 447-278 or 176-925.
Hence, the required elements are
^=60° 19' 13"] i?=119° 40' 47"
C=82° 26' 47" V or C=23° 5' 13"
0=447-278 I c= 176-925
SOLUTION OF TRIANGLES. 187
138. Case IV. — The three sides being given, to solve the
triangle.
tan4 = V(«-^>^«-«> -
9
s(s — a)
:. logtan2- = Hlog(s-??) + log(s-c)-logs-log(s-a)}.
Similarly,
75
logtan2=Hlog(s-c) + log(s-a)-logs-log(8-b)}.
From these equations A and B can be found, and C
from the equation
139. In this case, if more than one angle is to be found,
the formulse for the tangents of \A and \B are preferred
to those for the cosines or sines, as only four logarithms
have to be found, namely, those for s, s — a, s — h and
s — c. If the formulae for the cosines are used, six logar-
ithms must be found, those for s, s — a, s — 6, a, 6 and c ;
and six also if the formulae for the sines are used, namely,
those for s — a, s — 6, s — c, a, b and c.
140. Example 4— Given «=349, 6 = 521 and c=539 ; to find A,
B and C.
Here, s= 704-5.
log tan ^^ =\{\og{s - 6)+log(s - c) - log s - log(s - a)}
= ^(log 183-5 + log 165-5 -log 704-5 -log 355-5)
_ / 2-2636361 -2-8478810^
~*1 + 2-2187980 - 2-5508396/
^i/ 4-4824341 \
^V -5-3987206/
= |(T-0837135)
=1-5418567.
188 SOLUTION OF TRIANGLES.
T'5418747 = logtanl9° 12',
T-541 8567 = log tan 19" 1 1' 8",
l-5414678 = logtanl9° 11'.
4069:3889=60:8.
3889
60
4069 ) 233340 ( 57 3
20345
29890
28483
14070
12207
^4 = 19° 11' 57-3",
^=38° 23' 55".
Similarly, using the logarithms found above, we find
log tan f =1-8290603,
^5=34° 0' 16-1",
t A ^=68° 0' 32",
^]^'^ C=73° 35' 33".
-y'
\J\P ^1 = 38° 23' 55"]
^ i • V^ ^=68° 0'32"l-
h i ^ f) 0=73° 35' 33" j
v*^
Examples XVI. a.
Solve the triangles of which the following elements are
given :
1. a = 2992-95, 5 = 127° 54' 30", 0=33'^ 9' 10".
2. a = 5043-04, B = 84° 56' 14", (7= 58° 45' 33".
3. 6 = 7282-61, 0=28° 49' 5", ^ = 102° 40' 15".
4. 6 = 3572, c = 9147, ^ = 42° 15' 38"-
5. c = 3000, a = 1406, 5 = 120° 15' 40".
6. a = 304-532, 6 = 526-109, (7=78° 18' 44".
7. a = 5371-24, 6 = 2743-65, ^ = 49° 14' 30".
SOLUTION OF TRIANGLES. 189
8. a = 4857,
6 = 6104,
^ = 20° 19' 10".
9. a = 586,
6 = 987,
^ = 60° 25' 25".
10. 6 = 807-8,
c = 1162-4,
5 = 41° 88'.
11. 6 = 7412-5,
0 = 9182-1,
0=64° 12' 20".
12. a = 54,
6 = 48,
0 = 86.
18. a = 329-4,
6 = 451-7,
0 = 154-2.
14. a = 620- 124,
6 = 711-005,
Examples XVI.b.
0 = 932-147.
Solve the triangles of which the following elements are
given :
1. a = 4686-50,
5= 122° 37' 45",
0=28° 37' 50".
2. 6 = 4670-13,
(7= 151° 52' 85'',
^ = 18° 25' 15".
3. c = 6508-75,
H. 6 = 501-2,
^ = 66° 39' 55",
J5=25°32'15".
0 = 398-5,
^ = 68° 48'.
5. 0 = 50-38,
a = 68-4,
j5 = 94°17'.
6. a = 891-204,
.6 = 172-537,
0=104° 14'.
7. a = 548-28,
6 = 1051-87,
^ = 27° 12' 10".
8. a = 9621,
6 = 6758,
^ = 59° 40' 40".
9. a = 742,
6 = 824,
^ = 75° 10' 55".
10. 6 = 714-3,
0 = 958-2, .
0=87° 0' 4".
11. a = 2143,
0 = 4172,
^ = 25°1'14".
12. a = 200-4,
6=295-8,
0 = 811-1.
^18. a = 5102,
6 = 8074,
0 = 2314.
14. a = 5817-24,
6 = 345107,
0 = 2001-15.
Examples XVII.
1. In the ambiguous case, if a, 6, and A be given, and if
Oj, Cg be the values of the third side, then c^c^ = h'^-a^.
2. If one of the angles at the base of a triangle be 36°,
the opposite side 4, and the altitude ^o — 1, solve
the triangle.
190 SOLUTION OF TRIANGLES.
3. The angles of a triangle are 35°, 65° and 80°, and the
difference between the longer sides is 1000 ; find
the sides.
4. If^ = 6r, j5 = 37° and a-6 = 372, find a and 6.
5. Given two sides and the included angle, find the
length of the altitude on the third side.
If this altitude be greater than the third side,
then sin 0+2 cos C> 2, where G is the included
angle.
6. If 6, c, and A be given, shew that a may be deter-
mined, without finding B and C, from the formula
A
a — (b + c)sin ~ sec <p,
where tan <h = y^- cot ^ •
b + c 2
B—G d) A
7. Prove that tan — ^ — = tan^^ cot -^, where cos (f) = clb)
and apply this formular to find B and G, where
^ = 35°25'and6:c = ll:10.
8. If Q be an angle determined from the equation
cos 0 = {a — b)/c, prove that
A-B (a + b)sme , A-^B csinO
cos — ;r — = ^^ — —4= — and cos — -i — = — j='
2 2s/ab 2 2s/ab
9. In a right-angled triangle, the medians drawn from
the acute angles are a and /3 ; shew how to find
the angles.
Find the angles when a = 10, ^=14.
10. If c6s(5-0) = 31/32, and 6 = 50, c = 40, find a.
11. The tangents of the angles of a triangle are in har-
monical progression; given one of the sides and
the diflference of the first and third angles, shew
how to solve the triangle.
SOLUTION OF TRIANGLES. 191
12. Given the mean side of a triangle, whose sides are in
arithmetical progression, and the angle opposite to
it, investigate formulae for solving the triangle,
and find the greatest possible value of the given
angle.
Solve the triangle when the mean side is 542,
and the opposite angle 59° 59' 59".
13. Given the area, the perimeter, and an angle of a
triangle, shew how to solve the triangle.
14. Given the lengths of the three medians, shew how to
solve the triangle.
15. Given the lengths of the three altitudes, shew how to
solve the triangle.
§ 3. Practical Applications.
141. The application of the methods just explained
to the problems of Surveying forms the subject of this
section.
We have seen that, if three elements of a triangle,
including at least one of the sides, be known, we can
determine the magnitudes of the other three elements,
except in one particular case, when the solution is
ambiguous.
For the purposes of Surveying, therefore, we require
instruments : (1) for measuring the length of a line, called
a base-line or base, on the ground, and (2) for measuring
angles either in a horizontal or vertical plane.
192 PRACTICAL APPLICATIONS.
142. Measurement of Base-line. — If the district to be
surveyed be not extensive, and if no very great accuracy
be required, the base-line may be measured with the
ordinary surveyors chain, known as Gunter's chain.
This chain is 22 yards long, and consists of 100 links
joined together by rings, having at every tenth link a
piece of brass attached to assist in counting the number
of links.
In more elaborate surveys extending over whole
countries, the base-line may be five miles or more in
length, and the greatest attainable accuracy is required.
Steel chains may be used, or rods of metal, glass, well-
seasoned wood, etc. ; the temperature must be observed
throughout, and allowance made for the expansion or
contraction of the rods, etc., resulting from changes of
temperature. This correction may, however, be avoided
by the use of compound rods, made of two bars having
different coefficients of expansion and connected at their
ends by cross-pieces, on which marks are made that
remain at the same distance apart for all ordinary
temperatures.
. 143. The Theodolite. — The accompanying, figure re-
presents a simple form of theodolite. It consists of an
achromatic telescope mounted so that it can turn round
two axes, one vertical and the other horizontal, and pro-
vided with graduated arcs by means of which angles in
horizontal and vertical planes may be measured.
A, B, C and D are four plates, the two upper ones,
G and D, being in contact with one another. At the
bottom of the lowest plate A is a screw by which the
theodolite is fixed to the head of a tripod stand. From
PRACTICAL APPLICATIONS.
193
the plate G, a hollow axis projects downwards at right
angles to it, passes through the plate B, and ends in a
ball which works in a socket in the plate A. Inside
this axis, and co-axial with it, is another axis attached to
the uppermost plate D, called the vernier-plate, and
passing through the plate G. The theodolite can be
turned round as a whole about the outer axis, and can
be clamped in any position by the clamping-screw E,
which tightens a collar that surrounds the axis above the
plate B. When so clamped, a slow motion may be given
194 PRACTICAL APPLICATIONS.
to the instrument by the tangent-screw F. The inner
axis allows the vernier-plate D, and the telescope, etc.,
attached to it, to be rotated, while the rest of the
instrument is clamped by the screw E. By means of
the screw G, the vernier-plate can be clamped to the
plate below, and a slow motion can be given to the
former by the tangent-screw H. The four screws K
(of which only two are shewn in the figure) enable the
plate B and those above it to be set accurately level and,
consequently, one axis of the theodolite vertical. The
horizontality of the vernier-plate D is indicated by two
levels L fixed to it at right angles to one another. In
the centre of this plate is also fixed a compass M. The
edge of the plate G is graduated the whole way round to
every half degree, but angles can be read to one minute
by two verniers N on opposite sides of the vernier-plate,
exactly 180° apart.
The telescope T is carried by two F-supports 8, S,
attached to the ends of the bar R, which rotates about an
axis turning on the pillars, P, P. The axis is parallel
to the vernier-plate, and therefore horizontal when the
instrument is adjusted correctly. The bar R being at
right angles to this axis, the telescope turns in a vertical
plane w^hen the rest of the instrument is fixed by the
clamping-screws, E, G. A graduated semicircle Q is
attached to the bar P, and a slow motion can be given
to it by the screw Tf. The semicircle is graduated to
every half-degree, from 0° to 90° in either direction from
the middle point, and angles can be read to one minute
by the vernier X attached to the compass-bore M. A
level V is fixed to the telescope, and indicates when the
axis of the telescope is horizontal.
PRACTICAL APPLICATIONS. 195
In the common focus of the object-glass and eye-glass
of the telescope are placed two spider-threads, called
cross- wires, at right angles to one another, and adjusted
so that, when the intersection of the cross- wires appears
to coincide with the centre of an object, the direction of
the axis of the telescope coincides with the direction of
the object as seen from the observer.
The verniers N and X are short fixed graduated arcs,
constructed so that 30 divisions of the verniers are equal
to 29 divisions of the circles; the difference in length
between a division of either is therefore equal to -sVth of
a division of the circle. The first or zero division of the
vernier, marked with an arrow-head, indicates the num-
ber of degrees and half-degrees on the graduated circle.
If the arrow-head coincides exactly with a division on
the circle, say 47J°, then this is the required reading.
But i£ it be between 47J° and 48°, it is then noted which
divisions of the two arcs coincide. If the division of the
vernier be the 14th from the arrow-head, then the arrow-
head is \^ of 80' beyond 47 J°, and the correct reading is
47i°-f-14>r47°44^
In the above account the principal parts only of a
simple theodolite are described : no reference is made to the
details which are required for securing the correct adjust-
ments of the instrument, for it may be assumed that as
far as possible these adjustments are already made by
the maker.
144. In setting up the theodolite for use, the tripod is
firmly planted on the ground, its position being indicated
by a mark made below a plummet suspended from the
head of the stand exactly beneath the centre of the instru-
196 PRACTICAL APPLICATIONS.
ment ; and the legs of the tripod are arranged so that the
vernier-phite is roughly horizontal. The instrument is
then turned so that the two levels L are parallel to the
line joining two of the levelling-screws K ', and, by
means of these screws, the vernier-plate is set exactly
horizontal, the bubbles of the two levels being then in the
centres of their ranges. If the vertical axes have been
accurately adjusted, the bubbles of the levels will keep
their positions while the instrument is turned completely
round.
The object-glass and eye-piece of the telescope are then
adjusted for the distance of the object to be observed and
for the eye-sight of the observer ; both the object and the
cross-wires must at the same time appear well-defined.
145. To measure the horizontal angle between any two
objects, i.e. the angle between the projections, on the
horizontal plane through the observer, of the straight
lines joining the observer to the objects, we proceed as
follows :
Turn the vernier-plate round until the zero-line of the
vernier (indicated by the arrow-head) coincides nearly
with the zero-line (marked 360°) of the horizontal circle.
Clamp the vernier-plate by the screw G, and make the
two lines coincide exactly by the tangent-screw H.
When this is the case, the zero-line of the second vernier
should coincide with the line marked 180° of the hori-
zontal circle. Loosen the screw E, and turn the instru-
ment round until the intersection of the cross-wires
coincides nearly with the centre of one of the objects;
then clamp the instrument by the screw E, and make the
intersection of the cross- wires coincide exactly with the
PRACTICAL APPLICATIONS. 197
centre of the object by the tangent-screw F, Now, loosen
the clamp 0, and the rest of the instrument being still
fixed by the clamp E, turn the vernier-plate round until
the intersection of the cross-wires coincides nearly with
the centre of the second object ; clamp the vernier-plate
by the screw (r, and make the intersection of the cross-
wires coincide exactly with the centre of the object by the
tangent-screw H\ having previously, however, adjusted
the telescope, if the difierence of the distances of the two
objects is great compared with that of either. Read the
angles indicated by both verniers, and the mean of the
two readings will give the required angle.
If the instrument has been correctly adjusted, the zero-
lines of the vertical circle and the vernier X will coincide
when the vernier-plate has been set horizontal and the
axis of the telescope is also horizontal. Thus, the angle
of elevation of any object can be obtained by means of
the vernier X.
146. Trigonometrical Survey. — Let the length of a
base-line AB be measured on level ground, the ends A
and 5 being marked by flagstaffs or other prominent
and well-defined objects. c
Let (7 represent a similar
object on the same hori-
zontal plane with A and
By and let the angles
BAG, ABC be measured
with a theodolite. We D^
have thus sufiicient data for calculating the angle AGB
and the lengths of the lines AC, BC (art. 129) ; though, in
practice, the angle AGB would be measured from the
198 PRACTICAL APPLICATIONS.
station G to test the accuracy of the other measurements.
In a similar manner, if D represent another object in the
same plane, we can determine the lengths of the lines
AD, BD. Then, in the triangle AGD, knowing the
angle GAD and the sides AG, AD, we can find the length
of the line GD (art. 131). Now, proceeding to the points
represented by G and D, and selecting other suitable
objects E, F in the same plane as before, and measuring
the angles DGE, GDE, DGF and GDF, we can calculate
the lengths of the lines GE, DE, GF and DF ; and, from
these data, again, the length of EF can be found.
Proceeding in this manner, we may imagine a network
of triangles to be formed, increasing in size and spreading
over the whole surface of a country, the magnitudes of
the sides and angles determining the distances and bear-
ings of a series of conspicuous objects from one another.
This, briefly, is an outline of the manner in which a
trigonometrical survey of a country is carried out.
147. We have supposed, for simplicity, that the base
line AB is horizontal. This, however, is not necessary,
and, in practice, the base-line may be inclined at a small
angle to the horizon, but the requisite correction is easily
applied if the slope of the ground be known (art. 149).
Again, it will rarely, if ever, be the case that the
selected objects, G, D, E, etc., are in the same horizontal
plane with A and B. The angle GAB will not then be
the angle subtended at A by the line BG, but the angle
between the projections of the lines AG, AB on the
horizontal plane through A ; and the line AG represents,
not the actual distance between A and G, but the hori-
zontal distance between them, i.e. the projection of the
PRACTICAL APPLICATIONS. 199
line AG on the horizontal plane through A. Thus, the
figure of the preceding article represents, in this case, the
relative positions of the points A, B, (7, etc., as they
would be indicated on a map of the country.
148. Now, it is obvious that, if the lengths of the
sides of the triangle be appreciable fractions of the
radius of the earth, the triangles will no longer be plane,
but spherical, or rather spheroidal, triangles. In all the
great trigonometrical surveys that have been carried out
■ this is the case, the sides being sometimes as much as 50,
or even 100, miles in length. A detailed description of
such surveys must therefore lie beyond the scope of a
work on Plane Trigonometry.
If, however, the sides be short, say, not more than a
few miles in length, then, for all practical purposes, the
triangles may be treated as if they were plane triangles.
This is the case in the survey of the Mer de Glace and
tributary glaciers, executed in 1842 by Prof J. D. Forbes.
An account of this survey, though it was carried out
on a small scale and without many of the refinements
necessary in extensive operations, will illustrate some
of the more important features of a trigonometrical
survey.
149. Forbes' Survey of the Mer de Glace.— TAe Base-
Line. — The site chosen for the base-line was a road in
the valley of Chamouni, joining the villages of Les Praz
and Les Tines, and passing a short distance from the
foot of the glacier. At the time the survey was made,
this road was formed of dry, well-compacted gravel, and
its surface was apparently level, though in reality sloping
200 PRACTICAL APPLICATIONS.
upwards towards Les Tines at an average angle of 44'.
Opposite the foot of the glacier, the road for a distance
of 1000 yards is absolutely straight, and along this
portion of it the base-line NO (see map) was measured.
The two ends of the line were marked by nails driven
into the to(>8 of long pins of hard wood fixed in the
ground, and at each end is an object visible from at least
one other station used in the survey. The station N is
exactly at the eastern end of the beam which forms the
south side of a wooden bridge near Les Praz ; and, close
to the station 0, there is a solitary tree with its lower
branches lopped off.-
The length of the base-line was measured with a ten-
metre chain and a steel tape divided into English feet
and inches. It was found to be 91 chains and a fraction,
the fraction being approximately two-fifths of a chain,
but determined more accurately by the steel tape to be
26 ft 2'50 ins. Thus, the length of the base-line was
91 chains -h 26 feet +2*50 inches.
The chain, being compared with the steel tape, was found
to be 32 ft. 10*675 ins. long, giving for the length of the
base-line
2992-95 English feet,
a result shewn, by re-measurement of part of the base, to
be probably correct to within about an inch, or about
1/36000 of the length of the base.
The road, however, being inclined at an average angle
of 44' to the horizon, this length should be multiplied by
the cosine of 44', or 099991 81 ; but as this would result
in shortening it only by about 1/12000, the correction was
not applied.
PRACTICAL APPLICATIONS. 201
150. ThA Triangtdation, — The form of the glacier,
and the series of triangles by which it was deter-
mined, arc shewn in the accompanying map, which
is reduced from that of Professor Forbes. The sta-
tions, forming the angular points of the triangles,
and lettered /, L, F, G, 11, B, £, were chosen on the
rocks bounding the glacier and at some height above
it, HO that from each station two or more of the others
wam visible.
In many ways the survey was carried out under serious
disadvantages. " The walls of the glacier are excessively
rugged, often maccessible. The stations are difficult to
choose so as to be visible from one another, owing to the
intricate windings of the ice-stream and the enormous
height of the rocks. The fundamental triangulation
ijjust be carried up a valley, whose extremities, independ-
ent of mountains, differ in level by 4400 feet." The
triangles on these accounts ^re badly shaped and few in
number. In the triangle FOB, for example, two of the
angles are very small, and it is obvious that a very small
error made in the measurement of either of them would
'^'ive rise to disproportionately large errors in the lengths
of the calculated sides. A " well-conditioned " triangle
should be as nearly equilateral as possible, and in none
of the triangles employed in the survey are the sides
even approximately equal in length. With one excep-
tion, however, the three angles of every triangle are
irioasured, and in one only of the six other triangles does
the sura of the angles differ by more than a minute from
two right angles.
The third station /, forming with N and 0 the first
triangle, is a rock above the Chapeau, distinctly visible
202 PRACTICAL APPLICATIONS.
from both ends of the base-line. The observed angles of
this triangle were
NIO= 18° 55' 50''
/OiV= 127° 54' 0"
ONI^ 33° 8' 40"
179° 58' 30"
The sum of the angles of this triangle being 1' 30" less
than 180°, each angle is increased by one-third of this
amount; and a similar correction is also made in the
other triangles. Since iVO = 2992-95 feet, angle ION
= 127° 54' 30" and angle Oi\^/ = 33° 9' 10", we have,
therefore,
i\r/= 7275-78 feet,
70 = 504304 „
The fourth station L is a projecting mass of rock on
the ridge extending above the Montanvert towards the
Aiguille des Charmoz. From this point the stations /
and 0 could be seen ; the west end N of the base-line
being, however, invisible. The observed angles were
OLI= 3G°18'20"
LIO= 84° 56' 21"
XQ/= 58° 45' 40"
180° 0'2r
giving (since 70 = 5 043 04 feet),
70 = 8484-49 feet,
77=7282-61 „
The fifth station F is on the promontory of Les Echelets,
about 150 feet above the glacier, and from it the four
stations 7, 7, G and B were visible. The angles of the
triangle 777" are
PRACTICAL APPLICATIONS. 203
LFI= 48° 30' 15"
ILF=10r W 10"
FIL= 28° 48' 50"
179° 59' 15"
giving IF= 9485-56 feet,
Zi^= 4686-50 „
The sixth station G is marked by a pyramid of stones
on a large rock on the ridge of Tr^laporte ; it is about
300 feet above the glacier, and commands a view of the
three stations L, F, and B. Only two of the angles of
triangle LFG were measured, namely,
Zi^(^ = 122°37'45",
FLG= 28° 37' 50",
giving 6^i^= 4670-18 feet,
G^Z = 8208-28 „
The seventh station 5 is a pyramid of stones built on
the promontory of the Tacul, and is seen from five of the
other stations. The angles of the triangle FGB are
GBF= 11° 42' 0"
FGB=15r52'25"
GFB= 16° 25' 5"
179° 59' 30"
giving i?'5 = 108531 feet,
OjB = 6508-75 „
The eighth station H is at the foot of the Couvercle,
and is opposite station B. The angles of the triangle
GBH are
GBH= 66° 40' 10"
BHG= 87° 48' 5"
HGB= 25° 32' 30"
180° 0' 45"
204 PRACTICAL APPLICATIONS.
giving GF= 5980-79 feet.
£ir= 280801 „
The ninth, and last, station ^ is on a rocky promontory
high up the Glacier de L^chaud. The angles of the
triangle BHE are
BEH=- 21° 24' r
HBE= 67° 50' 40"
EHB= 90° 45' 40"
180° 0'2r
giving ^^=7127-97 feet,
J5i7= 7695-66 „
151. The principal triangulation being completed, the
sides of these triangles were then used as base-lines for
determining the positions of several subordinate points,
and the outline of the glacier was drawn with the aid of
compass and tape measurements from known points and
lines.
152. Measurement of Heights. — In Chapter V., two
simple cases of the calculation of the height of a tower or
other object above a horizontal plane were explained : the
base-line in the first being measured on the plane from
the foot of the tower, etc. ; and, in the second, in the same
straight line with the tower, the foot of the tower being
supposed inaccessible. We shall now explain the method
by which the height of a mountain or any other object
may be determined, the base-line being measured in any
direction and not necessarily on level ground.
PRACTICAL APPLICATIONS. 205
153. ^To find the height of a mountain.
(1) Let the length and bearing of the base-line AB be
measured on level ground, and
let G represent the summit of the
mountain.
From the two ends of the
base-line, the bearings of G are
measured. Let the directions of
these bearings be represented by
AD, BD ; these two lines meet
in a point D, vertically below (7,
and in the same horizontal plane with. AB (Eucl. XL 19).
At one end of the base-line, say A, measure the angle
of elevation GAD of the summit G.
Now, in the triangle ABD, the base AB and the angles
BAD, ABD are known ; and from these the side AD is
calculated. Also GD = ADi^^GAD.
This equation gives the height of the summit G above
the base-line AB, and, consequently, above the level of
the sea, if the height of the base-line above the same
level be known.
(2) Let the base-line be inclined to the horizon at a
known angle.
Let B and D now be the projections, on a horizontal
plane through one end A of the base-line, of the other
end of the base and of the mountain summit (7; so that
A, B and D represent the positions of the ends of the
base, and of the summit, as they would be depicted on
a map.
The same measurements are made as before, namely,
the length and bearing of the base, the bearings of the
summit from the two ends of the base, and the elevation
206 PRACTICAL APPLICATIONS.
of the summit from one end A ; and, in addition, the in-
clination to the horizon of the ground on which the base
is measured.
Thus, AB is now equal to the length of the base-line
multiplied by the cosine of its inclination to the horizon ;
and, precisely as in the first case, we find the height of
the mountain G above the horizontal plane through A.
154. Exam/pie. — In addition to making the survey of
the Mer de Glace described above, Prof Forbes deter-
mined the height of every station and of many of the
neighbouring summits of the Mont Blanc range by the
method explained in the preceding article. In this part
of his work it was necessary to take into account the
effects of atmospheric refraction and the curvature of the
earth ; but, in most xBases, both effects were eliminated by
observing, not only the elevation of one station above
another, but also the depression of the first station below
the second, and taking the arithmetic mean of the two
angles. We give one example, the height of the station G.
The line LF, determined from the triangle ILF to be
4686-50 feet long, is here the projection of the base-line
on the horizontal plane through F. The bearings of G
from the ends of the base-line are given by the angles
LFG and FLG, which are 122° 87' 45" and 28° 37' 50"
respectively, giving 4670-13 feet for the length of the
line GF.
The arithmetic mean of the elevation of G above F
and of the depression of F below G was found to be
4° 48' 15''.
The height Qi) of G above F in feet is therefore
4670-13 X tan 4° 48' 15';
I
PRACTICAL APPLICATIONS. 207
log h = log 4670-13 + log tan 4^ 48' 15"
= 3-6693290
+ 2-9245144
= 2-5938434 = log 892-51.
Hence, the height of the station G above the station F
is 392-5 feet.
In a similar manner, it was found that the height of F
above the Montanvert is 523-5 feet. The height of the
Montanvert above the observatory of Geneva was deter-
mined by barometric measurement to be 4960 feet, and
the height of the Observatory above the level of the sea
was known to be 1343 feet. Hence, the height of the
station G above the sea-level
= 392-5 + 523-5 -1-4960-1- 1343 feet
= 7219 feet.
155. Example. — The apparent dips of a stratum in directions
inclined to one another at an angle $ are a and f3, respectively ;
to obtain equations for determining the amount and direction of
the true dip.
Let OA and OB be the directions in which the apparent dips,
a and (3, are measured, A and B being Cj^
points in the same horizontal plane
with 0, and 0 a point in the plane
of the stratum vertically above 0.
Then, angle OAC = a and angle
0BC = f3.
Draw OB perpendicular to AB.
Then OD represents the direction of
the true dip, since AB is a. horizontal
line in the plane. Let 8 be the
amount of the true dip, and </» the angle made by its direction
with OA.
Now, 0D = OA cos (f) — OC cot a cos <^ /
= OB cos(^ -4>) = OC cot ^ cos((9 - </>),
PRACTICAL APPLICATIONS.
tan /? cos </) = tan a(cos ^ cos ^ + sin ^ sin ^),
cos ^(tan /? - tan a cos 6) = sin ^ tan a sin ^ ;
tan<^=^^"^~^^"^^^"^=(tan/?cota-co3^)cosec^. ...(1)
tan a sm ^
Also, tang=^=,^'^^ ^=tanasec</> (2)
OD OC cot acoa<j>
Examples XVIII.
[Note. — Unless specially mentioned, the height of the
observer's eye above the ground is nob to be taken into
account in the following examples. See also the note at
the head of Examples VII. A.]
1. AB is a line 1000 yards long ; B is due north of A.
At B, a distant point P bears N. 70° E. ; at ^, it
bears N. 41° 22' E. ; find the distance from A to P.
2. J. ^ is a base-line 200 feet long, measured close to
one bank of a river and parallel to it ; 0 is a mark
on the opposite bank; the angles BAG and ABC
are found to be 59° 15' and 47° 12', respectively;
find the breadth of the river.
3. The sides of a valley are two parallel hills, each of
which slopes upwards at an angle of 30°. A man
walks 200 yards directly up one of the hills from
the valley, and then observes that the angle of
elevation of the other hill above the horizon is 15°.
Shew that the height of the observed hill is 273 2
yards nearly.
4. A man, standing at the water's edge, finds the angle
of elevation of the top of a cliff to be a. He then
walks a feet directly up a beach, which slopes
upwards at an angle y, and, again measuring the
PRACTICAL APPLICATIONS. 209
angle of elevation of the top of the cliff, finds it to
be /3. Find the height of the cliff above the sea-
level.
Find the height when .a = 60 feet, a = 42°,
/5 = 65°, y = 8°.
Two distant spires, P and Q, are seen from the two
ends of a base-line AB, 1200 yards long, measured
on a straight and level road, the spires being on
the same side of the road. The angles FAB,
PBA, QAB and QBA are found to be 108° 14',
89° 5\ 27° 30' and 114° 20', respectively. Find the
distance between the two spires.
6. AB is a straight and level road, 1500 yards long ; C
and D are church spires on either side of the road.
The angles BAG, ABC, BAD, ABD are measured
and found to be 43°, 57°, 29° and 37°, respectively.
Find the distance between the spires.
7. From one end of a horizontal straight road, running
N.W. and S.E., and 2000 yards long, the top of a
mountain is observed at an altitude of 23° in a
direction E. 4° N., and, from the other end of the
road, it bears N. 37° E. Find the height of the
mountain above the level of the road, and its
horizontal distance from the road.
8. In order to determine the height of a mountain,
a north-and-south base-line, 1000 yards long, is
measured ; from one end of the base-line, the
summit bears E. 10° N., and is at an altitude of
13° 14'; from the other end, it bears E.46°30'N.
Find the height of the mountain.
9. To determine the height of a steeple, a base-line, 150
feet long, is measured on a road running due east
210 PRACTICAL APPLICATIONS.
and inclining upwards at an angle of 5°, the lower
end of the base-line being on the same level as the
foot of the tower. From this end of the base-line,
the top of the steeple bears due north, and its angle
of elevation is 1G° ; from the other end of the base-
line, the top of the steeple bears N. 28" W. Find
the height of the top of the steeple above the
ground to the nearest inch.
10. A ship, sailing uniformly and directl}'- towards a port
P, sights another sailing uniformly and directly
towards a port Q, and observes that the line join-
ing the two makes an angle a with its own
direction of motion. After a hours, the line joining
the two vessels points directly to P, and, after h
hours more, it points directly to Q ; and, at the
latter time, the distance PQ subtends an angle ^
at the first ship. Compare the rates of sailing of
the two ships.
11. On the side of a hill inclined to the south at an angle
of 20°, a road is made which slopes upwards in
the direction E. 15° N. ; if the road be a mile
long, find the difference in height between its two
ends,
12. The shadow of a cloud at noon is cast on a spot
1600 feet due west of an observer. At the same
instant, he finds that the cloud is at an altitude of
23° in a direction W. 14° S. : find the height of the
cloud and the altitude of the sun.
13. The elevation of a steeple standing on a horizontal
plane is observed, and at a station a feet nearer it
its elevation is found to be the complement of the
former. On advancing in the same direction h feet
PRACTICAL APPLICATIONS. 211
nearer still, the elevation is found to be double
the first ; shew that the height of the steeple is
{(«+^)-?}-
14. A person, walking along a straight road, observes the
greatest elevation of a tower to be a. From
another straight road, he observes the greatest
elevation of the tower to be /5. The distances of
the points of observation from the intersection of
the two roads are a, h, respectively; prove that
the height of the tower is
/ (^2_52 Y
\cot2^-cotV'
15. A person walks from one end J. of a wall a certain
distance a towards the west, and observes that
the other end B then bears E.S.E. He afterwards
walks from the end B a distance a(^2 + 1) towards
the south, and finds that the end A bears N.W.
Shew that the wall makes an angle cot "^2 with
the east.
16. A man standing on an elevation can just see over the
surface of a calm sea the top of a mountain, the
height of which he knows to be 1650 feet above
the sea-level, and the summit of which is at a
distance of 70 miles from his own position. What
is the height of his eye above the sea-level, assum-
ing the radius of the earth to be 4000 miles ?
17. An observer, whose eye is h feet above the surface of
a lake, determines the angle of elevation of a point
on a cloud to be /3, and the angle of depression of
the image of the same point to be a ; find the
212 PRACTICAL APPLICATIONS.
height of the cloud above the surface of the
lake.
Find the height of the cloud when A = 240 feet,
a = 30°, ^ = 17°.
18. Ay B, and C^are three points in a straight line. AB,
BC, and CA are 1000 feet, 2000 feet and 3000
feet respectively ; P is a point such that each of
the angles APB, BPG is 35°. Find the distance
AP.
19. A hill consists of two inclined planes sloping in
opposite directions at angles a and /3 to the
horizon. A cloud is driven with uniform velocity
and always at the same height in a line towards
the sun and at right angles to the axis of the hill.
The shadow of the cloud passes up one slope in t
seconds and down the other in f seconds. Find
the altitude of the sun.
20. At distances of 100 feet and 40 feet, measured in a
horizontal plane in the same straight line from
the foot of a tower, a flagstaff standing on the top
of the tower subtends an angle of 8° ; find the
length of the flagstaff.
21. A flagstaff stands on the top of a tower built on
a horizontal plane. A person observes the angles
subtended, at a point in the horizontal plane, by
the tower and the flagstaff; he then walks a known
distance towards the tower, and finds that the
flagstaff subtends the same angle as before. Find
the height of the tower and the length of the
flagstaff.
22. A person, walking along a straight road, observes the
greatest angle (a) subtended by two objects in the
PRACTICAL APPLICATIONS. 213
same plane with the road. He then walks a
distance a along the road, and the objects appear
in the same direction making an angle ^ with the
road. Shew- that the distance between the objects
2 a sin a sin B •
IS 7-.
cos a + cosp
23. A, B, and C are three consecutive milestones on a
straight road, from each of which a distant spire is
visible. The spire is observed to bear N.E. at A,
E. at B, and E. 80' S. at C. Find the distance of
the spire from^, and the shortest distance of the
spire from the road.
24. At each end of a horizontal base of length 2a it is
found that the angular altitude of a certain peak is
a, and at the middle of the base it is /3. Prove
that the height of the peak above the plane of the
base is a Ana An fi
V sin(/3 + a)sin(y8 — a)
25. The angles of elevation of a balloon are observed from
two stations a mile apart, and from a point half-
way between them, to be 60°, 30°, and 45° respec-
tively. Prove that the height of the balloon is
440^6 yards.
26. Two stars, A and B, are so situated that, when A
is due south and at an altitude of a degrees, B is
setting at a point /3 degrees W. of S. ; find the
angle subtended by the two stars.
27. A straight flagstaff, leaning due east, is found to
subtend an angle a at a point in the plain upon
which it stands a yards west of the base. At a
point h yards east of the base, it subtends an
angle /3. Find at what angle the flagstafi' leans.
214 PRACTICAL APPLICATIONS.
28. Two lines of straight railway, ABC, DEG, meet at
C, telegraph-posts being situated at A, B, D, E\
the angles DAE, DBE are each equal to a; and
the angles EAB, EBG are /8 and y respectively ;
shew th^t
BC=AB . ;'°^f°S°l^ix V
sin(y - ^)sin(a + /3+y)
29. Two vertical faces of rock, at right angles to each
other, exhibit sections of a stratum ; the dips of
the sections so formed are found to be a and /3
respectively ; if ^ be the true dip of the stratum,
and 0 its direction (measured from the vertical
plane corresponding to dip a), then
tan2^ = tan2a + tan2/3,
and tan 6 = tan /3 cot a.
30. A person, wishing to determine the dip of a stratum,
bores vertical holes at three of the angular points
of a horizontal square ; the depths of the stratum
at these points being a, h, c, and h the side of the
square, the dip of the stratum is
31. Find the amount and direction of the true dip of a
stratum, the apparent dips being 41° and 18° in
the directions N. 31° E. and E. 12° S. respectively.
32. Two lines traced on an inclined plane include an
angle a, and their inclinations to the horizon are
P and y. Shew that the tangents of their in-
clinations to a horizontal line traced on the plane
are
sin a sin /3 , sin a sin y
sin y - sin /3 cos a sin y8 — sin y cos a
PRACTICAL APPLICATIONS. 215
S3. Two lines inclined at an angle y are drawn on an
inclined plane, and their inclinations to the horizon
are a and /3, respectively. Shew that the sine of
the inclination of the plane to the horizon is
cosec y(sin2a + sin^/S — 2 sin a sin /3 cos y)'^.
34. Two planes are inclined to the horizon at angles a
and /3 in directions which make an angle 6 with
one another : find the direction and inclination
to the horizon of their common section.
35. A tight rope connects the tops of two vertical poles,
a and h feet high, respectively, placed c feet apart
in an east and west line. If the sun be due south
and at an altitude of a degrees, find the direction
of the shadow of the rope.
Find the direction if a = 35 feet, 6 = 24 feet,
c = 30feet, a=35".
36. Given the angular distances (0, a) of the sun from the
planes of the meridian and horizon, find the breadth
of the shadow cast by an east and west wall of
height n on the horizontal plane through its base.
37. A wall, 20 feet high, bears 59° o' E. of S. ; find the
breadth of its shadow on a horizontal plane through
its base at the instant when the sun is due south
at an altitude of 30°.
38. A man, walking along a straight road which runs in
a direction 30° E. of N., notes when he is due
south of a certain house. When he has walked
a mile further, he observes that the house lies
due west, and that a windmill on the opposite
side of the road is N.E. of him. Three miles
farther on he finds that he is due north of the
windmill. Find the distance between the house
216 PMACTICAL APPLICATIONS.
and the windmill, and shew that the line joining
them makes with the road an angle
39. A cloud, just grazing the top of a mountain a feet
high, is seen at an altitude a by a man at the
sea-level due south of the mountain. It is driven •
by the wind at the same height and with uniform
velocity, and, t seconds later, is seen by him at an
elevation /5 in a direction 0 east of north. Find
the direction of the wind and its velocity in feet
per second.
40. On a plane inclined to the horizon at an angle of 30°,
a circle of radius 10 feet is described, and a post
is placed at the highest point of the circle per-
pendicular to the plane. At one end of the
horizontal diameter of the circle the angular
elevation of the top of the post is 45°. Find
the length of the post.
41. A and B are two places 12 miles apart, A being due
north of B. An observer at A determines the
altitudes of the two ends of the visible path of
a shooting star to be 82° and 70°, and the azimuths
of the same points to be N. 23° E. and N. 59° E.
An observer at B measures the azimuths of the
same points to be N. 12° E. and N. 30° E. Find
the length of the visible path of the shooting star,
and the heights of its two extremities.
42. A beacon is due west of a lighthouse and three miles
distant from it. The channel of a river is given
by the condition that a vessel shall enter due
south of the lighthouse, at such a point that the
PRACTICAL APPLICATIONS. 217
lighthouse and beacon shall subtend an angle of
60° at the vessel, and shall continue to do so
until the beacon is north-west, when the channel
remains straight in the last direction in which the
vessel was sailing, until it is due south of the
beacon. Prove that the straight part of the
channel is ^3 + 1 miles long.
43. A curve on a railway, whose form is a circular quad-
rant, has telegraph posts at its extremities and at
equal distances along the arc, the number of posts
being n. A person in one of the extreme radii pro-
duced sees the^th and gth posts from the extremity
nearest him (from which his distance is a) in a
straight line. Find the radius of the curve.
44. A man, standing on a plain, observes a row of equal
and equidistant pillars, the tenth and seventeenth
of which subtend the same angles as they would
if they stood in the position of the first and were
respectively one-half and one-third of the height ;
shew that the line of pillars is inclined to the line
drawn to the first at an angle whose cosine is
nearly y%.
45. A man on a hill observes that three towers on a
horizontal plane subtend equal angles at his eye
and that the angles of depression of their bases
are a, a, a \ prove that, c, c\ and d' being the
heights of the towers,
sin(a — d') sin(a'' — a) , sin(a — aQ _ q
c sin a d sin d c" sin d
46. A vertical pole of height a stands on a plane inclined
to the south at an angle S to the horizon ; if the
angular distances of the sun from the planes of the
218 PRACTICAL APPLICATIONS,
meridian and horizon be 0 and a, find the length
of the shadow of the pole on the inclined plane.
47. Tvyo ships are sailing uniformly in parallel directions,
and a person in one of them observes the bearing
of the other to be a degrees E. of N. ; p hours
afterwards its bearing is ^ degrees E. of N. ; and
q hours afterwards it is y degrees E. of N. Prove
that the course of the vessel is Q degrees E. of N.,
where
tan ^— y^^QQsin(/3 — y) — gsinysin(a — ^)
p cos a sin(|8 — y) — 5 cos y sin(a — ^)'
48. A person in a balloon, which is travelling uniformly
eastward, and also rising uniformly, observes a
train travelling southwards. When it is seen in
the N.E., N., and N.W. its angular depressions are
a, )8, y, respectively ; shew that
tan a + tan y = ^^ 2 tan /3.
49. A person walking along a straight road observes that
the maximum angles of elevation of two hills on
the same side of the road are a, /3, and the distance
between the points of observation is a. Along a
road passing between them, inclined at an angle y
to the former, the maximum angles of elevation
are again a, /3, and are observed at points distant
h from one another. Find the heights of the
hills.
CHAPTER X.
APPLICATIONS TO THE GEOMETEY OF TEIANGLES,
POLYGONS, AND CIKCLES.
156. A knowledge of the following geometrical the-
orems, in addition to those given in Euclid, will be
assumed in this chapter.
The straight lines which bisect the sides of a triangle
ABC at right angles pass through the circumcentre (8) of
the triangle. The straight lines which bisect the angles
of the triangle, both internally and externally, pass
through the incentre (/) and the three excentres (/j, ig* ^s)-
From, the latter theorem, it follows that any angular
point, the incentre, and the excentre opposite that angle
lie on a straight line ; that the same angular point and
the other two excentres lie on another straight line ; and
that these two lines are at right angles. The altitudes of
a triangle, i.e. the perpendiculars from the angular points
on the opposite sides, pass through a point (0), which is
sometimes called the centre of perpendiculars, but gene-
rally the orthocentre. The medians, i.e. the lines joining
the angular points to the mid-points of the opposite sides
pass through the centroid of the triangle.
If the perimeter of the triangle be denoted by 2s, the
lengths of the tangents from the angular points to the
219
220 GEOMETRY OF TRIANGLES,
incircle are s — a, s — h, and s — c, respective!}'' ; and the
lengths of the tangents from each angular point to the
opposite excircle is s, the semi-perimeter of the triangle.
The lines joining the feet (D, E, F) of the altitudes
form the pedal triangle of the given triangle. Each
angle of the pedal triangle is bisected by the correspond-
ing altitude, and is equal to the supplement of twice the
opposite angle of the given triangle. Thus, the ortho-
centre of a triangle is the incentre of the pedal triangle.
Also, any triangle is the pedal triangle of that formed by
joining its excentres.
If the altitudes of a triangle be produced, the parts
intercepted between the orthocentre and the circumcircle
are bisected by the sides to which they are perpendicu-
lars. The parts of the altitudes intercepted between the
orthocentre and the angular points are double of the
perpendiculars from the circumcentre on the opposite
The nine-points-circle of a triangle passes through the
mid-points of the sides, the feet of the altitudes, and the
mid-points of the parts of the altitudes intercepted be-
tween the orthocentre and the angular points. Its centre
is the mid-point of the line joining the orthocentre and
the circumcentre, and its radius is half that of the circum-
circle of the given triangle.
157. To find the area of a triangle.
Let ABC be the triangle. Draw the altitude AD.
(1) In terms of two sides and the included angle.
Area of triangle ABG= \AD . EG
= i AB sin B. BO,
if B be acute or right.
POLYGONS, AND CIRCLES.
221
or lAB sin(7r — 5) . BC, if B be obtuse or right,
= |oa sin B, in every case.
Similarljr,
area = J6c sin A = \ah sin G.
(2) In terms of the sides.
Area of triangle ABG
= lea sin B
B . B
= eacos.^sm^.
y ca ^ ca
= /s/s{s — a){s — h){s — c) = S.
158. To find the area of any quadrilateral in terms
of its sides and the sum of two opposite angles.
Let a, h, c, d denote the
lengths of the sides AB, BC,
CD, DA, respectively, of the
quadrilateral ABGD ; s the
semi-perimeter; 2a) the sum
of the angles A and G; and
Q the area of the quadrilateral.
Let A=()0 + a, G = a) — a.
Then Q = AABD+ABGD
= Jac? sin(ft) + a) + J?>c sin(ft) — a),
.-. 2Q = (ad-{-bc)cos a sin o) — (6c — ad)8m a cos w (1)
Again, by art. 121,
BD^ = a^ ^-d^- 2ad cos(ft, + a) = h^ + c^- 26c cos(a) - a),
(6c — ad) cos a cos w + (ad + 6c) sin a sin o)
= i(62 + c2-a2-c^2) ^2)
222
GEOMETRY OF TRIANGLES,
Eliminating a from equations (1) and (2) by squaring
and adding, we get
{ad 4- 6c)2sin2ft, + {he - adf co?i'^w = 4Q2 + j(62 + c^^a^ - cV^)\
:. 4>Q'- = aW+h^c^-2ahcd{2 cos^a)-l)-i{b^+c^-a^-d%
.-. 16Q2 = 4(acZ+6c)2-(62+c2-a2-cZ2)2-l6a6c(Zcos2ft,
= (2acZ + 26c + 62 + c2 - a2 - c?2)
x(2acZ+26c-6'-c'+a'+(Z')-16a6ccZcos'a)
= (6+c+a-(^)(6 + c-(x+c^)
x{a+d+b — c){a+d — h+c) — 16ahcdcos^(jt).
Q=^{{s — a){8 — 6)(s — c){8 —d) — abed cos^o} .
Cor. 1. — If the quadrilateral be cyclic, ft) = 90° ; and the
expression for the area becomes
\/{8 — a){s — b){s — c){s — d).
Cor. 2. — Hence, if the sides of a quadrilateral be given
in length, its area is greatest when it is cyclic.
159. To find the area of a regular polygon inscribed
in a given circle.
Let ABC... be a regular poly-
gon of n sides inscribed in the
given circle, of which 0 is the
centre and r the radius.
Join OA, OB. Then angle
A0B = 2irln.
Area of triangle AOB
= lOA.OB^inAOB
= Jr^sin
area of polygon ABC. . . = ^^nr'^sin.
n
27r
n'
POLYGONS, AND CIRCLES.
223
160. To find the area of a regular polygon described
about a given circle.
Let ABC... be a regular poly-
gon of n sides described about
the given circle, of which 0 is
the centre and r the radius.
Let D be the point of contact
of the side AB. Join OA, OB,
OD. Then angle AOD = ir/n.
Area of triangle AOB^\OD.AB
= ^r . 2r tan
n
.'. area of polj^gon ABC. . . = nrHsna
n
161. To find the radius of the circumcircle of a
triangle.
Let R be the radius of the circumcircle of the triangle
ABC. Draw the diameter BA\ and join AV.
or
(1) In terms of a side and the opposite angle.
Since BCA' is a right angle,
BC=BA'BmBAV
= BA' sin A, if A be acute or right,
BA' sm{7r — A), if A be obtuse or right.
m:
224
GEOMETRY OF TRIANGLES,
a — ^R^m A, in every case.
7?_ a h c
(art. 120)
2sm^ 2sin5 2sma
(2) In terms of the sides.
Byart. 125, sin^=2>Sf/6c,
R=:ahcl4<8.
162. To find the radius of the incircle of a triangle.
Let / be the centre, and r the radius, of the incircle
of the triangle ABC. Draw
IG, IH, IK perpendicular to
the sides,
(1) In terms of the sides,
^ABC
= ABia-\-ACIA + AAIB
.-. S = \BG.IG+IGA.1H
-VlAB.IK
But IG = IH==IK=r,
S=r.i(a + b + c) = rs.
r = S/s.
(2) In terms of the angles and the radius of the
circumcircle.
BC = BG + GC = IG cot IBG + IG cot ICG,
' B G
cos 2 cos-^
sin- sm^^
5.0, G . B
cosg sin^ + cos^-sm-g
' . B . G
sin 2 sin-
POLYGONS, AND CIRCLES.
225
_ sin 1(^+0) '
. B . C
sm ^ . sin 7^
a sm - sin ^ 2E sm A sm -^ sin ;^
cos -^ COS g-
r = 4it sm ^ sm ^ sm ^.
163. To ^')i<i the radii of the excircles of a ti'iangle.
Let I^ be the centre, and r^ the radius of the excircle
opposite A ; rg and r^ the radii of those opposite B and (7.
(1) In terms of the sides.
AABC=AGI^A + AAI^B-ABT^G
S=iCA . I^H^-hiAB.I^K^-iBG. I^G^.
But JjG, = /i^i = I^K^ = r^,
.-. ^ = r^ J(6 + c — a) = ri(s — a),
.'. r^ = S/{s — a).
Similarly, r2 = >S/(s — 6), and
r^ = SI(s-c).
(2) In terms of the angles
and the radius of the circum-
circle.
BG=BG^+G^G
= GJ^cotG^BT^+G^I^coi Gfil^,
a = r^( tan^ + tan- j = '
^mlJB+G)
' B G '
cos— cos^
B G ^^ . . B G
a cos 77 cos ^ 2R sm A cos ^ cos ^
'2 2 2 2
cos
cos
226
GEOMETRY OF TRIANGLES,
r^ = 4ic sin ^ cos ^ cos ^.
Similarly,
Vc, = 4it cos » sm - cos -^ and i\ = 4i2 cos ^ cos -^ sm ly
164. jTo yiTicZ, -i^i any triangle, the distances between
(1) the circumcentre and the orthocentre, (2) the cir-
cumcentre and the incentre and excentres, and (3) the
orthocentre and the incentive and excentres.
Let S be the circum-
centre, 0 the ortho-
centre, and I the in-
centre, of the triangle
ABC.
(1) Since SA = R,
OA = 2R cos A, and
angle SAO = (90'' -B)
-(90° - C)-= C - B,
C therefore
SO^ = R\l -f- 4 cosM - 4 cos ^ cos(a- B)]
= R%1 - 8 cos J. cos B cos 6^).
A . B C
(2) Since SA =R, IA= r cosec — = 4i2 sin -^ sin ^,
and angle SAI=-^-{9Q'' -C) = \{G^B).
..2^.;.2^
J5 . G
:. SP = R'^-\QR^^\n^^m^^-2RAR^\n^f^m~co^\{G-B)
=:R^ — 2RAR sin -^ sin ;,^f cos — ^ —
= i22_2/?.4iisin|sin^cos^'^-,
Pt'^«
- 2 sin I sin ^),
POLYGONS, AND CIRCLES. 227
= R^ — 2RAR sin -^ sin ^ sin -^,
= R^-2Rr.
Again, if J^ be the excentre opposite A,
I^A = r^cosec ^ = 4it cos ^ cos -^.
.-. 81^' = R^-2RAR cos | cos ^(cos ^^ - 2 cos | cos ^),
= R^ + 2RAR sin ^ cos | cos |'
= i^2_{_2i^7V
(3) Since /J. = 7^ cosec ^ = 4i^ sin -^ sin ^,
0^ = 2jRcos^,
and angle /^O = ^-(90°-(7) = KC'-^)-
.-. OP
AT>2r 2A . A ' 2^ • 2<^ A A ' ^ ' ^ C-^l
= 4it^ cos^^ + 4sin^-;TSin^j^ — 4 cos A sm ^ sin ^ cos — ^^ I
A r,o\~ o A , A • oB • oC A A ' B . G 0 £
= 4it^ cosM + 4 sin^- sin^^ — 4 cos -4 sin ^ si n -^ cos -^ cos ^
— 4 cos A sin^ ^ sin^-^ I
= 4i?"^ cosM +4 sin^^ sin^^ — 4 cos A sin ^cos ^sin-cos-^
— 4f 1 — 2 sin^-^ j sin^- sin^-^ I
= 4!RMcos^A — cos A sin B sin C+ 8 sin^— sin^^ sin^-^ j
= 2r2 _ 4i^2cog ^ cos ^ g^g (^
Next, since A I^^r^cosec ~ = 4fR cos ^ cos — , we have,
similarly, 01^^ = 2r^^ — 4i?"^cos A cos 5 cos G.
228 GEOMETRY OF TRIANGLES,
165. The nine-points circle of a triangle tovxihea the
incircle and excircles (Feuerhach's Theorem).
Let 0 be the orthocentre of the triangle ABC, S
the circumcentre, F the nine-points-centre, / the incentre,
and /j the excentre opposite A.
We have to shew that IV^^R-r and l^V^^R + r^.
Since V is the mid-point of SO,
2IV^ = SP+IO^-2SV^
= E^- 2Rr + 2r2 - 4i22cos A cos B cos G
- JE2 + 4722COS A cos B cos (7
= i(R-2rf.
IV=lR^r.
Again, from the triangle I^SO,
2/^ F2 = 81^^ + 0/^2 - 2>Sf F2
= i^ + 2Rr^ -f 2ri2 - 4i22cos A cos 5 cos C
- Ji22 + 4i22cos il cos B cos a
= J(i2+2r,)2,
I^V=\R+r^.
Similarly, /g F= JiJ+rg and /g F= Ji? -}- rg.
Hence, the nine- points-circle touches the incircle and
three excircles.
166. To find the area of a circle.
Let regular polygons of n sides be described in and
about the given circle of radius r.
The area of the former polygon is ^nt^s^n. — , and of
the latter 'Mr2tan -. The area of the circle lies between
n
these values, whatever be the value of n, and therefore
lies between them when n is infinitely great.
^1'
POLYGONS, AND CIRCLES.
229
Now, when n is infinity,
limit of Inrhm— = limit of A'nr^— ( sin — -4-
= 7rr2 (art. 75),
and limit of 'n.r^tan - = limit of nr^-i tan --r- -
n n\ n n
n
the area of the circle = irr^.
167. To find the area of a sector of a circle.
Let 0 be the number of radians in the angle AOB
of the sector AOB, r the radius
of the circle. Then
area of sector A OB : area of circle
= angle A OB : 4 right angles
= e:27r,
area of sector = ■x-ttt'^
168. Example 1. — If / be the incentre of the triangle ABC, and
/i'j, /i'2, ^3 the radii of the circumcircles of the triangles BIC, CIA,
AIB, then li^ Li^R^ = 272V. .
(See figure of art. 162.)
By art. 1 61, BC= 2 /?isin BIG
B C
= 2^i8in(7r -
-2/2iCOS^-
2/2sin^ = 2/?iCOs|:,
i2i = 2/2sin4,
RiR2R3=SB^sm^ sin^sin^
2 2^
= 2i2V.
230 GEOMETRY OF TRIANGLES,
Example 2. — To fiud the area of a quadrilateral in terms of any
three sides and the two included angles.
Let AB=a, BC=b, CD = c, angle A BC = 6 &nd angle BCD=cf>.
E^ D
Draw BE parallel and equal to CD, and CF parallel and equal
to BA; and join AE, DF, BE and AF. Then ABCF, DCBE,
AEDF are parallelograms. Also, the triangles ABE, FCD are
equal, and A D bisects the parallelogram AEDF. Again, the angles
EBC, BCD are together equal to two right angles ; therefore, the
angle ABE ia the supplement oi 6 + 4>.
Now, twice the area of the quadrilateral A BCD
=par. ABCF+AADF-ADCF+p&r. DCBE- /\ABE-AADE
= par. ABCF-\-Y)a.r. DCBE-2AABE
= a6 sin ^ + he sin <f>-ac sin( 0-\-(f)).
.'. area of quadrilateral ABCD
= \ah sin ^ + ^6c sin </> - \ac sin( 6 + </>).
Example 3.— If the incentre of a triangle be equidistant from
the circumcentre and the orthocentre, one angle of the triangle
is 60°.
By art. 164,
0/2 = 2r2 - 4 A! '-« cos A cos B cos C,
and SP=R^-2Rr,
OP - SP =2r^ -4R^ COS A cos B cos C-R^ + 2Rr,
hence, using the relations
4 R C
r=4/2sin ^ sin „ sin- = ^(cos ^4-cos5+cos C— 1)
2i 2i 2t
and co'^A + cos^^ + co^C-\- 2 cos A cos B cos C= 1 , .
we get OP-SP= R\\-2cos A){\ - 2 cos B){\ - 2 cos C).
If 01= SI, one of the factors 1-2 cos J, l-2cos^, l-2cosC'
must vanish, and therefore one of the angles of the triangle is 60°.
POLYGONS, AND CIRCLES. 231
Example 4. — If the incircle of a triangle pass through the cir-
cumcentre and the orthocentre, the angles of the triangle are
|, | + cos-V2-i) and ^-co^-\J2-\).
Since the incircle passes through the circumcentre and the
orthocentre,
.*. the incentre is equidistant from these two points,
.•. one angle of the triangle must be -, by the previous example.
Let A =f .
3
Again, since the incircle passes through the circumcentre,
.'. the radius of the incircle is equal to the distance between the
incentre and circumcentre,
.'. 16/j:2sin2i sin2^ sm^-=R^ - SRhin ^ sin - sin -.
2 2 2 2 2 2
AD C
Now, 4 sin — sin — sin - = cos A + cos B + cos C - 1 ,
ju 2i 2i
. (cos ^1+ COS 5+ COS C-l)- = l-2(cos J+cos5+cosC-l),
. (cos A + cos B + cos C)2 = 2,
. cos A + cos ^ + cos C — ^/2,
. cos5+cosC'=v/2-i.
. 2cos^±f:'cos^-=^'=cos:^-<'^=V2-i,
.-. ^^=cos-\V2-i)and^=|,
.-. A=1, 5=^ + cos-Xv^2-|) and (7= J-cos--\^/2-|).
3 3 3
Examples XIX.
1. Find the areas. of the triangles whose sides are : (1)
15, 86, and 39 feet; (2) 198, 194, and 195 feet.
2. The sides of a triangle are 242, 1212, and 1450 yards,
shew that its area is 6 acres.
3. In the ambiguous case in the solution of triangles,
find the sum of the areas of the two triangles.
232 GEOMETRY OF TRIANGLES,
4. In the ambiguous case, find the difference between
the areas of the two triangles.
5. Find the area of a triangle in terms of two angles
and the adjacent side. If 5 = 45°, (7=60°, and
a = 2(^3 + 1) inches, shew that the area of the
triangle is 6 + 2^3 square inches.
6. Prove that the area of a triangle is
Ja2sin25 + J62sin2^.
7. Prove that the area of a triangle is
lid'coiA + 62cot B + c^cot G).
8. If a, h, c be the sides of a triangle, the triangle whose
sides are m(b+c), m{c+a), mia+h) will be equal
to it in area if
2m* = sin -^ sm - sin -^.
9. If the sides of a quadrilateral be 23, 29, 37, and 41
inches, respectively, the greatest area it can have
is 7 square feet.
10. Find the area of a cyclic quadrilateral in terms of its
sides, without deducing it from the general ex-
pression of art. 158.
11. li ABGD be a cyclic quadrilateral,
j^(ji^ (^^ + 6cg)(acZ + he)
ab+cd
12. If ABGD be a cyclic quadrilateral,
**"2 V(s-c)(s-d)-
13. The area of a quadrilateral in which a circle can be
inscribed is ijabcd . sin co, where 2co is the sum
of two opposite angles ; and, if a circle can be also
circumscribed about it, its area is Jahcd.
POLYGONS, AND CIRCLES. 233
14. If ABCD he a quadrilateral which can be inscribed in
a circle and also circumscribed about a circle,
, ^A he -I , o-D ah
tan^ ^ = — 7 and tan''-^ = — ^.
15. If a be the side of a regular polygon of n sides, and
-B, r the radii of its circumscribed and inscribed
circles,
16. The area of a regular polygon inscribed in a circle is
one-fourth that of the regular polygon of the same
number of sides described about the circle ; find
the number of sides in the polygons.
17. The area of a regular polygon inscribed in a circle is
to that of the circumscribed regular polygon of the
same number of sides as 3 : 4 ; find the number of
sides.
18. Compare the areas of regular decagons inscribed in,
and described about, a given circle.
19. Find the area of a regular dodecagon, the length of
whose side is a.
20. In any triangle, 4i^2sin A sin B sin G= 2S.
21. In the ambiguous case, the circumcircles of the two
triangles are equal.
22. If 0 be the orthocentre of the triangle ABC, the cir-
cumcircles of the triangles BOG, CO A, AOB are
equal to that of the given triangle.
23. In any triangle, cos(5 — C) + cos A = bc/2R^.
24. If S be the circumcentre of the triangle ABC, shew
from the areas of the triangles ABC, BSC, CSA,
ASB that
sin 2J. + sin 25-j-sin2C=4 sin J sin jB sin C.
234 GEOMETRY OF TIU ANGLES,
25. If S be the circumcentre of the triangle ABC, the
diameter of the circumcircle of the triangle BSC
cannot be less than the radius of the circunocircle
of the given triangle.
26. If the incircle of a triangle touch the sides in A\
B\ G\ the square of the area of the triangle ABC
is equal to the product of BA\ CB\ AC multi-
plied by their sum.
27. Prove (1) algebraically, (2) geometrically, that
r = (s — a)tan — = (s — 6)tan — = (s — c)tan ^,
, A '^ ^ B , G
rj = stan^, 7'2 = stanj^, ^3 = 5 tan-.
2i Ji Z
28. Find the radii of the circumcircle, incircle and ex-
circles of the triangle whose sides are 25, 52 and
63 inches.
29. Find the radii of the circumcircle, incircle and ex-
circles of the triangle whose sides are 25, 101 and
114 inches.
30. A sphere whose radius is 2 inches rests on three
horizontal wires forming a plane triangle, whose
sides are 3, 4 and 5 inches ; find the height of the
top of the sphere above the plane of the wires.
31. r7vv'3 = ^2
32. r^rr,+r^r^ + i\r^ = s\
33. 1+1+^=1.
n ^\ ^'3 ^'
34 ^^ = tan2^tan2ftan2^.
r^r^T^ 2 2 2
35. cot^^^^^i^.
POLYGONS, AND CIRCLES. 235
36. sm2'^ =
37. a = {r,-\-r,)J'':^. .
\r2r3
38. (7'i - r)(r2 - r)(r3 - r) = 4r2i2.
(X 6 7'3
40. 7\ + r2 + r3 — r = 4i^.
41. The sum of the reciprocals of the altitudes of a
triangle is equal to the sum of the reciprocals of
the radii of the excircles.
42. The sum of the radii of the two excircles of a triangle
which touch the side a produced is equal to
a cot ^.
43. If the sides of a triangle be in arithmetical progres-
sion, then will the radii of the excircles be in
harmonical progression.
44. Each altitude of a triangle is a harmonic mean
between the radii of two excircles.
45. The distances of the excentres of a triangle from
the incentre are a sec ^, b sec ^, and c sec ^.
46. The distances between the excentres of a triangle
A , B , G
are a cosec -^, 0 cosec ^, and c cosec ^.
47. If /g ^^d ^3 ^® ^^6 excentres opposite B and G,
48. The excentres of a triangle lie without the circum-
circle, and cannot be equidistant from it unless
the triangle be equilateral.
GEOMETRY OF TRIANGLES,
49. The distances of the orthocentre of a triangle from
its angular points are 2i2cos^, 2i2cosJ5, and
2R cos G, or a cot A , h cot B, and c cot G.
50. Find the area of the segment of a circle of 14 inches
radius, the arc of the segment subtending an
angle of 30° at the centre of the circle (7r = 3|).
51. Each of two equal circles passes through the centre
of the other; shew that the area common to both
is ia2(47r-3V3),
where a is the radius of either circle.
52. If 0 be the orthocentre of the triangle ABG,
OB . AB+OG . GA _0G . BG+OA . AB
BG ~ GA
OA.GA + OB.BG
AB
53. If /, /j, /g, Is be the incentre and excentres of a
triangle ABG,
lI,^ + IJ^ = II,^ + I,I,^=:IIi^-IJl
54. Also r3 . //, . 11^ . lis = I^' • I^^ ' I(^-
55. IfDEFhe the pedal triangle of a triangle, the lengths
of EF, FD and DE are i2sin2^, i^ sin 25, and
B sin 2C, or a cos A, 6 cos B, and c cos (7 ; and the
perimeter of the triangle DEF is
4i2 sin A sin B sin G.
56. The area of the pedal triangle of the triangle ABG is
2S cos A cos B cos G.
57. The radius of the incircle of the pedal triangle of the
triangle ABG is 2R cos A cos B cos G.
58. The area of the triangle formed by joining the
excentres of a triangle ABC is ahcj^r or sa/sin J..
POLYGONS, AND CIRCLES. 237
59. If 0 be the orthocentre and S the circumceatre
of a triangle, 80^ = 9R^ -a'-h''- c\
60. If 0 be the orthocentre of a triangle ABC, and DEF
the pedal triangle, shew, from the areas of the
triangles ABC, AEF, BFD, CDF, and DEF, that
Gos^A + cos^^ + cos^O + 2 cos A cos B cos C = 1 .
61. If P be a point on the circumcircle of the quadri-
lateral ABCD, the products of the perpendiculars
on the following pairs of sides, BC, AD ; GA, BD ;
and AB, CD; are each equal to
PA.PB.PG.PD
62. If (J) be the angle between the diagonals of a quadri-
lateral ABCD, its area is
63. li ABCD be a quadrilateral such that the lines joining
the mid-points of opposite sides are equal, then
ac cos(A -\-D) = hd cos( J. -f B).
64. If a regular pentagon and a regular decagon have the
same perimeter, prove that their areas are as 2 : ^5.
65. If P and Q be the areas of two regular polygons
inscribed and circumscribed respectively to the
same circle, and if P' and Q be the areas of the
inscribed and circumscribed regular polygons, in
the same circle, with double the number of sides,
then
l = Vp-J and 1 = 1(1+;^).
Hence, find an expression for the area of the
octagon circumscribed to a circle whose radius is r.
QiQ. If the radius of the circumcircle of a triangle be equal
to the perpendicular drawn from one of the angles
238 GEOMETRY OF TRIANGLES,
on the opposite side, the product of the sines of the
angles adjacent to that side is h
67. If 0 be the orthocentre of the triangle ABC, and K
the incentre of the triangle BOG, the radius of the
A
circumcircle of the triangle BKG is ^R cos -^•
68. ABC is a triangle inscribed in a circle of radius R,
and G, H, K are the mid-points of the arcs
BC, GAy AB\ prove that the radius of the incircle
of the triangle GHK is
,^ . B+G . G+A . A+B
4>M sin — 7 — sin — ^ — sin — -j — .
69. The internal bisectors of the angles of the triangle
ABG meet the circumcircle in G, H, K ; if S' be
the area of the triangle GHK, then
S'IS = Rl2r.
70. If p be the radius of the incircle of a triangle whose
sides are 6 + c, c+a, a +6, where a, h, c are the
sides of a given triangle, then
p^ = 2Rr.
71. / is the incentre of a triangle, and G, H, K its points
of contact with the sides. If p^, p.^, p^ be the radii
of the circumcircles of the triangles HIK, KIG,
GIH, respectively, then
72. /is the incentre of the triangle ABG ; R, R^, R^, R^
are the radii of the circumcircles of the triangles
ABG, IBG, IGA, TAB, respectively; prove that
73. If a, /3, y be the altitudes of a triangle.
POLYGONS, AND CIRCLES. 239
74 If I, m, n be the distances between the excentres of a
triangle,
Imn sin A sin 5 sin C= Sr^rg^'g.
75. If I^, ig, -^3 be the excentres of a triangle ABC, the
distances between those of the triangle /j/^a^s ^^®
8it cos — :r- , 8it cos — -. — and 8B cos —.--.
4 ' 4 4
76. In the ambiguous case in the solution of triangles,
the sum of the radii of the two incircles and of the
two excircles opposite the given angle, is equal to
twice the common altitude of the triangles.
77. In the ambiguous case, HA, a, h be given, and if
S, S' be the areas of the two triangles, the con-
tinued product of the radii of the incircles and of
the excircles opposite B, is equal to SS'.
78. The area of the triangle formed by joining the points
of contact of the incircle of a triangle with the
sides is 2rB'^lahc.
79. Find the radii of the circles which touch two sides of
a triangle and the incircle.
80. The radius of the incircle of a triangle can never be
greater than one half that of the circumcircle.
81. The rectangle under the segments of any chord of the
circumcircle drawn through the orthocentre is
greater than twice the rectangle under the seg-
ments made on a chord of the incircle drawn
through the same point by twice the square on.
the radius of the incircle.
82. Two circular sectors are of equal area, and the chords
of their arcs are equal ; their angles are as 2 : 1 ;
find the angles.
240 GEOMETRY OF TRIANGLES,
83. The circumference of a semicircle is divided into two
arcs, such that the chord of one is double that of
the other. Shew that the sum of the areas of the
two segments cut off by these chords : area of the
semicircle = 27 : 55. (tt = 3|.)
84. The altitudes of a triangle intersect in the point 0 ; if
Pv /°2' P3' P4» Pb> Pe ^® ^^^ radii of the circles taken
in order, inscribed in the six triangles of which 0
is the common vertex, then
PiP3P5 = PiPiPe-
85. AL, BM, ON, the medians of a triangle ABC, inter-
sect in G; if /Op pg- Ps' Pi> Pb> Pe ^® ^^® TS^d^x of the
circles inscribed in the triangles BGL, LGG, CGM,
MGA, AGN, NGB, then
Pi Pz P5 Pi Pi Pe
86. The points of contact of each of the four circles
touching the three sides of a triangle are joined ;
if the area of the triangle thus formed from the
incircle be subtracted from the sum of the areas of
those formed from the excircles, the remainder will
be double of the area of the original triangle.
87. If, in a triangle ABC, AG, BH, GK are cut off from
the sides AB, BG, GA, and respectively equal to
m . AB, m . BG, m . GA, and if R, R^, R^, R^, be the
radii of the circumcircles of the triangles ABG,
AKG, BGH, GHK, then
a?R^^ + y'R^ + cm^^ = (a2 + 6^ + c^)R%Sm^ Sm + l).
88. If 8 be the circumcentre, and / the incentre, of the
triangle ABG, then
a . A^^/= i2V(cos B ^ cos G).
POLYGONS, AND CIRCLES. 241
89. A triangle is formed by joining the points at which
the lines bisecting the angles of a given triangle
meet the opposite sides. Shew that the area of
the new triangle is to that of the given triangle in
the ratio of 2abc to {h-\-c){c-\-a){a-{-h).
90. If / be the incentre of a triangle ABC, IG, IH, IK
perpendiculars on the sides, p^, p^, p^ the radii of
the circles inscribed in the quadrilaterals AHIK,
BKIG, GGIH, then
Pi _|_ P2 _^ Ps = f ^
r-p^ r-p^ r-p^ r
91. F is the nine-points-centre of a triangle ABC, and
A\ B\ G' are the mid-points of the sides. If a, /3,
y be the angles subtended by VG, VA and VB,
respectively, at A\ B\ and G\ then
a cos a + 6 cos /3 -f- c cos y = 0.
92. If a\ h\ c' be the sides of the triangle formed by join-
ing the excentres of a given triangle ABG, then
g^ 6^ c^ ^ahc _ ^
a^ h'^ c'^ (Jbh'd
93. If the incircle of the pedal triangle of a given triangle
touch the sides of the former in A\ B\ G, then
B G G^A' A'B' a A -n n
^^77 = -7YT = -li^FT = " COS A COS B COS G.
BG GA AB
94. Through A, B and G are drawn straight lines A^B^,
B-fi^, G^A.^ perpendicular respectively to the sides
AB, BG, GA of the triangle ABG, forming the
triangle AJi^G^-, the triangle A^BJO^ is formed in
a similar way from the triangle A^B^G^; if AnBnGn
be the n^'^ triangle so formed, the radius of the
circumcircle of this trianofle is
\ 2 sin A sin BninG ) '
242 GEOMETRY OF TRIANGLES,
95. Two spherical surfaces, whose radii are p^, p^, cut at
an angle of 60°, and p is the radius of their circle
of intersection ; prove that
3^1 1 _ 1
V Pi^ P2 PlP'2
96. From a point at a distance c from the centre of a
circle whose radius is a, two tangents are drawn,
and a second circle is described touching the first
circle and the tangents, a third circle touching the
second and the same straight lines, and so on.
Shew that the sum of . the areas of all the
circles: the area of the first circle ={c — af-Aac.
97. On the sides BC, CA, AB of the triangle ABC are
described the three triangles A'BC, AB'C, ABC,
equal in every respect to the triangle ABC. Shew
that the sides of the triangle A'B'C are
aisjl + 8cos J. sin5sin G, h\/l + 8smA cos^sin C,
Cs/ 1 + 8 sin J. sin J5 cos G,
and that
area of A^'5'0' : area oiLABG= 3+8 cos^ cos^cosO : 1.
98. G is the centroid of a triangle ABG, and A', B\ G' the
mid-points of the sides ; if p^, p^, p^ be the radii of
the circumcircles of the triangles B'GG', CGA',
A'GB\ then
GA^j_pI^GBKpI^GG^ .^^1 a^+b^-\-c^
a2 62 ^-2 3'a^^ c^^*
9' 9 ' 9
Pi P2 PZ
99. Also,
2 ' 9 ' 2
P\ Pi Pi
100. The equation giving the length x of the diagonal
AG f)f the quadrilateral ABGD, is
POLYGONS, AND CIRCLES. 243
{x\ah + cd) - (ac + hd){ad + bc)Y
= 4 abed Gos^(o{{x^ -a?- b^)(x^ ~ c^- d'^) + 4a6ccZ sin-o)},
2ft) being the sum of two opposite angles.
101. If chords of the circumcircle of a triangle be drawn
through the points in which the line joining the
centres of the circumcircle and incircle meets the
incircle, the product of the rectangles under their
segments is equal to
102. The radii of the excircles of a triangle are the roots
of the equation
x^ - x\4R + r) + xs^ - rs2 = 0.
103. If the sides of a triangle be roots of the equation
x^ — Ix^ + mx — n = 0,
the altitudes of the triangle are roots of the
equation
SR^x^ - 4^mR^x^ + 2lnRx - oi^ = 0.
104. If 0 be the orthocentre of the triangle ABC, OA,
OB and OC are roots of the equation
x^ - 2(E+r)ic2+(r2 - 4>R^+s^)x - 2R{s^ - (r+2Rf] = 0.
105. A circle can be inscribed in a quadrilateral, three of
whose sides taken in order are 5, 4, 7; and the
quadrilateral itself is inscribed in a circle. Shew
that the sine of the angle between 'the diagonals is
Ss/70/67.
106. If 78 and 50 be the lengths of the diagonals of a
quadrilateral inscribed in a circle of radius 65 and
sin"^f be the angle between them, the sides of the
quadrilateral are llx/26, 5^26, 5^/26 and 19>v/26.
107. A circle touches two sides of a triangle and the
circumcircle, find its radius.
244 GEOMETRY OF TRIANGLES,
108. If S be the area of a triangle ABC, and >Si' the area
of the triangle formed by joining the points in
which the bisectors of the angles of ABC meet the
opposite sides, prove that
S' _ 2 sin J. sin 5 sin 0
S~(sin5+sin0)(8in(7+sin^)(sin^+sin5)*
109. If the incentre and circumcentre of a triangle be at
equal distances from one side, the cosines of the
angles adjacent to that side will together be equal tol.
110. A hexagon, two of whose sides are of length a, two
of length 6, and two of length c, is inscribed in a
circle of diameter d ; prove that
and that the difference between the square of the
area of the hexagon and the square of the area of
a triangle whose sides are a^2, 6^2, c^J'!, is
ahcd+\d\
111. If Xy y, z be the perpendiculars from the angular
points of a triangle on any straight line, then
a\x-y){x-z)^h\y-z){y-x)^-c\z-x){%-y)={'Lb.ABQ)\
if the proper sign be given to the perpendiculars.
112. If "p, q, r be the lengths of the bisectors of the
angles of a triangle produced to meet the circum-
circle, and ii, v, lu the lengths of the altitudes of
the triangle produced to meet the same circle, then
p\v — w)-{- q\w — u) + r\u — t^) = 0.
113. If ^, q, r be the bisectors of the angles of a triangle,
and p', q', r these bisectors produced to meet the
circumcircle, then
cos \A , cos JJ5 , cos \C _ 1,1,1
p q r a h c
and ^'cos J^-f^'cos J5+r'cos \G=a-\-h + c.
POLYGONS, AND CIRCLES. 245
114. On the sides of a scalene triangle ABC as bases
similar isosceles triangles are described, either all
externally or all internally, and their vertices are
joined so as to form a new triangle A'B'C'\ prove
that, if A'B'C be equilateral, the angles at the
bases of the isosceles triangles are each 30°; and
that, if it be similar to ABC, they are each
tan- ^^
115. A straight line AB is divided at C into two parts of
length 2a and 26 respectively. On A C, CB, and
^jB as diameters, semicircles are described so as to
be on the same side of AB. If 0 be the centre of
the circle which touches each of the three semi-
circles, shew that the radius of the circle is
ab{a-\-h)
and that its diameter is equal to the altitude of
the triangle AOB.
116. If, in a triangle, the feet of the perpendiculars from
two angles on the opposite sides be equally distant
from the mid-points of those sides, the other angle
will be 60° or 120°, or else the triangle will be
isosceles.
117. Find the relation which exists between the angles
of a triangle whose orthocentre lies on the incircle.
118. A triangle is formed by joining the feet of the per-
pendiculars from any point P on the sides of a
triangle ABC', if 8 be the circumcentre of the
triangle ABC, and S the distance of P from S,
shew that twice the area of the new triangle is
(E2_(52^siQ^sin5sina
246 GEOMETRY OF TRIANGLES,
Prove what it becomes when P is (1) at the
centre, (2) on the circumference of the circumcircle.
119. If twice the square on the diameter of the circum-
circle of a triangle is equal to the sum of the
squares on the sides, then the triangle is right-
angled.
120. The alternate angles of a regular pentagon are joined
by straight lines which form another pentagon ;
the alternate angles of this pentagon are joined,
and so on continually. Given a side of the first
pentagon, find the sum of the areas of all the
pentagons continued ad infinitum.
121. A circle of radius p touches externally three circles
which all touch each other externally, and whose
radii are p^, p^, p^ ; prove that
JP2±P3±P _{_ JPs±P2+_P + Jpi + P2 + P = JP1+P2 + P3
^ Pi ^ P2 ^ Ps ^ P ' '
If the first circle touch the other three and
include them all, find a similar relation between
the radii of the four circles.
If the three circles be each of radius a, the radii
of the other two circles are (2^3 ±3).,.
122. If /be the incentre, S the circuracentre, and 0 the
orthocentre of the triangle ABC, the area of the
triangle ISO is
— 2R^sin — ^ sin — ^ — sin — ^ — .
123. The triangle D^^i^ circumscribes the excircles of the
triangle A BC, prove that
_EF ^ FD _ DE
a cos A h cos B c cos G
POLYGONS, AxVB CIRCLES. 247
124. If c, c be the diagonals of a quadrilateral which is
incyclic and circumcyclic, D, d the diameters of
the circumcircle and incircle respectively, then
(P cc
125. Three circles, touching each other externally, are all-
touched by a fourth circle including them all. If
a, b, c be the radii of the three internal circles, and
a, |8, y the distances of their centres from that of
the external circle, respectively, prove that
\0G ca ao/ a^ b^ c^
126. Two points A, B are taken within a circle of radius
p whose centre is G. Prove that the diameters of
the circles which can be drawn through A and B
to touch the given circle, are the roots of the
equation
x\p^c^ - amf^m^G) - 2xpc^(p^ - ah cos C)
+ cHp^ - 2p^ah cos G+ a^¥) = 0,
where the symbols refer to the parts of the
triangle ABC.
Miscellaneous Examples. IT.
a.
If an arc of ten feet on a circle of eight feet diameter
subtend at the centre an angle of 143'' 14^22", find
the value of tt to four places of decimals.
If^+-B + 0=7r, then
sinM + sin25- sin2(7= 2 sin A sin B cos G.
248 GEOMETRY OF TRIANGLES,
3. Solv3 the equations : (1) cos 30 = cos 0,
(2) sia4e + sin20 = cosa
4. Find the cosines of the least and greatest angles of a
triangle whose sides are 7, 14, 15 ; and apply the
formula a^ =, ^2 _^ ^2 _ 26c cos J. to prove that, if the
straight line which bisects the vertical angle of a
triangle also bisects the base, the triangle must be
isosceles.
6. It is observed that the altitude of the top of a mountain
at each of the three angular points J., 5, (7 of a
horizontal triangle is a ; shew that the height of
the mountain above the plane of the triangle is
Jatanacosec^.
6. Shew that four times the area of a triangle is
62sin2a+c2sin25;
and interpret the result geometrically.
1. 8(cos^a + sin^a) = 5 + 3cos4a.
2. cos-iff+2tan-H = sin-H.
3. Find the conditions under which it is possible that the
expressions sin(a + /3)cos y and sin(a + y)cos/3 may
be equal.
4. If the sines of the angles of a triangle are as 13 : 14 : 15,
then the cosines are as 39 : 33 : 25.
5. If the sides of a triangle be 4219, 5073, 3104, find the
greatest angle.
6. Shew that (1) r= ^^^'^^^ •
/2> 1+1+1= 1
^^ hccaah Mr
POLYGONS, AND CIRCLES. 249
y-
1. If, in a triangle, each of the angles J. and 5 is double
of the third angle G, then
A+B A + B-{-G A^-B+G
cos 7^ COS = COS* -. •
2 0 4
2. If^ + 5+(7=27r, then
sin 2J.+sin 2i?4-sin 2(7= — 4sin^ sin J5 sin G.
3. Solve the equations: (1) sin 80 = 2 sin 0,
(2) tanO-|-tan20 = tan3a
, T- . . 1 a^ + y^ l + cosM-5)cosO
4. In any triangle, ^ , o=i— ; t-j — t^t »•
•^ * a^ + c^ 1 + cos( J. — C/)cos 5
5. When the sun is 20° E. of S. and at an altitude of 25°,
the shadow of the top of a church spire falls at a
point A on the level ground on which it is built.
At a point B, 60 feet north of A, the bearing of
the top of the spire is 15° E. of S. Find the
height of the spire.
6. The sides of a triangle are 11, 90, and 97, find its area,
and the radii of its circumcircle, incircle, and
excircles.
S.
1. cos^a + cos^f J + a j + cos^f I" + a j + cos^f -^ + a J = 2.
2. tan-Xl + r.H^)-tan-i(l+r^^i.r) = tan-i^^,^^J,^^.
3. If ABG be a triangle^ and sin 2A, sin 25, sin 2(7 be in
arithmetical progression, then tan A tan (7=3.
4. If ABO be a triangle, find cos G, having given
sin^ m , tan J. p
-. — D = — and D= •
sm B n tan B q
250 GEOMETRY OF TRIANGLES,
5. If ^=49^5' 30", 5 = 64** 15^20", and 6" = 5127, find C
and a.
6. lip, q, r be the lengths of the perpendiculars from the
circumcentre of a triangle to the sides, then
4^^ + ^ + ^^ = ^.
\p q rJ pqr
€.
1. Shew how to construct an angle when its sine is given,
and apply to the construction of an angle whose
sine is 3/(2 + ^5).
o
2. lfA+B+G=-^^, express cos 2A 4-cos 25+cos 2(7-1
as a single term.
8. Find the general values of the limits between which A
lies, when sinM is greater than cos^J..
4. If a, /3, y be the lengths of three straight lines AB, BC,
CD, and they be so placed that A, B, C, D are on
a circle whose diameter is ^D ; then will the
length oi AD be the positive root of the equation
a;3 - a;(a2 + /32 + y ) - 2a;5y = 0.
5. A tower stands on a slope inclined at an angle a to
the horizon. At the foot of the slope, directly
beneath the tower, the angle of elevation of the
top of the tower is 2a, and a feet further up the
slope it is Za. Find the height of the tower, and
the distance between the base of the tower and
the toot of the slope.
G. Between two concentric circles lies a series of circles,
given in number, each of which touches the two
nearest of the series and also the two concentric
POLYGONS, AND CIRCLES. 251
circles ; find the ratio of the areas of the con-
centric circles. If the number of the series of
circles be six, the area of the outer of the con-
centric circles is nine times that of the inner.
1. sin3a + sin3(12(r-ha) + sin3(240°H-a)= -f sinSa.
2. ABCDE is a regular pentagon, 0 the middle point of
the arc AE of the circuracircle ; if a be the radius
of the circle, shew that
(1) OB-OA = a,
(2) OA.OB = a\
3. If sin a — a cos a = a, find the value of tan a.
4 ABC is a triangle, and K is the middle point of AB ;
AD is drawn perpendicular to BG cutting GK in
L ; prove that AL is equal to
ah sin G
a~\-hcosG'
5. If ct = ^3-l, 5 = ^3 + 1, and J. = 15°, solve the tri-
angle.
6. In any triangle, R = ^ \\ ^^^/ ^ , ^ . ■.
Y].
1. A ring, 10 inches in diameter, is suspended from a
point one foot above its centre by six equal strings
attached to its circumference at equal intervals.
Find the cosine of the angle between two con-
secutive strings,
2. If^-f5+C=7r, then
2cos(^ + 40)sinU + 20) + sin2(5~a)
= -4sin0sin(5-C')sin(5-2C).
252 GEOMETRY OF TRIANGLES,
3. Prove that cos 9° = iVs + Jb + \Jb - ^5,
sin 9° = iV3 + V5- J V5£75,
cos 27° = W5"4-V^+ J V3^V^',
sin 27° = \Jh + V5 - i V3 - Jh.
4. Solve the equation
sina + sin(0-a) + sin(2e + a) = sin(^ + a)H-sin(20-a).
5. A church tower BCD with a spire above it, stands on
a horizontal plane, B being a point in its base and
G being 9 feet vertically above B. The height of
the tower is 289 feet and of the spire 35 feet;
from the extremity JL of a horizontal straight
line BA it is found that the angle subtended by
the spire is equal to the angle subtended by BG\
prove that AB is 180 feet, nearlj^
6. If a', h\ c' be the sides of the triangle formed by the
external bisectors of the angles of a triansie, then
"^ 6 "^ c " r
h'-^ c'^ abc
1. 2.an-i(J^tan|) = cos-<^ + ^^"^\
\Va + 6 2/ \a + o cos x/
2. ABODE is a regular pentagon and 0 any point on the
arc AE of the circumcircle. Prove the formula
cosa + cos(72° + a)+cos(72°-a) = cos(36° + a) + cos(36°-a)
and apply it to shew that
OA + OG+OE=OB-{-OD.
3. Eliminate 0 between x = acos(0 :\- a) and y = h cos(0 + /3).
4. 1( {a^ + h^)sm(A-B) = (a^-h^)sm{A+B), the triangle
is either isosceles or right-angled.
5. If a : 6 = 379 : 214 and C = 40" 24' find A and B.
POLYGONS, AND CIRCLES. 253
6. AD, BE, OF, the altitudes of a triangle ABC, are
produced to meet the cireumcircle in d, e, /;
shew that
^DEF : Adef : AABG= 2 : 8 : sec ^ sec 5 sec C.
1. Find the cosine, sine, and tangent of the angle between
two faces of a regular tetrahedron ; also of half the
angle between two adjacent faces of a regular
octahedron.
2. If ABC be a triangle, and if 1 — cos J., 1 — cos 5,
1— cosO be in H.P, then sin^, sinj5, sin (7 are
also in H.P.
3. The number of grades in an angle of a regular polygon
is to the number of degrees in an angle of another
as 5 : 3 ; find the number of sides in each, and shew
that there are only three solutions.
4. Solve the equation tan 80 = 5 tan 0.
5. Find the smaller value of c, having given ^ = 10°,
a = 2308-7, 6 = 7903-2.
6. In any triangle, dR^ is not less than a^ + h'^ + c^.
X.
1. Given (p) the sum of the three tangents, and (q) the
sum of the three cotangents, of the angles of a
triangle ; find an equation whose roots will be the
three tangents.
2. ABCDEFG is a regular heptagon, 0 the middle point
of the arc AG of the cireumcircle; if a be the
radius of the circle, shew that :
(1) OC-OB+OA=a,
(2) OA .OB . OC=a^
234 GEOMETRY OF TUT ANGLES,
3. Find the relations which must exist between a, /5, y in
order that
(1) tan a + tan/8+tany = tanatan^tan y.
(2) tan /3 tan y 4- tan y tan a + tana tan/3=l.
4. ABC being a triangle, express
a . h c
cos A cos B cos C
in a form adapted to logarithmic computation.
Find the numerical value of the expression when
a = 1000, A = 35° 4', jB= 10° 30'.
5. All vertical sections of a hill from the base to the
summit are alike, and consist of two equal arcs of
equal circles, of which the lower has its convexity
downwards, and the upper has its convexity up-
wards, the highest and lowest tangents being
horizontal ; shew that a person who goes right
over the hill traverses a less distance than one
who goes half round it.
6. Lines drawn parallel to the sides of a triangle ABC
through the excentres form a triangle A'B'C.
Shew that the perimeter of the latter triangle is
,^ .A B n
4iK cot cot -^ cot » .
fi.
1. Find two regular polygons such that the number of
their sides may be as 3 : 4, and the number of
degrees in an angle of the first to the number of
grades in an angle of the second as 4 : 5.
2. If X satisfy the two equations
x^+a^-2xacose = b^, x^+a^-2xacos(60° - e) = c\
then 2^2^a2 + 62 + c2±4^3.>Sf, where >Sf denotes the
area of the triangle whose sides are a, h, c.
POLYGONS, AND CIRCLES. 255
8. Solve the equations: (1) tan 0 + sec 20 = 1. .
(2) tan - hix + J sec " '^hx = -y.
4. If P be any point within a triangle ABC, prove that
cotPi?(7.AP5(7+cotP(7^.APa^ + cotP^5.AP^j5
is independent of the position of P.
5, Prove that three times the area of a triangle is equal
to
where I, on, n are the lengths of the medians.
V.
1. The angles x and y vary subject 'to the relation
smx = ksiny, where Jc is a constant greater than
unity. Shew that, as x changes from zero to a
right ancjle, continually increases, and find
"= ° ' tan^/ ^
the values of this expression, when x = 0, and
when x = ~.
„ T-,. , ^ sin a sin 0 .t, - . a sin a sin 0
2. It tan (/) = ^ — , prove that tan 0 = : ^^.
^ cos t^ — cos a cos </) ± cos a
3. Solve the equations :
,-. V sin 0 I cos 0 1
cos 0 — cos a sin a — sin 0 sin(a — 0)
(2) lOcos0 = 2cosec0 + seca
4. Two lines of length f, q are drawn from the same
point, and are inclined to one another at an angle a.
From their extremities perpendiculars are drawn
to each of them : find the area of the parallelogram
so formed.
5. Each of three circles within the area of a triangle
touches the other two and also two sides of the
256
triangle ; if a be the distance between the points
of contact on one of the sides, and 6, c be like
distances on the other two sides, prove that the
area of the triangle, of which the centres of the
circles are the angular points, is equal to
IT.
(Geometry of the Quadrilateral.)
1. In any quadrilateral
g^sinM - chm'^C hhin^G- d%m^A
m\\A+B) "^ 8m\B+C)
2 sin AmuG , j j^ i dx
= -.— 7-i y^^-^—y-n — TvN (cd cos i) — a6 cos B).
8in(^+i^)sin(i^ + 6y ^
2. If a, b, c be three sides of a quadrilateral, 0 being the
angle between a, h, and (/> that between h, c ; also,
if ah cos 6 = bc cos (p = ac cos{0 + 0) ; shew that the
quadrilateral is inscribable in a circle of which the
fourth side is the diameter.
3. If, in a quadrilateral, a + b==c + d, the difference be-
tween the areas of the triangles ABC and CD A is
equal to
area of quadrilateral x -.—i:~ r^l
sin h{B + I))
4 If the sum of the opposite angles of a quadrilateral be 2ft),
and if the angle between the diagonals be a, then
. 2 _TiQ{(8 — a)(8 — b){s — c)(s — d) — abed cos^o)}
tan a (^2 _ 52 4. ^2 ^ ^2y2 •
5. The length of the line joining the points of intersec-
tion of pairs of opposite sides of a cyclic quadri-
lateral is
(ad+bc)^(ab + cd)^bd{c^ - a^f -h acQ)' - d''f]h
(b^'-d^Xc'-a^)
CHAPTER XI.
HYPEEBOLIC FUNCTIONS.
169. Circular Functions in relation to the Sector of
a Circle. — Let a point move from X on the circumference
of a circle, whose centre is 0, to
the position P ; let ^ be the
number ot radians in the angle
XOP, A the area of the sector
XOP ; then, if a be the radius
of the circle, we have
2A
A = ^a^O, and therefore 6 = —^'
Hence,
cos 6 = cos
2A
2A
„, sinO = sin ^
and so on for the other circular functions, i.e., we may
regard the circular functions as functions of a sector of a
circle, and the results obtained will be identical with
those for angles measured in radians, provided that the
unit of area, in terms of which the sector is measured,
is the square whose diagonal is the radius of the
circle.
The sense of the sector XOP is the same as that of the
angle XOP, and is denoted by the order of the letters.
R 257
258
HYPERBOLIC FUNCTIONS,
Thus, for all positions of X, P, and Q on the circumference
we have
sector XOQ = sector ZOP+ sector POQ.
170. Definitions of the Hyperbolic Functions. — Let
a point move along the curve from the vertex A of one
bi-anch of a rectangular hyperbola, whose centre is 0 and
semi-axis equal to a, to the position P ; let J. be the area
of the hyperbolic sector A OP, and let u= — ^ » ^-^ ^^^ ^'^ ^^
the measure of the sector AOP, the unit of measurement
being the square whose diagonal is the semi-axis.
Take OF a line making an angle of 90** in the positive
sense with the transverse axis OAX, and let OM. ON be
HYPERBOLIC FUNCTIONS. 259
the projections of OP on OX, OY respectively, then the
r«atio
OM : OA is called the liyperholic cosine of u,
ON : OA the hyperbolic sine of u,
ON : OM the hyperbolic tangent of u,
OA : 0-M the hyperbolic secant of u,
OJ. : ON the hyperbolic cosecant of tt, and
Oif : OiV the hyperbolic cotangent of u.
The abbreviations for these hyperbolic functions are
cosh ^6, sinh u, tanh Uy sech u, cosech u, coth u..
Inverse Hyperbolic Functions.— If a; = coshu, then
we write inversely, as in the case of the circular functions,
u = cosh " '^x. Similarly, we denote the other inverse func-
tions by sinh-^a;, tanh-^aj, etc. The symbol 'cosh~^aj'
may be read: 'the sector whose hyperbolic cosine is x,
the unit in terms of which the sector is measured being
the square whose diagonal is the semi-axis.
If u be determined from the equation a; = cosh u, where
a; is a given number greater than unity, u is a two- valued
function of x, the values being equal in magnitude and
opposite in sense ; we define cosh~^a; as the positive value
oi 11.
Similarly, sech'^a; is defined as the positive value of n,
which satisfies the equation a; = sechu, x being a given
positive number not greater than unity.
The sign of each of the other inverse functions is the
same as the sign of x.
Hence, each of the quantities cosh~^a;, sinh~^a;, tanh~^a;,
sech~^a;, cosech -^aj, coth~^aj is a one- valued function
of X.
260
HYPERBOLIC FUNCTIONS.
171. Elementary Relations between the Hyperbolic
Functions. — We have, by definition,
cosh it-=
and seen u = — ^,
Similarly,
and
OA
sech u =
cosech u =
coth u =
OM'
cosh It'
1
sinh u'
1
tanh u'
.(A)
Again, by definition.
tanh. = g5
So also.
coth 16 :
OA ' OA coshtfc"
coshu
(B)
sinh u
From the property of the rectangular hyperbola, we
have (Taylor's Elem. Geom. of Conies, art. 53)
OM^-PM^ = OA\ I
OM^-ON^=OA^
fOMV_fONY_,
\0AJ \0AJ ^ '
Similarly,
and
cosh^u — sinh% = 1.
1 — tanh^u = sechht,
coth^u — 1 = cosech%.
•(C)
172. To determine the value of any Hyperbolic Func-
tion in terms of any other.
The formulae of the last article furnish five inde-
pendent relations between the six hyperbolic functions,
from which, in the same manner as in the case of the
HYPERBOLIC FUNCTIONS.
261
circular functions, we may deduce the value of any func-
tion in terms of any other. The hyperbolic cosine and
secant of any sector are, hj definition, positive ; the hyper-
bolic sine, cosecant, tangent and cotangent of a sector
are all positive or all negative. Hence, we must write
r--^o ^ 1 . Vl + cosech%
cosh u=-\- V smhm + 1 ; cosh u=± ■ — ;t- >
the upper or lower sign being taken according as cosech u
is positive or negative; and so on. The values and
proper signs are given in the following table :
cosht^
= c.
sinhu
= 8.
tanh w
sechM
= 05.
cosech u
cothw
= 2.
c
1
s
t
~ c
s
1
c
1
**-
1
s
t
+ ^
+ ^
s
1
t
-Jl-x'
t
±sf\-x^
+ 1
X
z
1
X
^sli-x"
X
1
y
-^l-x^
y
±s/z^-l
I
z
z
173. Even and Odd Functions.— From the definitions
and figure of art. 170, we have immediately
cosh( — u) = cosh u,
and sech( — u) = sech w ;
thus, the hyperbolic cosine and secant are even functions
of u.
262
HYPERBOLIC FUNCTIONS.
Also, sinh( — u) = — sinh u,
tanh( — u)= — tanhif,
cosech( — }i)=— cosech u,
coth( — u) = — coth u,
i.e. the hyperbolic sine, tangent, cosecant, and cotangent
are odd functions of u.
174 If P, Q, R, S he points taken in order on a branch
of a rectangular hyperbola, and if QR and PS be
parallel, then will the sectors POQ, ROS be equal ; and,
conversely, if the sectors POQ, ROS be equal, then will
QR and PS be parallel.
Let QR be produced to meet the asymptotes in Q\ R',
and let PS be produced to meet them in P\ S'.
Since the intercepts on any chord between the curve
and its asymptotes are equal {E.G.G. art. 50), therefore
HYPERBOLIC FUNCTIONS. 263
the curvilinear areas PP'QQ and BR'B'8 can be divided
into an infinite number of pairs of equal strips by draw-
ing chords parallel to PSj and therefore these areas are
equal.
We have also ^OQQ' = ^ORR\ (Eucl. I. 38)
and AOPP' = AOSS\
... AOQQ' + PP'Q'Q - AOPP' = AORR + PR'S' 8 - AO;Sf>S',
i.e. sector POQ = sector RO 8.
Hence, conversely, by a reductio ad ahsurdum, it
follows that, if the sector POQ = the sector R08, then
will QR and P8 be parallel.
175. If POQ, R08 he equal sectors of a rectangular
hyperbola, and if p, q, r, s he the projections of the points
P, Q, R, 8 on an asymptote, then will
0p:0q = 0r:0s.
Since the sectors POQ and R08 (see figure of art.
174) are equal, it follows that the chords QR and P8
are parallel (art. 174),
hence Op:Oq = PP':QQ'
= 88':RR
= 8s:Rr.
But, since R and 8 are on the hyperbola, we have
Or. Rr = Os . 8s, (E.G.G. art. 49)
and, therefore, 8s:Rr = Or : Os.
Hence, 0p:0q=0r:0s.
176. If RV he an ordinate to any diameter OQ of a
rectangular hyperhola, and if u= rTAT — > ^^^^ ^^^
OA'^
, OF , . , VR
cosh n= jy^ and smn u^-z^-
264
HYPERBOLIC FUNCTIONS.
Let P be a point on the curve such that sector
2y .4 OP = sector QOR; let a, p, q,
T be the projections of A, P, Q,
R on the asymptote OT \ then
by art. 175, we have
Op'.Oa^^Or.Oq.
But, since OM and MP are
equally inclined to the asymp-
tote,
OM-^MP_Op
OA Oa
AM
Similarly, since 0 V and VR are equally inclined to the
asymptote (E. G. C, art. 54),
OV+VR^Or
Oq
_OV±VR
OQ
Hence,
OQ
OM+MP
OA
In like manner, since Pp :Aa = Rr: Qq,
OV-VR
(art. 175)
we have
/. by addition,
and, by subtraction,
OM-^MP
OA
OM
OA'
MP
OA'
OM
OQ
qv
OQ'
VR
OQ'
MP
But, by definition, cosh u = -y-^, and sinh u = ^yj;
, 07 , . , VR
cosnu = -rj^, and smh u= jj^-
HYPERBOLIC FUNCTIONS.
265
177. To "prove that
cosh (if ■\-v) = cosh u cosh v + sinh u sinh v.
2A0P
Let P. Q be points on the curve such that u= ^.^
±POQ
Draw QV, the
ordinate of Q to the dia-
meter PV, and draw VL
perpendicular to OA and
VW to Qi\^.
Since lAOP = lVQW
{E. G. a, art. 54), the tri-
angles QFF and 0PM are
similar.
Now,
AM L N
^, - , OiY OL + LN
cosh(u+^) = (^=-^X-
^a¥ qL_ MP_ VW
OA' OM OA MP'
But,
and
OX OF ,
W=OP = ^^^^^'
FIT FQ . ,
(art. 176)
cosh (it + ?;) = cosh u cosh 'y-f- sinh u sinh ^...(D)
In like manner,
. ,, ^ ^ NQ MP LV.OM WQ
smh(^ + ^) = ^=^.^+^.^
^MP OV OM VQ
OA' OP OA'OP
= sinh u cosh v + cosh usinh v (D)
HYPERBOLIC FUNCTIONS.
178. Hence we have
sinh(u + i;)
i&.n\\{u + v) =
cosh(^ + v)
sinh u cosh ^;+cosh u sinh v
cosh u cosh '?;+sinh u sinh v
, ,, n i. 1./ . \ tanh t/, + tanh -y
and th erefore tanh (u-\-v) = q— -7 — r-'-— — ^— .
1 + tanh u tanh v
Again, putting v = Uy we get
cosh2u = cosh2ifc4-sinh% ^
= 2 cosh% — 1
= 2sinh2u+l.
sinh 2^ = 2 cosh u sinh u,
2 tanh u
,(D)
tanh 2it
l + tanh^tt"
.(E)
179. Since the geometrical proof of art. 177 is inde-
pendent of the sense of u and v, we have, by changing v
into —V and attending to the results of art. 173,
cosh(u — v) = cosh n cosh v — sinh u sinh v,
sinh(u — v) = sinh u cosh -y — cosh u sinh -y, I _ m^
. , , , tanh u — tanh V
tanh (16 — ?;) = :; — 7 — \ 7 — r— •
^ 1 — tanh u tanh v
From the addition formulae
cosh(u + 'y) = cosh u cosh y+sinh u sinh v^
cosh(u — v) = cosh 16 cosh v — sinh i6 sinh v I
8inh(i6 + i^) = sinh 16 cosh v + cosh u sinh v I
8inh(i6 — -y) = sinh i6 cosh v — cosh u sinh i;J
we obtain, as in the case of the circular functions,
2 cosh 16 cosh V = cosh(i6 + v) + cosh(i6 — vy
2 sinh 16 sinh v = cosh(i6 + -y) — cosh(i6 — v)
2 sinh u cosh i; = sinb(u + v) + sinh (u — v)
2 cosh 16 sinh v = sinh(K, -\-v) — sinh (u — v)>
whence, putting ;S for 16 + ^; and D for u — v, we have
,(F)
HYPERBOLIC FUNCTIONS.
267
cosh >Si + cosh Z) = 2 cosh — ^ cosh — ^
cosh 8 — cosh D = 2 sinh — ^ — sinh — g—
sinh >Sf + sinh D — 2 sinh — ^ — cosh — ^
sinh >Sf — sinh D = 2 cosh — ^^^ — sinh — ^ —
.(F)
180. Gudermannian Function. — Let P be a point
on a rectangular hyperbola,
FM the ordinate of P, MT
a tangent to the auxiliary
circle, P and T being on the
same side of the axis.
^, .„ 2 sector J. OP
ihen, II u =
and
e= LAOT,
. OM OM
we have sec6> = jyn — q-j>
and .'.
Hence,
and
Also, since
. ^ I ta 1 x/Oi/2-0^2 ^p
tan 0 = s/sec^O — 1= y^-, = tt-t-
OA OA
sec 0 = cosh u^
tan 0 = sinh. uJ
0 tan 6 ^ , 1 u sinh u
tan - = IT- 7^, and tanh ^ = .r- ^ — .
2 l + sec0 2 1 + coshu
tan ^ = tanh -^
•(G)
(G)
The angle 6 is called the gudermannian of u, and the
relation between 0 and ^t is written
^ = gdit, or u = gd-'^0.
268
HYPERBOLIC FUNCTIONS.
181. Curves of the Hyperbolic Functions. — It will
be shewn at a later stage that
cosh'M, = l + n7 + 7T+ ...ad inf.
11 ^ 1L
and sinh u = u+.-jr-{-r-=+ ...ad inf.
[3 [5 -^
From these series the values of cosh u and sinh u for given
values of u may be calculated. Then, by division, we
obtain the values of the remaining hyperbolic functions
of u. If 6 = gdu, we may employ the equation
sec 0 = cosh u and a table of natural secants to find 0
when u is known.
Table of Approximate Values of Hyperbolic
Functions.
d
sec^
cos^
tan^
COt0
8in^
cosec 6
u.
gdu.
cosh u.
sechu.
sinh u.
•00
cosech u.
tanh u.
coth u.
0-0
0°
1-0!)
1-00
00
0-00
00
0-2
11°
ro2
•98
•20
4-97
-20
507
0-4
22"
108
•92
•41
2-43
•38
2-63
0-6
32'
119
•84
•64
157
•54
1-86
0-8
42°
1-34
•75
•89
1-13
•66
1-51
10
50"
1-54
•69
1^18
•85
•76
1-31
1-2
56°
1-81
•55
1-51
•66
•83
1-20
1-4
62°
2-15
•46
1-90
•53
•89
1-13
1-6
67°
2-58
•39
2-38
•42
•92
1-09
1-8
71°
3-11
•32
2-94
•34
•95
1-06
2-0
75°
3-76
•27
3-63
•28
•96
1-04
30
84°
10-07
•10
10-02
•10
•995
1-005
40
88°
27-29
•04
27-27
•04
•999
1-0007
50
89°
74-74
•01
7473
•01
•9999
1-0001
00
90°
CO
0-0
00
0^0
1^0
10
By aid of the above table we may readily draw the
curves representing the functions.
HYPERBOLIC FUNCTIONS.
269
i:^^
O Co
270
HYPERBOLIC FUNCTIONS.
Curves of the Hyperbolic
Tangent and Cotangent
Tangent •
Cotangent
jxf
Q
/
^
I
/
^
u
-6
'»
•I
-pcf
O
e'
/
-3
Curve of the Guderniannian
[
HYPERBOLIC FUNCTIONS. 271
Examples XX.
L Shew that the area included between a branch of a
rectangular hyperbola and the asymptotes is
infinitely great.
2. 1-
sinh^a , „ /., tanh^a
sinh 2fl3 ^ , J sinh ^x , ,
4-. — r^^ — r^ = tanh x, and — r-s r = coth x.
cosh2a; + l cosh2£C — 1
^ . 1 "-, 2 tanh a;
5. sinh 2x =
6. cosh 2x =
1-tanhV
1 + tanh^o;
1 — tanhV
1 + tanh X
7. cosh2ir + sinh2cc-^ ^ , .
1 — tanhir
8. 2 cos^iT cosh-y + 2 sin^a; sinh^i/ = cos 2x + cosh 2y.
9. 2 coth 2cc - coth x = tanh a).
10. tanh «;+ tanh y= ^'f t^' + y) .
•^ cosh a? cosh 2/
11. cosh(a;+2/)cosh(i:c — 2/) = cosh2cc + sinh22/
= cosh^y + sinh^ic.
12. sinh(aj+ 2/)sinh(aj — y) = sinh^cc — sinh^^/
= cosh^ix; — Gosh^y.
18. cosh{x+ y + z)
= cosh X cosh y cosh 0 + S cosh a; sinh y sinh 0,
sinh(aj + 2/ + 2;)
= sinh X sinh 2/ sinh 2; + 2 sinh a? cosh 3/ cosh z.
, ^ , , , . , tanh £C tanh y tanh 0 + S tanh x
14 tanh(a;+2/ + «) = i + vfanh ytanh^ '
272 HYPERBOLIC FUNCTIONS.
16. cosh 3tt = 4 cosh% — 3 cosh -M,
sinh 3t6 = 4 sinh^u + 3 sinh u.
tn J. 1 « tanh% + 3tanlia;
1 + 3 tanh^a;
17. If 0 = gd'M<, then
cos 0 = sech u, sin 0 = tanh u,
cot 0 = cosech u, cosec Q = coth -a.
18. cosh - la = sinh "Vo^^.
19. sinh-ia= ±cosh-i/\/a2+l, the upper or lower sign
being taken according as a is positive or negative.
20. tanh-ia+tanh-i6 = tanh-i:^^.
1 + ah
21. If PT, the tangent to a rectangular hyperbola at P,
meet the radius OQ in T, then
PT , , 2P0Q
^ = tanh-^^.
22. If from i\^, the foot of the ordinate of a point Pon a
rectangular hyperbola, NQ be drawn to touch the
auxiliary circle at Q, then the bisector of the angle
AOQ will bisect the hyperbolic sector AOP.
23. If
24. If
-^2:- = 1, and y^ + -r^- = 1,
sin^aj cosmic ' cosh^y sinh^^/
then a^ = sin^a; cosh^y and y8^ = cos^a; sinh^^/.
u -y 1
sin 2a; si nh 2y cos 2a; + cosh 2y
then u^+v^+2ucot2x = '[,
and u^ + 1;2 — 2^ coth 2y= —1.
25. (cosh u+sinh u)(cosh v+sinh v) = cosh(tt+i;)+sinh(u+v).
26. (cosh Uj+sinh u^Xcosh Ug+sinh u^). . .(cosh Un+sinh. Un)
= cosh(u^+U2+. . .+Un)-\-smh(u^+U2+. . .+Un).
27. If n be any positive integer,
(cosh u + sinh u)^ = cosh nu + sinh nv,.
HYPERBOLIC FUNCTIONS. 273
28. If t'= 2 cosh u, Vn = 2 cosh nu, prove that
Apply this formula to shew that
v^ = v^ — 2v,
29. sinh - ^a - sioh - 1& = sinh " \as/b^ + l - bs/cf^).
•^0 ^^^-if^^^^O-^^'^^^^^\ . X i/tan0-tanh0
• ^^ Vtan20-tanh2^7'^^^'' VtanO + tanh0
= tan - i(cot 0 coth 0).
31. Prove geometrically that
. , . , - tanhu+tanhv
tauh(u+iO=rT-i — tt'— I — T— •
1 + tanh'i^tanhi'
CHAPTER XII.
INEQUALITIES AND LIMITS.
§ 1. Inequalities.
182. In Chapter VI. (arts. 74, 76) the following in-
equalities have been proved, 0 being the number of radians
in an acute angle : sm$<6< tan 0,
0-2
cos 0 > 1 — -^ ,
4
183. If 6 he the number of radians in an acute angle,
sme>e-^,co8e<l-^ + -^, and tsin6>e + ^-
(1) Let the arc AB subtend
an angle of 6 radians at 0,
the centre of a circle of any
radius (a) ; let G be the middle
point of the arc AB, D that of
the arc AG, E that of the arc
AD, and so on. Then, OG cuts
AB at right angles in N.
Now, the area of the segment
AGB — area of sector A OB - area of triangle A OB
= ^-(0-sin0).
274
J
INEQUALITIES. 21 h
Also, since the area of a circle is the limit of the area
of a regular polygon inscribed in the circle, when the
number of its sides is infinitely great (art. 166), it follows
that the area of the segment ACB is
= ^AGB+''1^ADG+^''^AED+ adinf.
Now, AAa5 = J^5.(7i\^ = asin|.a(l-cos|)
= 2a%n-sin222<2a^-2*y ^'ir^
.,., ...i,c<.5(|)-.,|,«*
^^^AED < I. • ^', and so on.
2* lo
a20V. .1.1. \ . .a^ 4
16 '3'
-(0-sin0)<-^(^l+-2 + ^^+-->^-e-<
e-sin0<-^.
0
sin e> 0-77-
o
(2) cos0 = l-2sin2|
<^ JO e^v
V4 48^23047'
<l-^+^.
2^24
(3) tan0=^,
cosO
2''"24
276
INEQUALITIES.
%,e.
2 "^24
, by division.
Now, since ^ ^ ^,
^ 07 05
8" ~ 72' °^ 72^^ ~ ^^^' ^^ P^s^^^^® J
and l-|V~,or^V{(6-a7-12}, is positive;
tana>0+|-.
o
184. In the accompanying figure are drawn the curves
of sin 6, 0'-\6^ and 0 — ^6^, between the values 6 = 0 and
TT
0 = — ; and the figure thus represents graphically the
id
inequalities we have been considering. It also shews
how closely the expression Q — \Q^ may be regarded as an
INEQUALITIES.
277
approximation to the value of sin 0, so long as Q is an
acute an^le.
The dotted line is the curve of 0 — f
e^an
expression which may be shewn to be greater than sin 0,
while Q lies between 0 and -^.
185. To shew that sinh x>x> tanh x,
and
cosh a? > 1 + '
If A denote the area of the sector, we have
sinh x:x: tanh x
_PM. 2A .PM
OA'OA^'OM'
_PM. 2 A .AL
OA'OA^'OA'
= iPM.OA:A:iAL.OA,
= triangle J. OP : sector AOP
: triangle A OL.
sinh x>x> tanh x.
Also,
cosh a? = 1 + 2 sinh^-,
>l + 2
I.e.
>^+r
186. Example 1.
angle,
-If 0 be the number of radians in an acute
sin d >
> ''' vers I
Let AOP be the given angle containing 6 radians, AOB a right
angle, APB an arc of a circle with centre 0.
Draw PM perpendicular to OA, and, with if as centre, and MP and
MA as radii, describe the quadrantal arcs PN and AL.
278
INEQUALITIES.
Then (art. 68), arc PiV> arc ^P> arc AL.
Z-PM>BxcAP>l- AM,
sin6>>^>^versa
O AT
"When ^ = ^, these three quantities are ultimately equal, for the
arcs NP. AP and AL coincide in the limit with the arc AB.
Example 2. -
greater than 1.
If ABC be a triangle, 8 sin— sin— sin— is not
2 ^ ^
(1) We have sin — sin— = \{
B-C B+C'
cos — - — — COS —
= i(co.^-sin|).
and this is greatest, for a fixed value of A, when B=G.
Hence, if any two of the three angles A^ J?, C be unequal, we
ABC
can increase the product sin — sin — sin by making the two angles
2 2 2
equal without changing the third angle.
It follows that the product sin - sin sin— is greatest when the
triangle is equiangular, in which case it is equal to sin^ 30°, or \.
ABC
8 sin — sin - sin -^ ^ 1-
2 2 2
(2) Oeometrical proof, — If S be the circumcentre, and / the in-
centre, of the given triangle,
.S72 = /22_2i?r (art. 164).
2r:j>i2,
INEQUALITIES. 279
Z A A
8sm:^smf siii^4-l.
Z " Z
Example 3. — If ABC be an acute-angled triangle,
sin ^ + sin ^4- sin (7>cos A +cos jB+cos C.
sin ^ + sin ^ -1- sin C— cos ^ — cos 5 - cos C
= sin^-sin(|-^^ + sin B - sin^|-Z?)+ sin C - siuf^ - C\
= 2cos^|sin(^-j)+sin(^-|)+sin(C--)}.
Now, sin a + sin /? + sin y - sin(a + /5 + y )
= 4sintosinr+^sin^,
z z z
whatever be the values of a, /?, y ;
.•.sin(^-j) + ri„(iJ-|) + sm((7-^)
.77,.- /5 + C Tr\ . IC+A jr\ . /A+B t\
and each factor of the last expression is positive, since all the angles
A, B, C are acute.
sin J. + sin B + sin C - cos A — cos B — cosC is positive,
sin ^ + sin ^ + sin (7 > cos ^ + cos ^ + cos C.
Examples XXI.
1. If 0 be the number of radians in an acute angle,
tan e> 30-2 sin a
2. Draw the curves of cos 0, l-JO^ and l-^O'^ + ^O^
f between the values 0 = 0 and 0 = -^-
3. If m and n be integers,
. TT
n sm —
n
7r> : — ^.
TT TT TT
cos ,,— cos -,— ... COS ^~—
271 4?i 2"*'^i
280 INEQUALITIES.
4 If ABC be a triangle, 8 cos ^ cos jB cos O 1.
5. The sum of the cosines of the three angles of a
triangle cannot be less than 1 or greater than IJ.
6. If ABC be a triangle,
sin2^+sin25+sin2C
< 2 sin jS sin (7-f 2 sin C sin ^ + 2 sin A sin B.
A B
7. If ABC be a triangle, the least value of tan^-j^ + tan^^
Q
+ tan2- is unity.
8. The sum of the acute angles which satisfy the
equation cos2a + cos2y8 + cos2y = l is less than tt.
9. The sum of the positive angles which satisfy the
equation cos2a + cos2/3 + cos'^y = 2 is greater than ^.
10. If ABC be a triangle,
tan^^ tan2-4 tan^^tan''^— H-tan^— tan^^
is always less than 1 ; and, if one angle approach
indefinitely near to two right angles, the least
value of the expression is J.
] 1. The value of the expression
V(a'cos-^^ + hh\n^<f>) + ^(a^in^^ + ^^cos^)
is intermediate to a + 6 and ^(2a^ + 26^) ; also, the
value of the expression
11 4
is intermediate to — [-7- and
12. The geometric mean of the cosines of n acute angles is
never greater than the cosine of the arithmetic
mean of the angles.
\
INEQUALITIES. 281
13. If 6 be the number of radians in an angle less than
two right angles,
. 2 sm -
sm 6 2
<
2+COS0 „ , e
S + cos^
Hence, prove that
3 sin e
>
2 + COS0
14. If cos{a + 0)-\-m cos 0 = n, shew that n^ cannot be
greater than l + 2mcosa + m^.
15. If (J), cf), yfr, xj/ be all acute angles, and 0 be greater
than i//-, and sin 0 = /x sin 0', sin i/r = ^ sin yj/, where
/x > 1, then ^ — i/r > <^' — >//■'.
16. sin(cos Q) < cos(sin 0), for all values of 0.
17. If -A^C be a triangle,
. „B . „G , . Ju . 9^ , . ^A . J^
sm'^g sin^— + sm^— sm^ „ + sin^- sm'^^
Li A Z Z Li li
is not less than
iV (sinM + sin^i^ + sin-^O).
§ 2. Limits.
187. We have already defined a limit, and illustrated
the definition by finding the limits of cos 0, sin 0, etc., for
the values 0 = 0 and 0= „ (arts. 18. 19). It has also been
shewn that, if Q be the measure of an angle in radians,
the limits, when Q is zero, of sin 0/0 and tan 0/0 are both
282 LIMITS.
unity (art. 75). We express these results briefly by the
notation :
Xt(cos0) = l, Lt{^\ne) = l, Lt.(sme/e) = l,etc.
e=Q a_lL e=o
The following important algebraical propositions will
be frequently used :
and Lt(l-^Y = \
where e is the sum of the infinite series
1+1+1+1+
188. // A and B he functions of a quantity 0, luhose
limits for a given value of 6 are known, to find the limits,
for the same value of 0, of A±B, AxB, A^B and A^.
Let a and b be the limits of A and B for the given
value of Oy and, when 6 has any other value, let A=a + x,
and 5 = 6 + 2/, where x and y ultimately vanish as Q
attains the given value.
(1) Lt{A±B)^Lt{a^x±{h-{-y)) = a±h '
= Lt.(A)±Lt.(B).
(2) Lt.(A X B) = Lt.{{a-\-x) x (6 + 2/)}
= Lt.{ah+hx-\-ay+xy)
= ab = Lt(A)xLt{B).
Hence,
Lt(A xBxCx...) = Lt.{A) x Zt.(B) x Lt{C) x . . . .
(3) ^=^-^'
LtiA)^Lt.{^xLt{B), by (2),
LIMITS. 283
(4) Suppose a > 0, and therefore, for values near to a,
A>0; then A^ = (e^''s Ay = e^iog a ^
Now Lt.(B log A) = Lt.B. Lt.Qog A), by (2)
= h log a,
since log ^ is a continuous function of A,
B log A = h log a-\-z, where z ultimately vanishes.
Hence J.^ = e&ioga+z^g&iogaxe^.
but, ultimately, e^ = e^ = 1,
189. To find the values of
^sinhcc\ ,^^/tanhcc'
Lt\ ) and Lt\-
:0\ X / a;=0\ ^
It has been shewn (see art. 185), that
sinh x'.x'. tanh x = triangle A OP : sector A OP : triangle A OL.
Now, triangle A OP : triangle AOL=OP:OL = OM : OA.
Let P move along the curve towards A. Then, ulti-
mately, M coincides with A and OM is equal to OA.
Hence, the triangles AOP, AOL, and the sector AOP
(which is intermediate to them in area) are ultimately
equal,
190. To find the values of
284
LIMITS.
Now, Z«. {(l-sin2|)«-^T} = l,
by the theorem referred to in art. 187, and
, / sm -
71=00 ^^ ^^ n = oo
2n\ a
\ n
n
= 0,
therefore, by art. 188,
i,(eos°)"=(lY = l.
n=«,\ w ve/
Again, since
sin-<- <tan-,
n
sm —
a
sm
71 n
a
—<1, and cos— < —
^ a
a
71
or
sm
C0S°<_J'<1,
71
This is true for all values of n,
a\n
= 1.
Cor. — In the same way, it may be shewn that
/ a\'^^ -°^ ( aS^
• X^. I COS - ) —e 2 Xf. I cos - ) = 0, etc.,
n=oo\ ^/ n=^\ "Tl/
for Lt (""- sin2 '') = "', and Lt ("^ sin^-') = oo , etc.
n=oo\2 71/ 2 n:=xV2 7?/
J
1 ^2
!a\8inh2"
n
) n
1/
sinh2.
LIMITS, 285
191. To find the values of
I sinh -\
Lt. ( cosh - ) and Lt. ' •
n=ooV W n=co\ a j
We have
n
XI (cosh-T = Z^. (l+smh2-)'
71 = 00 ^ ^Z 71 = 00 ^ '^/
= Lt. ifl+sinh^^
n = oo [\ 71
1
Now, XI (l+sinh2-)^^^^'" = 6 (art. 187),
71 = 00 ^ 71''
and Xt(^sinh^«) = xJW!^
therefore, by art. 3 88,
Lt fcosh-y = eO = i.
Affain, since sinh — > — > tanh -,
n n n
sinh —
cosh- > > 1,
n a
= 0.
This is true for all values of n,
286 LIMITS.
sinh
Lt\ Vl
n=cp\ ct
n
Cor, — In the same way, it may be shewn that
Lt. ( cosh - ) = e 2 and Lt ( cosh — ) =00, etc.,
for Lt (-^ sinh^- ) =^, and Lt ( — sinh^- ) = go , etc.
n=ooV2 nJ 2' n=«\2 nJ
Examples XXII;
1. Prove that the ultimate vahae of the ratio sin 0 : 6,
when 0 is zero, may be made equal to any quantity
whatever by adopting a suitable unit-angle. Find
the unit-angle in order that the ultimate value
of the ratio may be tt.
2. Find the value 0^ Lt(sinpO/ta.nqO).
6 = 0
3. If sin 0 = 2 sin (0 — ^), find the limit of smO/sm<p,
when 6 and 0 apj^roach to tt.
4. Find the value of Zl{(7r-20)/sin 20).
V=2
5. Find the value of X?^.|(tane-l)('l-tan|)|.
.. -D ,x. ^ T^ { ^ ^ ^^ sine
6. Prove that Lt (cos ^ cos -^^... cos -^j = —^.
/ Q rfh'^
7. Prove that Lt cos - + sin — =6^0.
-Find the value of X^.(cos 0)0"".
/ 6=0
LIMITS. 287
9. If u = (cossa^y^°«^°'-<
-ii
shew' that X^.(u) is either 1, e srs^ or 0, according
as k is less than, equal to, or greater than, 2.
10. What is the ultimate value of the logarithm of the
sine of an angle with the tangent of the angle as
base, when the angle is diminished indefinitely ?
11. Regular polygons, each of n sides, are inscribed in
and 'described about each of a series of equal
circles, the number of which is n times n. Prove
that the difference between the sum of the areas
of the inscribed and the sum of the areas of the
circumscribed polygons, when n becomes infinite,
is equal to ttM, where A is the area of one of the
circles.
12. Two circular arcs have a common chord and lie on the
same side of it ; prove that, if the radii approach
indefinitely near to equality, the quotient of the
area contained between the two arcs divided by
the difference of their lengths is ultimately equal
to the radius of either of them.
13. If a and h be positive quantities, and if a-^ = ^{a-{-h),
\ = {a-p)^, a^= lia^ + h^, h^=={aj)-^^, and so on;
prove that a^ = b^ = -^ ;
cos
■ r
b
and shew that the value of tt may be calculated
by means of this theorem.
CHAPTER XIII.
SERIES.
§1. The Addition Formulae Extended.
192. Notation. — The symbol l>GrSn-r is used as an
abbreviation for " the sum of all possible terms that can
be obtained from n angles, each term being the product
of the cosines of r angles and the sines of the remainin":
{n — r) angles." Thus, with two angles A^, A^, we have
(72 = cos A-^coa A^, 110^8^ = cos A^ sin ^2+ cos A^ sin A^,
/S'2 = sin J.isin^2 5 ^^^» ^^^^ three angles J.^, A^, A^, we
have
Cg = cos A^coa A2COS A^,
SCg/Sj = cos J.2C0S J-gsin J.^ + cos J.3COS ^.^sin A ^
+ COS J-^cos J-gsin A^,
1,0^82 = cos ^jsin A2sin ^3+ cos ^2sin A^^'m A^
+ cos ^gsin A-^sm A 2,
Sq = sin J-jSin J.2sin A^.
The number of terms in llCrSn-r is the number of
combinations of n things taken r at a time, or
— ^ , ,^ / — ^^ — -• This number is denoted by
1 .2.3. ...r -^
the symbol (n)r.
288
ADDITION FORMULAE EXTENDED 289
193. Formulae for the Cosine and Sine of the sum of
11 Angles. — To ijvove that
and
We know that Q,o^(A^ + A^=^C^ — S2y
^m{A^^Al) = ^GX.
and that co&{A^-{- A^-\-A^ = G^ — Y.G^Sc^,
sin(A, + ^2 + ^) = 2aA->Sf3.
Assume that
COS(^l + ^2+...+^„_l) = a„_i-S(7«_36f2 + 2Cn_6/Sf4-...,
sin(^, + ^2+"- + ^n-l) = 2C„_2^1-SCn-4^3
+ E(7n_6'Si6— ....
Then we have
C0S(^l + ^2+--+^n-l + ^n)
= C0S(^1 + J.2+ . . . +^7i-l)C0sJ.„
-sin(^i + ^2+ ••• +^«-i)sin^n
= ((7„_i-SC,_3<Sf2+2a,_5^4-...)cos^„
- (2C,_2^Sfi - Ea,_4*Sf3+ . . .)sin ^„
= Gn-\CO^ An — (^Gn-AQ0^An-{-^Gn-2Si^\nA^
+ (2(7n-5>S^4COS J.„ + E(7«_4/Sf3Sin J-n) — . ..
= Gn — 2C^_ 2^02 + 26.^_4>S4 — ...
Also, ^m{A^+A^+...-{-An-i+An)
= ^m{A^+A^-\-...+An-i)(to^An
+ cos(^i + J.2+...+J.„_i)sin^n
= (EC„_2^1-2C,_4^3 + 2C„_6^5-...)COS^„
+ {Gn-i-J.Gn-A+ ...)^inAn
= (2C„_2^iCOs ^n+ C„_isin ^„)
— (E(7«_4>Sf3Cos J.,i + 2(7„_3>Sf2sin J.„)
+ (26\_6>Sf5Cos ^„ 4- 20„ _ s^Sf^sin J.„) — . . .
= 2C7,i_lOl — ZjGn _303 + 2Cyi_ 5^5 —
Thus, if the formulfe be true for {n — 1) angles, they are
290 ADDITION FORMULAE EXTENDED.
also true for n angles. But we know that they are true
for two and three angles, hence they are true for any
number of angles.
Gov. 1, — Hence, it follows that
Dividing the numerator and denominator of this frac-
tion by C„, and writing ^T^ for "the sum of all possible
terms that can be obtained from n angles, each term being
the product of the tangents of r of the angles," we have
tanu,+^.+...+^„)=Yii:iysl!i:;:-
Got. 2. — In like manner, it may be shewn that
cosh (Uj -h 1^2 + • • • + '^«) = ^^^n + ^Ghn - 2^A'2 + 2C/l„ . Sh^^ + ... ,
sinh(iti-|-U2+ . .. -f-tf'n) = I'Ghn-iShi + 'EChn-sShs
+ 2GK-,8h+...,
tanhK+tt,+ ...+^.)= i + ^ThJ^Th,+ ... '
where 'EGhrShn-r stands for " the sum of all possible
terms that can be obtained from n sectors, each term
being the product of the hyperbolic-cosines of r of the
sectors and the hyperbolic-sines of the remaining (n — r)
sectors " ; and XThr for " the sum of all possible terms
that can be obtained from n sectors, each term being the
product of the hyperbolic-tangents of r of the sectors."
194. To prove that
2"C0S JjCOS J.2 ... cos J.n = 2cOS(± J.j±^2- ... ±^n-
We know that
2 cos ^icos A^ = cos(i4i + J-g) +cos(^i — J.2),
and therefore that
ADDITION FORMULAE EXTENDED. 291
22cOsJ.^COsJ.2 = COs(+^l + ^2) + ^<^s( + ^l~^^)
+ C0S( — ^l + ^2) + C0s(-J.i — J.2),
or 22n2(cos^ ) = 222Cos( ± A^ ± ^2).
where 1X2(008 J.) is an abbreviation for "the product of 2
factors of the type cos J.,"
and S22COs(± ^.^±^2) ^*^^ "the sum of 2^ terms, each
term being the cosine of one of the 2^ angles ( ±^^ ± A^!'
This formula may be shewn to be true for any number of
angles.
Assume that
2"-in,,_l(C0S^) = S2n-iC0S(±^l±^2±---±-^n-l).
Multiplying each side of the equation by 2 cos An, and
expressing each double product on the right side as the
sum of two cosines by aid of the formula 2 cos a cos /3
= cos(a + /3) + cos(a — /3), we get
2^nn(C0sJ.) = E2nC0s(±J.i±^2±---±^n-l±^«)-
Thus, if the formula be true for ('^ — 1) angles it is also
true for n angles. But we know that it is true for two
angles, hence it is true for any number of angles.
Got. — In like manner it may be shewn that
2*^nn(C0sh U) = S2nC0sh( ±Wj ± 1^2 ± . . . ±Un)-
195. Formulae for the Cosines and Sines of Multiple
Angles. — By art. 193, we have
cos(^i+^+...+^i)=c,,-sa^_2/Sf2+2C„-A-....
Let each of the angles A^, A^, A^, ..., An be equal to 0,
then each term in 'ZCn-rSr becomes cos^"^0 sin*'^, and the
number of such terms is -^ \ <^ ^ — ^ or in)^,
\. L. 6 ...r ^
2(7,,_ A-= Wrcos"-'*^ sin'-a
Hence, we get
cos 710 = cos^0 - ('njgcos^ - ^0 sin^O + ('Ji^cos^^ - ^Q sin^^ - ... (1 )
292 ADDITION FORMULAE EXTENDED.
So also, from the equation
we deduce the equation
sin7i0=('M)iCOS»*-iasin0-(7i)3COs«-30sin3e+ (2)
Cor. 1. — Dividing (2) by (1), we get
1 - {n\%d.n^e + {n)^t2in^e -...'
Cor. 2. — From art. 193, Cor. 2, it follows, in like man-
ner, that
cosh nu = cosh**i^ + (ti)2C0sh*^ - ^u sinh%
+ ('?i)4Cosh" - *u sinh*tt + . . . ,
sinh nu={n\cosh.'^-'^ii sinh u + (7i)3COsh**"^i6 sinh%+ ...,
ta.nhnu== Witanhu+W3tanh%+...
1 + ('^i)2tanh% + (rij^tanh^ + . . .
196. To prove that, when n is even,
2" - i cos'^e = cos nO + {n\QO^{n - 2)0 + (n)2Cos(n - 4)0
+ ... + iWn.
2
and that, when n is odd,
2« - icos»0 = cos nO + (7i)iCos('M - 2)6 + ('m)2Cos('7i - 4)0
+ ...+Wn-lCOS0.
2
By art. 194 we have
2"nn(cOS^) = S^^COS(±^l±^2±---±-4«)-
Let A^ = A2= ...=An = 6, then the left side becomes
2"cos"0. On the right side collect the 2" terms into
n
groups. Take as the (r + l)th group, where r not > ^, all
the terms in which the expression for the angle con-
tains r and only r plus signs, or r and only r minus signs.
Each term of this group will be equal to cos {n — 2r)0, and
the number of terms in the group will be 2{n)r, except in
ADDITION FORMULAE EXTENDED. 293
the case for which n is even and r = ^, when the number of
terms will he {n)^ only, the half-group which contains -
2 ^
and only ^ plus signs being also the half-group which con-
tains - and only ^ minus signs.
Thus,thefirstgroup = 2cos')i^,the second 2(t^)iC0s(7i — 2)0,
the third ^{n)^Q)^{n — '^)Q, and so on, the final group when
n is even being (7i)„, and when n is odd 2(?i)„_icos0.
Hence, dividing by 2, we have
2n-icos"0 = COS nO + (^)iCos(7i - 2)0 + ... + \{n)n.
2
or 2'^ - ^cos'^O = COS nQ + (71)^008(^1 - 2)0 + ... + (7i)h-iC0S 0,
2
according as n is even or odd.
Cor. — From art. 194, Cor., it follows, in like manner,
that 1^ - ^cosh"i6 = cosh nu -h {n)^Q^{n — 2)u -f- . . • ,
the last term on the right side being
\{n)n or (7i),^^iC0sh u,
2 2
according as n is even or odd.
197. To express 2^"^sin*^0 as a series of cosines or sines
of multiples of 0.
By art. 196 we have
2/1 - icos^O = cos nO + (n\cos{n — 2)0 -f {n)2C0s(n — 4)0 -f . . .
Change 0 into ., — 0, and reject from each angle the
multiple of a right angle.
There are four cases, according as n is of the form 4m,
294 ADDITION FORMULAE EXTENDED.
when n is called an even even-number, since such num-
bers occupy the even places in the series 2, 4, 6, 8 ...,
of the form 4m + 1, an odd odd-number,
of the form 4m +2, an odd even-number,
or of the form 4m +3, an even odd-number.
Observing that cos( r'^-A ) is equal to cos A,ainA,-coaA,
or —sin A, according as r is even even, odd odd, odd even,
or even odd respectively, we have the following results —
n even even, 2^-^sin"0
= cos nO — (n\cofi(n — 2)0 + (n)2Cos(n — 4<)0
2
n odd odd, 2^-hm^O
= sin nO — {n)-^sm(n — 2)0 + (n)2sm{n — 4)0
-\-,..-\r{n)n-ismO,
2
n odd even, 2^-hm^O
= — cos nO + (n)jCos(n — 2)0 — {n)2C0s(n — 4)0
+ ...-{-i{n)n,
2
n even odd, 2'^-hm^O
= — sin nO + (n)^sm(n — 2)0 — {n)2sin{n — 4)0
+ . . . + ('^),^isin 0,
2
the last term being positive in every case.
The method of this article depends on the periodicity
of the circular functiono, and is not, at the present stage,
applicable to hyperbolic functions.
198. When n is not very large the series for 2""^cos*^0
and 2""^sin"0 may be conveniently obtained by aid of
ADDITION FORMVLAE EXTENDED. 295
Pascal's Triangle,
1
2
1
8
3
1
4
6
4
1
5
10
10
5
1
6
15
20
15
6
1
7
21
35
35
21
7 1
8
28
oQ
70
56
28 8
etc.
Taking the coefficients from the Triangle, and observing
that a coefficient which occurs once only in a row is to
be halved, we get
cos 0 — cos 0,
2cos2e = cos20+l,
4cos30 = cos30 + 3cos0,
8 cos*0 = cos 40 + 4 cos 20 + 3,
16 cos50 = cos 50 + 5 cos 30 + 10 cos 0,
32 cos60 = cos 60 + 6 cos 40 + 15 cos 20+10,
64 cos70 = cos 70 + 7 cos 50 + 21 cos 30 + 35 cos 0,
128 cos80 = cos 80 + 8 cos 60 + 28 cos 40 + 56 cos 20 + 35,
etc.,
and
sin 0 = sin 0,
2sin20=-cos20 + l,
■ 4 sin30 == _ sin 30 + 3 sin 0,
8 sin40 = cos 40 - 4 cos 20 + 3,
16 sin50 = sin 50-5 sin 30+10 sin 0,
32sin60= _cos60 + 6cos40-15cos 20 + 10,
64 sin70 = -sin 70 + 7 sin 50 -21 sin 30 + 35 sin 0,
128 sin80 = cos 80- 8 cos 60 + 28 cos 40- 56 cos 20 + 35,
etc.
296 SERIES OF PO WEliS
199. Example. — Express siii''^ cos^^ as a series of sines of multi-
ples of 6.
By art. 198, we have
16 sin5^ = sin 5^-5 sin 3^+10 sin 6.
Multiplying by 2 cos ^, we get
32 sin^^ cos $= 2 sin 5^ cos ^ - 5 . 2 sin 3^ cos 6^+ 10 . 2 sin 6 cos 6
=sin Q9 + sin Ad-b sin 4^-5 sin 2^-f 10 sin '2.9
=sin 6^-4 sin 4^+5 sin 19.
Again, multiplying by 2 cos 9., and proceeding as before, we get
64 sin^^ cos2^=sin 7^-3 sin 5^-fsin 3^ + 5 sin ^,
and, again,
128 sin6^cos3^=sin 8^-2 sin 6^-2 sin 4(9 + 6 sin 29.
§ 2. Series of Powers of a Cosine or Sine.
200. From the series of art. 195, namely,
cos nO = cos^a - (7i)2Cos" - ^0 sin^O + (t^^cos" - *^sin*0- . . . ,
and
we may, by substituting 1 — cos^O for sin^^, obtain series
for cos nO and sin nO/sin 0 in powers of cos 0 only. The
general form of the coefficients may be determined in
this manner, but the method adopted in arts. 201 and 202
is somewhat simpler, and depends only on the elementary
identities
sin('7i + 1)0 -\- sin(n — 1 )0 = 2 sin nO cos 6
and sin('M -\-l)0 — sm(n — 1)0 = 2 cos nO sin 6.
The series may be arranged either in descending or
ascending powers of cos 0, and we may also obtain similar
series in descending or ascending powers of sin 0. It will
OF A COSINE OR SINE. 297
be seen hereafter that, except in the case of the series in
descending powers of cos 0, the form of the series will
vary according as n is even or odd. There will accord-
ingly hQ fourteen series of this type in all. In arts. 201
and 202, we investigate the two series in descending
powers of cos Q ; in arts. 203 and 204, we obtain, by
re-arrangement of the terms of the series in descending
powers of cos 6, four series in ascending powers of cos 0.
By changing 0 into ^ — 0 we may without difficulty
deduce four series in descending powers of sin 0, and
four in ascending powers of sin 0.
The group of series here considered is an important
and a natural one ; but the order in which the series are
taken, and the method of demonstration, are to a great
extent arbitrary. The series in descending powers of cos 0
are the simplest of the group, since their form is the
same for even as for odd values of n, and for this reason
they are here made the fundamental ones ; the expan-
sin nQ
sion for — ; — ^r- is taken before that for cos?i0 because the
sin t^
coefficients of the terms of the equivalent series are simpler
for the former.
201. To prove that
^^^ = (2cose)"-l-(7^-2X(2cos0)^-3 + O^-3)2(2cosef-5
-... + (-iy(n-r-l)r(2Gosey'-^'-'^+....
Let Un = sin nO/s'm 0, v=2 cos 0.
From the identity
Bin{n + 1)0 + sin(7i — 1 )0 = 2 sin nO cos 0,
it follows that Un^l = UnV — Un-l (1)
and from the definition of ^(^ and U2 that
298
SERIES OF POWERS
u.
1,
and u^ = v.
Hence, by the use of (1), we have in succession
etc.
The numerical va^
following table : —
ue of the coefficients is shewn in the
Uc
u.
u.
Un
10
etc.
If these coefficients are read obliquely from left to right
downwards, we obtain the coefficients of a binomial series,
and this law of the coefficients is general, since the
process of formation of the successive numbers by aid
of (1) is the same as that employed in forming the
OF A COSINE OR SINE. 299
binomial coefficients, the (r + l)th number in any oblique
line being the sum of the (r+l)th and rth numbers in
the preceding oblique line.
Now, the coefficient of the (r4-l)th term of Un is the
(> + l)th number of the {n — l)i\i row in the table, and
therefore the (r+l)th number of the {n — T — \)th. oblique
line, and consequently is equal to {n — r—\\.
Thus, we get
+ (-iy('M~r-l),i;"-2^--i+...,
or
sm iiif)
-^—^ = (2cosa)" - ^-{n-2\(2coHey - ^ + (71-8)2(2 cos 0}" -'-,..
sm (7 "
+ (-l)^(7i-r- 1)^2 cos 0f-2'-i+...
Cor. — In like manner, it may be proved that
'^Sr^= (^ '"^^ u)-l-(.^~2X(2 cosh ^)"-3
+ (^-8)2(2 cosh uy-^- ...
+ (- 1/(72 -r- 1)^2 cosh i(,)"-2»-i+...
202. To prove that
2 cos nO = (2 cos Oy^ - ^^2 cos 0)" - 2 + -^^^^2 cos 0)" - ^- . . .
From the identity
sin(7i + 1)6- sin(7i - 1 )0 = 2 cos nO sin 6,
it follows that
^ ^ sin (71 + 1)0 sin(7i — 1)0
2 cos Old = — \ ' ^ ^ — —^^,
sm 6 sm 0
and, by art. 201, we have,
!in|+l)^ = (2 cos Br-{n-lU2 cos 0)"-
+ (n-2\(2 cos ey-*- ... + {-iy{n-r),{2 cos e)»-='-|- ■ ■■
300 SERIES OF POWERS
and
+ (-l7(?i-r-lX_i(2cos0)"-2^+...
Now, (n — 1)1 + 1 = '^,
(7i-r-l)...(ii-2?- + l)
"^ 1.2.3...(r-l)
_ 7i(?i — r — l)(7i — 7' — 2) . . .(ti — 2r + 1)
~ 1.2.3...r *
Hence,
2cosrie = (2cos0y^-7i(2cose)^-2+^^:^^?l^\2c^^
J.2.3...r
Cor. — In like manner, from the identity
Sinn 26 smhu
and from art. 201, Cor., it follows that
2 cosh nu = {2 cosh u)" — ')i(2 cosh u)" - ^
+!<!L^)(2coshu)-*-...
+ (-l)."0^-^-lX^->-2)-(^-2>-+l)(2co3hu)»-^-+-
203. To prove that, when n is even,
= n cos 0 ^—r^ — ^cos^^ H — ^^ r^ cos^0 — . . .
and that, when n is odd.
OF A COSINE OR SINE. 301
^-^^ -ST
By art. 201 we have
^^ = (2cos^)^-i-(7i-2X(2cos0f-3.
that for
equal to
am (7
+ (- lX(?i-r- 1)^2 COS 0)"-2»--H
When n is even, the last term of this series is
which 91 — 2r — 1 = 1, orr=^ — 1, and is therefore
1.2.3...g-l)
— -1
= ( — 1)2 71 cos 0;
the last term but one
, ,t-(^')i(i--)-
,4
— — (2cos0)^
1.2.3...g-2)
= -(_l)l-!^gzf)eos3e,
the last but two
= (-!> r (2cos0)=
1.2. 3. ..(1-3)
and so on.
302 SERIES OF POWERS
Hence, when n is even,
sin nO
sin0
= (-1)2 j ncosO — ^ — ^-cos^6+— j^ -^cos^^-. . . k
and therefore
. .^.1+1 sin tie
^ ^^ sin 0
^ n(n^-2^) 3^ , 71(7^.2-22X712 -42) ,.
[3 [5
Next, let n be odd, then the last term of the series for
— ; — ^ in descending powers of 2 cos 6 is that for which
n — 1
71 — 2r— 1 = 0, or ?'= , and is therefore equal to
n—1 71—3
1.2.3...^
(-1J
the last term but one
n + l 71-1
n-3 2 2 ■' «-l772__-|2
= (-1)"^-^ ^ ^(2cos0)2= -(-1)— !L-J:-cos2(9,
1.2.3...^ ^-^
the last but two
71+3 71 + 1 „
n-5 2 ' Q •••^
= ( - 1)^ ^;r3T(2 cos e)*
1.2.3...^V^
/ ,,^-^H^'-82)(7l2-l2)
= (-1) ' 1.:^ 3.4 ^Q"^>
hence, when 7i is odd.
and so on
OF A COSINE OR SINE. 303
sin nO
sinO
= (_]) 2 |1_____COS20 + ^ j'^- - ^COS^0-...|,
and therefore
^~^^ TirTe
= 1 r^-- cos^O + —ri — - cos*0 — . . . .
[Z [4
Cor. — In like manner it follows from art. 201, Cor., that
when n is even,
/ |/|+i sinhjTiu
^ sinhi6
, '71(7^2-22) , - , 7^(n2-22)(^2_42)
= n cosh It ^ — ^coshm -\ — ^ rj~ ^coshm - . . .
[3 [5
and that, when n is odd,
, _ .VlzI sinh nu
(_1) 2 .
smh u
^l-^^C0sh2.+ (-^-^y-^^0sh%-....-
|2 [4
204. To 'prove that, when n is even,
( - 1)2 cos -TlO = 1 - ,— COS20 H ^ , ^ COS^O - . . . ,
and that, when n is odd,
n-l
( — 1) 2 COS 110
. n{n^-l^) „. , n(n^-V)(n^-S^) ,.
= ncosO ^^-^^r ^ cos^O + ^ -^ '^cos^O - . . ..
[S_ |5
By art. 202, we have
2cos7i0 = (2cos0)^-~(2cos0)"-2 + ^^^^^:::^^
304 SERIES OF PO WERS
If n be even, the last term of this series is that for
which r = ^, and is therefore equal to
^e-i)e-2)...
(-1)' ;; (2cos0)'>=(-i)2.2,
1.2.8...|
the last term but one
« ,™-2-l2-V-^ « n^
1.2.3...g-l) L2
the last but two
, ..■,.'(i^')i(;-o-',. ,,
= ( - 1)22 . — ^-^ -^ COS^0,
and so on ; hence, when n is even,
2cos7i0 = (-l)^2.|l-^cos2e+^?^^^^^cos*0-...|,
and therefore
( - l)^cos ne=l-^ cos^e + ^'^'^'^~ ^'^ cos^^ - . . . .
Next, let n be odd, then the last term of the series for
2 cos nO in descending powers of 2 cos 0 is that for which
-, and is therefore equal to
2
n — 1 ti — 3
n.
9
(-l)V ? ^ (2cos0) = (-l)''2 .7i.2cose,
1.2.3...^-
OF A COSINE OR SINE.
the last term but one
71 + 1 n — 1 ,
1.2.3...^
= -(-1)2 -A_^ ^.2 008^0,
the last but two
71 + 3 n + 1
= (-1)'^ t-T-(2cose)^
1.2.3... -~-
, ^ «-Zi 7l(ri2 -12) (^2 _ 32)
= ( - 1) 2 — ^^ ^ i 2 cos^O,
and so on ; hence, when n is odd,
2 cos 710
= (-1) 2 2J71COS0 — ^-T^ — ^cos^O+-^ r^^ ^cos^0-... K
and therefore
n-l
( — 1) 2 cos 710
. 7l(n2-12) ^^2_12)(^2_32)
= -^1008 0 ^— rs — -^008^0+-^^ -iP ^cos^O— ...
Cor. — In like manner from art. 202, Cor., it follow^s that
when 71 is even,
( — 1)2 cosh Tiu = 1 — — cosh^it H ^- -^cosh% — . . . ,
\2 [4
and that, when oi is odd,
M-l
( — 1) 2 cosh Tilt,
;= 71 cosh u ^— , rr ^cosh^^t + -^ r-^^ ^cosh ^u-,.,
|3 \o
u
306 SUMMATION OF SERIES.
By changing 0 into f,— ^ we may derive from the
series of arts. 201-204 four series in descending powers
of 2 sin 0, and four series in ascending powers of sin 0,
but we cannot, at the present stage, apply this method to
the series of powers of the hyperbolic cosines.
§ 3. Summation of Series.
205. To find ike sum of the cosines of a series of angles
in arithmetical 'progression.
Let a be the first angle of the series, /3 the common
difference, then the (r + l)th term of the series of cosines
is cos(a + r^). We have
2cos(a + r/3)sin| = sin{a + (r+i)/3}-sin{a + (r-i)/8}.
. Putting r = 0, 1, 2, . . . {n — 1) we get
2 cos a sin 'I = sinf a + ^ ) — sinf a — ^ j,
2 cos(a + /3)sin ^ = sinf a + -~ j — sinf a + ^\
2cos(a + ^i^.)5)sin| = sin{a + (n-J)^}-sin{a + 0i-f)/3}.
Hence, by addition,
2sin^{cosa + cos(a + ^)+cos(a + 2^) + . . . +cos(a -\-n-l.^)}
= sin{a + (-^ — J)/5} - sinf a - ^ )
= 2cos{a + (7i-l)f}sin^^,
SUMMATION OF SERIES. 307
and therefore
cosa + cos(a + /3) + cos(a + 2/3) + ...+cos(a + 7i-l./3)
sin I
This result gives the following Rule : —
'' To find the sura of the cosines of a series of n angles
in an arithmetical progression whose common difference
is p, multiply the cosine of the average value of the
angles by the ratio sin -^ /sin ^."
The average value of the angle is readilj^ obtained by
taking half the sum of the first and last angles.
The result may be written in the form
S cos(a + 7^/3) = cos 1 a-\-{n — iy^\ .sin -7j-/sin^.
Cor. — In like manner it may be shewn that
2 cosh(a + 7'^) = cosh j a-{-{n — l)'^\ . sinh -^/sinb ^.
206. To find the sum of the sines of a series of angles
in arithnetical 'progression.
Let a be the first angle of the series, /3 the common
difference, then the (r + l)th term of the series of sines is
sin(a + rP).
We have
2sin(a + r/3)sin| = cos{a + (r-J)/3}-cos{a + (r + J)/3}.
Putting ?' = 0, 1, 2 ...(71 — 1), we get
2 sin a sin ^ = cosf a ~ 9 ) ~ cosf a + ^),
308 SUMMATION OF SERIES.
2sio(a+/3)sin| = cos(a + f)-cos(a + ^),
2sin{a + (7i-l)^}sin| = cos{a+0i-f)/3}-cos{a+(7i-i)/3}.
Hence, by addition,
2sin|{sina+sin(a+/3)+sin(a+2;8)+...+sin(a+7i-l/3)}
= cos(^a - ^ j - cos {a + ('^?' - J)/?}
= 2 sin j a + {n — 1)^ [sin -^,
and therefore
sina + sin(a + ;8) + sin(a + 2^) + ... + sin{a + ('>r-l)^}
3in j a + (71 - Vf^ Uin -2^
sm
This result gives the following Rule : —
" To find the sum of the sines of a series of n angles in
arithmetical progression whose common difference is /3,
multiply the sine of the average value of the angles by
the ratio sin -|^ /sin ^."
The result may be written in the form
2 sin(a + r/5) = sm j a + ('^ — l)^- f • sm -^/ sm ^.
Cor. — In like manner it may be shewn that
'"lr'sinh(a + r/3) = sinh|a4-('^-l)5| • sinh ^^-/sinh |.
SUMMATION OF SERIES. 309
207. Many series may be summed by aid of the method
or results of the two preceding articles.
The following list of difference-forms will be found
useful : —
cot r^ — cot X = cosec x.
tan X — tan ^ = tan= sec x.
tan 36 - tan 9 = 2 sin 0 sec Sa
tan-i(l + r.?Tl)-tan-Xl+^^i.r) = tan-L-^j^^2-
cot 33 — 2 cot 1x — tan x.
tan 2cc — 2 tan x — tan^cc tan 1x.
2 coth 2x — coth x = tanh x.
sin a
cotra — cot(7^+l)ct
sin7'asin(r+l)a*
sin^
cos(a + r — l/3)cos(a + ^'/3)
tan(a + t/3) - tan(a + r - 1 /5) =
2sin|
cosec 2x — l cosec x = —. — ^i —
cosec X — cosec Sx = 2 cos 2x cosec Sx.
cosec^a:; — cosec^^a^ = 8 cos 2x cos^o) cosec^3fl3.
sin^O — 2 sin^j^ = 2 cos 0 sin^^-
cof^2aj — I cot^ic = I tan'^a; — h.
2n
tSiii~hi(n-\-l) — tan"^(?i — lW = tan"^, . , ^ — tt~^*
cosec^o; — 2 cosec-2ic = 2 cos 2x cosec22a;.
tan 8^ — 3 tan x = S sin^x sec Saj.
sec rO sec(?' + 1 )0 — sec(7' — 1 ) 0 sec rO
= 2 sin 0 sec(r- 1)0 tan rO sec(r ^- 1)0.
310 SUMMATION OF SERIES.
The list may easily be extended by the observation or
invention of the reader. The discovery of the difference
form, by means of which a given series may be summed,
furnishes indirectly a valuable exercise in the manage-
ment of trigonometrical formulae.
208. Example 1. — Sum the series
tan x+i tan l+^L^tan 1 + ... +-L, tan g^.
"We have tan x=cotx-'2, cot 2;r,
1 , X \ ,x ,
- tan - = - cot - - cot .j;,
2 2 2 2
lx.37l,.rl,.r
22^^^22=P'"'p-2'"*2'
1 -^ ^ 1 cot-A,,
271-2'
gSTitan 2;^^ = ^^^cot-^j-^^2^
Sn=-^ cot -^j - 2 cot 2x.
Example 2. — Sum the series
tan^a tan 2a + i tan^Sa tan 4a + ... + J^, tan22"-^a tan 2"a.
•6 2**-!
We have tan^a tan 2a = tan 2a - 2 tan a,
\ tan^2a tan 4a = ^ tan 4a - tan 2a,
— tan24a tan 8a = — tan 8a - ^ tan 4a.
-1-. tau^2'»-^a tan 2"a = -^ , tan 2"a - JL tan 2"-^a.
2n-i 2"-^ 2"~2
Sn = o^zY tan 2"a - 2 tan a.
Example 3. — Find the sum of n terms of the series
tan-.2+tan-.^^ + tan-'^g + tan-'j^^ + ...
SUMMATION OF SERIES.
311
We have tan~^-
2r
tan
.^ r(r+l) -(/'-!>
= tan- V(r + 1 ) - tan-\r - l)r.
Putting r = l, 2, 3 ... w we get
tan-^2 = tan-U.2-0,
tan-i ^— =tan-i2 . 3 - tan-U . 2,
1+3.4
tan"
27i
:tan-^7i(7i + 1) - tan-^^t - l)n,
>S'„=tan-^?i(?i + l).
Example 4. — Sum to n terras the series
cos*^ + cos*2(9+ 008*36'+ etc.
We have 8 cos*.r = cos 4.^; + 4 cos 2^ + 3,
. • . s'^s' cosV^ = Ycos 4r (9 + ^1. cos 2r (9 + 3?^
r=l r=\ r=l
= cos2(?i+l)^sin2?i^/sin2^ + 4cos(?i+l)^sin7i^/sin^+3n,
2 cosV^= 1 cos 2(w + l)^sin 2n(9/sin 2(9
+ ^ cos(?i + 1 ) ^ sin ?i^/sin B + ^n.
Examples. — Let Aq, A^, A2...An-\ be n points symmetrically
ranged on the circumference of a
circle whose centre is 0 ; then shall
the sum of the projections of OA^^
OAi...OAn-i on any line OX be
equal to zero.
Let LXOAQ=a,
lAqOAi = lA^OA^ = etc. = /?,
then 7i/3 = 277.
Also, if r be the radius of the circle,
the sum of the projections of OA^,
OA^...OAn-i on OX
= r{cos a + cos(a + 13) + cos((x + 2/3) +
= rcos-(a + (w-l)^}sin -^/sin^
^ 2 -* 2 / 12
= 0, since sin — - = sin tt = 0.
^^«.z
+ cos(a + 7i-l . /S)}
312
GONVEROENCY AND
Example 6. — If a regular polygon of n sides be iuacribed in
a circle, and if I be the length of the chord joining any fixed point
on the circle to one of the angular
points of the polygon, then
where a is the radius of the circle,
and m any positive integer less than n.
Let 0 be the fixed point, J ,. a vertex
of the poly on, LOCAQ=a.
Then
27r>
0Ar=2a Bin Ua+r.^\
Now
( - 1)'"(2 sin </)f~ = 2 cos 2m</) - (2m)i2cos(2?n -2)cf>+ ...+(- l)"'{2m),„
r=o *- \ n J
-(2m)i2cos(m-l)(a + r?^)+... + (-l)'«(2mV}.
But iim<7i the sums of the cosines vanish by art. 205,
= na — -^'
Qrny
§ 4. Convergency and Continuity of Series.
209. In dealing with an infinite series of terms we
inquire, in the first instance, whether the series is con-
vergent, i.e., whether, however great the number of terras
CONTINUITY OF SERIES. 313
may be, their sum is a finite quantity tending to some
fixed limit; and, in the second, whether the series is
continuous, i.e., whether an indefinitely small change in
the variable involved in the series produces an in-
definitely small change in the limit to which the sum
approaches as the number of terms is continually in-
creased.
The following classification and terminology will be
adopted.
Infinite series are either convergent, oscillating, or
divergent.
A convergent series is either absolutely convergent, or
semi-convergent.
A convergent series has already been defined as one
the sum of whose terras is a finite quantity tending to
some fixed limit, however great the number of terms may
be. Tf the terms of such a series are all positive, or if
the series remain convergent when the terms are made
positive without change in their numerical value, the
series is said to be absolutely convergent.
Thus ^ + iT + ^ + To + • • • ^or all values of x,
Li Lt l£
and 1+iccos O + o^^cos 2O + i^^cos30+... when x<\
are absolutely convergent series.
If a series is convergent, but does not remain conver-
gent when all its terms are made positive, it is said to be
semi-convergent.
Thus, 1-I + 5-1 + -
and cos-^ + ^cos -^-hgCos -g--F...
are semi-convergent series.
314 CONVERGENCY AND
If the sum of a series never exceeds a certain finite
quantity, however great the number of terms may be, but
at the same time the sum does not tend to a fixed limit,
the series is said to oscillate.
Thus, 1-1 + 1-1 + 1-...,
and cos -^j + cos ., + cos -^ + . . .
O tJ o
are oscillating series.
If the sum of a series increases without limit with the
number of terms, the series is said to be divergent.
Thus, 1 + 1+I+1 + ...
is a divergent series.
For the fundamental Theorems on Convergency and
Divergency the reader is referred to Todhunter's Algehxi,
chap. XL., or C. Smith's Treatise on Algebra, chaps.
XXI., XXV.
With respect to continuity, it should be observed that
the continuity of an infinite series is not a necessary con-
sequence of the continuity of its several terms, for the
sum of an infinite number of indefinitely small changes in
the terms may be a finite quantity.
210. If c&o, ttj, a^...he a series of constantly decreasing
'positive quantities, and if Lt an = 0, and /3 be not equal
to zero or any multiple of ^ir, then will the series
a^cos a + aiCos(a + /8) + a^C0B{a + 2^) + . . ,
be convergent.
Let 8n = a^cos a + a^QO^{a + /5) + . . . + anCOs(a + n^),
CONTINUITY OF SERIES. 315
then 2^m^Sn = aA sin fa + ^j-sin («-§)[
-\-aA sin (^a + ^j-sin (a + |)|
+ (Xn{sin(a + 7i + i . p)-^m{a-\-n-\ . /3)}.
Therefore
2 sin ^AS^ji+ctoSin (a - ? ) — ajisin(a + '?v+i • jS)
^-{an-i-an)^m{a-n-l.^) (1)
Now ((Xq - a^ + («! - ag) + . . . + (a„_ i - an) ="%- o^n, and,
by hypothesis, Lt an = 0,
n=oo
.-. the series (a^ — a^) + (ttj - Wg) + • • • is convergent.
Also, since ao>ai>a2..., its terms are all positive.
Hence, the series
{aQ-aj)sm\^a + ^J + {a^-a2)8m{a + -^j + ...
whose terms are numerically less than the corresponding
terms of (a^ — a^) + ((Xj - cig) + . . . is also convergent.
Hence, observing that the limit of ansin{a + n + ^. ^)
is zero, and that sin ^ is not equal to zero, we conclude
from (1) that Sn tends to a finite limit as n is indefinitely
increased, and therefore the series
a^cos a + aiCos(a + P) + a.2Cos{a + 2^) + . . . ad inf.
is convergent.
Co7\ — Similarly, it may be shewn that if a^, a^, ofo . . . be a
series of constantly decreasing positive quantities, and if
316 CONVERGENCY AND
Lt. a„ = 0, and /3 be not equal to zero or any multiple of
'Z-TT, then the series
a^siu a + aisin(a + ^) + a2^in{a + 2^) + . . . ad inf.
is convergent.
211. // the series aQ+a^x + a^^+ ...ad inf. he abso-
lutely convergent for all values of x not greater than
some fixed quantity r, then for all values of x less than r
the series will he a continuous function of x.
First, suppose x and each of the coefficients a^, a^,
a^... positive.
Let ccj, x^^ be two adjacent values of x less than r, of
which a?! is the greater,
let ^j = a^ + a^x^ -f a^fc^ + ...ad inf.
and ^2 = %+ a-^x^ + a^x^ + ...ad inf.,
then we have to prove that, when x^ — X2_ is indefinitely
small, so alscJ is 8^ — 82-
By subtraction, we get
8^-S2 = a^{x^-X2)-\-a2(x^^-X2^') + ...+an(x^''-X2'')+...
Now -1^^2"^^«-l_,_^n-2^^^n-3^^2_^_^ _|.^^«-l^
1 2
and therefore, since a?i>a?2> we have
■a?.
<vx^^~'^ and >iia;2*^"^
Hence, S^-'82< (x^ — ccg) (a^ + 2c/2a5i + SagCt^^ + . . .
and > (iCj — a?2)(^i + 2a2a'2 + ^a^x^^ + . . .
Now, the ratio of the (n + iy^ to the n^^' term of the
series a^ + 2a^ + Sa^x^ + . . . + nanX^ ~ ^ +
_n-{-l an+iX
~ n an
r l\an+ix^
\ n) an
CONTINUITY OF SERIES. 317
and, by sufficiently increasing n, this ratio can be made
as nearly equal to ""'"^ as we please, and therefore, if x
be less than r, this ratio can be made less than _^!±2_ ^ g^
an
less than the test ratio of a convergent series ; therefore
the series a-^-{-2a^x-\-^a^x^-\- ... is convergent when x<t.
Hen ce, % + 2a^x^ + ^a^x-f + ...ad inf.
and a^ + ^a^x^ + ?ta^x.^ -\- ...ad inf.
are finite quantities.
Therefore, when x^ — Xc^ is indefinitely small, 8-^ — Sc^ lies
between limits which are also indefinitely small, and,
consequently, 8-^ — 82 is indefinitely small, hence the
series a^ + a^cc + ctgCt;^ + . . . ad inf. is a continuous function
of X for all values of x less than 7\
Secondly, if some of the terms of the series be positive
and some negative, we can arrange all the terms of the
same sign into a group, and apply the theorem just
proved to each of the groups separately. Then, since the
change in each group is indefinitely small when that in x
is indefinitely small, the algebraical sum of the two
changes will aiso be an indefinite!}'" small quantity.
Hence, the theorem holds for negative as well as for
positive values of %, ^g' ^3 • • • ^^^ ^•
212. The argument of art. 211 is not valid when x^^Vy
for the series aj + Sctgr+Sagr^H-... may be divergent.
For example, let r=l, and let the series «! + ag + %+•••
be J2 + 22 + 32+--- >
then a^ + 2a2r + Sa^v^ + . . . becomes t + :^ + « + • • •.
318 CONVEROENCY AND
and this is divergent. Thus, the reasoning of art. 211
does not sliew that tlie limit of the series
when X increases up to unity, is the series
The theorem of the following article may be applied in
such limiting cases.
213. If the series a^-^- a^ + a^-\- . . . ad inf. he convergent,
and if x be less than unity, then the limit of
aQ + ajX + dgic^ + . . . ad inf.
as X approaches the value unity will be equal to
aQ-{-a^-\-a.2+...-ad inf.
Let s = aQ+a^+a2+... ad inf.,
S = aQ+a-^x + ^2^^ + • • • ^^ '^V-»
and let x = l—h, where h is a. small positive quantity,
then we have to shew that
Lt.{s^S) = 0.
h=0
Let 8 = ao+^i+... + an+?'n,
S=aQ-\-a^x+...+anX''-\-Rn,
then s — S
= a^(l-x) + a^{l-x^) + ...-han{l-x'') + rn-Rn
= (\-x:){a^+a2{l+x)+aQ{l-hx+x'^)+.. +an{l-\-x-\-...+x'^-'^)]
+r„-R„
= h{a^-\-a2+a^+...+a„
+x(a^ + a^-\-...+a„)
h(u^ + u^ + u^-{-...-\-u„) + r„-B„,
CONTINUITY OF SERIES. 319
where Ur = x^' ''^{ar + ar+i + . . . + a„),
hence, if U be the average value of the quantities
we obtain the result
s — S=hn U+ Tn — Rn-
Now, in consequence of the convergency of the series
C6o + ai + «2+---
and of the limitation of the value of x to numbers :t- 1,
it follows that each of the quantities denoted by Ur is
hnite (or indefinitely small) however great n may be ;
and therefore that U the average value is finite (or in-
definitely small) ; also as n increases indefinitely, Vn and
Rn diminish without limit.
Let h = ~j, then s — >Sf = — [-Tn — Rn-
Now let h diminish without limit, and n increase without
limit, then Lt{s-8) = 0.
h=0
Cor. — If the limiting value of x for which the series is
convergent be R, where R is any fixed number, the limit
o^ ctQ + a^x + a^pc^ -}-... as x increases up to R will be
aQ+a^R + a^R^ +....
For if we put bn for cinR"', and p for x/R, we may write
the series in the forms
and 60+61 + fe2+---,
and apply the theorem of the present article.
214. The argument of the preceding article depends
on the conditions that a^ + ag+.-. + ttn is finite for every
value of n, and that Vn and R^ vanish for any indefinitely
320 CONVERGENCY AND CONTINUITY, ETC.
great value oi n. In Example 1, all these conditions are
satisfied ; in Example 2, some of them only.
Example 1. — When x increases up to the value 1, the linjit of
the series -^ - „ + a" ~ • • • ^^ infinitum = 1-- + -^ — ... ad inf.
2 3 2 3
Here ai + a2 + ... + a„ is less than 1 and greater than \ whatever
n may be ; also Tn is numerically < -, and therefore a foi'tiori
n-{- 1
Rn is numerically < ; hence, the conditions of art. 213 are
satisfied, and the proposition is true.
Example 2. — Consider the series
\-x-\-x^-x^+... ad inf. (1)
in relation to the series
1-1 + 1-1 + .. . adinf (2)
The series (1) is convergent for any value of x, if x<\, and the
limit to which it converges is - — — ; hence, as x approaches the
value unity, the series (1) approaches the value ^.
The series (2) oscillates, its sum being 0 or 1 according as the
number of terms is even or odd.
The equation s- S=hn U-\- ?•„ - Rn
still holds; a-^-^-a^-V ...■\-an is always finite, and therefore C^ is
finite ; hence, if h = -^., the term hn U becomes indefinitely
small, when h becomes indefinitely small. But r„ does not vanish,
nor does Rn vanish, for, with the assigned relation h= -^, the
{n + iy-^ term of (1) is still finite, since
Thus, we are unable to infer that
Lt.(s-S)=0.
h=0
INFINITE SERIES FOR THE COSINES, ETC. 321
5. Infinite Series for the Cosines and Sines of x in
ascending powers of x.
215. To prove that
X ■, ^ . / ^x^^'
2r
cos« = l-|+^-... + (-l)'j2^+(-l)'-+'iJ,
where 0<R<
2r+2
If n be any even integer we have, by art. 1 95,
cos nO = cos«e - --^^^^cos^- 20 sin^O
+ ... + (_ l)'^^|^^^|^^cos--0 sin-e
+ ••• + 1 ^^i.2...(m-l)m^ '^'
the number of terms in the series being o + 1-
Hence, cos n0 = cos«0{l _-«-l^-^)(ta^y
, , . - ,,'yi0('^0 - 0). . . (ti0 - 2^^^! . 0)^tan 0V>-
+ ...+(-1) l.2.3...2r \ e J
+ ... to f^ + lj terms h.
In this equation, let nO remain constant and equal to
X, and let n be indefinitely increased, and consequently
0 indefinitely diminished ; then, since
Lt. cos«0=il(cos-Y = l, (art. 190),
we have, by art. 188,
322 INFINITE SERIES FOR THE
,,/' a;(aj-e)/tan0\2
-h...i-^ i) i.2.3...2r \ e )
+ ... to f^ + lj terms K
Let kr denote the absolute value of the ratio of the
(7* + l)th to the rth term of this series,
then k r; (^'-^^--^ ■ e){x-iT-l . eVtan fly
then k,-Lt (2r-l)2r V d /* '
Hence, for &i\ finite values of r, kr= „ ._i\9 » *id,
for values of r such that - is finite, ^r<7o tto"' until
at the end of the series, when ''"=-^.
7 _ 20.6 /tan0\
n ""^^^'"'^^(2r-l)2r'
~2'
2 2a;2
{n-'l)n\ 6 J {n-l)n^
It follows that the terms of the series for cos x increase
in absolute value with r so long as {2r — l)2T<x^, and
that, from and after the greatest term, each term is less
than the preceding term, and that the terms ultimately
vanish. Also, the terms alternate in sign. Hence, we
have for all finite values of r such that (2r—l)2r>x\
cos.=i-|+|-...+(-i)^+(-ir'ij,
where R = a. series of diminishing terms, alternately
positive and negative, and therefore R is numerically less
than the first of these terras, viz..
x(x-e)...(x-2r+l. 0)/tan e\2^+2
1.2.3...(2r+2) \ 0 J '
COSINES AND SINES. 323
and therefore, since
we have, a fortiori, R < r-r -.
' -' |2rH-2
Cor. — In like manner, it may be shewn from the for-
mula of art. 195, Cor. 2, that
x^ x^ x^^
where i2 = a series of diminishing positive terms, of which
the first is less than j^ — —^ and the ratio of each to the
|2r + 2
and.-. . -^<[-2^q:2/r'"(2r + l)(2r+2y'
^.2r+2
or R<
|2r{(2r+l)(2r+2)-a;2}-
216. To prove that
/y.2r+l
where 0<R<
[2r-hl
If n be any even integer, we have, by art. 195,
sin 710 = 71 cos"-iO sin 0-^^^7i^^o^^^cos^"^^ sin30+ ..,
+(_ i)r-i<^zi%:%r:!r+?)cos'.-^+i0 sin2-'e+ ,
1. , A , o . . . i^r -^ i^
. , ^.Vl-1 n(n — l)...S .2 /> • „ i/^
+<- ^>^ i.2:..(».-2)(^-ir ^ '"^ ^ ^'
the number of terms in the series being -^.
324 INFINITE SERIES FOR THE
Hence,
-20)/tan0y
smn0 = cos-0|n0.-g 1,2.3 \-e-) +••
■*"^ ^ 1.2.3...(2r-l) V e /
+ . . . to ^ terms K
Let 710 = Xy and let a; remain constant while n is
indefinitely increased, and therefore 0 indefinitely di-
minished ; then, since
Lt cos'^0= Lt. (cos ^Y= 1 (art. 190),
we have, by art. 188,
r^ ( tane a;(a;-0)(£C-20)/tan0V ,
sin a?
a;(a3-0)...(a;-2r-2.0)/tan0Y^-i
"^^ ^ 1.2.3... (2r-l)" V 0 /
+ . . . to ^ terms !-.
Let kr denote the absolute value of the ratio of the
rth to the (r — l)th term of this series, then
, _ , (g; - 2r - 3 . 0)(a; - 2r ~ 2 . 0) /tan QV
'^^-r. (2r-2)(2r-l) \ 0 J'
and therefore, for all finite values of r, kr=
(2r-2)(2r-l)
and, for values of r such that - is finite, A;y<-
Sl
71 ' ^ (2r-2)(2r-l)'
when ^=9^,
, 20.20 /tan^Y^ 3.2.a;^
'^'~(ri-2)(n-l)\ 0 ) {n-2){n-\)v?
7h
until at the end of the series when ^= «,
COSINES AND SINES. 325
It follows that the terms of the series for sin x increase
in absolute value with r so long as (2r— 2)(2r— l)<a;^
and that, from and after the greatest term, each term is
less than the preceding term, and that the terms ulti-
mately vanish. Also, the terms alternate in sign.
Hence, we have for all finite values of r such that
(2r-2)(2r-l)>i:c2
sin^ = a)-|+... + (-iri^^+(-iyE,
where jR = a series of diminishing terms, alternately posi-
tive and negative, and therefore R is numerically less
than the first of these terms,
and therefore, since Lt ( -^— ) =1,
we have, a fortiori, R < .^ -. •
Cor. — In like manner, it may be shewn from the
formula of art. 195, Cor. 2, that
0? /p2r-l
sinha; = a; + ,-^+... + ,- =-|-jK
(3 \iT-\
where R <
L2r-l(2r.2r+l-a;2)
/y%Z /yi4 /yiD
217. The series l±^+^±^+...
x^ . x^ , x^
and ^±_+_±_+...
are absolutely convergent for all values of x, and there-
fore, by art. 211, they are continuous functions of x for
all values of x.
326 INFINITE SERIES FOR THE
It will be observed that the series for cos x and cosh x
contain even powers only of x, a result in agreement with
the known theorem that cos a; and cosh a; are even
functions of x. Similarly we observe that the series for
sin X and sinh x contain odd powers only of x.
The series for cos x and sin x are equal to these func-
tions for all values of x, and therefore the series must be
periodic, i.e., they must converge to the same limit when
X has values differing by a multiple of 27r, a result that
may be verified by actual computation.
Thus, to take a simple case, let x have the values
^ and -y^, then we must have
1
(fo)7(i+(fo)yii--'^-/-
^-r^^k^-mk-.'^^i^i
MO
Working to two places of decimals we get
first series =l-05 + -00- ... =-95,
and second series
= 1 - 21-76 + 78-93 - 114-52 + 89-01
-43-05 + 14-19-3-39 + -62--09 + -01--00 +
= 183-76 -182-81 =95.
218. To prove that cosh a; = J (e^^ + e " *)
and that sinh x = ^{^—e-^).
By arts. 215, 216, we have
x^ aj*
cosh a; = 1 +777 + r-r + . . . ,
[2 [4 .
and sinh a; = a;+,— + .—+...,
[3 [5
COSINES JiND SINES. 327
cosh oj + sinh a; = 1 +a;+— + j-^ + . . .
and coshic — sinncc = l— cc+r^ — ro + ---
Hence, cosha; = J(e*+e-«),
and sinhaj = |(e^— e-^^).
It is to be observed that in the exponential values of
cosh a? and sinho; the single arithmetical value of e* is
always to be taken ; thus, if cc = J, then e* = + s/^i or if
a;=-i then e^ = ^^^
219. To 'prove that, if Q — gd u, then will
0'
. = logtang+|)
By art. 21 8, e^ = cosh u + sinh u.
But cosh u = sec 6, and sinh u = tan 6 ;
,, /I , X /I 1 + sinO
e« = secO + tan 0 = —^ — ^r—
cos 6
= 0 . 0 = *^^4+2>
cos « — sin ^
u = logtan(|+|).
220. The hyperbolic cosine and sine might have
been defined by the equations cosh aj = J(e*+e"*) and
sinh a; = J(e* — e~^) or the equivalent series.
From these definitions we at once obtain the results
cosh!53 + sinh£c = e* and cosher — sinh cc = e~*,
hence, by multiplication, we have cosh^aj — sinh2aj=].
828 INFINITE SERIES FOR THE
Again, cosh(a;+2/) = K«''"^^+6"''"^) = JC^'"- e^+e"*-^'^)
= cosh X cosh y + sinh x sinh y.
Similarly, the other formulae relating to the hyperbolic
functions may be established. Thus, we obtain a purely
algebraical treatment of the hyperbolic functions.
It will further appear in Part III. that the circular func-
tions might be defined and their properties investigated
in a similar manner, without any reference to geometry.
221. By aid of the exponential values of coshoj and
sinh X, series involving these functions may frequently be
reduced to known algebraical forms.
Example 1. — To prove that
cosh a: + cosh(:r +y) + cosh(a;+ 2y) + . . . to w terms
=co8h(^+^y ) sinh ^/sinh |. (See art. 205.)
We have cosh(.2: + r^/) = |(e*+'-^ + e"*"*^),
's" cosh(a; + n^)=^*^ 2~V+'^ + ^ '^S'e-*-'^.
r=0 r=0 r=0
Hence, hy the formula for the sum of n terms of a geometrical
progression,
= coshf ;r + ^^^^— y jsinh ^ /sinh-|.
COSINES AND SINES. 329
Example 2. — Find the sum of the series
^inh u + n sinh '■lu -h ~ ^sinh 3w + . . . to (?i + 1 ) terms.
1 . ^
Let AS'=the sura of the series,
then 2AS'=e"+?ie2« + ^^^^^)e3M4., , to {n+l) terms
_|e-«+ne-2«+'^^^"-^V3«+... to (w + 1) termsj,
and therefore, by the Binomial Theorem,
2>S = e"( 1+ e")" - e-"(l + e-"r
= 2sinh^|+lV. (2cosh|V.
aS'= 2"cosh"| sinh^l + 1 V
Examples XXIII.
1. Find the sum of n terms of the series
cos 0 + cos 20 + cos 30+ ....
2. Sum the series
siu a + sin 3a + sin 5a + . . . to n terms.
3. Prove that
TT , 37r , Stt , Ttt , Qtt 1
cos Y^ + cos ^ + cos jY + cos Yi + cos YY = 2'
and that
27r , 47r , Gtt , Stt , lOx 1
cos — + C0S jy + cosyy + cosyy + cos-y^= -^•
4. Sum the series
008(71— 1)0 + cos('M — 2)0+ cos('M — 3)0+... to 271 terms.
5. Sum the series
cos a — cos(a — /3) + cos (a — 2/3) — ...+( — l)"'Cos(a — n^).
330 SERIES.
6. Sum the series
sin Q cos 30 + sin 20 cos 60 + sin 40 cos 1 20 + . . . to n terms.
7. Find the sum of the series
cosec0+cosec20+cosec40+... to n terms.
8. Sum the series
tan^ + 2tan2J[ + 22tan2M + ... + 2'^-itan2«-i^.
9. Find the sum of the series
tan0sec20+tan20sec220+... + tan2«-i0sec2«0.
10. Shew that
3«sin |, - sin 0 = 4|sin3|+ 3 sin3|+ . . . + 3*^- isin^l^j-.
11. Sum the series
8in20+sin2204-sin230+...+ to n terms.
12. Shew that
cos*g + cos*-g- + cos*-g + cos*-g- = 2-
13. Expand cos^0sin20 in a series of cosines of multiples
of0.
14. Prove that when m is an odd integer
r . m2-l . „ , (m2-l)(m2-9) . . "I
8minx=m\ sma;- ^sm^fl;4- . 9. i a, k sm^a;-... .
15. Shew that when 71 is a positive integer
^ — 172"*^''^+ 1.2.3.4 *^"^"--:^^^-
16. Find all the vahies of 0 determined by the equation
sin 0 + sin 30+sin 50+... +sin(27?,-l)0
= cos 0 + cos 30 + cos 50 + ... + cos(2'?i - 1)0.
17. Sum the series
sin0 , 2 sin 20 , 22sin 2^0 ,
+ ^ S7^ — T + ^ S7?i — T + ... to 71; terms.
2cos0~l 2cos20-l ' 2cos 2^0-1
SERIES. 881
18. Shew that
cosO — cosf 0 + 3-) + cosf 0+-^j — cosf^+n?)
+cos(e+^)-cos(0+|^) + cos(e+^) = O.
Interpret the equation geometrically. (See Ex. 39.)
19. .AO is a diameter of a circle of radius unity, AP^ any
arc of the circle. If arc J.P„ = ti.arc AP^, and
chords OPi, OP«_i, OPn, OPn+i are drawn, shew
that the formula
2cos(7i+l)0 = 2cos7i0.2cos0-2cos(n-l)0
may be written in the form
OPn+l=OPn.OP,-OPn-l.
Prove the formula geometrically by aid of Eucl.
VI. D.
20. A series of points are distributed symmetrically round
the circumference of a circle. Shew that the sum
of the squares of their distances from a point on
the circumference is twice that from the centre.
21. Find the sum of the series
cos h COS f- COS [-••• + cos^ :
n n n n
22. If ^ = -^, shew that
lo
cos0 + cos3^ + cos50+... + cos 110 = J.
23. Find the sum of the series
cos a + cos 3a + cos 5a + ... + cos(27i — l)a.
24. If 71 be a positive integer, and sinJ0=^— , shew that
cosJ0 + cos|0 + cos|0+... + cos — ZL^Q = n^m7i6.
332 SERIES.
25. Find the sum of
sin 2a + sin 5a + sin 8a + ... to 7i terms.
26. Find the sum of
cos2a + cos^2a + cos23a+... to n terms.
27. Sum to n terms
cos3a4-cos3(a + ^) + cos3(a + 2/3)+....
28. Sum to n terms
sin^a + sin2(a + ^) + sin2(a + 2^8) + . . . ,
and hence find 12+22+32+.. .+^2.
29. If n be an even integer, prove that
sinti^
COS0
= (2sme)»-i-^(2sine)''-H^^^^|^— \2sin0)''-5-...
30. If n be an even integer, prove that
n
(—1)2 2 cos TlO
= (2 sin QY- ^(2 sin 0)^-2 + ^-^\2 sin 0)^-*- ...
31. If ti be an odd integer, prove that
^ = 1 j-5— sin20 + ^ f} -^sm*0— ....
008 0 [2 [4
32. Prove that
, X ,, X X , , X X , 7'/..
tan ^ sec a; + tan ^^sec ^ + tan ^sec ^-\-...aa inj. = tan x.
33. Sum the series
1, a.lj a,li. a.
to 71 terms, and also to infinity.
SERIES. 333
84. Find the sum of
1
cosacos(a + )8) cos(a + )8)cos(a + 2^)
+-
cos(a + '3^ — 1 . /8)cos(a + n/3)
35. Sum the series
^ 1 2.1 ,, , 2.2 ,, , 2.3
144.12.^2^ 2H22 + 2^ 3*4-32 + 2
2 4
+^^^"V + 4H2"*"-- ^ '^ *®^°^^'
36. Sum the series
^sec2|+-2sec2|+^sec2|+... to n terms.
37. If n be even, find the sum of the products of the
sines of pairs of angles equidistant from the
beginning and end of the series
a, (a + ^), (a + 2/3) ... (a + ^i^=a/3).
38. From a point within a regular polygon perpendiculars
are drawn to all the sides : find the sum of the
squares on these perpendiculars.
39. If A^,. A^, A^... A^n+i be the angular points of a
regular polygon inscribed in a circle, and 0 any
point on the circumference between A^ and A^n+i)
prove that the sum of the lengths of OA^, OJ.3,
0 J. 5, ... OAin+i will be equal to the sum of OA^,
0^4, 0^...0^2n.
40. If P1P2' ^2^z> • • • PnPi he equal arcs round the cir-
cumference of a circle, and if P^M^^, P^^i • • • Pn^n
be drawn perpendicular to any diameter ABy
shew that the arithmetical mean of the rectangles
AM^.BM^, AM^.BM^,...AMn.BMn is half the
square on the radius of the circle.
334. SERIES.
41. If 71 be an odd integer, prove that
^ ^ COS0 ^ 1 ^ ^
+ <^-y^-^>(28me)n-»— ■
+(-i/^-^-if-;-2)-(^-2^)(2sine)n-v-i+....
42. If n be an odd integer, prove that
(-l)V2sin7i0
== (2 sin er- 1:(2 sin 0'*-2+^?:^^i:i?l(2 sin 0)^-*-
+(-i)-"<-"7y;3(;";^-+^>(2siner-^+....
43. If 71 be an even integer, prove that
COS0 1 |3^ |5
44. If 71 be an even integer, prove that
cos nQ = l — Tg-sin^^ H — ^ - — ^sm^0
If Iz
7l2(7l2-.22)(7l2-42) . „. ,
45. Prove that 2 2 sin(pa + g^)
p=l g=l
_sin|masin|?i^sin|{(m + l)a + ('y^+l)i3}
~ sin la sin J/3
46. Sum the series
cos 20 cosec 30+ cos 60 cosec 3^0+ cos 180 cosec 3^0+ . ..
to n terms.
47. Sum the series
2 cos 0 sin2-+ 2^003 ^ sin2-2+ 2^cos^sin2^^3+ ... to 7i terms.
SERIES. 336
48. Sum the series
cos W cosec220 + 2 cos 2^0 cosec2220 + . . .
+ 2«-icos2'^0cosec22»^a
49. Find the sum, to n terms, of
sinO . sin 20 . sin 30
+
cos 0 + cos 120 ' cos 20 + cos 220 ' cos 30 + cos 320
50. Sum the series
1 1
sin^a; sec 3a; + ^ sin^3fl3 sec Z^x + -^ sin332ic sec 3^ic + . . .
to n terms.
51. Sum the series
tan 0sec 20 + sec 0 tan 20sec30+sec 20tan 30sec 40+ ...
to n terms.
52. A regular polygon of n sides is inscribed in a circle,
and from any point on the circumference chords
are drawn to the angular points ; if these chords
be denoted by c^, ^2' • • • ^» (beginning with the
chord drawn to the nearest angular point, and
taking the rest in order), prove that the quantity
^1^2 ' ^2^3 + • • • + ^n - l^n ^rfil
is independent of the position of the point from
which the chords are drawn.
53. From a point 0 a straight line OA is drawn, making
an angle «(<— xt) with a fixed straight line AB^
and n other straight lines OA-^, OA^, ... OA^ are
drawn to it making the angles A OA^^, AfiA^, ... all
equal and each equal to a ; if B^, B^... Bnhe the
radii of the circles circumscribing the triangles
OAA^y OA^A^ ..., find the value of
i^i + jRg + -'^s + • • • + ^-
SERIES.
54. Shew that
and give the coefficient of 0^".
55. From the equation
2 sin ra — sin ^(n + l)a sin J^ia/sin Ja
r=l
deduce the sum of the first n natural numbers, and
also the sum of their cubes.
56. Sum to n terms the series
JL . ^ J. . Z . o
57. Find, by aid of the exponential value of the hyper-
bolic sine the sum of the series
sinh u + sinh(u + v) + sinh(u + 2v) + ... to n terms.
58. Sum the series
(n-l )cos e+{n- 2)cos 26 + (n- 3)cos 3^
+ . . . + 2 cos(n -2)0+ cos(7i - 1)0.
59. Sum the series
(I **" i) +(i**° p) +-+(^ta°|i)''
and shew that the sum to an infinite number of
terms is
(tana)2 a'^S
60. The sum of cosecaj + cosec 2ic + cosec 2^33 + . . . + cosec 2" " ^a;
is zero i^ ^ = o^i— f> w, and n being integers.
61. Find the sum, to n terms, of
sm^O cosec 2^0 + 2 sm^20 cosec 2^0
+ 22sin222^ cosec 2^0+.. ..
62. Sum the series
sin 0 sec S^ + sin 3^ sec m+. . . + sin 3«-i0 sec 3*^^
SERIES. 337
63. Find the sum, to n terms, of the series
COS d Q0&\ cosec"-^ + cos W cos^— cosec^-^
+ COS 3^0 cos^' gj cosec^-^ + -
64. Find the sum of all the values of cos(j0a + g'|8), where
f and q may have any positive integral values
between 0 and ?i — 1.
65. Shew that
sin{a- 271^} +sin{a- 2(71- l)/3}+sin{a- 2(71-2)^}+...
+ sin{a + 27i/5}=^^sin(27?, + l)^.
^ sin/5 '
66. Prove that
vi=M n=N p~P
S 2 E ... cos(ma+7i/5+_29y4-...)
m-O 71=0 p=0
= cosi(ifa+i\r/3 + Py+...)sinK^+l)asinJ(iV'+l)/3
X sin J (P + 1 )y . . . X cosec Ja cosecJ^S cosec Jy . . . .
fiX g ~ *
67. If (p{x) denote ^ _^, shew that
6 ~)~ 6
9n+l -I
9!>(a;) + 2(p(2x) + 220(22^^) + . . . + 2"0(2«aj) : ^ -^
9^(2-+ia)) 0(^)-
68. Prove that the sum of the series
- log tan 20 + "2 log tan 2^6-{-...to n terms
= log(2sin20)-l^log(2sin 2^+i0).
Zi
69. Find the sum of
cos -^+ cos —^ — I- cos -^ + ... to 71 terms ;
and apply the result to prove that, if one angle of
a triangle be n times another angle, the side
opposite the former angle is less than n times the
338 SERIES.
side opposite the latter angle, where n is any
integer.
70. Straight lines whose lengths are successively pro-
portional to the numbers 1, 2, 3, ...,n form a
rectilineal figure whose exterior angles are each
equal to — . If a polygon be formed by joining
the extremities of the first and last lines, its area is
— ~ ^ cot — h-r^ cot - cosec^-.
24 . n 16 n n
CHAPTER XIV.
FACTOES.
§ 1. Fundamental Theorem on Trigonometrical
Factors.
222. The object of this chapter is to shew that the
resolution of trigonometrical expressions into factors can,
in a great number of cases, be made to depend imme-
diately on a single fundamental theorem.
The fundamental theorem may be enunciated as
follows : —
Ifvn denote any one of the functions 2 cos nx, 2 cosh nx,
or x^^—, and if Un denote the function 2 cos na, then
X
Vn — Un
= {Vi-2cosa}|vi-2cos(« + ^)||vi-2cos(a + -;^)|...
to n factors.
The theorem may be written in the form
r=n-l C / 27r\l
^^71 — ^11= n i'^i — 2cosf a + r . — jk
where 11 \v. — 2 cosi a-\-r . — )}■ is an abbreviation for
r=o I ^ \ n/j
339
340 FUNDAMENTAL THEOREM ON
" the product of the n factors obtained by assigning to r
the values 0, 1, 2, ... (n— 1) in the expression
Vj — 2cosfa + r . — j.
It will appear in the following articles that the proof
of the fundamental factor theorem depends only on the
elementary facts that the functions 2 cos tix, 2 cosh nx and
a;"+ — severally obey the law
X
'^in^n ^^ '^m+n "i I'm - nj
and that 2 cos na obeys the same law and is also a periodic
function of nay of period 27r, so that for cos na we may
substitute cos(7ia + 27r), cos{na + 47r). . . or cos('7ia + r . 27r),
where r is any integer.
223. If Vn and Un he two functions, each of which obeys
the law f(m)xf(n)=f(m-{'n)+f(m — n)for all integral
values ofm and n, and if Vq=Uq, then will Vi^--u^ be a
factor of Vn-Un.
We have VnV^ = Vn+i + Vn- 1,
Vn+l^VnV^-Vn-i.
So also, Un+1 = UnU^ — Un-i ',
.'.Vn^l-Un+i = {Vn-Un)v^-]rUn{v^-U^)-(Vn-i-Un-\)...{l)
Hence, if v^ — u^ is a factor of Vn — Un for any two con-
secutive values of n^ it is also a factor for the next higher
value of n.
But, from (1),
v^ — u^=^ {v^ — u^v^ + u^{v-^ — 'W'l) , since Vq — Uq = Q\
v^ — u^ is a factor of -y^ — u^ and v^ — u^
and therefore of v^ — u^, and therefore, by successive in-
ferences, of Vn — Un.
TRIGONOMETRICAL FACTORS. 341
224. Since
2 cos mx . 2 cos nx = 2 cos(m + n)x + 2 cos('m. — n^x,
2 cosh mx . 2 cosh nx = 2 cosh(m + n)x + 2 cosh(m — ^1)33,
»d (a=».+l.)(cr» + l) = (^"'■^»+^J + (^'"-'•+^0.
and since, when yi = 0, each of the functions 2cos?ia;,
2 cosh nx, and rc^+ — is equal to 2, we see that 2 cos na?,
2cosh7icc, or x^ + — may be substituted for Vn or u,i in
the theorem of art. 223 ; thus, we infer that
cos a — cos /5 is a factor of cos na — cos n^,
cosh X — cos a of cosh, nx — cos na,
cosh cc — cosh y of cosh nx — cosh ti^/,
03 H 2 cos a of 03'^+— -—2 cos Tia,
X x^
and so on.
225. // Vn denote any one of the functions 2 cos nx,
2 cosh nx, 0;"+ — , then will
X
r=n-l C / 27r\1
Vn — 2Q,o^na=^ 11 j-yi — 2 cosf a + r — j\.
By the theorem of art. 223, we know that v^ — 2 cos a is
a factor of -yri— 2 cos na.
Hence, if ?' be any integer, v^ — 2cosia + r'^] is a
\ n '
factor of -Vyj— 2 cos (71a + r. 2ir),
i.e. of I'n — 2 cos na.
Assigning to r the succession of values 0, 1, 2,...{n — V)
we obtain n factors of 'y^ — 2 cos na, and these factors are,
342 FUNDAMENTAL THEOREM ON
except for special values of a, all different, hence we may
write
Vn — 2cos7ia
= X{Vi — 2 cos a}\ v^ — 2 cosf aH — -j \.-'
x{..-2cos(«+(2^-)}
where X has to be determined.
From the formula Vn+i = VnV-^ — Vn-i, it follows, by-
repeated inferences, that Vn is an integral function of v^
of the n\h. degree, and that the coefficient of v^ in the
value of Vn in terms of v^ is unity, hence we get X = 1, and
therefore
Vn — 2cos'^a
Jn;{.,-2cos(„+.^)}.
r=0
Cots. — cos nx — cos na
= 2^-i{cosa; — cosajMcosaj — cosfa + — H
X ] cos a; — cosf a + — j [ . . . to n factors
_2n-i n jcosa; — cosfa+r—jk
cosh nx — cos na
= 2"~^{cosh £C — cos a} j cosh x — cosf a H j f
X I cosh X — cosf a H — - ) [. . . to ?i factors
r=n-lf / 27rM
_2n-i jj -^coshaj — cos(a + r — ) h.
r=o I \ n/j
TRIGONOMETRICAL FACTORS.
X'^-\ - — 2 008 710
X^
= \x-\ 2 COS a M ^ H 2 cos ( a + —
y X ) y X \ n
x\x-\ 2cosf a + — j L.. ton factors
='l[~'|aj+i-2cos(a + r— )|.
226. Demoivre's Property of the Circle. — If
Aq, A^, A^ .., An^x he n points ranged symmetrically on
the circumference of a circle whose centre is 0, P any
point in the plane of the circle, then shall
PA,\PA,\PAi..,PA:_,
= 0P2"- 20P« . O^o^^cos nPOA^^ 0A;'\
Let
OP = X, OAq = a, /. POAq = a. 4
By art. 225, we have, writing
X p
-- for X,
a
'=^-Hx a . / 27r\\
= n \'-\ 2cos( a+r — jK
,.=0 W X \ n/j
multiplying by a'^x^ we get
a;2n - 2a;"a"cos na + a^"
='^~n'|x2-2a;acos(a + r^)+a4 (1)
From the triangle POAr we have
PA,' = x^- 2xa cos(a + r^) + a\
and therefore, from (1),
PAS.PA^.PAl.. P^„li = a;2«-2a;Wcos7ia + a2«
344 • FUNDAMENTAL THEOREM ON
or
PA^ . PA^ . PAl . . PAl, = OP^^' - 20P^' . OA^cos n POA^
■hOA;\
Core. — If the angle POAq = 0, i.e., if P lie on a radius
through one of the n points, then
PAo .PA^.PA^.., PAn-i = OP^ - OA^.
If the angle POAq = -, i.e., if P lie on the bisector of
the angle between radii through two consecutive points
of the system, then
PA(,.PA^.PA^...PAr,.i = OP''+OA^.
These results are known as Cotes s Properties of the
Circle.
227. The wide applicability of the fundamental factor
theorem arises from the consideration that by assigning
special values to x, or a, or both, and making elementary
transformations, we can deduce an indefinite number of
factorial forms. Thus, if factors of cos na are required,
we may write v,i = 2cos(?ia + 7r), then v^^Un becomes
— 4cos7ia, and we obtain a factorial form of cos-na;
or, if factors of sin-nO be required, we may put x = 0,
a = W, then Vn — Un becomes 2 — 2 cos -ti. 20 = 4 sin^nO, and
we obtain a factorial value of sin^nO.
In deriving particular cases in which the factors are
given from the general theorem, the necessary substitu-
tions and transformations may often be inferred from
a careful consideration of the number of factors, and of
the limits between which angles involved in the product
lie. A geometrical representation of such a series of
angles as that denoted by a + r — will be of service;
TRIGONOMETRICAL FACTORS, 345
thus, if XOAq be the angle a, and if r range from 0 to
27r .
(n — 1), a-\-r — will denote a series of angles obtained
by drawing radii to n points A^, A^,A^, ...,An-h succeed-
ing one another at equal distances on the whole circum-
ference of a circle ; while, if we have such a series of
angles as a, a H- — , a H , . . . , a + ^^ — , the representative
points Aq, A^, A^, ..., An_i are n points on a semi-circle.
228. Example 1. — If n be a positive integer, shew that
sinw(/) = 2«-^sin(/)sin('(^ + !^)sin('(^+^V..sin(<^+^!i::l^y
By the fundamental theorem,
cosn6-cos7i.2cl> = 2''-^ U ]cos ^-cos( 2</) + r— ) [.
Let ^=0, then
l-cos7i.2(/) = 2"-i'^'n |2sin2^<^ + r-U,
Extracting the square root we get
sin?i</) = 2'*-i'^~n |sin^(^ + r-U.
The positive sign is taken for all values of <^, since
if 0 <n(f)<7r , sin ntf) and all the factors are positive,
if TT <ncf)< 27r, sin ncf) and the last factor only are negative,
if 2ir<n(f)< Sir, sin n(f> is positive, and the last two factors
only are negative^ ; and so on.
Example 2. — Prove that, when n is an odd integer,
:r" + l=(^+l)/'.v2_2^cos-+ yA^2_2^cos^^ + l
x(^2_2a;cos^i:i^7r + lY
"We have, by art. 225,
I r=n-l f I / 27r\"\
^" + -„-2cos?ia= n -^ A- + --2cos(a+r — ]}.
346 FUNDAMENTAL THEOREM ON
Multiply by af*y and let na=7r, then
r—w— 1 f 9r4- 1 ^
(^" + 1)2= n {a^-^XCOB^^-^TT+l}.
r=0 '^ n i
Since n is odd, there is a middle factor, viz., that for which r= t_.
This factor = a^'-^ - 2^ cos tt + 1 = (^ + 1 f,
and the factors equidistant from the middle factor are equal,
hence (^"+l)^ = (^-2xcos- + lWar^-2^cos— +1^...
Ix^ - ZX cos ^^TT + 1 ) (a?+ 1)2,
and therefore, since :*?'*+ 1 and :r + l have the same sign, we obtain
the result
^"+l = (jp+l)f^2_2A^cos-+lVa;2-2^cos — + l)...
X Ix"^ - ^x cos — — TT + 1 y
Example 3. — Shew that
coalcos H^ ... cos(?2jdk=(::i)lzi
n n n 2""^
By the fundamental factor theorem, we have
cosm^-coswa=2'"-^ 11 -jcos ^-cosf a + r--j j-.
Ijetm=2n, 6='^, a=0, then
»-=2n-l ^ A.,r"^
cosw7r-l=22'»-i n -^-cos — y
=0
since the number of factors is even.
= 2-' n {cos^}.
Hence, <rl)lll=cos?r. cos?2:...cos(?^:il)f.
22n-l
=cos- . cos
n n
Example 4. — Qj^, Q^ . . are n points ranged symmetrically round
a circle of radius a and centre 0. P is a point such that OP=c,
FOQi = 0. If c = a cos - sech (j>, prove that
71
TRIGONOMETRICAL FACTORS. 347
2aVsiii"-(cosh «</> + cos nO)
sin QiPQ^ . sin Q<,PQz ... sin Q^PQi
Draw QiNj^, $2^2 perpendicular to
OP, then
A^iP$2 = ^OQ,Q., + AOPQ^ - AOPQ2,
.'. PQ^.PQ^. Bin Q,PQ.,
= a cos - . 2a sin - + ( a iV, - §2^2)^
n n
= ac|2sin-cosh^ + sin^-sin(^+— H
= 2ac sin — -{ cosh </> - cosf ^ + - ) }■•
Similarly,
PQ2 . PQz sin $2^^3= 2ac sin --! cosh (/> - <
and so on.
Hence, {PQ^ .PQ^... PQnf sin Q^PQ^ . sin Q^PQ^ ... sin QnPQx
= 2"a'*c"sin"-'^ fl I cosh d) - cos f ^ + - + r-^ ) ) .
But we know that
{PQx .PQ.2... PQnf = a''' - 2c«a"cos %^ 4- (r%
and that 2'»-i''~ff '{cosh </> - cos^ (9 +^ +r^) |
= cosh w<^-cos(?i^ + 7r)=cosh 7i<f) + cos nO.
2a"c'*sin'*-(cosh 7i<f) + cos nO)
•t)}
848
PRODUCTS.
o o •n-.^j,,^*^ *^ zi sin 71^ , sinh-MU
§ 2. Products for cos nO, — r—^, cosh nu,
n sin 0'
71 sinh 16*
229. To prove that
cos7id = cos"0"ff'[l4— ^^^
tan
27Z,
COS 710 -cos 710 = 2*^-1 n jeos^-cosr^+r-^U.
We have, by art. 225,
cos Tiff) — cos nO = 2^-
Let nd) = n0+'7r, and therefore (h = 0-\--,
^ n
then - 2 cos 716 = 2^- ^'~li 'jcos^^ + '^) - ^^^(^ + ^— ) }
on i^'^-iV'/o • fn 2r+l \ . 2r-l 1
In this identity put 0 = 0,
then -2 = 2«-i'^~n'|2si]
r=0 I
.'. by division,
2r+l . 2r
sm -7^ — X . sm
271
2n
M.
r=n-l
cos 710 = n
r=0 -
sin0
. 2r + l
tan — t:: TT
+ COS0
271
or
COS710 = COS"0 IT
r=0
1 +
tan0
. 2r+l
tan^ — TT
271
It should be observed that, as r passes through the
series of values 0, 1, 2...(?i — 2), (ti— 1), the angle
-^ — TT passes through the series of values ^, ^, ~ ...,
2n ^ ^ 2n 2n' 2n '
( TT — 2^J, ( TT — ^ ), and that, the nearer r is to the middle
PRODUCTS. 349
2'r+l
of its series of values, the less does the angle — 7^ — tt differ
from a right angle, and therefore the more nearly is the
factor 1 H — equal to unity ; hence, in estimating
tan — ^ — TT
2-71
the value of the product we must remember that the
factors at the beginning and end of the series have the
greatest weight.
The reader will find it useful to represent the factors
by a rank of equidistant ordinates, the corresponding
abscissae being the values 0, 1, 2...('7i — 1) of r. The
Corollaries of arts. 229-232 may be illustrated by folding
the rank of ordinates so as to bring the extremes together.
Cor. — Taking together factors equidistant from the
beginning and end of the product, and observing that,
7?/ — 1
when n is odd, the factor for which r= — — is equal
to unity, since — ^ — tt is then a right angle, we get
the results :
ft
n even, cos n6 = cos^O U J 1 —
.=0 I .....2r4-l
tan^-
2n
n odd, cos nd = cos'^0 U J 1 -
r=0
tan^^ TT
'2n
230. To prove that
sin 710 ^ i/=A"^fi , tan^
I n
850 PRODUCTS.
We have, by art. 225,
Let 710 = nQ + ^, and therefore 0 = 0 + g- ;
and let Tia = 7i0 — ^ , and therefore a = 0—^\
then
-2 sin ne = 2"-' n"{cos(0+^) -eos(0+*^x)}
= 2»-/n;{2.n(.+3sin?^.},
-2sin7i0 ^,, /=W-^f • (^ , ^7r\V=;W^ . 2r-l 1
SmO r=l I \ '^/J r=0 I 271 J
In this identity put 0 = 0, then since Xt -^ — 7r = n, we
have -2'M = 2^-i H Uin — [ H ■^2sin^^x[;
r=l I '?«' J r=0 I 2W J
therefore, by division,
sin'7i0_^=Jl-^r sinO
'7^sinO
I I hcos^l,
"^ I tan — I
I tan — I
or — ^-p^=cos^-iO n J1+-
-Tismt^ • ,.=1 i
^- n
We observe, as in art. 229, that the factors near the
middle of the product approach to unity, and also that,
til
if n be even, the factor given by 'r = - is equal to unity.
A
Got. — Taking together factors equidistant from the be-
ginning and end of the product, we get the results : \
PRODUCTS. 351
-yz, even, — ^ — 7; = cos^"^t? II I
7ismt^ ,.=1 -\ x...9'^
n-l
71 odd, — ^7, = cos^-iO n 1 —
^ n
231. To prove that
We have, by art. 225,
r=n-lf / 27rM
cosh, nv - COS na = 2^-'^ H scosht' — cos(a + '?"- — •
r=0 I ^ '^/J
Let 71a = TT, and therefore a = -,
r=7i-l r 27'+ 1 1
then cosh 71V+ 1 = 2^-1 11 ■{ cosh v — cos -ttK
2 cosh2!|^=2^-i'^rf '|2smh2^ + 2sin2?^x|.
Let ^ = 11, then
r=n-ir 2r-4-l 1
cosh27iu = 22^-2 n jsinh%+sin2-^^7rK
In this identity put u = 0,
then 1 = 2^^-2 H W^^-^^^h
.-. by division, cosh^nu = II J I H ^ -, I
1 «'°^;r'^/
362 PRODUCTS.
And, by a simple transformation,
- , sinh% , „ f, , tanh^ti)
cosh^Tiu = cosh^^it n IH ft-^ I
Gov. — Taking together factors equidistant from the
beginning and end of the product, and observing that
cosh nu is positive, and that each of the factors is positive,
we get the results : —
n even, cosh nu = cosh**u II 1 1 +
r=0
{^ tanh^ 1
n odd, cosh oiu = cosh"i6 11 1 +
^ ^ tanh%
232. To prove that
\nsmhuJ .=, I ^^^,r^^
We have, by art. 225,
r=n-l f / ^_
cosh Tiv — COS 71a = 2*^-1 11 jcoshi; — cosf a + r—
r=o I \ n .
Let a = 0, then
cosh'/i-y— 1 = 2""^ n i cosh i; — cos r — \,
2 sinh2'^'" = 2»^-i^~lf ' J2 sinh2^ + 2 sin2M.
PRODUCTS.
353
Let Q = i^, then
smh2rm = 22«-2 n -^smh^i^ + sin^— [,
— ^-.j = 2^" - 2 n i sinh% + sm^ — \.
\ smh u / r=i L "^i J
In this identity put u = 0, then since Lt. . , — = 7i,
M=0
we get 7i2=22«-2 n isin2-^k
therefore, by division, ( — ^-, — ) = 11 J 1 H 1,
-^ Vti smh uJ r-\ 1 • 2'^7r I
^ n ^
and applying the transformation
^ , sinh^t^ 1 9 /-, , tanh^wN
1+ -2/3 =cosh2i6( 14-— — 2^-)
sm^O \ tan^O /
to each of the factors, we have finally
Cor. — Taking together factors equidistant from the
beginning and end of the product, and observing that
sinh nu and sinh u have the same sign, and that all the
factors of the product are positive, we get the results : —
n even,
n odd,
n _ ,
sinh nu , ^ , ~|,
— r-T— - = cosh"-% n
71 smh u r-i
1 +
tanh^u
tan^ —
tanh^u
smh nn . , tt 1 1 .
— r--r— - = cosh'^-iu n Jl +
71 smh u ^^1 \ ^^^2T2L
n
354 INFINITE PRODUCTS.
§ 3. Infinite Products for the Cosines and Sines of x-
233. To j)rove that
and to find limits between which Fr lies for a given
finite value of r.
By art. 229, Cor., we have, for any even value of n,
Let n6 = x, and let x remain constant while n is in-
definitely increased, and consequently 6 indefinitely
diminished. Then
Lt cos«0 = Lt (cos -)" = 1 ; (art. 190)
n=oo n=oo ^ n/
and, so long as r is finite,
2r+l \2
tan2a
cos nO = cos'
n=co,__o^r-\- 1
-l(2r+l)xj '
hence, cos», = (l-^Xl-3^,j ... (1-^^--^^^)F,
where /', = it/l ^^\(l- *^^'' ^
to (rt— '^) factors.
INFINITE PRODUCTS. 365
We shall now shew that 1> Fr> 1 — > tt— o for all
^ (r-l)7r2
values of r such that (2r + 1)^^ > x.
A
Whatever finite value x may have, we may take r
«j. 2t4-1
such that (2r + 1)^ > x, and therefore —z^ — ir > 6. There-
foie, from and after this value of r, every factor of Fr
h positive and less than unity, and therefore jP,.< 1.
Again,
^->1- --;2rTr + , ,2r+3 +-to(^-r)terms I
therefore, a fortiori,
^ ^ , f tan^e . tan20 , . ("n \ , 1
, /27itan0Vr 1 1 J. fn \^ 1
^'•^- > 1 - (-^^- j l(-2^Tl?+(2^^T3P^- *" W -^) *^^^^|-
tan —
Now, Lt.{ni2ijnQ) = Lt ,x = x',
n=oo n=o3 X
n
^^ (2r+l)2 + (2r + 3)2+---
^(2^2+(2r+2)2+-"
4l(r-l)r^r(r+l)^'"'r
<4
*For if Oi, ttg, ... a^ be positive, and each < 1, then (l-ai)(l-a2)
>!-(«! + a2)> ^^^ therefore (1 -%)(! -a2)(l -a3)>(l -aj + ag)!! -aj)
> 1 - (tti + ttg + a^) ; and therefore, by successive inferences,
(l-ai)(l^a2) ...(l-a^)>l-(ai + a2+...+a^).
356 INFINITE PRODUCTS.
UJ. 11 1_4. \
<4\r-\ r"*"r r+1 J'
1 1
and therefore, <
Hence Fr>l-
4 r— 1
(r-l)7r2
Thus, we have 1 > i^^ > 1 -
(r-l)7r2
And since the latter limit can be made as nearly equal
to 1 as we please by sufl&ciently increasing r, we may
write cosa; = (^1 — ^j(l "svA^ ""svj - «^ ^V-
234. To prove that
sin X (^ x^
=(.-5)(>-^)-('-i.)^.
X
and to find limits between which Fr lies for a given
finite value of r.
By art. 230, Cor., we have, for any even value of n,
=--1
sinnO . ,^ ^ [^ tan^^ ^
nsinO
= cos"-i^ 11 1
-^ 1 tan^^
I n .
-]■
Let nO — x, and let x remain constant while n is
indefinitely increased, and consequently 0 indefinitely
diminished.
. X
sm-
Then Lt (n sin 6) = Lt. x = x,
■n=oo TO=oo ^
n
and Lt. cos^ - ^^ = X^. ( cos - ) =1,
n=oo n = oo ^ "'' •
also, so long as r is finite,
INFINITE PRODUCTS. 357
n \ nl \n J
Hence, ?i^^ = (l-5)(l-^J...(l-^>„
/ tan^e \ / tan^e \
to f— — r-1) factors.
n=oo
iC"
We shall now shew that l>Fr>l ^ for all values
Tir-
of r such that (r + l)7r > x.
Whatever finite value x may have, we may take r such
that (r+l)7r>cc, and therefore —>Q. Therefore,
from and after this value of r, every factor of F^ is
positive and less than unity, and therefore Fr<l.
A • 1P^^ { tan^e , tan^^ ,
n . n
to f ^ — r — Ijtermsl,
, ^. . „ , r tan^e , tan20 .
.-., <.fomori, ^,> 1-1^3^-^, + ^^^-^^+...
^^ f:^ — r — ijtermsl,
_, /?itan0\2r 1,1,
to f^ — r— Ijtermsi.
368
INFINITE PRODUCTS.
Now, £1.(71
n—oo
tan
^tan0)= Lt £C = a3;
n
and
(r+l)^' (r+2)2+--
^r(r + l) ' (r+lXr+2) ' -
i.e.
^ ^+1 • r + 1 r+2^*"'
and therefore, < -•
r
Hence
^'^l-rl^-
Thus, we have 1 > i^^ > 1 — —^.
And since the latter limit can be made as nearly equal
to 1 as we please by sufficiently increasing r, we may
write
235. To 'prove that
and to find limits between which F^ lies for a given
finite value of r.
By art. 231, Cor., we have, for any even value of n,
. ''X^fi , tanh%
cosh Tiu = cosh" 16 ii j ^ -\ iyZTTT
INFINITE PRODUCTS. 359
Let nu — Xf and let x remain constant while n is in-
definitely increased, and consequently u indefinitely
diminished.
Then Lt cosh'^i^ = Lt ( cosh - j = 1 ; (art. 191)
X
n.
and, so long as r is finite,
J. tanh^u
2r + l
_ i^tanh^YI 2n '^ | ^ u s^' ^ | 2x
-^ ^ ^Uan^^n2r+lJ 1(2.+ 1K.
2^ / \ 2?i
Hence, cosh x
where F,
ff-
TT^ JV'^S'VJ"'V'^{2r-l)V.
T^ , ^ , tanh% \ /^ , tanh% \ , fn
2n /\ 2n I factors.
We shall now shew that 1 < Fr<e^''-^^'^\
Since every factor of F^ is positive and greater than
unity, we have at once F^>\.
Again, i^^ < e '^ ^n 2^ j
r_tauh2^ + tanh2u to(r^-r )termsl
.:, a fortiori, Fr<e^ '^'' / \ 2n / J^
i.e. <^^ "" "^ l(2;-+l)2^^(2r+3)2 V2 ' / i
*For if ttj, a^ ... a,„ be positive, then l+ai<l+— V-— + -.af^ inf.,
2.e. <e''i ; and therefore (l + ai)(l + a2)...(l+aw)<e«i+«2+--+«OT,
360 INFINITE PRODUCTS.
tanh -
Now Lt. (n tanh u) = Lt .x = x;
n=oo n=ao X
n
and -^ — 7-TT9+7^ — riv:9+---<
(2r+l)2 ' (2r+3)2 ' •"^4 r-1
Hence Fr< eC'-DTS.
Thus, we have 1< i^^ < e(r-i)7r2.
And since the latter limit can be made as nearly equal
to 1 as we please by suflSciently increasing r, we may
write
cosh a; = (l + -^)(l + pfij^^ + 52^2) • • • «^ W-
Cor. — In like manner it may be proved that
sinhic A , ic^\A , x^ \ /-, . a;^ \_,
where l<Fr<e'''^'',
and therefore = n (l+-?-o).
236. Example 1. — Prove that
J=2 2 4 4 6 6 ^^ ^. . /pallia's Theorem).
2 13 3 5 5 7
We have «_l£^=fl -?Vi _ JlVi _ ^ \ ....
6 \ irVV 2VA 3V2r •
Let 6=^, and therefore - = -,
_1.3 3^ 5/7
2.2' 4.4' Q.Q""'
7r224466 ,..
2133557 -^
INFINITE PROD UGTS. 361
EXAMPLT5 2. — Prove that the sum of the products of the recipro-
cals of the fourth powers of every pair of positive integers is
3847r8
Equating known values of — ^— , we have
u
Taking logarithms, we get
Expanding by the logarithmic series, and equating coefficients of
e^ and ^, we get
r^ I [5^2 ((3)2/
and 2^=4^{-|+l_l_+_|_^^-__|^+^^^^^^
Hence 22-\. A=(4y-4 = 4.^(r + -^"r— 1
=i^.l92,
^|9
2I i = 3847r8
r4 s4
L519
Example 3. — Prove that
1 + 3+6 10
The n^^ term of the series =M^L±1)
(27i+l)4
64V"^ Vi
. 4.71^ + An _!/• 1 1 ^
'8(271+ 1)^ 81(271 + 1)'^ (27r+Ty/"
3G2 INFINITE PRODUCTS.
we we «>s^=(l-|?)(>-|?)(l-^^)....
and coa^=I-,^+,^-....
Equating coefficients of ff^ and ^ in these expressions, we get
V^^2:' b'' 8'
and i . i + 1 .1 + 1.14-. =Z!..
X2 g-^-T-p 5-^^32 -52 42j4'
hence, ^ +1 +1 + ... = (!LY_ gJEl^'^l.
1* 3* 54 V8/ 42|4 96
Therefore the given series = l{(^^ l,)-(^- 1,)}
64\ 12/
Example 4. — To prove that
sin (9 see , . .
— TT— = cos - cos -, cos - . . . aa iwr.
^ 2 4 8 -^
We have sin (9 = 2 cos - sin
2 2
„ cos— i sin-;
2 22 22
^ (9 (9.6'
2COS2,co823sm-
= 2«cos^cos2^,cos|3...cos|sin|,
'''" ,-S=^^^|-^2'-«2-3---2--
Now, Lt. 2''sin -^ =Z^. S • ^= ^'
2" n-=co ^
2"
and consequently
sin <9 ^ e 6 . . .
__=cos - cos^-jCOSga ... ad mf.
FACTORS. 363
Examples XXIV.
1. Prove that, when n is an even integer,
x^-\-l = \x^ — '2.x COS —+\\\x^ — 2x cos f- 1 ) • • •
2. Prove that, when n is an even integer,
a;^_l = (a;2_l)(^ic2-2cccos — + lV£c2_2^cos — + lY..
x(^2~2^cos^^^ + l)
8. Prove that, when n is an odd integer,
x'^ — l={x — V)\x^ — 2ic cos — + 1 jf a?^ — 2fl? cos — + 1 j . . .
4. Write down the factors oi x^-\-\,x^ — l, x^ + \,x^+l,
x' — 1, x^-\-\,x^^ — \; illustrating each case by the
division of the circumference of a circle.
5. Shew that, when n is even,
2 /
cos -Jia = n / 1 ^ , ^
sm" „
zn
n
r=^ - 1
and — -. — = cosa 11 /I-
sin^ —
n
6. Shew that, when n is odd,
n—5
' 2
cos7ia = cosa 11 / I
'■=0 1 sin^'^.
2^1
364 FACTORS.
and fiEL^^J'iTA-^.
7. Shew that
. X sinj 0 + (-^i — 1)— [ = cos 4^ — cos Tif 0 + 2 )•
8. Shew that, if n be odd,
(--l)~2"sin'n,0 = 2"-ism0sin(^+— )sin('0+— v..
xsin(9!> + ^i:^.^).
9. If 71/3 = 27r, prove that
cos a cos(a + /3)cos(a + 2/3). . .cos(a + 7i — 1.8)
(-l)Y nir \
10. Prove that
11. If w be odd, shew that
cos 719!) = ( - 1)^" 2»» - 1 cos (f) cos(d) + '^)cos((f> + — ) . . .
xcos( 0H tt)-
12. Shew that, when n is even,
2^ sm-sin — sm — ...sm^ ^ =1.
71 7i 71 71
13. Shew that, when n is odd,
/ o^^ -^ Stt 57r (7i-2)7r
jjn = 2 2 cos -^ cos ^r— cos ;:i— . . . cos — •
^ 2n 271 271 2n
FACTORS. 365
l^. Find the value of
.7r.27r.87r . (n—V)7r
sm — sm — sm — ... sm^
n n n n
where 71 is a positive integer.
15. Shew that, when n is even,
J^ . 7r . 3x . 5x . (n~l)7r ^
16. If a = -r/ — r^iT> prove that
4(71 + 1) ^
cosec a cosec 5a cosec 9a ... to 7^+1 factors = 2"^2.
T7 -P +1. . 1 1 3 5 7 9 11 , . .
17. Prove that ^ = ^.^.g.g.^.j^...ac^ tT./.
18. Sum the series
l + 32+52 + f2+--^^^V->
and 1 + 34+54+^+... ad inf.
19. Shew that the sum of the squares of the reciprocals
of all positive integers is -^ .
20. Prove that ji-2+^+3ig+^ + ...=g.
„, T, ,, , 3 5.7 11.13 17.19
21. Prove that - = -g^. ^2^ .^g^
22. Find the sum of the products two together of
1 1 i i
12' 32' 52' 72""
23. Find the sum of the products three together of
I i i 1
12' 22' 32' 42 " "
24. Shew that p-^+^-^+...=|^-
366 FACTORS.
25. Sum to infinity the series -^-\--ki — f""!!" + • • • •
^^ „ 2 10 26 50 , . .
, 5 17 37 . . .
^ = ^4' 16' 36 •••^^'^•^•'
then will 4a2-62=4.
27. If ^ = 2n^x> prove that
2"cos Ocos 20 cos 220 ... cos 2«-i0 = l.
28. Prove that
2^ V2 J{2+J2) V(2V(2+V2)) J(2+J{2+J(2^J2)))
TT 2 ' 2 * 2 ' 2
29. Find to n terms the sum of the series
log(l + cos 0) + \og(l + cos I) + log(l + cos I) + . . . .
30. Prove that
(2cos0-l)(2cos20-l)...(2cos2«-i0-l) = ^^^"^"^/.
^ ^ 2 cos 0+1
31. Prove that
(l_tan^f)(l-ta.^|)(l-tanf,)...a^i./. = ii-,.
32. From the identity sin0 = 2sin-sin— ^— , deduce in
succession :
(i.) sin0
^„ ■ . e . e + TT . 0 + 27r . 0+(^-l)7r
= 2^-ism-sin sin ... sm — -^^ ^—
p p p p
where p=2^*;
(ii.) sin0
= 2^ " ^ sin ( sin^ sin^- ) ( sin^— — sin^- ) . . . ;
p\ p p/\ p p)
FACTORS. 367
(iii.) sin 6
= e(l-5)(l-2&)(l-3&)-«'^^™/-
33. A^, A^y A^ ... A^n+i, are the vertices of a regular
polygon inscribed in a circle of radius a, OA^+i is
a diameter, prove that
OA^.OA^.OA^...OAn = a^.
34. A^,A<^,A^...A2n are the vertices of a regular polygon
inscribed in a circle of radius a, 0 is the mid-point
of the arc A-^A^n, shew that
OA^.OA^.OA^...OAn = s/^
a'
35. AB is a diameter of a circle, Qq any point on the
circumference ; if Q^, Q^, Q3 • • • Qn be the points of
bisection of the arcs AQq,AQj^,AQ2 ..., prove that
36. Shew that if A-^A^A^ ... A2n, B^BJB^... B2n be two
concentric and similarly situated regular polygons,
then
PA^.PA^...PA2n-l _ PB,.PB,...PB2n-l
PA,.PA,...PA2n ~PB^.PB,...PB2n '
where P is anywhere on the concentric circle
whose radius is a mean proportional between the
radii of the circles circumscribing the polygons.
37. Prove that
sin 910 = 2" " ^ sin ^f cos ^ — cos — j f cos r/> — cos
X (cos ^ — cos^ ^- 1.
38. From the equation
sin?i0 = 2^^-i'"ff^|sin(0-|-^)|,
n
368 FACTORS,
deduce, when n is even,
(-l)^sin'n,0 = 2»-i'r['Vcos(0+^-^)|,
and'n;{tan(04-3} = (-l)^
39. Prove that
-(''+-l)=7-^(-+?)(-?J(-^S--
40. Deduce the factorial expression for cos0 from that
for sin Q by aid of the formula cos 0 = f> . ^-
•^ 2sm6
41. Prove that 1 —cos a is a factor of
71 sin('?i + 2)a — (371 + 2)sin(7i + l)a + (371 + 4)sin na
— (71 + 2)sin(7i — l)a ;
and find the other factor.
42. If 71 be odd, shew that sin 7i0 + cos 7i0 is divisible
either by sin 0 + cos 0 or by sin 6 — cos 6.
43. For what integral values of n are sin nQ and cos nQ
divisible by cos 0 ?
44. From the formula
sin(7i +1)0+ &in{n — 1)0 = 2 sin nO cos 0,
shew that sin7i0 is divisible by sinO for all
integral values of n.
45. Prove that if 0 = -— and — = 7: ,
271 <p 0 TT
, 2sm^esm^2e...sm%n-l)0_<l)
sin'^0 sin'-^^^ . . . sin2(?i — 1)0 ~ 6'
46. If 04-0 + i/r = 7r, shew that
^^^2n+1^^2n+l.|-^2n-fn
I 271+1 /
n=or (27i+l)V2"*" (27i+l)V (2r^+l)Vr
71 = 00 r
2 (-1)'
CHAPTER XV.
APPEOXIMATIONS.
§ 1. Approximations and Errors.
237. In many theoretical calculations, it is not always
necessary that the result should be perfectly accurate, an
approximate value being sufficient for the purpose in
view. In all calculations based on measurements (which,
however carefully made, cannot be free from error), an
approximate result only can be obtained; and it is
useless giving to that result an appearance of greater
accuracy than the character of the measurements will
allow.
For example, if we wish to find the square of 30001
correctly to four places of decimals, the term ('0001)^ may
be omitted from the equation
(3-0001)2 = 9 + 2 X 3 X 0001 + (0001)2,
and we obtain 90006, a result which differs from the
true value only by -00000001.
Again, if x and y be quantities so small compared
with a that their products and second and higher powers
may be neglected, we have approximately : —
{a + x){a + 2/) = ^2 + a(a; + y),
(a + a^)-..(a + 2/)=l+^,
2 a 369
870 APPROXIMATIONS AND ERRORS.
X _x
a±x~ a
V(a^±a,) = a±^,
cos(a ± ic) = cos a + a; sin a,
sin(a±a;) = sin a±x cos a ;
X in the last two cases being the measure of an angle
in radians.
Definition. — If, in measuring or calculating the magni-
tude of any quantity whose real magnitude is a, an
error x be made, the absolute error is x, and the relative
error is xja.
From the third of the above approximations, it follows
that the relative error may be taken as either - or —7—.
'^ a a±x
238. Example 1. — If A be the length of the chord of a circular
arc, B that of the chord of half the arc, the length of the arc is
approximately equal to {8B-A)ld.
Let 6 be the number of radians in the angle subtended by the
arc at the centre of the circle, a the radius of the circle : then
A = 2a sin _ = 2a( - - -—-\-— — - .
2 V2 48^3840
2 \2 48 ' 3840
and 5=2asinf=m-^ + ^^^-...),
Let \ and /n be determined by the equations
These give X= -1, ^=|, ^^,^X+^,I^= -^,
8B-A /, e^ _^
= /l-
3 V 7680
where I is the length of the arc.
I
APPROXIMATIONS AND ERRORS. 371
Hence, approximately,
;=(8J5-J)/3.
This is known as Huyghens' approximation to the length of a
circular arc. (See also Example xxv. 23.)
The closeness of the approximation may be shewn by finding the
length of an arc which subtends an angle of 45° at the centre of a
circle whose radius is one mile. We have
^ = 2 X 63360 X sin 22°30' = 2 x 63360 X 0-3826834
=48493-64 inches,
5=2 X 63360 X sin 11°15'=2 X 63360 x 0-1950903
= 24721-84 inches.
The values, using Huyghens' approximation, give for the circular arc
a length of 49760-36 inches.
The real length is 63360 x 3-14159265 x J or 49762-83 inches.
The absolute error in this case is therefore a little less than 2^
inches, and the relative error about 1/20,000.
Example 2. — The height of a tower standing on level ground
is determined by measuring a base-line from the foot of the tower,
and the angle of elevation of the top of the tower from the other
end of the base-line. If a small error be made in observing this
angle, to find : (1) the absolute error in the calculated height of the
tower, and (2) the angle of elevation for which the relative error is
a minimum.
Let CB represent the tower, BA the base-line. Let A be the true
angle of elevation of the top of the
tower, and B radians (represented by
CAC) the error made in observing
this angle, the error being either in
excess (as in the figure), or in defect,
of the true angle.
Let a be the true height of the
tower, X the error in the calculated
height due to the angular error 6, c the length of the base-line.
Then, a=ct3iii A, a+x=cta.n{A +6),
"^ :tan(^-f^)-tan^= ^^" ^
c cos(^ -f ^)cos A
= ^sec2u4, neglecting 6'^ and higher powers of 6,
,r=^csecM.
372 A PPROXIMA TIONS A ND ERRORS.
Agaiii) the relative error is xja^ which
^ec&ed^A^ e ^ J'd
c tan A cos A sin A sin '2, A'
Hence, for a given error 6 made in observing the angle of eleva-
tion, the relative error in the calculated height of the tower is least
when sin 2 J is greatest, i.e. when the angle of elevation is 45°.
239. In any right- angled triangle the number of
degrees in the smallest angle divided by 172 is very
nearly equal to the smallest side divided by the sum of
the other side and twice the hypotenuse. (Ozanam's
Formula.)
In the right-angled triangle ABC, let G be the right
angle, and A the smallest angle ; let A be the number of
degrees, and a the number of radians in this angle, so
that a = 7rjl/180 = 3J./l72, approximately.
"N" ^ _ csin J. _ sing
6 + 2c~2c + ccos J.~2H-cosa
= ^, approximately,
^ 2"
= l = lf2^ approximately
This proves Ozanam's formula, when A is not large.
Writing / for the fraction
^(2 + cos^)
sin J.
we see then that, for small values of A, J does not differ
greatly from 172. In the following table, the value of J
is given to three places of decimals for every five degrees
from 0' to 45*' ;^
APPROXIMATIONS AND ERRORS. 373
A.
J.
A.
/.
0°
171-887.
25°
171-923
5°
171-887.
30°
171-962
10°
171-888.
35°
172-026
15°
171-892.
40°
172128
20°
171-.902.
45°
172-279
The degree of approximation may be shewn by solving the
triangle in which 6'= 90°, c=4156, a =2537.
We find 6 = 329r8, and, by Ozanam's formula,
- ^^ 2537x172 ^436364^3^0 3^. ^g.
3291-8 + 4156x2 11603-8
The correct value of A is 37° 37' 17", so that the absolute error in
this case is only 59".
240. Any three elements of a triangle (except the three
angles) being increased by given small amounts, to find
the consequent changes in the other three elements.
Let the increments of a, b, c, A, B, C be denoted by
^y y, z, 0, 0, V^, respectively, 0, (p, and yfr being measured
in radians.
Then a sin B = b sin A,
and (a + x)sm(B + 0) = (6 + y)sm(A + 0).
Now, since all terms of the second and higher degrees
of X, y, z, 0, (p, \lr may be neglected in comparison with
the first,
(a 4- x)sm{B + 0) = (a + x){sm 5 + 0 cos B)
= asin J5 + a;sin5 + 0acos5.
Hence, making use of the first of the above equations,
the second may be written
ic sin 5 + 0a cos B = y sin A-\-Qb cos A.
And, dividing the left side of this equation by a sin 5
374 APPROXIMATIONS AND ERRORS.
and the right side by the equal expression 6 sin J., we get
--h0COt5 = ^ + 0COt2l,
a 0
or --dcot^=^-0cot5.
a b ^
Similarly, from the equation a sin (7= c sin ^, we obtain
6 cot A = — yfrcotC.
|-dcot^=|-0cot^ = ^--V^cota (1)
Again, A+B+C^tt,
and A + e-\-B+(p + G-{-\[r = 7r,
e^<p+xlr = 0 (2)
Hence, if any three of the six quantities x, y, z, 0, <p,
xfr (except 0, <p, yj/) be given, the other three can be
determined by means of the three equations (1) and (2).
There are obviously four cases to be considered, cor-
responding severally to the four cases in the solution of
triangles : — (1) given x, <p, \[r, to find y, z, 0; (2) given
y, z, 6, to find x, <f>,-^; (3) given x, y, 0, to find z, <l),\lr;
(4) given x, y, z, to find 6, 0, xfr. (See Examples xxv.
13-16.)
241. The last proposition may be proved geometrically
as follows : —
Let the triangle ABC be changed into the triangle
A'B'C by changes in any three elements (except the
three angles).
Draw BL parallel to B'A\ AN and BP perpendicular
to FA\ AQ parallel to GA\ and AM and QR perpen-
dicular to GA\
Let LABL = <f), LACM=\lr, and BB' = x; then y = A'M,
a.nd z = A'N+PB'.
APPROXIMATIONS AND ERRORS. 375
A
\
C B B
Since ^ and x/^ are very small angles, we have AL —
C(f) and AM=h\lr', also, if sin(5 + 0) be multiplied by
any of the small quantities conside'red, we may replace
sin(-B+0) by sin J5, etc.
Now, y = MA' = AQ + RA'
^ANco^Qc{A + e) + RQcoi{A+e)
= AL cosec J.+5Pcosec^+J.if cot J.
= c0 cosec A+x sin(5 + 0)cosec J. + 6\/r cot ^,
2/ sin J. = c^ + 6i/r cos ^ + aj sin 5 (1)
Again, z = A'N-^PE = A'Q + QN^PB'
= QRco^Qc{A + e)+ANcoi{A-\-6)+BFco^{B+cl>)
= 6i/r cosec -4 + C0 cot A +£c sin B cot J. +aj cos B,
z^inA — h-yp--\-C(l)Q,o^A-\-xs,m.{A-\-B)
= 6\/r + c0 cos J. + a? sin (7 (2)
Also, as before, 0 + ^ + \/r = O, and, by means of this
equation, equations (1) and (2) may be transformed to
the equations (1). of the preceding article.
376 APPROXIMATIONS AND ERRORS.
Examples XXV.
1. It -~vj = TTTTT. then 0 = 4 24 , approximately.
2. If 0 be nearly a right angle and n>l, then
(sm dr= — , ., , ; Tv-i — 7i, approximately.
3. If 0 be less than a radian, then
^_^/3-3cos0\^
^-"^VS + cos^y *
very approximately, the approximate measure of
the error on the left-hand side being 07^^80 radians.
4. What value should be given to the constant m, in
order that the formula
/I tan^+msin^
^= — THi^: —
shall be the best approximation to the number of
radians in a small angle Q in terms of its sine and
tangent ?
5. If u — esinu = t;, where e is small, shew that, if
powers of e above the first may be neglected, we
have tan| = (l + e)tan|
6. If u = ^ 4- ^ sin n^ where e is small, then
u = 0 + e sin 0 + terms of the second order,
t6 = 0-|-esin0-|-^sin20H-terms of the third order,
u = 0 + e sin 0 + ^ sin 2^ + ^ (3 sin 30 — sin <^)
+ terms of the fourth order.
7. Solve the triangle in which c = 5793, a = 1489, G= 90^
by means of Ozanam's formula.
APPROXIMATIONS AND ERRORS. Til
8. If, in a triangle ABC, ^ = 30°, c=l, and a = 250, find
approximately the number of minutes and seconds
in the other angles.
9. If, in a triangle ABC, c, A and B be given, A and B
being measured in radians and very small, find the
other sides, and shew that
a-\-h = c{l-\-lAB).
10. If, in a triangle ABC, a, h and C be given, when
C='7r — 0, 0 being measured in radians and very
small, find c, A and B in terms of a, b and 6,
11. If a parallelogram, formed by four jointed rods, be
slightly deformed, find the change in its area.
12. In the ambiguous case in the solution of triangles, if
a, h and A be the given parts, and there be a small
error y in the value of b, prove that the error in
either of the corresponding values of c is
y(G cos A — b)
c — b cos A
13. If, in a triangle ABC, a, B and C be increased by the
small quantities x, <p and yfr, find the resulting
changes in b, c and A.
14. If b, G and A be increased by the small quantities
y, z and 0, find the resulting changes in a, B and G.
15. If a, b and A be increased by the small quantities x,
y and 0, find the resulting changes in c, B and G.
16. If a, b and c be increased by the small quantities x,
y and z, find the resulting changes in-^, B and G.
17. If a triangle be solved from the observed parts
(7=57°, a = ;^6, 6 = 2, shew that an error of 10" in
the value of G would cause an error of about 3"'66
in the calculated value of B.
378 APPROXlMATIOyS AND ERRORS.
18. The sides of a triangle are observed to be a = 5, 6 = 4,
c = 6, but it is known that there is a small error in
the measurement of c ; examine which angle can
be determined with the greatest accuracy.
19. Three vertical posts are placed at intervals of one
mile along a straight canal, each rising to the same
height above the surface of the water. The visual
line joining the tops of the two extreme posts cuts
the middle post at a point 8 inches below the top.
Find, to the nearest mile, the radius of the earth.
20. On the top of a spire is an iron cross, the length of
whose arms is a, and whose plane lies east and
west. A person standing due north* observes that
the horizontal and vertical arms subtend small
angles of a and /3 radians respectively. Find the
height of the tower and its distance from the
observer.
21. Ay B, G are three given points in a straight line ; D is
another point whose distance from B is ascertained
by observing that the angles ADB, CDB are equal
and of an observed magnitude 6 ; prove that the
error in the calculated length of BB, consequent on
a small error S in the observed magnitude of Q is
_ 2ah{a+hfs.m 6
{a^-\-b^-2abcos2e)^' '
approximately, where a and h are the distances
between A, B and B, G respectively.
22. The side c and the angles A and B, of a triangle ABC,
are measured, but, on measuring the angles, equal
small errors are made. If the resulting relative
errors of the sides a and b be equal, shew that the
triangle must be isosceles.
APPROXIMATIONS AND ERRORS. 379
23. If A be the length of the chord of a circular arc, B
that of the chord of half the arc, and C that of the
chord of a quarter of the arc, the length of the arc
is approximately equal to
^-40^ + 256(7
45
Find the absolute error in the length of an arc
determined by this formula, the circle being one
mile in radius, and the angle subtended by the arc
at the centre being 45°.
24. Shew that the area of a small segment of a circle is
very nearly f base X height, and that the error is
very nearly Q'^J^O of the area, where 0 is the num-
ber of radians in the angle subtended by the arc at
the centre.
Calculate the numerical value of this fraction
when the angle contains 5°.
25. The sides of a triangle, a, h„ c, are increased by small
amounts x, y, z; shew that the radius of its cir-
cumcircle is increased by
^cotAcotBcotG{xsecA-^y8ecB+z&ecC).
26. If, in a triangle, a, h and B be given, and if the true
value of the angle exceed the measured value B by
-a small angle of 6 radians, the cube and higher
powers of which may be neglected, shew that the
diameter of the circumcircle is
h cosec B[l - 0 cot B + iO^cotW + cosec^B)].
If the square of 6 may be neglected, how is the
third side affected by 0 ?
f
w
380 THEORY OF PROPORTIONAL PARTS.
§ 2. Theory of Proportional Parts.
242. We shall now examine the principle stated in
Chapter IV. for finding the logarithm of a number, a
circular function of an angle, or its logarithm, when the
number or angle lies between two given in the tables.
The approximate nature of this principle, which is
known as the principle of proportional parts, will be
evident from the following geometrical illustration for
the case of the sine of an angle.
Let OL, OM and ON, measured along a line OX, be
^ proportional to the angles 6,
e + S, and e + S\ S and 6' being
-, small. From L, M and N draw
LP, MQ and NR perpendicular
to OX, and proportional to the
sines of 6, Q+S and 6-\-S' ', so
lif N H that P, Q and R are three points
near to one another on the curve of the sine. Draw PK
perpendicular to RN, cutting QM in H. •
Now, according to the principle of proportional parts,
we must have
siD(0+^)-sin 0 : sin^O +S')- sin 6 = 8 : S\
i.e. QH:RK=PH:PK,
i.e., since the angles PHQ, PKR are equal, PQR must
be a straight line.
Hence, all that is implied in the principle in this case
is that if we take two points very near to one another
on the curve of the sine, the part of the curve between
them does not differ from a straight line.
We shall suppose the tables to give the logarithms to
THEORY OF PROPORTIONAL PARTS. 381
the base 10 of all numbers from 1 to 100,000 correct to
seven places of decimals, and the circular functions and
their logarithms to the same number of places for every
minute from 0° to 90°.
243. Logarithms of Numbers. — Let n be any number
and S a number small compared with n. Then
\og{n + ^) - log ^ = log(l + -) = M log.(l + -
S 1 ^2 1 ^
where fj. is the modulus of the common system of loga-
rithms.
Now, if S be so small compared with n that the squares
and higher powers of Sjn may be neglected, we have
\og{n -\-S) — log n = jjiSIn, approximately,
i.e., the change in the logarithm varies approximately as
the change in the number.
In the case in which the principle of proportional parts
is usually applied, we have n<t 10,000 and ^ = 1 ; also
since />t = -43429448, i.e. < |-, we have
^ < 7 • Va9 ^-e- < -0000000025 ;
2n^ 4 10^ '
hence, to at least 7 places of decimals, we have
\og{n-\-§)- \ogn •.\og{n-\-\) — \ogn = S'. 1,
S being <1.
To find the smallest number n whose logarithm can
be obtained correctly to 7 places of decimals by means
of the principle of proportional parts, S being 1, we have
^' = •0000001,
and 91 = 1474, approximately.
382 THEORY OF PROPORTIONAL PARTS.
Again, if the difference between the logarithms of two
consecutive numbers, each containing six digits, be
•0000100, then a difference of 0000001 between the
logarithms of two numbers lying between the former
pair will correspond to a difference of "01 between the
numbers ; i.e., given the logarithm, we can in such a
case find the corresponding number correctly to two
places of decimals. Now, w^hen yu. l/7i= 00001 we have
71 = 43429. Hence, if the logarithm be given, we can
find the number correctly to two places of decimals if
the number be less than 43429, and correctly to one
place of decimals if the number be greater than 43429.
244. Circular Functions. — We shall examine the prin-
ciple of proportional parts fully in the case of the sine,
and briefly in the cases of the other circular functions.
Sine. — We have
sin(0 + (5) — sin 0 = cos ^ sin 5 — sin 6(\ — cos S)
= Scose-iS^sme-i6^cose+....
The ratio of the third term of this series to the first
being — ^<5^ and therefore very small, the third and suc-
ceeding terms may be neglected.
The ratio of the second term to the first is — ^S tan 0.
(1.) If tan 6 be not great, i.e. if 0 be not nearly 7r/2, this
ratio is very small, and therefore the second term may be
neglected in comparison with the first, and the above
equation becomes
sin {6-\-S)— sin 6 = Scos 0, approximately,
i.e. when the change in the angle is very small, the change
in the sine of the angle varies approximately as the
change in the angle. Hence
THEORY OF PROPORTIONAL PARTS. 383
(2.) If, however, 0 be very nearly 7r/2, the ratio
- \S tan 0 may be finite. If this be the case, the second
term cannot be neglected in comparison with the first,
and the above equation becomes
sin(^ + ^) — sin 0 = ^ cos 0 — J^^sin 6, approximately ;
consequently, the change in the sine does not vary as the
change in the angle ; the change in the sine is said to be
irregular.
Again, the second term, though it cannot be neglected
in comparison with the first, is less than the first, for the
greatest value which J(5tan0can have is J^tanf^ — ^j
or J^/tan S^ and this is never greater than J (art. 74) ;
also, Sco^O is vei:}" small compared with 6, since 6 is
very nearly 7r/2 ; hence, sin(0+^) — sin 0 is very small
compared with S, and the change in the sine is said to
be insensible. A small change in the sine will there-
fore correspond to a great change in the angle ; in other
words, several successive angles differing by 1" will have
the same tabular value of the sine, and, consequently,
an angle cannot be found exactly from its sine when it is
very nearly a right angle.
Eeferring to the figure of art. 89, it will be seen that, if POP' be
a small constant angle, the difference between P'M' and PM, and
therefore the difference between sin J. OP' and sin ^ OP, diminishes
as the angle AOP increases, and becomes infinitely small, i.e. insen-
sible, when AOP is nearly a right angle.
245. Cosine. — The cosine of an angle being the sine of
its complement, it follows that, when the change in the
angle is very small, the change in the cosine varies
approximately as the change in the angle, except when
the angle is very small; and then the change in the cosine
becomes irregular ; in this case it is also insensible.
384 THEOR Y OF PROPORTION A L PA RTS,
These results may also be deduced from the equation
cos 0 — cos(^ + ^) = ^ sin 0 + \^co^ 0, approximately.
Tangent. — We have
tan(0-4-^)-tand
_ sin^
""cos(0-f^)cosO
= sec20 tan S(l- tan 0 tan ^) - 1
= sec2a(^+~+...)(l + (5tan0+^tan2a+...)
= 8 sec2^ + ^Han 6 sec^O +S^{i+ t&n^e.jsec^O + . . . .
As before, the third and succeeding terms may be
neglected in comparison with the first. The ratio of the
second term to the first is ^tan^. Hence, when the
change in the angle is small, the change in the tangent
varies approximately as the change in the angle ; except
when the angle is nearly a right angle, and then the
change in the tangent is irregular. It is never insen-
sible, for S sec^O is always > S.
Cotangent. — The cotangent of an angle being the tan-
gent of its complement, it follows that, when the change
in the angle is small, the change in the cotangent varies
approximately as the change in the angle ; except when
the angle is small, and then the change in the cotangent
becomes irregular ; it is never, however, insensible.
These results may also be deduced from the equation
cot 0 - cot(0 + ^) = <5 cosec^e - S^cot 6 cosec^^,
approximately.
Secant— ^We have
sec(0 +S)-sece = S tan 6 sec ^ + ^( J + tan20)sec 0,
approximately, the third and succeeding terms being
small in comparison with those that are retained.
THEORY OF PROPORTIONAL PARTS. 385
The ratio of the second term to the first is (5(JcotO+tan 6).
Hence, when the change in the angle is small, the change
in the secant varies approximately as the change in the
angle, except when the angle is small or nearly a right
angle. If the angle he small, the change in the secant is
irregular and insensible; if the angle be nearly a right
angle, the change in the secant is irregular but not
insensible.
Cosecant. — The cosecant of an angle being the secant of
its complement, it follows that, when the change in the
angle is small, the change in the cosecant varies approxi-
mately as the change in the angle, except when the angle
is small or nearly a right angle. If the angle be small,
the change in the cosecant is irregular ; if the angle be
nearly a right angle, the change is irregular and insensible.
These results may also be deduced from the equation
cosec 0 — cosec(0 + (5) = ^ cot 0 cosec Q — 6\^ + cot20)cosec Q,
approximately.
246. Logarithms of the Circular Functions.— >Sfme. —
We have
log sm(0 + (5) - log sm 0 = log — ^T^^-g—
= log(cos ^ + cot 0 sin S)
= log(l+^cot0-i^2_ __)
= fi{S cot e - i^2cosec20 +...),
fx being the modulus of the common system of logarithms,
and the third and succeeding terms being neglected in
comparison with those that are retained.
The ratio of the second term to the first is
— ^S cosec^^ tan 6 or —S cosec 20, which is small unless
Q be small or nearly a right angle.
2 b
386 THEORY OF PROPORTIONAL PARTS.
Hence, when the change in the angle is small, the
change in the logarithm of the sine varies approximately
as the change in the angle, except when the angle is
small or nearly a right angle. If the angle be small, the
change in the logarithm of the sine is irregular but not
insensible ; if the angle be nearly a right angle, the
change is irregular and insensible.
Cosine. — When the change in the angle is small, the
change in the logarithm of the cosine varies as the change
in the angle, except when the angle is small or nearly a
right angle. If the angle be small, the change in the
logarithm of the cosine is irregular and insensible ; if the
angle be nearly a right angle the change is irregular.
These results may also be deduced from the equation
log cos e - log cos(0 +S) = jjl8 tan 6 + iimShec^e,
approximately.
Tangent — From the two preceding results we have
logtan(O + ^)-logtan0 = logsin(0 + ^)-logcos(^ + ^)
— (log sin 6 — log cos 6)
= juiS(cot e + tan 0) - ifxS^cosec^O - sec^^) + . . .
= 2jj.S cosec 20 - 2iuLS^cosec 26 cot 20+....
Now, cot 20 being great when 0 is small or nearly
a right angle, it follows that, when the change in the
angle is small, the change in the logarithm of the tangent
varies approximately as the change in the angle, except
when the angle is small or nearly a right angle. In both
these cases the change in the logarithm of the tangent is
irregular but not insensible.
Cotangent. — The same results are true for the loga-
rithm of the cotangent.
Also, since
log cot 0 - log coi(0 -\-S) = log tan {0 + S)- log tan 0,
THEORY OF PROPORTIONAL PARTS. 387
it follows that the diifereDces for any small change in
the angle are numerically the same in the logarithms of
the tangent and cotangent of the angle.
Secant and Cosecant. — Lastly, since
log sec(0 -\-S) — log sec Q = log cos 0— log cos(0 + S),
and log cosec(0 + ^) — log cosec d = log sin Q — log sin(0 + (5),
it follows that the results for the logarithms of the secant
and cosecant are the same as for those of the cosine and
sine, respectively ; also that the differences for any small
change in the angle are numerically the same for the
logarithms of the cosine and secant and for those of the
sine and cosecant.
247. If f{0) denote any circular function of an angle 6^ or its
logarithm, we have seen that, in every case,
f{d+^)-f{e)=A^+B^^+...,
where A, B, etc., ar^ functions of 6 but not of 8.
If B8^ be not small compared with A8, the change in the circular
function or its logarithm is irregular. If A8 be small compared
with 8, the change is insensible.
248. When the Angle is given. — In determining the value of any
circular function of an angle or its logarithm, by means of the prin-
ciple of proportional parts, we have to take into account only the
irregularities in its change.
The largest difference-angle with which we have to deal is just
less than one minute, and therefore the irregularity in the change
of the function may affect the seventh place of decimals if B8^ be
not less than '0000001, where 8 is tlie number of radians in one
minute. The limiting value of the angle 6 is given by the equation
B=-ooooooix(15§ooy.
249. When the Circular Function or its Logarithm is given. — The
accuracy of the calculated value of the corresponding angle to the
nearest second depends both on the insensibility and the irregularity
in the change of the function.
388 THEORY OF PROPORTIONAL PARTS.
The change in the function will be insensible for two angles
differing by one second, if Ah be less than '000000 1, where S is the
number of radians in one second. The limiting value of the angle
6 is given by the equation
^ = •0000001x548000
IT
The calculated value of the angle may, on account of the irregu-
larity in the change of the function, differ from the true value, if
ji?82 be not less than -^ of ^8, where S is the number of radians in
one minute, this being just greater than the largest difference-
angle with which we have to deal. The limiting value of the
angle 6 is given by the equation
i?_ 10800
A GOtt*
Miscellaneous Examples. III.
a.
1. If ^+5+ C=7r, prove that
sin3^ sin(5 - C) + sin35 sin((7- ^) + sin3(7 sin(^ - 5) = 0.
2. If^+5+a=f, prove that
cosec A cosec B cosec G—cotB tan G-cotC tan B
— cot G tan ^ — cot ^ tan G— cot A tan 5— cot B tan A = 2.
3. If.4 + 5+a+i) = 2'7r, prove that
cos J^ cos JD sin J5sinJC— cos JJ5cos JCsin hA sin JD
= sin i(A-{-B)sin ^{A + C)cos i(A +Dy
4. Prove that
S cos 2a sin(/3 - y) + 2 sin(;8 - y) . S cos(/3 + y) = 0.
5. Prove that
S cos 2a cot J(y - a)cot J(a - /3) = 2 cos 2a + 22 cos(/3 + y).
6. Prove that
2 sin ^ sin y sin(/? — y)sin(3a+/3 + y)
+sin(a+/3 + y) .nsin(^-y) = 0.
MISCELLANEOUS EXAMPLES. 389
TT 1 1
1. Prove that ^ = 2tan-^^ + tan-ij=.
2. Prove that •-- = tan-i^ + tan-i-+tan~i-.
3. Prove that :r = 2 sin" ^—7-rr — sin ~i-
4. Provethat2 tan~i- — tan-i^ = j,2tan-ij, + tan-^^ = 2-,
5 1 TT
2tan-i— — tan-i^^=— , and, generally, that
2tan-i^"-' + (-irtan-i— = ^, where l,^\-^2
are successive convergents to ^2.
5. Prove that
tan-^=,=tan"i7i-tan-^K.tan-^-7Tr=tan-^— r-tan-^Ts,
7 2 3' 41 12 17
tan"^^^ = tan-^j=^ — tan-i^, and, generally, that
tan-i — = tan-^ tan~^ , where 1, —,—,...
are successive convergents to ^2.
6. Prove that tanh-^-^ = tanh-i^ — tanh-i->,, tanh'^Tr;^
17 0 12' 99
= tanh-i„ — tanh-i— , and, generally, that
tanh-i = tanh-^ tanh"^ — , where 1, — ,
P-2n+l q2n 'p2n ?i
— ,... are successive convergents to ^2.
390 MISCELLANEOUS EXAMPLES.
y-
1. If in a triangle the median which bisects the base c
is perpendicular to the side 6, then
2tan^ + tana=a
2. In any triangle S- + 6i2=2i2. 2—.
T-^ Ob
3. If a, P, y be the lengths of the lines joining the feet
of the altitudes of a triangle, then
a- 62+^2- 2a6c *
4. Shew that the line joining the middle point of BG to
the middle point of the perpendicular from A on
BG, makes with BG an angle whose cotangent is
cot B <- cot G.
5. Find the inclination of the line joining the centres of
the inscribed and circumscribed circles of a triangle
to its base.
6. If straight lines be drawn through the vertices of a
triangle, bisecting the exterior angles, and if A be
the area and P the perimeter of the original
triangle, and A', P' the corresponding quantities
in the new triangle, shew that
4AA' = Pa6c,
and PP' = 4AY cos ^ + cos ^ + cos ^ j.
8.
1. In any quadrilateral figure whose diagonals intersect
at right angles, if S and D be respectively the
sum and difference of two opposite sides, and S\
ly the sum and difl'erence of the other two oppo-
site sides, then
MISCELLANEOUS EXAMPLES. ^91
2. In any triangle, whose perimeter 2s is given, the value
of Rr^^^ is greatest when the triangle is equi-
lateral, and is then equal to ^.
3. A polygon of Zn sides, which are a, b, c successively,
repeated n times, is inscribed in a circle; if the
angular points be A, B, G, D, E, etc., and the
radius of the circle be denoted by r, prove that
AC'^ = I ac + 2hr Hm-]U}C + 2ar sin -]-r-(ab-\- 2cr sin- j,
with similar expressions for BD and CB.
4. The circumference of a circle is divided into twelve
arcs, whose lengths, taken in order, are in arith-
metical progression, and the first six together form
a quadrant. Shew that the area of the polygon
bounded by the chords of the twelve arcs is
— cosec fz-x, where a is the radius of the circle.
o 72
5. If the diameters AA\ BB\ CC of the circumcircle of
a triangle ABC cut the sides in D, E, F respec-
tively, then
ad^be^cf~b:
^^^ 11^^ WE'^ GT^ ¥R ^^^^ '^ ^^^ ^ ^^^ ^^ ^^'
6. A flagstafi" on the top of a tower is observed to subtend
the same angle (a) at two points in a horizontal
plane on the same line through the centre of the
base, whose distance from each other is 2a, and
an angle /3 at a point half way between them.
Find the height of the flagstaff.
MISCELLANEOUS EXAMPLES.
1. One of the aogles of a plane triangle is 60°, and the
sides including it are in the ratio of 3 to 5 ; find
the tangents of the other angles.
2. A, B, G are three points in a straight line such that
AB and BG each subtend an angle of 80° at a
point P. If AB = a, BG=c, shew that the differ-
ence between AP and GP is , ->
3. From a point P in the side AG of a, triangle ABG, a
line is drawn bisecting the triangle and making
an angle 6 with AG, shew that
2AP^
cot 0 = -Tn Tri COSCC A—COtA.
AB .AG
4. The alternate angular points of a regular pentagon are
joined by straight lines. Find the length of a side
of the pentagon formed by these lines, and shew
that the radius of the circle circumscribing it is
as/^ — ^ls/^f where a is the side of the original
pentagon.
5. Through the angular point C of a triangle ABG a
straight line GPQ is drawn, on which are let fall
the perpendiculars AP, BQ ; prove that
PQ = AP cot B~BQ cot A.
6. A regular pentagon and a regular hexagon are in-
scribed in a circle, so as to have an angular point
in common, and the other adjacent angular points
are joined ; shew that the perimeter of the figure
so formed is 4rsin 18° sin 15°cosec3°, where r is
the radius of the circle.
MISCELLANEOUS EXAMPLES. 393
t
1. A circle whose centre is I and radius r is inscribed
in the triangle ABC, and touches the sides in
D, E, F. Circles whose radii are r^^ r^, r^, are
inscribed in the quadrilaterals AEIF, BFID^
CD IE; shew that
2. In an isosceles triangle a series of circles is inscribed,
the first of which is the inscribed circle of the
triangle, and the others touch the preceding one
and the two equal sides of the triangle. If the
sum of the areas of the circles is equal to four-
thirds of the first, find the ratio of the sides of the
triangle.
3. A triangle is divided into two parts by a line through
one of the angular points A, such that the circles
inscribed in the two parts touch the dividing line
in the same point. Shew that 0, the inclination of
the dividing line to the opposite side, is given by
the equation
cot (p = jf tan ^ ~ tan ^^j.
4. A plane polygon whose sides are a^, a^, a^..., and area
A, is divided into triangles by joining its angular
points with a point, at which the sides subtend
angles 0-^, 0^, 0^ If the centres of the circum-
circles of these triangles be joined in order, the
area of the polygon thus formed is
i{A-i^{a^cote)}.
894 MISCELLANEOUS EXAMPLES.
6. From a point A outside a circle two lines of equal
length AB, AC, drawn to the ends of a diameter
BG, and the circumference m D, E -, if BE, CD
meet in 0, the area of the quadrilateral
ABOC= ^BC^cot BA C. Compare that part of the
area of the circle which is outside the triangle
ABC with that which is included within it.
6. PX and PY are two fixed right lines meeting at an
acute angle P. On PX two fixed points B and C
are taken. Shew that, if A be the point on PF at
which BC subtends the greatest angle possible,
and that when P is a right angle, this reduces to
PC-PB
sin B AC:
PC+PB'
1. If a = -y- and x = cos a, shew that 8x^ + 4!X^ — 4a3 — 1 = 0,
and that the other roots of the equation are cos 2a
and cos 3a.
2. If a = ~ and x = sma, shew that a^-'^x^ + ^ = 0,
and that the other roots of the equation are sin 2a
and sin 4a.
3. If a = -i^, then cos a + cos 2a + cos 3a = — J,
cos a cos 2a + cos a cos 3a + cos 2a cos 3a = — J,
cos a cos 2a cos 8a = J.
MISCELLANEOUS EXAMPLES. 395
27r
4. If a = ^, then sin a + sin 2a + sin 40 = 1^^7,
sin a sin 2a -f- sin a sin 4a + sin 2a sin 4a = 0,
sin a sin 2a sin 4a = — Jx/'^-
27J-
5. If a = -=-, then cos^a + cos22a + cos^Sa = f ,
sin^a + sin22a + sin24a = |.
6. If ABGDEFG be a regular heptagon inscribed in a
circle of unit radius, then will
e.
1. Prove that one solution of the equations
x^ = a^ + ay, y^ = a^-\-xy
X y a
sin — sm -y sm y
and find the other solutions.
2. Prove that the equations
ay+a^ = x^, xz-\-a^ = y^, yz + a^ = z^,
are satishea by -. — rr- = -r-^^ = — — t- — - — ,
'' sm 2a sin 3a sm 4a sm a
when c(=q, and solve the equations.
8. Shew that one root of the equation ^x^ — 4a;2 — 4;^? + 1 = 0
is cos -=, and find the other roots.
4. Shew that sin = is a root of th'e equation
x^ + ^x''-^^ = 0,
and find the other roots.
396 MISCELLANEOUS EXAMPLES.
6. If o. = jxy then will sin a + sin 13a = —J,
and sin a sin 13a = — J.
6. Prove that
(a: — 2 cos -R- )( ^ — 2 cos -r- )( !» — 2 cos -v )( ^ — 2 cos -^j
=a;H2a;8-a;2_2a;+l.
1 . Prove that cc = tanh X'\-\ tanh^a? + ^ tan \\^x + ...ad. inf.
2. Prove that 2(cos ^ + J cos^^ + \ cosM + . . .)
= cos2— - sin2^ +-(^cos*2 - sill 9 y
■^iG^^'2 "^'"'2 ) + ••••
3. Prove that 2«cos Q cos 20 cos 2^0 .. . cos 2*^0
= cos 0 + cos 30 + cos 50+ ... +cos(2'^+i - 1)0.
4. Find the sum of n terms of the series
log(l + cos 0) + \og\l + cos 2) + \og{\ + cos^aj + • . • •
5. Prove that
( j = 1- sin^Jic-cos^Ja? sin^Jaj-cos^liccos^Jicsin'^Ja;-....
6. Sum to infinity the series
1. Tr.l, TT.l, TT,
22^^^22'^23 23"^2^ 2'*"^*'"
X.
1. If /i be positive, and 0 < 0 < ^, then
sin0 sin(0 + ^)
~0~ ~0+r"'
MISCELLANEOUS EXAMPLES. 897
. . x^
2. Find the limit of , when x = 0.
1 — cos mx
3. Find the limit of ^^ , when Q = 0.
4. If the unit of measurement be a right angle, find the
limit of ^3 , when 0 = 0.
5. Find the limit of ^^ -, when x = 0.
SlW^X
6. Three mountain peaks A, B, G appear to an observer
to be in a straight line when he stands at each
of two places P and Q in the same horizontal line ;
the angle subtended hy AB and BG at each place
is a, and the angles A QP, GPQ are (p and \[r,
respectively. Prove that the lieights of the
mountains are as
cot 2a+cot i/r : J(cota+coti/r)(cota+cot0)tana : cot2a+cot0,
and that, if QB cut AG in D,
AG= GD . sin 2a(cot a + cot i/r).
PART III.
COMPLEX QUANTITY.
"Every combination of symbols can be explained, and everything
explicable is a line of definite length and direction, and every such
line can be represented byp + gV-1." — De Morgan.
CHAPTER XVI.
COMPLEX NUMBERS.
250. On the Representation of Positive and Negative
Numbers by Straight Lines. — We know that if lengths
measured along a straight line, from a point in the line as
an origin, be denoted by positive numbers, then lengths
measured from the origin along the line in the opposite
direction will be denoted by negative numbers.
Conversely, all positive numbers may be represented
by lengths measured in one direction along a straight line,
and all negative numbers by lengths measured from the
same origin in the opposite direction. «
In the same manner, quantity of any kind may be
represented by lengths measured along a straight line by
taking a unit of length to represent a unit of the quan-
tity considered. Thus, if angles are the quantities con-
sidered, a radian may be represented by an inch, and
398
COMPLEX NUMBERS. 399
radians may then be added or subtracted by operating
upon their representative lengths.
All real number or quantity may accordingly be repre-
sented by length measured along a straight line ; and, in
the case of quantity which can be conceived as existing
in opposite conditions, if one of these conditions be repre-
sented by lengths along the line in one direction, the
other condition will be represented by lengths measured
in the opposite direction.
In some cases these opposite conditions of quantit}?- are
inconceivable, for example, a negative number of lbs. of
matter, or a negative number of ergs of energy are in the
fullest sense of the word ' impossible ' quantities.
251. On the Direction of Number. — Hitherto, a nega-
tive number has been obtained by measuring a length
from the origin in the negative sense of the straight line
along which numbers are represented.
The same result may be obtained by measuring the
same length from the origin in the positive sense, and then
rotating the length about the origin through two right
angles in a plane containing the line ; thus, a negative
number is the positive number equal to it in magnitude
turned through an angle of two right angles. We may
accordingly substitute for the symbol — a the equivalent
symbol {a, tt), where a denotes the magnitude only of the
number and tt the number of radians through which
rotation has taken place.
This mode of representing negative numbers suggests
an extension of the idea of number. From the origin we
may take a length along the line of positive number, and
* then rotate the length about the origin through any given
400 COMPLEX NUMBERS.
angle. The line thus obtained is called a vector, the
number it represents a complex number ; the magnitude
of the line is called the modulus of the number, and the
angle through which rotation has taken place the ampli-
tude of tlie number. The vector or number is fully
denoted by the double symbol (a, a), where a is the
modulus, and a the amplitude of the number.
The modulus of a given complex number a; is a one-
valued quantity and is denoted by mod(a;). The ampli-
tude is many-valued, its values forming a series of angles
of constant difference 27r, and is denoted by Amp(.'z;).
The principal value of the amplitude is that value which
is greater than — tt and not greater than tt, and is denoted
by amp(aj).
It will be observed that
Amp(aj) = 2n7r + amp(aj),
where n is any integer, positive or negative.
The successive values of the amplitude obtained by
assigning to n in the equation
Amp(cc) = 2n7r + amp(a;)
the values 1, 2, 3,... are called the 1st, 2nd, 3rd, ... positive
values, while the values obtained by putting n— —1,
— 2, —3,... are called the 1st, 2nd, 3rd, ... negative values.
Positive and negative numbers are special cases of com-
plex numbers, e.g., -|-3 = (3, 0) or (3, 27r) or, generally,
(3, 2n'7r) where n is zero or any positive or negative
integer ; and — 5 = (5, tt) or (5, 2n+l . tt) where n is zero
or any positive or negative integer. Thus,
mod( + 3) = 3, Amp(-|-3) = 27i'7r, amp(-f-3) = 0,
and
mod( — 5) = 5, Amp{ — 5)={2n + l)7r, arap( — 5)=^.
COMPLEX NUMBERS. 401
252. The Addition of Complex Numbers.
OZ, OB represent any complex numbers
{a, a), (h, P) respectively, and if from
A a line AC he drawn equal to OB,
and inclined to OX the line of positive
number (called the primary axis) at
the same angle as OB, then the com-
plex number represented by 00 is 6 X
called the sum of the complex numbers (a, a), (b, ^) ; or,
if (c, y) be the number represented by 00, then
{a,a)+(6,/3) = (c,y).
By making a and /3 equal to 0 or tt we obtain the sum
of two positive or negative numbers as defined for arith-
metical or real algebraical quantity.
253. The order of addition of two complex numbers
is indifferent.
Complete the parallelogram OAGB (see fig., art. 252),
then, by Eucl. I. 34, BC is equal and parallel to OA, there-
fore, by the definition of addition, the sum of OB and OA
is also represented by 00 ; hence
OA + OB^OB + OA,
i.e., the Commutative Law in Addition holds for complex
numbers.
The addition of two complex numbers may be repre-
sented in three ways : —
(1.) 0^ + 0^ = the third side of a triangle whose other
sides taken in order are equal to OA
and OB in magnitude and direction ;
(2.) OA + 0^ = the diagonal through 0 of the parallelo-
gram whose sides are OA, OB.
402 COMPLEX NUMBERS.
(3.) Oil + Oi? = twice the complex number represented
by the median through 0 of the
triangle OAB.
254. The Multiplication of Complex Numbers.— In
considering the addition of complex numbers, it was
necessary to extend the definition of the operation of
addition ; for multiplication no such extension is required.
We employ the following definition used in arithmetic and
real algebra :
Def.— To multiply one number by a second, we do to
the first what is done to unity to obtain the second.
255. With this definition we can prove that The 'pro-
duct of two complex numbers is a complex number whose
modulus is the product of the moduli of the factors, and
whose amplitude is the sum of the amplitudes of the factors.
Let (a, a) and (b, /3) be two complex numbers, then
shall (a, a)x(6, /3) = (a6, ^+;g).
To multiply (a, a) by (6, /3) we do to (a, a) what is done
to unity to obtain (b, /B).
Now, to obtain (b, (3) from unity, we multiply the unit
by the number 6, and rotate the resulting length through
an angle ^.
Hence, to multiply (a, a) by (b, /3) we multiply the
length of (a, a) by the number b, thus obtaining a length
ab, and then rotate the length ab through an angle (3
from its former direction, thus finally obtaining a length
ab making an angle a + /3 with the primary axis, i.e. a
complex number whose modulus is ab, and whose ampli-
tude is a + /3.
Hence, (a, a) X (b, /3) = (ab, a + ^).
Cor. (1, a)xa = (a, a).
COMPLEX NUMBERS.
403
256. The order of multiplication of two complex
numbers is indifferent.
We have shewn that
(a, a) X (h, /3) = (ah, ^+2) (art. 255)
and that (6, /5)x (a, a) = {ba, 13 + a) „
Now, ah = ha SLYid a+/3 = ^-ha,
therefore (ah, a + /3) and (ha, /3 + a) are one and the same
complex number, and therefore
(a, a) X (h, 0) = (b, /3) X (a, a),
i.e., the Commutative Law in Multiplication holds for
complex numbers.
257. The Distributive Law in Multiplication. — To
prove tha^t
(u-\-v)xw = uxw+vxw,
where u, v, lo are complex numbers.
Let OA=u, AB = v,
then OB = u + v.
Let m = mod(w), 0 = amp(^t;).
In OA take a length Oa equal to
m. OA, and from a draw ah parallel
to AB and equal to m.AB; then, by
similar triangles OAB, Oab, Ob is
in a straight line with OB, and the
length Ob = m. OB. Euel. YI. 6.
Next, rotate the triangle Oab' about 0 without change
of size or shape through an angle 0 to the position OA'B\
then, by the definition of multiplication, we have
OF'='OBx(m,e) = OBxw,
UT = qA X (m, 0) = OA X w,
and A7B' = ABx(m,e) = ABxw.
But 6B = 0A'+A'B\ (Def. of Addition)
404
COMPLEX NUMBERS.
OBxw=OAxw+ABxw,
or (u+v)xw=uxw-{-vxw,
i.e., the Distributive Law in Multiplication holds for
complex numbers.
258. The Grouping of Complex Numbers in Addition
iC and Multiplication. — Let u, v, %v
be complex numbers ; then shall
{ii+v)-\-w = u-\-{v+iu)
and {uxv)xw = ux{vxw).
(1) Let OZ, AB, BG represent
the numbers u, v, w respectively,
then,
{u+v)+w = (OA-\-AB)-^BG
(Def. of Addition)
O X
= OB+BG
and u-h(v-hiu) = OA-{-(AB-\-BG)
= qA+AG
= 00;
(u-\-v)+w = u+(v+w).
(2) Let a, a be the modulus and amplitude of u; b, ^
of V ; and c, y of w.
Then (uxv)xw = [(a, a) X (b, /5)] X (c, y)
= (ab, ^+^) X (c, y) (art. 255)
= (abc,a + /3 + y),
and ux{vxw) = (a, a) x [{b, /3) x (c, y)]
= (a, a)x(bc,/3+y)
= (a6c, a + /3+y).
Therefore {uxv)xw = ux(vxw).
Thus, the Associative Law in Addition and Multipli-
cation holds for complex numbers.
COMPLEX NUMBERS. 405
259.' Conjugate Complex Numbers. — Def.— If two
complex numbers have equal moduli, and amplitudes
of equal magnitude and contrary sense, the complex
numbers are said to be conjugate to one another.
Or, in symbols, (a, a) and {a, — a) are conjugate com-
plex numbers.
Addition of Conjugate Numbers.— Let OA, OA' re-
present two conjugate complex num- ' ^^a
bers {a, a), (a, — a); join AA', and
let AA' meet the primary axis in N,
then, by elementary geometry, ON ^
bisects AA' at right angles.
Since N is the mid-point of AA'^ a!
we have '0A + 0A'=20N, (art. 253)
(a, a) -f (a, — a) = (2(x cos a, 0) = 2a cos a.
In like manner, it may be shewn that
(a, a) — {a, —a) — {a, a) -|- (a, tt — a) = 2la sin a, |^J.
Multiplication of Conjugate Numbers. — We have
(a, a) X (a, — a) = (a^, a — a) = (a^, 0) = a^,
or, the product of two conjugate numbers is a number
whose modulus is the square of the modulus of each of
the numbers, and whose amplitude is zero.
If a = l, we have
(1, a)x(l,-a) = (l,0) = l,
or conjugate complex numbers of unit modulus are re-
ciprocal to one another.
260. Powers and Roots of Complex Numbers.—
Powers. — By art. 255 we have
(a, a)x(6, ^) = (a6, ^+^),
hence, by repeated application of the rule, we get
(ttp ai) X (a^, ag) X . . . X (a„, an) = (a^a^ ...an, ai + a2+...+a„).
406 COMPLEX N UMBERS.
Let the moduli a^ a^,-.. an be each equal to a, and the
amplitudes aj, 02... an be each equal to a, then we get
(a, a)" = (a^'r^a),
i.e. the nih. power of a complex number is a complex
number whose modulus is the nth. power of the modulus
of the original number, and whose amplitude is n times
the amplitude of the original number.
Roots.— Since (a, aY^ia"^, no) it follows that (a, a)
is an nth. root of (a**, no), or, putting r for ct", and 0 for
na, and denoting the arithmetical nth root of r by Jijr,
we see that
(^r, - j is an nth. root of (r, 0).
Again, since (r, 0) = (r, 0+2m7r), where m is any in-
teger, we see, further, that
(jijr, -) is an nth root of (r, 0).
By giving m the series of values 0, 1, 2, ...n — 1 we get
as nth roots of (r, 0) n complex numbers no two of which
have the same amplitude, hence n distinct nth roots of
any complex number exist.
We may assign to m integral values other than those
of the series 0, 1, 2 ...n—1, but we shall not in this way
obtain any additional roots. For, if we put m = rn+8
where r is any integer and s an integer of the series
0, 1, 2...n-l then
[:^r, —^-) = [^r, -^^+ 2rxj = [^r, -^^),
Thus the nth root of a given complex number x is an
-^-valued quantity. The many-valued nature of the root
is indicated by using double brackets, thus ((x)y.
COMPLEX NUMBERS. 407
We define as the 'principal nih root of x that root
whose amplitude is the nth. part of the principal value
of the amplitude of x, and we denote this principal
root either by {xY or a;" or ^cc.
Fractional Indices. — Let n = -, where p and a are
positive integers prime to each other.
Then, defining the -th power of a number as the pth
power of the ^th root of the number, and using double
brackets as before to denote the many-valued-ness of the
fractional power, we have
{{r,6)Y = {{{r,e)ff
If 0 be the principal value of the amplitude of (r, 0) we
p
get the principal value of ((r, 0))q by putting m = 0, and
the 1st, 2nd, 3rd,... positive and negative values by
putting m = l, 2, 3, ... or —1, —2, —3, ... respectively.
We shall obtain the same values of ((r, 0))q, but, in
general, in a different order, by giving to m the series of
values 0, 1, 2,...q — l in the expression l^^^,— —),
provided that - is in its lowest terms.
Square Roots of - 1. — The two square roots of (1, tt)
are (l, |) and (l, -g.
Now, (1, 7r)=-l,
therefore the square roots of — 1 are the complex numbers
408 COMPLEX NUMBERS.
f 1, ~j and f 1, — r), tlie former being the principal value
of the square root, aijd in consequence that denoted by
The symbol i is used as an abbreviation for the com-
plex number f 1, ^j or \/—l.
Since f 1, ^j and f 1, —-) have equal moduli and
opposite directions,
0'-|)=-(^-|>
thus, the two square roots of —1 are i and —i where
Cube Roots of 1. — The three cube roots of (1, 2?i7r)
-(i.o).(i,^)(i,-|).
Therefore the cube roots of 1 are 1, (l,-^) and
/ 27r\ . . . \ ^/
(l, — S-), the first being the principal value of the cube
root, the second the first positive value, and the third
the first negative or second positive value of the root.
If we use the symbol o) as an abbreviation for the com-
plex number f 1, -^ V then f 1, -^j =0)^, and the cube roots
of unity are 1, w, w^.
It will be observed that (a))22 = (l, ^) = (l, ^) = co;
thus either of the roots co, or' is the square of the other.
COMPLEX NUMBERS. 409
Negative Indices. — Let n= —m where m is positive.
Then, defining a negative power of a number as the recip-
rocal of the corresponding positive power of the number,
we have
(a, aY = (a, «)""*= --i^=^ — ^ — -,
^ ^ ^ • ^ {a,aT' (a"*, ma)
= (^^>-^a) (art. 259)
= (a", na).
261. The Resolution of Complex Numbers. — Jl ^2/
complex number may he expressed as the sum of two
complex numbers having given amplitudes not differing
by a multitude of x.
Let OP = any complex number B^
(r, 0), and let OA, OB make any
given angles a, /3 with the primary
axis OX.
Draw PN parallel to OB to meet q
OA in iV, then
OP=ON+NP,
i.e., (r,e) = m',a) + (NP,fi}.
It may be shewn that ON or mod (OiY)= . }^ — -^,
•^ ^ ^ sm(/3-a)
and that NP or mod(:^)=:^^l'^^'^.
^ ^ sin(/3 — a)
Let a = 0, /3=Q, so that 0^ coincides with OX the
primary axis, and OB coincides with a line OY, making
an angle ^ in the positive sense with OX (called the.
secondary axis), and let x, y denote the lengths ON, NP ;
410 COMPLEX NUMBERS.
then OP=ON+NP,
or . (r, e) = (x, 0)-\-[y, '^) = x+iy
= rcos 6 + i. rsin 0
= r{cofi 0+i sin 0).
Def. — If a complex number (r, 0) is expressed in the
form x-\-iy, the term x is called the real part, and the
term iy the imaginary part of the complex number (?', 6).
The expressions " real part " and " imaginary part " are
to be regarded as conventional expressions only. Con-
sidered as abstractions, all numbers, positive, negative,
and complex alike, which obey their definitions and laws
of combination, are equally real, or, if we will, equally
imaginary.
Considered as applied to things that can be counted,
or to quantity that can be divided into parts that may
be counted, positive numbers are always real ; when the
quantity is of a nature such that it can be conceived as
existing in opposite conditions, negative numbers are
real ; and, when the quantity has the attribute of direc-
tion, complex numbers are real. Thus (3, ^] feet-per-
second, 5^ poundals, or an impulse denoted by a + ib
units of impulse, are as real expressions as 10 shillings,
or 30 miles-per-hour.
Further, when applied to quantity having direction,
such as distance, velocity, acceleration, force or mo-
mentum, complex numbers furnish the direct and
complete representation of the object, and the results
obtained by the study of complex numbers regarded as
abstractions may be transferred to concrete quantity
having direction in the same manner, and with the same
COMPLEX NUMBERS. 411
degree of confidence, as is the case when the properties
of arithmetical numbers are applied to examples dealing
with concrete magnitude.
The term "complex number" is often used to denote
a mixed quantity a-\-h\/ — \ partly real and partly
imaginary, the sign x/ — 1 and the sign of addition being
unexplained .symbols subject to the laws of real algebra.
In the treatment here adopted the word "complex" refers
to the double nature of the number, i.e., its modulus or
arithmetical magnitude and its amplitude or directional
magnitude. The mode of representing a complex number
as the sum of two parts, one along the primary axis (the
real part), the other along the secondary axis (the imagi-
nary part), is for many purposes of the greatest value ;
but it is to be regarded as one among an infinite number
of similar modes of resolving a complex number into
component parts.
Equivalent Forms of Results. — The following pairs of
equivalent forms of results already established are to be
noted : —
(l,a)x(l,/3) = (l,i+;8),
(cosa + isina)(cos^ + ^sin/5) = cos(a + /3) + ^sin(a+/3);
(1, a)x(l, -a) = l,
(cos a + i sin a)(cos a — -i sin a) = 1 ;
(1, a)« = (l,^a),
(cos a + '^ sin a)'* = cos na + i sin iia (n integral) ;
ct + 2r7r^
((1, «))"- = (l,
n
((cos a + ^ sm a)r = cos h i sin ,
where '^^ is a positive integer, and r = one of the numbers
0, 1, 2...^^i^l;
412 COMPLEX NUMBERS.
((l,a))-(l,^^±f^>
((cosa + isina))' = cos^«+^-^""+isin^"+^-^"",
where 7' = any one of the numbers 0, 1, 2, ... g^ — 1.
Demoivre's Theorem. — The statement that, when n is
any real number, positive or negative, integral or frac-
tional, cos 710+'?^ sin 710 is a value of (cos 0 + -^ sin 0)**, is
known as "Demoivre's Theorem." The theorem was
given by Demoivre in the form
where a? = cos 0, Z = cos nO.
262. Trigonometrical Formulae derived from the
properties of Complex Numbers. — In the demonstra-
tions of the preceding articles we have assumed the Com-
mutative and Associative Laws for the addition and
multiplication of real quantities (ab = ba, a + l3 = P+a,
art. 256, and ab.c = a .be, a + /3 + y = a+j8 + y, art. 258),
the theory of parallel straight lines, and of similar
triangles. It should be noticed that Eucl. I. 47 has not
been used. The theorem that the sides of a triangle are
proportional to the sines of the opposite angles, employed
in art. 20 1, is an immediate consequence of the definition
of the sine of an angle.
Without further assumption we can derive all the
fundamental properties of the circular functions. From
the relation (1, a) x(l, — a) = l we get
(cos a -h "^ sin a) (cos a - i sin a) = 1,
and, therefore, distributing the product on the left hand
side, cos^a -|- sin^a = 1.
COMPLEX NUMBERS. 413
From (1, a)x(l, /3) = (1, a + ^) we get
cos a cos j8 — sin asin/5 + i(sin a cos /5 + cos a sin ^)
= cos(a + )5) + ^sin(a+)8),
and, therefore, since a complex number can be resolved in
two given directions in one way only, we have
cos(a + jS) = cos a cos /3 — sin a sin /3,
and sin(a + /3) = sin acoS|8 + cosasiny8.
Similarly, from the continued product
(l,a)x(l,^)x(l,y)X...X(l,X)=(l,a-|^+7+TrTX),
we get the general formulae
COS(.ll + ^2+---+^„)=0„-2a_2^2+--->
sin(^ + ^2-f...+^n) = 2>SfiOn-l-2lSf3C«-3+....
And from the equation
(cos a-{-i sin a)'^ = cos na + i sin na,
n being a positive integer, we have
cos?la = cos*^a ^j — ^^— cos'*~2asin2a+...,
„ 1 n.n — l.n — 2 . „ « o ,
sm na = n sm a cos'^-^a .. ^ ^ sm^a cos^~^a+ . . . .
Since the fundamental laws of algebra have been
shewn to hold for complex numbers, we may substitute
such numbers in any algebraical identity dependent on
these laws. Then, resolving the complex numbers along
the primary and secondary axes, we derive (in the case in
which the modulus of each number is unity) two trigo-
nometrical relations connecting the angles which represent
the amplitudes of the numbers.
For example, in the identity a^—¥ = (a — h) (a^ +ah + 6^)
let a = cosa-|-isina, 6 = cos/5 + ^sin/5;
414
COMPLEX NUMBERS.
then we obtain the trigonometrical identities
cos 3a — cos 3/3
= (cos a — cos ^)(cos 2a + cos a + /S + cos 2/3)
— (sin a -- sin ^) (sin 2a + sin a + )8 + sin 2^8),
and sin 3a — sin 3/3
= (cos a — cos /3)(sin 2a + sin a + /8 + sin 28)
+ (sin a — sin ^)(co3 2a + cos a + ^8 + cos 2/3).
263. The following examples will further illustrate the
use of complex numbers : —
Example 1. — Shew that the sum of the nth powers of the five
fifth roots of unity is either 5 or 0.
The fifth roots of unity are the five complex numbers of unit
modulus, and having amplitudes 0, a, 2a, 3a, 4a, where 5a = 27r.
The ?ith powers of these have unit modulus, and, rejecting
multiples of Stt, the amplitudes
0,
0, 0, 0 when
a, 2a, 3a, 4a
is of the form 5m,
„ „ 5m + 1,
„ „ 5m + 2,
„ „ 5m + 3,
5m +4.
2a, 4a, a, 3a „ „
3a, a, 4a, 2a „ „ ,
4a, 3a, 2a, a „ „ ,
In the first case the sum of the ?ith powers = 5.
In the other cases the sum is zero, since it is in each case the sum
of five complex numbers of unit modulus whose directions are
symmetrically distributed about the origin.
Example 2. — Demoivre's pro-
perty of the circle. See art.
226.
Take OP as primary axis.
Let LPOAi = e,
LPOAr=e+(r-i)^-^;
n
let OP=r, and a = radius of the
circle.
We have 0A^-\- A^=OP= r,
COMPLEX NUMBERS.
415
.-. {ArP-rr={-moArr={-\r{a, e+{r-\f^y
i.e, ArP is a root of the equation
(^-r)"-(-l)'V,^^)=0-
The absohite term of this equation is (-l)"[r"-(a", n6)\
hence A^ . A^ . . . I^= f" - (a'^ n 6).
Let ar be the image oi Ar with respect to the primary axis, i.e. a
point such that Ara^ is bisected at right angles by OP^ then by
changing the sign of 6 we get
a[P .^ ...aj'=f'-{a\ -nO).
Multiplying, and putting A^P .arP^p?., where p^ = mod(Jr^)
= mod(a^/^) we get
^,,2n_2aV^cosw6'+a^".
PiW
Example 3, — If ^i, ^2» ^3) ■•• -4„ be 7i points at equal distances on
the circumference of a circle of centre 0 and radius a, and if P be
any point and OP=r, and A^OP=$; prove that the sum of the
angles which A^P, A^P, A^P ... A^P make with OAi is an angle
1 , J. • r'^sinw^
whose tangent is y,
r"cos nO — oT'
Take OAx as the primary axis.
We have OA^ + A^P = 0P_ = (r, 6),
A^P-{r,e)=-OAr,__
.'. [ArP-{r,e)f = {-mOA;)-
=(-l)V,
.*. ArP is Si root of the equation
[.r-(r, (9)]'^-(-l)^a" = 0.
The absolute term of this equation is (-1)"[(^', Oy^-a^],
hence A^.A^\..A^=:(r, 6y-a"=r''(cosnd + ismn0)-a'''
Let pr=mod{ArP), ^^=amp(J^/^),
th en P1P2 '■• Pn(cos 2^ + ?' sin 2^) = r'*(cos w ^ + 1 sin w ^) - a'*.
416 COMPLEX NUMBERS.
pip2 . . . p„cos 2<^ = r"cos n6 - a**,
and P)P2'-' PiM^ 2^ = r**sin nO,
r**sin nO
and therefore tan 2^ =
Example 4. — A man walks on a plane in such a manner that,
when he has passed over a distance a in a straight line, he always
changes his path through the same angle a in the same direction.
Shew that when he has done this n times, his distance from his
starting point is a sin ^ /sin ~, and that this distance makes an
angle (n—l)^ with his first path.
Let OAxA<:i... An be the path. Take
OAx as the primary axis. We have
OZn =(«, 0) + (a, a) + (a, 2a) + ... H-(a,^L a)
= a(l + cosa + cos 2a+... + cosM- la)
+ ia(sin a + sin 2a + . . . + sin n - la)
cos(w - \)% sin ^ sin '^-^ sin(w - 1)^
= a +za
• a -a
Sin — sm -
2 2
or modi{OAn)=a -. Amp(OJ„)=(w- 1)-.
|cos a + 0) cos^ a + ^) + w^cos^ a + ?^) V
= (- j Icos na-\-bi cos( wa + ^) + w^cos(^7ia + -r ))■
Wehavea,= -1, (l, |) or (l,-|).
If 0)= - 1, each side of the equation vanishes.
COMPLEX NUMBERS. 417
If o) = (l, I), take Oi;;i=(l, p, ar2=(l, ^), 02 = (1, a),
a5=(l,-a).
Draw the chords ALB, BMP, BNQ
at right angles to OX, 0(0i, Oin^ re- -^
spectively.
Then,
cos a + (0 cosf a + ^ J -f- w^eosf a + -^ j
='OL^OM+ON
'=^{OB+OA + OB + OP+OB+OQ)
=1 OB, since ^+0^+0^=0
=Kl,-a).
In like manner, writing na for a, we have
cos ?ia + w cos(^ wa +^) + w'-^cosfwa-l- ^)=f (1,- wa).
But [|(l,-a)]« = (|r(l,-^a),
|cos a + 0) cos^ a + ^ j + w2cos( a + — H
= (f r"^| cos wa + 0) cos(7^a + - ") + a)2cos(wa + ^ ) I .
If (o=M,-^ j, each side of the equation = (§)"(], na).
Example 6.— If
^ = cos a + 1 sin a, B=cos/3 + i sin /?, (7= cos y + 1 sin y,
express ^^^4±^ in the form P+^i, and prove that
^ _ _ 4 sin ^(/? + y - 2a)sin |(y + a - 2/3)sin K« + jS - 2y)
1 + 8 cos(/3 - y)cos(y - a)cos(a - ;8)
We have
^(7+(7^ + ^^_^cos(;8+y)+...+^•{sin(^+y) + ..■}
A^ + B'+C'^ cos2a+...+^{sin2a+...}
^ [cos(jg + y) + . ■ ■ + 1 {sin(^ + y) + . . . }][cos 2a + . , . - z{sin 2a + . . .}]
, (cos2a+...)H(sin2a+...)^
Denoting this expression by P+ Qi, we have
2d
418 COMPLEX NUMBERS.
^ _ [8in(^ + y) + ■ . ^[cos 2a +...] - [co8(/3 + y) + ■ ■ .][8in 2a -t- . . .]
^ (cos2a+...)H(sin2a+...)^
_ 8in(;8+y -2a)+two similar terms ^ _4n{sin ^(/? + y-2a)}
~ 3 + 2 cos 2( j8 - y) + two similar terms 1 + 8n{cos(^ ^ )}
Example 7. — To prove that
cos 2a/sin ^~ i^ sin ^!-l2 sin ^5lZ_ -f. three similar terms
'2 2 2
=88m«+%r_+S.
Supposing the trigonometrical identity to be deduced from an
algebraical one, we may determine the latter as follows : —
Let a, h, c, d stand for complex numbers cos a + i sin a, etc., '
then a-6 = 2isin^(cos^+tsin?±i?|
(a - 6)(a - c)(a - c^; = - 8i sin ^^ sin ^5^ sin ^(1, 1+^),
2i Ji Jt
where «=^(a + /? + y + 5) ;
also, Va6c5=(l, s\
\/ara ^ (1, -2a)
a{a-h\a-c\a-d) .g^'sin ^"/^sineir^rsin^
2 2 2
_ • ^' cos 2a + sin 2a
"8sin«-:;^sin±Z2sin^*
2 2 2 ^
But 1 =(l,-,)=co8^±to±i-isin^±^+Z±S.
Hence the trigonometrical identity will be true if we can shew
that —. ,, , ^ ^ w T. + 3 similar terms = - . ,
a(a-6)(a-cXa-o?) ^ahcd
i.e. that 2 ,. ■ ^ ■ ■ -j.-\- \-.=^.
a{a - o){a - c){a - a) abed
This algebraical identity is readily obtained by assuming
:v{x-a){x-b){a;-c)(x — d) a: — a x-h x~c x-d x
determining A^ B, C, 7), E^ and substituting their values in
A-\-B^C+D^-E=0,
COMPLEX NUMBERS. 419
Examples XXVI.
1. Prove that amp((X + 6i) is equal to tan"^-, tan"i-4-7r,
or tan"^ — tt, according as a is positive, a negative
and b positive, or a negative and h negative,
respectively.
2. If a = cos A+i sin A, b = cos B+i sinB, c = cos C+isinC,
where A, B, G are the angles of a triangle, then
abc= —1.
8. If a = y^, then
[(1. a) + (l, 2a) + (l,4a) + ip
= 2[(1, a) + (l, 2a) + (l, 8a) + (l, 4a) + (l, 5a) + (l, 6a)] + i.
4. If (X = (l, a), then 2cosa = a+-, 2isin a = a , and
a^ — 1
i tan a = o , -,-
5. If a = (1, a), then 2 cos 2a = a^ +-^, and 2^ sin 2a = a^ — s.
6. If a = (l, a) and n be any positive integer, then will
1 . 1
2cos7ia = a^+— :, and 2isinna = a'^ — — .
a a
7. Find the simplest form of (<^o^^+^-}^'''^)\
(cos 2/ — V — 1 sin yy
8. Find the real and imaginary parts of the expression
(cosa+x/— 1 sina)(cos/3 + /v/ — 1 sin^)
(cosy + x/ — 1 siny)(cos(5+/v/ — 1 sin^)
9. Find the simplest form of the expression
(cosg + V—lsinay
(sin;8 + v/-lcos/3/
420 COMPLEX NUMBERS.
10. Determine the simplest form of (cos 0 - V --1 sin Oyo
(cos a + V — 1 sin a)^^
11. Find the value of (-l + V^)H(-l->v/^)^
12. Apply De Moivre's theorem to express the real and
imaginary parts of (a+6/v/ — 1)**, when n is any
integer. Find the value of
(1 + ^33)10 4. (i_ ^113)10,
13. Shew that
p
{m+nj -lY+(m-nsJ -\y = 2(m2 + n^f cos pO.
where 0=amp(m+'>ix/— 1).
(^cos|^-x/^sin^j
14. Simplify the expression -.
(^cos^ + x/-lsin^J
15. Find the cube roots of V— 1.
1 6. Find the four values of (( - 1 + J'^J^)%
17. Exhibit the four fourth roots of 1 + ^ — 3.
18. Find the three values of ((l + x/-'T))i
19. Find the three cube roots of >^3+i.
20. Find the five values of the expression (( '^ ~ )]"•
21. Express by De Moivre's theorem all the values of
((-1))^^
22. Reduce ^^^ to the form A+Bs/~^. where
a+o
a = cos a + /v/ — 1 sin a, and 6 = cos ^8+ V — 1 sin /3.
23. If cCy = cos^+ V— Isin J, prove that, the product
being continued to infinity,
X^^^^ ... = cos TT.
COMPLEX NUMBERS. 421
24. Prove that
(sina; + V— 1 co^ xY = Qo^ ni-^ — x) + s/ — 1 &mn(^—x\
25. Calculate in a form free from imaginary quantities the
value of
[cos 0 — cos ^ + x/ — l(si n 0 — sin 0)]**
+ [cos d — cos (p — \/ — l(sin 6 — sin 0)J\
26. Prove that (^ + ^y"' is reducible to the form
p{cos6-\-iiiin0), and find the values of p and 0.
27. Apply De Moivre's theorem to shew that, if
l/{a+bs/-l)-\-'^(a-hs/-l) = 2l/{a^ + ¥).cosie.
28. Prove that the expression (a-\-ihy'^+{a+ih'y' is
reducible to the form JS(cos0 + isin0), and find
the values of M and </>.
3 1 1 1
29. Shew that —---„ = -—; f-^; h-, 77-, where a, 8
1+x^ l-^x 1 — ax 1 — px '^
are the imaginary values of (( - 1))^, and deduce, by
writing .t = cos 20 + V— 1 sin 20, that
3 tan SO = tan e - cot^O + ^) - cot^O - ^).
30. If ^, 5, 0 be the angles of a triangle, then
2cos3^ + 3 = 2cos^.2(cos2^+cos^)
— S sin J. . 2(sin 2^ — sin ^),
and 2 sin 3 J. = 2 cos ^ . S(sin 2A — sin A)
+ 2 sin ^ . 2(cos 2^ + cos ^).
422 COMPLEX NUMBERS.
31. If 0) be an imaginary cube root of —1, prove that
i cos a + ft) cosf a + ^ j + w^cosf a + -^) f
X |cos^+ft,cos(/3+|) + a)2cos(^4-^)}
= f |cos(a + i8) + ft) cos(a + /3 + 1) + a)2cos(a + )^ + ^)},
and deduce the value of
jcosa + ft)COsfa + ^j + ft)2cosfa+-i^H .
32. Shew that the roots of the equation
{{a-\-h)x-{a-h)y'^{a + h-{a-h)xY
Tit ... Ttt
a cos %b sill —
are the values of , where r has
a cos f-it)sm —
n n
any positive integral value between 0 and n — 1
inclusive.
33. Determine the values of x from the equation
(cos A +x sin J.)(cos B+ x sin S) =cos(J. -{-B) + x sin(^ + 5).
34. Find m in order that (cos 0 + msin 6)^ may be equal
to cos nO+m sin nO for all integral values of n.
35. If, in the identity
1 ^ 1 1
(x — a){x — h) {a—b){x—a) (a—b)(x — by
cos 26 + s/^ sin 20, cos 2a + \/^l sin 2a, and
cos2/3 + x/ — lsin2^ be written for x, a, b, re-
spectively, obtain the trigonometrical identity
resulting from equating the real parts of the two
expressions which are identical.
4
COMPLEX NUMBERS. 423
36. From the identity
obtain by writing for a, cosa + isin a, and similar
substitutions,
sin(a-/3)sin(y-^) = sin(a-^)sin(y-^) + sin(a-y)sin(/5-^;.
37. If (r, 0) = (1, a) + (l, /3l prove that r = 2cos'^'^,
0 = ^ J^. Hence shew that
cos a + cos /3 = 2 cos ^' cos — ^- ,
and that
sin a + sin /3 = 2 sin ^^ cos ^^.
38. If a = -S-, then cos a + cos 2a + cos 4a = — J,
and sin a + sin 2a + sin 4a = J v^''- '
CHAPTER XVII.
SERIES OF COMPLEX NUMBERS.
264. Finite Series. — Let Uj^+U2+u^+...-\-Un be any
series of complex numbers, and let Sn denote the sum of
n terms of the series.
If all the terms of the series have the same amplitude,
the vectors representing them form a straight line, of
length equal to the sum of the moduli of the terms, and
inclined to the primary axis at an angle equal to the
common amplitude of the terms ;
or, if /Sfn=(r, 0),
then r = the sum of the moduli of the terms,
and 0 = the common amplitude of the terms.
If the terms have not the same amplitude, the vectors
representing them do not form a straight line, and it
follows, by Eucl. I. 20, that the modulus of Sn is less than
the sum of the moduli of the terms.
In this case, if /Si„ = (r, 6), then r has a value less than
the sum of the moduli, and dependent on the values of the
moduli and amplitudes of the terms, and 0 has a value
also dependent on these moduli and amplitudes.
265. Definitions. — If the sum of the first n terms of
a series of complex numbers tends to a limit S of finite
424
SERIES OF COMPLEX NUMBERS 425
modulus and fixed amplitude, when the number n is
indefinitely increased, the series is said to be convergent,
and 8 is called its sum.
If the modulus of the sum increases without limit as n
is indefinitely increased, the series is said to be divergent.
If each term of a series of*complex numbers is expressed
in the form x-{-iy, the conditions of the definition of con-
vergency will be satisfied if the real series 'Zx and Zy are
each convergent ; and if one or both of the real series 2cc
and 22/ be divergent, the series of complex numbers is
divergent.
If the series whose terms are the moduli of the
terms of the original series is convergent, the original
series is said to be absolutely convergent. (See art. 266.)
If the series of complex numbers is convergent, and
the series of moduli of its terms divergent, the original
series is said to be semi-convergent.
If the modulus of the sum does not increase without
limit as n is increased indefinitely, and the sum does not
tend to a limit of finite modulus and fixed amplitude, the
series is said to oscillate.
The following are examples of oscillating series.
Example 1.— Consider the series whose wth term is +(l,~\.
Let 0AiA2A^A^A^ be a regular hex-
agon, then if OAi make an angle ^ with
the primary axis, the sides of the hex-
agon taken in succession, and repeated A^^
continually, represent the terms of the
series, and the sum of n terms is
zero, all, ^> ^^ ^' or OA^, ^^ ^^ — ^
according as n is of the form '^
6m, 6m+l, 6m + 2, 6m + 3, 6m + 4, or 6m + 5, respectively.
426 SERIES OF COMPLEX NUMBERS.
The series consequently oscillates^ and has any one of six distinct
values, each of finite modulus and fixed amplitude.
Example 2.— Consider the series obtained by placing in a circle
of unit radius a succession of chords A1A2!,
A2A3, ^3^4, ... of lengths hhh ""
We have
OA„=OA, + A^A2+A2A3+...+An-iAn.
Now the modulus of the sum OAn is unity
for all values of ?i, but in consequence of
the divergency of the series 1+^ + ^+...
the point An does not tend to any fixed
point on the circumference when n is indefinitely increased.
The series consequently oscillates, and has any one of an infinite
number of values, each of unit modulus.
266. A series of complex numbers is convergent when
the semes of moduli of its terms is convergent.
Let OAn represent the sum of "the first n terms of the
series of the complex numbers.
Then, since the series of moduli of the
terms is convergent, and since the modulus
of the sum is less than the sum of the moduli
of the terms, therefore mod (OAn) is finite,
however great n may be. Next, let AnA^
represent the sum of the m terms immed-
iately following the first n terms. Since
the series of moduli is convergent, it follows
the modulus of AnA^ can, by sufficiently in-
creasing n, be made as small as we please, and this
however great m may be; and therefore OAm can be
made to differ in modulus and amplitude from OAn by
as little as we please. Hence, the series is convergent.
SERIES OF COMPLEX NUMBERS.
427
267. Example 1. — Consider the
series
^-^+^-"-<^dinf.,
where ^ is a complex number.
The test ratio of the series of
moduli =i?^i^^^, and this can
be made as small as we please
by increasing n, hence the series
is absolutely convergent.
The diagram represents the
first four terms of the series,
when x^^sl^-'ri or (2,^), and
shews the rapid convergence of
the series after the second term,
the vector OAq representing the
sum of four terms, and approxi-
mately the sum to infinity.
Example 2.— The diagram re-
presents the series
-V.2 /»,4 ^6
H- — -U- 4- — -4-
(2+|4+^_ + 7
when x=J'i-^i, the vector OA
giving the sum of four terms,
and approximately the sum to
infinity.
268. If %, a^, a^... an he a series of constantly de-
creasing ^positive quantities, and if Lt. an = 0, and if /3
be not equal to zero or a multiple of 27r, then ivill the series
a,{l, a) + ai(l, a + ^) + a^{l, a + 2/5)-f...
be convergent.
It has been shewn in art. 210 that with the given
conditions each of the series
a^cos a i- aiCos(a + /3) + a^cos(a + 2/3) + . . .
428 SERIES OF COMPLEX NUMBERS
and aQsina + ai8in(a + /8)+a2sin(a + 2^)+...
is convergent, hence the series
ao(l> a) + «i(l, ^Hh8) + a2(l, a + 2/3) + ...
is also convergent (art. 265).
269. If the series aQ-\-a^X'\-a.fy^-\- ...ad inf., where
aQ, dj, a2, ... are real quantities and x a complex number,
be absolutely convergent when mod{x) = R, it will be a con-
tinuous function of x for all values of x such that
mod(x) < R
Let X = (r, 6) where r < R.
Then aQ-{- a^x + a^'^ -\- . . .
— aQ+a^r cos 6 + a^r^cos 20+...
+ i (a^r sin Q + a^r'^sin W+...)
= G+iS, say.
First, let 6 remain constant while r changes. Then
each term of the series G is numerically not greater than
the corresponding term of the series
and, by hypothesis, this series is absolutely convergent,
therefore G is absolutely convergent, and consequently G
is a continuous function of r so long as r<R (art. 211).
Similarly, the series /Sf is a continuous function of r ;
therefore G + iS, or a^ + a^x + a^^ + . . .
is a continuous function of mod(ic) so long as iaod(x) < R.
Next, let r retain a constant value less than R, and let
6 change from 0^ to 0^, where 6^ < 62-
Let G^=aQ + a{r cos 0^ + a2r2cos 20^+...,
(72=^0+ a^r cos O2 + a^r^cos 2^2 + • • • >
then
G^ - C2 = air(cos Oi - cos ^2) + a2^2(cos20i - cos 20^) +....
Suppose that a^, a^, a^,... are all positive, then since
cos 0 ~ cos (/)' is numerically less than 0-0' (art. 89), we
SERIES OF COMPLEX NUMBERS.
429
see that 6\ ~ G^<{e^ - e^){a{t^-\-'ia^T'^+2,a^r^+ . . .),
and, as in art. 211, it may be shewn that for any fixed
value of r < -R, the series
a^t' + ^a^f^ + SagT^ + . . .
is convergent and therefore finite.
Therefore G^ ~ G^ diminishes indefinitely with Q^ - 0^
or 0 is a continuous function of 6.
The result follows, a fortiori, if the coefficients a^, a^, otg,
etc., are not all of the same sign.
Similarly, the series >S' is a continuous function of 0,
therefore 0+ iS, or a^ + a-^x + a^p^ + . . .
is a continuous function of amp(ic) so long as mod(aj) < R.
Combining these results, we see that
is a continuous function of x for all values of x such that
mod (a?) < R.
Geometrical Illustration.
Let OP-=x,
CE=aQ+ap-^acfc^+....
Then if P move continuously from P to P' within a
circle of radius P, 8 will move continuously to a new
position S' ; and if P and P' are indefinitely near to one
another, so also are 8 and 8'.
430 SERIES OF COMPLEX NUMBERS
270. If the series aQ-\-a-^x + a^^-\- ... ad inf., where
a^, a^, ttg . . . are real quantities and x a complex number,
he convergent when x has a value (1, a), then the limit of
aQ+ai(r, a) + aj^r, a)^+... ad inf. as r increases up to 1
ivill be
aQ+a^(l, aJ+a^O-, a)^+... ad inf.
By hypothesis, the series
ao+ai(l, a) + a2(l, af-\-...
is convergent ; therefore each of the real series
aQ + (Xjcos a + a^coB 2a + ...
and a^sin a + agsin 2a+...
is convergent.
Now the series
<^o + ^iO'' a) + a2(?', a)H...
= (Xq + a^r cos a + cv'^cos 2a + . . .
+ i{a^r sin a + agT^sin 2a+...}.
But, by art. 213, the limits of
^0 + <^i^' ^^^ " + agT^cos 2a + . . •
and a^r sin a + a^r^sin 2a + • . • ,
as T increases up to unity, are
% + ^iCOS a + dgcos 2a + . . .
rind a^sin a + dg^in 2a + . . .
respectively ; therefore the limit of
% + ^i(^'' «) + «2(^'> a)2 + . . . ,
as r increases up to unity, is
ao+ai(l, a) + a2(l» a)2+....
Cor. — If the limiting value of x, for which the series is
convergent, be (R, a) where R is any fixed modulus, the
limit of aQ + a^{r, a) + a^ir, a)^ + . . . ,
as r increases up to i^, will be
aQ + a^(R, a) + ttgCi^, a)^ + . . . .
SERIES OF COMPLEX NUMBERS
431
For if we put hn for anR^, and p for rjR, we may write
the series in the forms
and 60 + hip^ a) + \{p, a)^ + . . . ,
and apply the theorem of the present article.
Geometrical Illustration.
Let
0^=(1, a),
OP = (r,a\r<l,
CS=aQ + a^(r, a) + a^{r, af+...,
then as P moves up to _p, 8 moves up to s.
271. Series involving the cosines and sines of angles in
arithmetical progression may be reduced to an algebraical
form by the use of complex numbers.
Thus, if (7=aoCosa + aiCOs(a + /3) + a2COs(a + 2/5)+...
and S = a^sin a + aisin(a + /3) + a2sin(a + 2^) + . . . ,
and if i:c = cos a + isin a, 2/ = cos iQ+'i sin /5,
then G+iS = a^ + a^xy ■^■a^x'ip'-^- ... .
If the sum of the algebraical series
a^x-\-a^xy-\-a^x\p'-\- ...
is known, the values of C and B may be found by resolv-
ing the sum of the series of complex numbers into com-
ponents along the primary and secondary axes.
432
SERIES OF COMPLEX NUMBERS
As simple examples of this method we may take series
leading to geometrical progres-
sions. Examples of binomial,
exponential, and logarithmic
series will be found in succeed-
ing chapters.
The accompanying diagram
shews geometrically the nature
of the process.
Here
Mo m;m, c s=on,+n^n, + n^n^+,..,
272. Example 1. — Find the sum of the series
cos a + cos(a + j8) + cos(a + 2^) + . . . ton terms,
and sin a + sin(a + /8) + sin (a + 2^) + . . . to n terms.
(See arts. 205, 206.)
Let C and S denote the sums of the series, and let « = (1, a),
6 = (1, ^). Then
C+iS=a+ab-\-ab^+... + ab''-'^
_a(l-6«)
1-6 •
This expression=^"^^°^+^'"^^"Xl-cosn/3-zsin7i/?)
l-cos/5-^sin^
(cosa+isina)(sn !^-^cos'^) . 2sin^
(sin^-icos^V 2sin^
V 2 2/ 2
(cos a + z sin a)( cos ^ + 1 sin ^ )sin -^
Tcos^+isin^ jsin^
=^cos(a+7^^|) + ^sin(a+w^|)Jsin^^sin|;
SERIES OF COMPLEX NUMBERS. 433
and >S'=sin(a + 7^^|)sin^/sin|.
Example 2. — Sum to infinity the series
cosa + ^cos(a + /3)+^2cos(a + 2^)+...
and sin a + .r sin(a + j8) + ^2sin(a + 2^) + . . . ,
when a^ is less than unity.
Denoting the series by C and S, and the complex numbers (1, a)
and (1, /3) by a and 6, we have
C+iS=a+x . ah+x"^ . ab^+... .
This series is absolutely convergent, since mod(6.r) < 1,
C+iS--
Hence, C+iS=-.
l-bx
cosa + isina
1—.V cos y8 — ix sin (3
_ (cos a + ^ sin a)(l - .a; cos /3 + ix sin /3)
1-2X008/3 + x^
_ COS g - :r cos(a - (3)-\-i sin a - 2^ sin(a - ft) .
\—2x cos ft + x^
jy_ cosa-^cos(a-^)
~ l-2.rcos/3 + ^2~'
and ^_sina-.ysin(a-/3)^
1 -2^cos^ + .<:^^
Example 3. — Sum to infinity the series
cos 2^ + cos ^ cos 3^4- cos^^ cos 4^+...
and sin 2^ + cos ^sin3^+cos2^sin4^+
Let C and S denote the series, and let ^=(1, ^)=cos ^+isin ^.
Then C+i'S'=^2+^cos^+^*cos2^+... .
If cos ^ is numerically less than 1, i.e. if 6^=mr, this series is
convergent, and we have
C+ iS= ^^ = —
1-^cos^ l-cos^^-icos^sin^
^ x'^ _ (1, 26>)
8in(9(sin6>-^■cos<9) gin ^(1,^3^)
ihZ+e)
TT ,
2 ^ -sin 6+1 coa 9 , . ,
1+ 1 cot
sin 6 sin 0
2e
434 SERIES OF COMPLEX NUMBERS
Hence, if d^nir, C= - 1, and >S'=cot $.
If d = mry we have C=H-l + l + ... = oo, and >S'=0 ; thus in each
case there is discontinuity when ^=0, tt, Stt..., or — tt, -27r —
The curve of the first series is a straight line parallel to the axis
along which $ is measured, with a series of isolated points at
infinity ; that of the second series is the cotangent curve with a
series of isolated points on the axis of 6 corresponding to $=0,
e=Tr, ^ = 27r, ..., (9=-7r, ^=-27r, ....
The fact that cos20+cos ^ cos 3^+ cos^^ cos 4^+... ac/m/l is equal
to - 1 for any very small value of 0, say one-millionth of a second
of angular measurement, while when ^=0 the sum of the series
is infinite, may serve to shew that theorems such as those of arts.
211 and 213 are not self-evident truths.
Examples XXVII.
1. Shew that, if x be any complex number, each of the
series
'^+^ + i5 + -
is absolutely convergent.
2. If ic be a complex number, the binomial series
\-\-nx-\-^-—^ — x^+. . . ad mf.
is convergent for all real values of n, when
mod(a;) < 1.
3. If a? be a complex number, and oi real but not a posi-
tive integer, the series
l + nx+^'^~^x^+...adinf.
is divergent when mod(a:;) > 1.
SERIES OF COMPLEX NU3IBERS 435
4. Find the values of the oscillating series
5. Find the limit of
where x — ii\ ~), when r increases up to \m\ty.
r. n l^ . sina , sin 2a , sin 3a , j.^ m +««^o
6. Sum the series --^ — I — ^ — I — ^ — h... to 10 terms.
7. Sum the series
cosa + a;cos2a + i:c'^cos3a+... ad inf., when x<l.
8. Sum the series
ccsin a — cc^sin2a + a3^sin3a — ... ad inf., when x < 1.
9. Find the sum to n terms of
cos 0 sin 2^ + cos20 sin W + cos^O sin 40+ ... .
10. Find the sum to n terms of
cos a + x cos(a + P)+ x^cos(a + 2/5) + • . .
and sin a 4- aJ sin(a + ^) + aj^sin(a + 2^) + . . . .
11. Given the sum to n terms of the series
cos a + cc cos(a + j8) + cc^cos (a + 2/3) + . . . ,
deduce the sum to infinity when x < 1.
12. Given the sum to infinity of tlie series
sin a + a? sin(a + 1^) + a;2sin(a + 2^) + . . .,
deduce the sum to n terms.
13. Prove that the sum ofn terms of the series
^^cosa^cos|a^cos3«_^_ is equal to 0,
cos a cos^a cos'^a
if na = 7r.
14. Sum to infinity
sin 0 cos 0 + sin20 cos 20 + sin^O cos 80 + . . .
and sin 0 sin 0 + sin^O sin 20 + sin^O sin 30 + ... .
15. Sum the series
-L sin e + l sin 20 + 2-^ sin 30+ . . . ad inf
436 SERIES OF COMPLEX NUMBERS
16. Sum the series
sin 45°sin 0+sin245''sin 20+siii345°sin 30+ . .
to n terms.
17. Prove that, if A, <1,
l^h?
(l-/i)2cos2|-|-(l+^)2sin2|
= l + 2^cosaj+2^2cQg2a;+... ad inf.
18. Sum to infinity
cos 0 cos 0 + cos20 cos 20 + cos^^ cos 30 + . . .
and cos 0 sin 0 + cos^^ sin 20 + cos^^ sin 30+ ... .
%
CHAPTER XVIII.
THE BINOMIAL THEOKEM.
273. If ii be a positive integer, x any complex number,
then
1.2 \r\n — r
The proofs usually given for real values of x depend
only on the fundamental laws of algebra, and therefore
hold for complex values of x.
274. If m, n be any real numbers, x any complex
number such that ■mod(a;)< I, and if the infinite series
- , , m(m — 1) 9 ,
l+maj+-Y-2— ^ +...
be denoted by /(m), then
f{m)xf{n)=f{m+n).
The proof given for real values of x, by aid of Vander-
monde's theorem,* depends only on the fundamental laws
of algebra and on the absolute convergency of the series
/(m) and f{n), and therefore holds for complex values of x
if mod(£c) < 1.
* C. Smith's Treatise on Algebra, art. 279.
437
438 THE BINOMIAL THEOREM.
275. Positive Fractional Index. — From the equation
/(m) xf{n)=f{m + n), we deduce, as in the case when x is
real,
{/(p)*-<'«)' ■
p and q being positive integers, and mod(ic) < 1.
Hence, it follows thaty M j is equal to one or other of
the values of the g- valued quantity ({l-\-x)y. We shall
shew that this value is the principal value, or, if
1 +x=(p, (p) where — tt < 0 < tt, then
For any given value of-, provided only that mod(a;)< 1,
we know that the series denoted by^f -j is absolutely
convergent, and that it is a one-valued, continuous fiinc- U
tion of X ; or, geometrically, if OF=yy^\ then ^ is at a fl
finite distance from 0, has a single fixed position when x
is given, and moves continuously for any continuous
change in x.
Let OG=\,GP = x,
then 0P = l+x, length OP = p, lGOP = 0.
Since mod(a3) < 1, P cannot lie outside a certain circle
with centre G and radius less than 1, and since 0C=1, 0
is evidently outside this circle.
Hence, p or OP never vanishes, and (f> or lGOP lies
between limits greater than — ^ on the one side and less
than ^ on the other.
THE BINOMIAL THEOREM. 439
Make the angle COQq = --^, and the length OQ^ — pi, and
let OQq, OQy OQ2, ... OQq-i be a system of lines of equal
lengths symmetrically distributed about 0. Xhen the
vectors OQq, OQ^, OQ^, ... OQ^.i represent the q values of
p
((l + a?))*^, OQq representing the principal value.
The statement thaty f- j is equal to one of the values
of ((1 -{■x))q may now be expressed geometrically by saying
that i^ coincides with one of the points Q^, Qi, ... Qg-i; and
we have further to shew that F coincides in all cases
with Qq.
As P moves continuously, the length OP or p changes
p
continuously, and therefore pi, or each of the lengths
OQq, OQ^, ... OQq-i changes continuously, and, siuce OP
never vanishes, therefore OQq, OQ^ ... OQq^i never vanish,
and therefore F cannot pass through 0 from one of the
points Qq, Qi, ... Qq-i to another.
440 THE BINOMIAL THEOREM.
Again, as P moves continuously, the angle COP or
0 changes continuously, and therefore — or lCOQq
changes continuously.
Hence, each of the points Qq, Qi,...Qq-i moves con-
tinuously, and therefore in no way can F pass from one
of these points to another.
Hence, if we can shew that for any one position of the
point P, F coincides wnth a particular point of the system
Qo' Qv ^2' ••• Qq-if ^^^^ fo>' ctll positions of P within the
limiting circle, F will remain coincident with that
particular point of the system.
Let P be at G, then p = l, and ^ = 0,
length OQo = 1, and lGOQ^ =^ = 0,
i.e., Qq is at G.
Also, since, when P is at G, x = 0, thereforeyf-j = 1,
and therefore 0F= 1, i.e. F is also at G.
Thus F and Qq are together when P is at G, and there-
fore F coincides with Qq for all positions of P ; i.e., in all
cases for which mod(£c) < 1, we have
OF=OQq,
/(p = ;(cos^^H.isin^^).
276. Negative Index. — Let n he a, real negative
number commensurable with unity, and equal to — m, say.
Then/(n)=/(-m) = J^^,
since /(-m)x/(m)=/(-m+m)=/(0) = l;
THE BINOMIAL THEOREM. 441
f(n) = -- ^—r-. ^ (art. -273 or 275)
= p- "^(cos m</) — i sin m^)
= p^(co9 n<l) + i sin n(f)).
277. Example. — If n be any real number, and - - < 6 <^. then
4 4
will"-gi^=l-^(f-J)tan^^+^^^^-^X^-^)i^-^Jtan4^-...,and
cos^^ 1.2 1.2.3.4 '
«^^^^=^tan^-^(V ^XV^>tan3^+....
cos"^ 1.2.3
The principal value of ((cos ^+tsin ^))" is cosw^+isin?i^, since
6 lies between - tt and tt ; therefore the principal value of
((l+*tan^))"=^+*^.
cos"^ cos"^
But, since - 1< tan ^ < 1 , we have, by the binomial theorem,
the principal value of
((1+2 tan d)f = 1 + n(i tan 6) + ^^^~}\i tan Ofi-....
Hence, costi^ -sin
cos'*^ cos''^
Equating real and imaginary parts, we have
cosnd^ 1 _ n{n-l) ^^2^ + n{n - l)(n - 2Xn - S) ^^^^^ _
cos**^ 1.2 1.2.3.4 ■■■'
and E£i^^n tan ^-^<^-^)^V^) tan3^+... .
cos^t* 1.2.3
Examples XXVIII.
1. Find the modulus and amplitude of the sum of the
series l+nx+ '^y"^ x^ -{-... ad inf., when n = }-
and x= — .
4
442 THE BINOMIAL THEOREM.
2. Draw the vectors representing the first three terms of
the series l + j(|:) + ilri)(|:y + ..., and the
vector representing the sum to infinity.
3. Sum the series
cosa + ^cos2aH — ^. ^ ^cos3a+... 1 0(71+1) terms,
and
71/(71/ '~' 1^
sin a +71 sin 2a + \ 0 sin3a+... to (ti + I) terms.
4. Shew that
7l/(7h ~~" X ^
sin2a+'nsin5a + --Y-2--sin8a+...to('n + X)terms
. 371 + 4 /_ 3a V
= sin-^-a.(2cos^j.
5. Shew that, if tan ^ = a; sin a/(l + ic cos a), then
77 (7) —" X I
l-\-nx cos a + -y-^-^ cos 2a+ . . . to (71 + X) terms
I
fx sin aV ^ ,
( ^ — TT ) cos 7?.^, and
\ sin ^ / '
^sin2a+... i
(x sin a\** • /.
= ( ---^) sin7ia
\ sin 0 /
7?a;8ina+ ^.j ^ ^sin2a+... to 71 terms
6. Sum the series
cos 710 + 7^ cos (71 — X )0 cos 0
71(^71 — X ^
+ \ ^ C0S(71 — 2)0 cos 20 + ... + COS 710.
7. Prove that in a triangle, where a is less than c,
cosji^ If, , a „,«-(«+ l)a^ oD
i
THE BINOMIAL THEOREM. 443
8. If 0 < ^, shew that
4
r==o l??4-2r 1
cos*^0cos^0= S (-1)^-^ ^ -tan^^a
r=o ['yi-1 [2r
9. Shew that, if 0 < t. then
4
cos^e sin '^^0 = 7Z tan e - '^('^ + 1X^ + ^) tan^O
+ 1.2.3.4.5 tan0-....
10. Shew that
mo fi/3/1 ,1 5.8 1 5.8.11.14 1 \
cos 10 =VJ|1 + 33-3;-^ -3-0+ 3.4.5.6 •39--I
and hence calculate the value of cos 10° to three
places of decimals.
CHAPTER XIX.
THE EXPONENTIAL SERIES.
278. For all values of x, whether real or complex, the
series
is absolutely convergent, and therefore also a continuous
function of x. The series is also one-valued, since each
of its terms has one, and only one, value for a given value
of ic.
This absolutely convergent, one-valued, continuous
series is called the exjponential series, and is denoted by
the symbol exp(a;).
279. If X and y he any nv/nihers. real or complex, then
exp(a?) X exp(2/) = exp(aj + y).
In the product of the absolutely convergent series
11 I? \L
the term of the (m + ny^ degree is
^m+n ^m+n-1 y ^m+n-2 y2 ym+n
[m+n [m+n — l'\l \m-{-n — 2' [2 '" \m+n
444
THE EXPONENTIAL SERIES. 445
or
1 L«+n+2!^a;«.+n-w("^+^X;^+^-l)^^+n-y^.,
\m+n\, U: I?
+«|,
and, by the binomial theorem, this is equal to ^ — ^ — ,
' -^ ' ^ \m-\-n '
which is the general term of exp(a; + 2/); we have therefore
exp(ir) X exp(2/) = exp(a;+2/).
Cor. — By repeated applications of this theorem we have
exp(aJi) X exp(aj2) x . . . x exp(fl?„) = q^^{x^+x^+ . . . +Xn).
280. If X he a real number, then
exp(a;^) = cosa) + '^sinaj. 'i^ -l*^^-:' <^''*-'^ i"'
We have, for real values of x,
x^ x^
cosa; = l-r- + ^--... (art. 215)
[2 [4
and smx = x-^+%-..., (art. 216)
therefore cobx = 1 + ^j^+^^+ ...
and i8mx = (xi) + i^+^^^X-^"->
[3 [5
therefore, by addition,
[2 ^
cosx + i8mx=l + (xi) + ^-^+^^+...,
and therefore
exp(a;'i) = cos ii! + ^ sin x.
Cor. — Since
exp(a;i) = cos x + i sin x
and exp( — a5i) = cos( — fl?)+isin( — a?)
= cosx — ismx,
446
THE EXPONENTIAL SERIES.
and
cos a; = -{exp(a;i) + exp( — ici)}
sin X = --.{exp(a;i) — exp( — a;i)}.
281. Tfx+yi be any comj^lex number, then
exp(a; + yi) = exp(£c)(cos y+isiu y).
For by art. 279, exp(aj+2/i) =exp(a;) x exp(2/i),
and by art. 280, exTp(yi) = cos 2/ + i sin y,
therefore exp(aj + yi) = exp(a;) (cos y-\-ismy).
From this theorem, we see that exp(x + yi) is a complex
number whose modulus = exp(aj) and whose amplitude is y.
Since cosy = co8(y-{-2n7r) and smy = s'm(y-\-2n7r), it
follows that, if z = x + yi, exp(0) is a periodic function of
s, and that its period is the imaginary quantity 27ri.
282. Example 1. — Represent geometrically the value of
e.p(2+|.).
We have exp(2) = 7"4 ajDproximately,
e.p(2 + |i) = (7-4, I).
Make ^vOP=1, OP =7%
o
)■
then
0F=exp\2 + '^t
Example 2. — Sum the series
1+- cos ^+ A cos 2
} \1
and
^ + — coa36+...adinf.,
^ sin ^+ i sin 20 + ,4 sin 3^+ . .. ao? m/
li (2 L^
Let (7 and >S' denote the sums of the series,
and let a: = cos d+isin 6,
THE EXPONENTIAL SERIES. 447
then (7+^•>S=l+-^+^+^ + ...
lJi \1 If
=exp(^)
= exp(cos d + i sin ^)
= exp(cos ^)[cos(sin 0) + ? sin(sin d)\
C=exp(cos ^) . cos(sin ^),
and /S'= exp(cos d) . sin(sin B).
Examples XXIX.
1. If cc be a real number, prove that
exp(a;) = cosh X + sinh x.
2. Shew that
exp(a; + yi) = (cosh x + sinh aj)(cos y + i sin y).
3. Calculate the values of exp(l) and exp(2) each to four
places of decimals. Verify the result by substitut-
ing the values in the equation
exp(2) = exp(l) x exp(l).
4. Calculate, to three places of decimals, the values of
exp(J) and exp(f). Verify by substituting in the
equation exp(f)-i-exp(J) = exp(l).
5. Represent by vectors the values of
exp(^^-|-ij and exp(^2~|)'
6. Sum the series
II n /I . sec^O sec^O ..^ , , . ,
l + sec0cos04- |t) cos 20+ cos 30+... adinf.
and
secOsin^H — ^r— sin20H — r^— sin S0+... ad inf.
il If
7. Find the sum of the series
- , X cos 0 , a;^cos 20 a?^cos 30 7 . /.
IH ^ 1 — Y~2 — 12 3 "^•" ^'^•^*
448 THE EXPONENTIAL SERIES.
8. Prove that
csm 0+ ^-^+y-2-gH- ... ac« in/
= exp(c cos 0)sin(c sin 0).
9. Sum to infinity the series
, csin(a + i8) , c2sin(a + 2^) ,
sinaH ii — ^— H yy 1-....
10. Find the sum of
sm A + sin(^ + J5)cot Q H — ^ .^ ' -\-...ad inf.
l±
11. Find the sum of
, . e sin20 . 02 sin30 ^e^^m^B , , . .
l + (l-^i^ + g^I^ + 5 sin^e +...acZtn/
12. Sum to infinity the series
cosaj+acos2ic+ +
1 . ^
13. Shew that
1 , zi n , cos20cos2O , cos^0cos30 , ^7 ._/.
1 +COS 0co:4 e + ^— ^ + ""i~2T~ •^*
sin 20^
= exp(cos20)cosf — - — j
0
~''^'-
CHAPTER XX.
LOGAEITHMS OF COMPLEX NUMBERS.
283. Def. — If y = exip(x), where x is any number, real
or complex, then x is called the logarithm of y. The
logarithm of y, thus defined, is denoted by Log 2/, the
capital letter indicating that the logarithm is many-
valued (see art. 284).
When X is & real commensurable number the above
definition is equivalent to that obtained by substituting
e^ for exp(a;), provided that we restrict e'^, when x is not
integral, to its arithmetical value. These real logarithms
to base e of arithmetical numbers are called Napierian
logarithms, being closely connected with the logarithms
of sines calculated by Napier. It will be assumed that
the fundamental properties of Napierian logarithms are
known.
When X is Si complex number, or when the value of e*
is unrestricted, the definitions are not equivalent, for (1)
when cc is a complex number e'^ is an undefined and
therefore a meaningless expression, and (2) when i:c is a
fraction —, where p and q are integers, e* has q values,
while ex-p(x) has one value only.
For example, if ^e denote the positive value of the
2f 449
450 LOGARITHMS OF COMPLEX NUMBERS.
square root of e, one value of ((e))^ is — ^e, and accord-
ingly, with a base-index definition of logarithms, J is a
logarithm of — mJc to the base e ; while with the inverse-
exponential definition given above it will appear that
- /^e has an infinite number of logarithms, and that no
one of these is J.
284. To 'prove that, if (r, 6) he any complex number,
Log (r, e) = logr+(e+2n'7r)i,
ivhere log r denotes the Napierian logarithm of the
positive number r*
Let Log(r, 0) = x-\-yi,
then, by definition, (r, 6) = exp(£C + yi).
But exp(aj + yi) = exp(a^)[cos y+i sin y],
(r, 0) = exp(fl?)[cos y + i sin y],
hence, r = exp(a;) or a; = log r,
and y = 0+2n7r,
Log(r, 6) = log r + (0 + 2mr)i.
285. Def. — If r be the modulus and 6 the principal
value of the amplitude of any complex number x, then
\ogr+6i is called the principal value of Logo; and is
denoted by logic.
Thus we may write
log(a +bi) = i log(a2 + 62) + Qi
where 6 = amp(a + bi) ;
or, if a is positive,
log(a + 6i) = I log(a2+?)2) +^ tan'^- ;
a
* For a geometrical discussion of the exponential function and the
inverse exponential, or logarithmic, function, see Chrystal's Algebra,
chap, xxix., § 19.
LOGARITHMS OF COMPLEX NUMBERS. 451
if a is negative and h positive,
log(a -[■hi) = l log(a2 + 6^) + i tan " ^—\-iri;
and if a and h are both negative,
log(a + 6i) = J log(a^ -\-h^)-\-i tan - ^- — 7ri.
The following particular cases may be noticed :
logl = 0, log(-l) = 7ri
log^ = |i, log(-^)=-|^.
286. If a he any number real or complex, and x any
real number commensurable with unity, then will
{{a)Y and exp(a; Log a)
have the same values ; and the principal value of {{a)y
will be equal to exp(cc log a).
Let a = {r, Q), where — tt < 0 :!> tt, then
Log a = log r + (0 + 2ii7r)i, (art. 284)
and therefore
exp {x Log a) = exp {a?[log r + (0 + 2n7r)i] }
= exY){xlogr){cosx(0-}-2n'7r) + i8mx{0-\-2n'7r)].
But, by the properties of Napierian logarithms,
exp(a; log r) = exp (log r^) = r^,
.*. exp(a; Log a) = r*{cos x{6-\-2n7r) + ism x{0-{-2n'7r)}.
Again, we have
{(a)f = r'^{cosx(e + 2n'7r) + is'mx(e-h2n7r)}. (art. 260.)
Hence, ((«'))* = exp (a? Log a).
Putting n = 0, we see that the principal value of
((a))*, or a^, =exp(£cloga).
462 LOGARITHMS OF COMPLEX NUMBERS.
Example. — In illustration of the method of the foregoing proof
we will shew that ((9))* and exp(^ Log 9) have the same values.
The values of ((9))* are +3 and -3.
Also Log 9 = log 9 + 2w7ri,
exp(^ Log 9) = exp{^(log 9 + 2w7n)}
= exp(^ log 9) (cos nir + i sin rnr)
= 3(cos nir + i sin mr).
Assigning to n the values 0, 1, 2, 3, ... - 1, -2, -3, ... we get as
the only values of exp(| Log 9) the real numbers 3 and - 3.
Hence, ((9))* and exp(^ Log 9) are equivalent symbols.
287. If z he any number, real or complex, such that
mod(z) < 1, then will
2 3
log(l+«) = 2;~|+|-...acZm/
By the Binomial Theorem, if ic be a real number com-
mensurable with unity, we know that the principal value
of ((1+2))-
= l+a,.+^^)«H^<^^^t|^^ + .... (1)
Also, by the last article, the principal value of ((1+0))*
= l+a;log(l+0) + -^^'-{log(l+«)F+ (2)
Therefore the series (1) and (2) are equal to one another.
The series (1) is convergent for all real values of x,
provided that mod(2;) < 1, and the series (2) for all values
of X, provided that log(l+0) is finite, which is always
the case when mod(0) < 1. Hence, we may equate co-
efficients of the powers of x in (1)* and (2).
* The argument is incomplete. It has not been shewn that the series
(1) remains convergent when its terms are re-arranged according to
powers of x ; or, if convergent, that it converges to the same limit as
before. For a complete treatment see Chrystal's Algehra, chap. xxvi.
§§ 32-.S5, chap, xxviii. § 9, chap. xxix. § 22.
LOGARITHMS OF COMPLEX NUMBERS. 453
Equating coefficients of x, we have for all values of z
such that mod(2;) < 1,
\og{l + z) = z-t^t- ...adinf.
Cor. — Since Log(l+0) = log(l -\-z) + 2n7ri, we have
z'^ z^
Log(l+z) = 2n7ri-{-z — —+— — ... ad inf.
when niod(2;) < 1.
288. If mod{z) — \, and Amy{z)=\=^{2n + l)Tr, then will
log(l + ^)=^-|'+|'- ... ad inf
Let a = Amp(0).
We know, by art. 2G8, that the series
(1, a) + J(l. ^Hh§) + W . ^+2^) + . . . ac^ inf
is convergent, provided that P=\=2n7r.
Let /3==a + 7r, then the series
(1, a) - i(l. 2a) + 4(1, 3a) - . . . acZ inf,
i.e., the series z — ^z'^+^z^— ... ad inf,
is convergent, provided that a=\=(2n + l)Tr.
Hence, by art. 270, the series
z^ , Z^ 7 . /.
z--^+--...adinf
is a continuous function of z up to the limit when
mod(0) = l, provided that Am-p{z)=\=(2n-\-l)7r.
Also log(l+2;) remains finite, and is a continuous
function of z under the same conditions.
Therefore, since the equality of log(l+0) and the series
s;— -^+^— ... holds, however nearly mod(2;) approaches
to unity, it will also hold when mod(:5) = l, provided that
Amp(2;)=i=(27i + l)'7r.
454 LOGARITHMS OF COMPLEX NUMBERS.
289. The geometrical interpretation of some of the
symbols used in arts. 287 and 288 may be noticed.
Let z={r, 0), l+z = (p, <f>\ then, since log(l+;$;) is
the principal value of
j^ N^ Log(l+2;), 0 cannot
be numerically greater
than TT.
Take^Oa=l, GP = z,
then 0P = 1+^. By
hypothesis, r or mod(;2)
< 1, therefore 0 lies outside a circle whose centre is C
and radiusr. If P move round the circle from A to A'
in the positive sense, 0 increases from 0 to tt, or, gene-
rally, from 2mr to {2,n + V)ir, and ^ increases from 0 to
a maximum value sin-V, which it attains when OP
touches the circle, and then decreases to 0.
If P move round the circle from A' io A in the posi-
tive sense, 6 increases from tt to 27r, or from — tt to 0,
or, generally, from (27i — l)7r to 2mr, and ^ is negative
and increases numerically from 0 to a maximum value
sin~V, and then decreases numerically to 0.
It will be seen also that p decreases continuously from
OA to 0A\ i.e. from l-\-r to 1 — r, and then increases
from 0^' to OA.i.e. from 1 — r to l-j-r; hence log/) is
always finite.
The limiting case, in which r=\ should be carefully
considered. A' then moves up to, and ultimately coin-
cides with, 0.
As P moves from A to 0, 6 increases from ^nir to
(27i-|-l)7r, and 0 from 0 to ^. As P passes through 0,
r
LOGARITHMS OF COMPLEX NUMBERS. 455
TT
(t) changes suddenly from +-^ to — — , and then increases
from — ^ to 0 as P moves onward from 0 to A. During
the same period p changes from
2 to 0, and then from 0 to 2.
The excepted case of art. 288,
in which Amp (2;) = {^n + Ijtt, q\
is that for which P coincides
with 0.
The discontinuity of the am-
plitude 0 as P passes through 0 gives rise to some inter-
esting results in the values of certain infinite series derived
from the logarithmic series. See art. 293, Examples 1, 2, 3.
290. If z he any number, real or complex, such that
mod 2; < 1, then will
Z^ 03
log(l-0)= -^- 2~ 3~--* ^^ ^''^•^•
Changing z into —0 in the equation
log(i+«)=^-i'+J-...,
we obtain the required result.
Co7\ — If mod(0) = l, and Amp{z)=\=2mr,
then log(l -2;)= -0-- ---....
If, as in art. 289, CP = z,
and if PC meet the circle
again in Q, then OQ = l—z
or (p, (p), and the changes
in 0 and p may be dis-
cussed as before.
The geometrical interpre-
tation of the limiting case when mod(0) = l should be
considered.
456 LOGARITHMS OF COMPLEX NUMBERS.
291. Gregory's Series.— //a; he a real number behveen
— I and +1, both limits included, then will
/y.3 ^6
t8in-^x = x— — +——... ad inf.
o 5
Under the given conditions with respect to x, we have
\og{l+xi) = xi-i(xiy+i(xi)^- ...,
and log(l —xi)=—xi — i(xi)^ — i{xif - ...,
log(l + xi) — log(l — xi) = 2i(x — ix^+\x^— ...).
. Now 1+xi and 1—xi are conjugate complex numbers,
having the common moduhis +s/T+x^, and the ampli-
tudes tan"^a; and — tan"^a; respectively; hence
log( 1 + ici) — log(l — xi) = \og\/l-\-x^-\- i tan - ^x
- logVl +x^+i tan " ^x
= 2iiaii~^x,
tan -^x = x — Jcp^ + \x^ — ...ad inf.
Cor. — Since Ta,Ti-^x = n7r + tsLn-^x, we have for values
of a; between —1 and +1, both limits included,
Tei,n~^x=n'7r-\-x — ix^+^x^—...,
where n is to be so chosen that T&n'^x — nTr may lie
between — ^ and + ^.
292. Numerical Value of tt.— By aid of Gregory's
Series the numerical value of tt may be obtained to any
required degree of accuracy. The following methods
may be noticed : —
(1) In Gregory's Series, let x = l, then
4 3^5 7^-*
(2) Let a? = — 7H,then
LOGARITHMS OF COMPLEX NUMBERS. 457
6~V3\ 8 3^5 32 7 33^"
and therefore 7r = 2^3(^l-^- 3 + 5 '32-7 '33 +
(3) -^ = tan-i^ + tan-ii (Euler's formula).
4 2 3"23"^5'25 •••
^3 3 33^5 3^ ••••
TT 1 1 .
(4) ~ = ^tsji-^~ — iQXi-^^^ (Machin's formula).
4
-4^1-1.1 + 1.1-
~A5 3 53^5 55 •••
\239 3 (239)3^ "V
(5) ^ = 4tan-i--tan-y^-+tan-i^
(Rutherford's formula).
293. Example 1. — Sum the series
cos ^ + ^ cos W+^ cos 3^ + . . . ad inf.^
and sin ^ + ^ sin 2^ + ^ sin Z6+ ... ad inf.
Let C and S denote the sums of the series, and suppose that Q
is not an even multiple of tt.
Let .r=cos ^+^sin ^=(1, 0.
Then (7+ iS=x-\- \x^ + |a^ + . . . ad inf.
= -log(l -x\ since O^^nir.
Now, l-^=l-cos^-isin^,
mod(l - x) = v^(l - cos ^)2 + sin2(9 = ^2 - 2 cos ^ = 2/y^sin2?,
458 LOGARITHMS OF COMPLEX NUMBERS.
and amp(l -x)=^ tan~M ~^^"^), since 1 - cos ^ is positive,
\ J. — " COS \j/
= -tan-i(cot|),
hence, log(l -a;)=logf 2\/sin2_ j -i tan-M cot ^\
C-\-iS= -log(2.^^^) + ?-tan-i(cot|),
C= - log(2A/sin2| V and ^=tan-^(cot |).
Since tan~M cot - ) = tan-M tan^- ^) = n7r+J--, where n is so
chosen that n7r + ^-^ lies between -^ and ^, we may write
^=W7r + ^--, with the above condition with respect to the vahie
of n.
If ^=2?i7r, we have immediately from the series, C=xf, S=0.
Example 2. — Sum the series
cos <^ - ^ cos 2<^ + ^ cos 3<f>- ... ad inf.,
and sin </> - ^ sin 2<^ + ^ sin 3^ - . . . ad inf.
In the results of Example 1, put 6=7r-c{), then, provided that
^>=|=(27i+ I)7r, we have
cos </) - ^ cos 2(^+^ cos 3<^ - . .. =log( 2A/cos2r V
and sin^-i^sin2<^+^sin3<^-... =tan-*(tan^j
where n is so chosen that - - < ^t + ^ < ^.
If <^=(27H-l)7r, then, immediately from the series,
cos ^-\ cos 2(/) + ^ cos 3i/) - . . . = - 00 ,
and sin<^-^sin2^+^sin3^-... =0.
LOGARITHMS OF COMPLEX NUMBERS.
459
The curves of the series in Examples 1, 2, 3 are represented in
following figures.
y = cos x+ f cos 2x-\-j cos 3x * —
y =cos x--fcos 2X + -^C0S JX
y =COS X + jCOS JX + jCOS JX+ —
There are no isolated points.
.'-'O
• y = sin X -t- -f-sin 2x ■*■ -j- sin jx* -—-■
TTtere are isolated joints along Ox at -ztt, O, 2T, etc.
-y= sin X --j-stH 2x -t-j sinjx-
TTiere are isolated points along Ox at-tr, IT, Sir, eto.
-y = sin X ■t--i-sin jx *■ 4-sin jx^r.
There . are isolated Points along Ox at -2Tr, -it, 0, v, 2ir, etc.
460 LOGARITHMS OF COMPLEX NUMBERS.
Example 3. — Sum the series
cos ^+^ cos 3^ + ^ cos 5^+...,
and sin ^+i sin 3^+^ sin 5^+....
If ^=|=»7r, we have by examples 1 and 2,
cos ^ + ^ cos 2^+ J cos 3^+... =-log('2A/sin2|\
and cos^-^cos2^+^cos3^-... =log(2A/cos2|V
.*. by addition,
cos ^ + ^cos3^+^ cos 5^+ ... = -^ log(A/tan2|Y
If d=mr, the series =±(1+^ + 1 + ...) =±oo.
Again if ^=i=W7r, we have
sin6^ + ^sin2^+Jsin3^+... =tan-Mcot|V
and sin ^-^ sin 2^+^ sin 3^-... =tan-Mtan|j,
.'. by addition,
8in^+^sin3^+isin5^+... =^{tan-^(cot |) + tan-i(tan ^)]-
the upper or lower sign being taken according as tan - is positive
or negative ; thus from ^=0 to 9=ir^ the series = ^, from ^=7r to
d='2.Tr its value is-^* and so on. If d = mr, we have immediately
sin^ + ^sin3^+... =0.
Example 4. — Supi the series
7i sm a + -^sm 2a + — sin 3a + . . . aa inf.^
2 3
where n<\. If ^ is the sum of the series, shew that
sin ^=wsin(^+a) and that taxi(e+^=\±^ tan %
V 2/ 1-n 2
^2
Let ^=7isin a+ — sin2a+...
LOGARITHMS OF COMPLEX NUMBERS. 461
^=yiCOSa + — cos2a+ ...,
and let a = cos a + i sin a.
Then </) + 6'^ = wa + ^^-^ 4- — ^ + . . .
= -\og{l-na)
= -log{(l - 71 COS a)2+(n sin a)2}* -itan-^f -?isina\
°^ ■^ Vl-7iC0sa/
= - log\/l-27^eosa+w2 + ^ tan-^f _^i^HL^\
\l-wcosa/
and.-. • ^=tan-^f-Ji!HL^y
From this result we see that
sin ^ . (1 - ?i cos a) = cos ^ . 71 sin a,
and therefore sin ^ = n sin( ^ + a).
We have, further,
sin(^ + a) + sin^_l+7i
sin(^ + a)-sin ^ 1 — n'
whence tan((9+^Hl±^tan^.
V 2J 1-71 2
Example 5. — If sin ^=72'Sin(^+a) where n<l, find an expansion
for ^ in a series of ascending powers of n.
Since the equation is unaltered by the addition of any multiple
of TT to 6, we may consider 6 to lie between — ^ and ^.
2 A
Let ^ = (1) 0), a = {lj a), and /. a:r=(l, ^+a),
then 2^ sin ^ = .r - - and 2^ sin( 0-\-a) = a.v- —
a: ax
Substituting in the given equation, we have
1 / 1\
x--=n{ax- — I,
X \ ax)
1--
and therefore x'^=
1—na
Taking logarithms we have
26>i + 2r7r?'=log(l --) -log(l
462 LOGARITHMS OF COMPLEX NUMBERS.
In this equation r must be zero; for, if ainp(l--J = ^, and
therefore amp(l - wa)= - ^, the right-hand member of the equation
reduces to 2<^i (art. 285) ; also, since 1 - ?i cos a is positive, <^ lies
between - ^ and ^, and consequently, with the restricted value of
2 2
0^ the value zero is the only admissible value of r.
Expanding the logarithms we obtain the equation
and .-. ^=7isina+^sin2a-|-^8in3a+... .
Example 6. — If tan a=7i tan ^, where w > 1 , then will
a+nr = ^ + wisin2/3+^sin4^-f-^sin6^+...a<]?Mi/-.
where m=^?l^I— .
71+1
Let a=(l, a), 6=(1, /?),
then itana=^-^- and ttan^=,.,"~ .
a^+1 ^ 6H1
Substituting these values of tan a and tan /?, the given equation
may be written
a^-l_ 62-1
a2+l VTi*
1 -!?^
whence a^=h^ -^.
1 -mo^
Taking logarithms we have
2az + 2r7rj = 2/3^ + log^l - p) - log(l - w62)
2
or a + r7r = /?4-msin 2^4- — sin 4^+... adinf.
LOOARITHifS OF COMPLEX NUMBERS.
463
294. Short Table of Napierian Logarithms, or Real
Logarithms to Base e.
No.
log.
No.
log.
No.
log.
1
0-00
16
2-77
2
0-69
17
2-83
200
5-30
3
110
18
2-89
300
5-70
4
1-39
19
2-94
400
5-99
5
1*61
20
3-00
500
6-21
6
1-79
600
6-40
7
1-95
700
6-55
8
2-08
30
3-40
800
6-68
9
2-20
40
3-69
900
6-80
10
2-30
50
3-91
1000
6-91
11
2-40
60
4-09
1100
7-00
12
2-48
70
4-25
1200
7-09
13
2-56
80
4-38
14
2-64
90
4-50
15
2-71
100
4-61
Examples XXX.
1. Find the values of Log(l + V — 1), and represent them
geometrically.
2. Find the values of Log( — 20), and represent them
geometrically.
3. If \og{l+iidMa) = A-\-Bi, prove that J. = log sec a,
7r TT
and find -B, having given ~ « < « < o*
464 LOGARITHMS OF COMPLEX NUMBERS.
4. li\og{x-\-iy) = a-\-ip, prove that
x^ -\-y^ = e^a, and y=xt&u /3.
5. Prove that Log( — ^e) = J + (2'/^+l)'7^^, and represent
the logarithms by vectors.
6. Shew that the expressions
Log(a + hi) and J log(a2 + 62) + Tan " i-
are not equivalent.
7. Prove that Log(cos 0+i sin 0) = {0-\-2nTr)it where n is
any integer ;
and that log(cos 6+i8mO) = (0+2n'7r)i, where n is
an integer so chosen that —tt < (O + S-nTr) > ir.
8. Shew that Logxy = 'Logx-\-Logy-\-2mri, where n is
an integer.
9. If a and /3 are the principal values of the amplitudes
of two complex numbers x and y, then will
log xy = log X + log y + 2r'7ri,
where r= —1, when a-\-^ > tt,
r = 0, when — tt < a + ^S 4»7r,
r=l, when a + fi :^ — tt. .
10. From the series tt = 2^3(l -|- i+^ • i- ...),find the
value of TT to two places of decimals.
11. Find the value of tt to three places of decimals by
Euler's formula.
12. Find the value of tt to five places of decimals by
Ma chin's formula.
13. Find the value of tt to ten places of decimals by
Rutherford's formula.
14. Trace the curve
2/ = cosa7 — J cosS-Tj + ^cos 5rc— ... ad inf.
LOGARITHMS OF COMPLEX NUMBERS. 465
15. Trace the curve
2/ = sin a; — J sin ^x + i sin 5aj — . . . ad inf.
10. If msin(m0 + O) = sinm^, where m < 1, then
0 = sin0 + imsin20 + Jm%in3O+....
17. Sum to infinity, when x<\,
x^m e + lx^BmW-\-\xhmW-\-.., .
1 8. Shew that, \i x<l,
tan \— 7T = a:;sin0 — -^sin 26+-^smS0— ....
l-\-xcosO 2 3
19. Sura the series, when m<l,
m sin^a — ^mhm^2a + Jm^sin^Sa — ... ad inf.
20. In any triangle, if a > 6, shew that
h b^ b^
log c = log a - - cos 0- ^2^08 20- — 3C0S 3C- . . . .
21. In any triangle, if & < c, shew that
j5 = -sin^ + Hsin2^ + J^sin3^ + ....
c c c
22. In any triangle, if 6 < c and a<c, shew that
, b bcosA — acosB, b'^cos 2A — a^cos 2B
'°Sa= c + 2^5
, fe^cos SA — a^cos 35
"^ 3c^ "^
23. If cot y = cot X + cosec a cosec x, shew that
2/ = sin aj sin a — I sin 2a;sin^a + 1 sin Sa^sin^a — . . . .
24. Sum the series
cos ^ sin 0 + J cos^O sin 20 + J cos^0 sin 80 + . . . ad inf.
25. Shew that cot-i(l + cot0 + cot2e)
. sin 0 . ^^ sin20 , . ^^ sin^O .
= sm 0 . — sm 20 . — ^- + sin 30 . — -^ ...ad %nf.
26. Sum to infinity
cos 0 — J sin 20 — J cos 30 + J sin 40
+ i cos 50 — ^ sin 60 — ... .
2g
466 LOGARITHMS OF COMPLEX NUMBERS.
27. Prove that
\ log sec^a; = sin^ic — J sin^2fl? -h i sin^Sa; — . . . .
28. Sum the series
cos^e - i cos220 + J cos^se -...ad inf.
29. Sura the series
sin^flj — J sin^Sa; + \ sin^Sa? — ...ad inf.
30. If 0 be an angle whose cosine is positive, expand
log cos 0 in a series of cosines of multiples of Q.
31. Expand log(l — 2acos0+a2) in a series of cosines of
multiples of 0.
CHAPTER XXL
COMPLEX INDICES.
295. The expressions {{A)y and A^ have been defined
for those cases in which A is any number, real or complex,
and 5 is a real number commensurable with unity. It
has been shewn that under these conditions the values of
((A)y are identical with those of ex-p{B . Log A). We
now extend the meaning of ({A)y by the following
definition : —
Def. — For all values of A and B, real or complex, the
expression exp(5. Log^) is denoted by ({A)y.
296. To reduce ((A)y to the form x-\-yi.
LetA=(a,a),B=ib,l3)>
then, by definition, ((A)y =exp(5. Log^)
= exp[(6 cos ^ 4- ^^ sin /3)(log a-\-a-\- 'Im-rr . i)]
— exp[6 cos P log a — 6 sin /3 . a + 2m7r
+ ^(6 sin ^ log a + 6 cos yS . a + 2m7r)]
= exp(6 cos /5 log a — 6 sin ^ . a + 2m7r)
X [cos(6 sin ^ log a + 6 cos |8 . a + ^mw)
+ i sin(6 sin ^ log a 4- h cos /3 . a + 2m7r)],
an expression of the ^ovm x-\-yi.
467
468 COMPLEX INDICES.
Hence, {{A)y is a many- valued function of A and B,
whose modulus = exp (b cos ^ log a — 6 sin /3 . a -f 2m7r),
and whose amplitude
= 6 sin ^ log a+6 cos /5 . a + 2m7r+ 2ti7r.
Def. — The value of {(A)y obtained by putting m = 0
in the above result is called the ^^rmcipa^ value of {{A)y,
and is denoted by {Ay or A^.
Thus, 7l^ = exp(6cos/81oga— 6sin/3.a)
X [cos(6 sin ^ log a+ 6 cos |8 . a)
+ i sin(6 sin ^ log a -b ^ cos /3 . a)].
297. The formula of the preceding article is cumbrous,
but the method used may readily be applied to any
particular case, as in the following examples.
Example I. —To find the values of ((\/-l)) ~^-
By definition, ((\/^))^^=exp(V^ LogV'^l).
Now, -vrri = (l, |), therefore LogV^=|*'+2^T2,
... ((x/-^))^=exp[z(|i + 2.m)]
= exp(-|-2n7r).
Let n=0, then
the principal value of ((\/^))'^^ = expf -^j
= 1-000 -1-571 + 1-234 --646+ -254 --080+ -021 --005 + -001-...
= -208 approximately.
Example 2.— To find the values of ((e))^' where ^ is a real
number.
By definition, we have ((e))^* = exp(^/Loge)
= exp[^2J(log e + 2w7^^)]
=exp[-2^27r.^+(9i]
=exp( - 2^?7r . ^)(cos ^+?"sin 0).
COMPLEX INDICES. 469
Let M = 0, then e^*=cos ^+*sin ^=exp(^^).
Since the principal value of ((e))^* is exp(^i), S^ is often used in
the place of the one- valued symbol exp((90, or in place of
cos ^+isin d.
Thus, we may write cos^= — — ,
and sin 0=^ % .
2^
298. The symbol {{A)y having been defined for complex
values of A and x, we may examine the consequences that
result from the identification of a logarithm of a number
with the index of the power to which a given base must
be raised to obtain the number.
We have the following definition : —
Def. — If A,B,xhQ any numbers, real or complex, such
that one value of {(A)Y is equal to B, then x is called a
logarithm of B to the base A.
We express this relation by the equation x = Log^i^.
It will be shewn in the following articles that a; is a
two-fold many-valued function of A and B; and that
Jjog^B is identical with Log B as previously defined when
A = {e, 0), so that Log 5 is a particular case of Log,, 5.
It has already been shewn that when 5 is a real posi-
tive number the Napierian logarithm of 5 is a value of
Log 5.
299. To express hog^B in the form x + yi.
If Log^B = x-\-yi,
then, by the definition of the last article, we have
((A)Y+^^ = B.
Let A = (a, a) and B = (6, /3).
470 COMPLEX INDICES.
By tho definition of art. 295,
= exp{(ic+2/t)Log^}
= exp{(a;+2/^)(log a+a + 2m7r . i)}
= QX^(x\oga—y.a + ^mir+i{y\oga + x. a + 2mx)},
or {{A)y'^y^ is a complex number whose
modulus = exp(a; log a — y.a + 2mir),
and amplitude = y log a + a; . a + 2m7r.
But mod(5) = 6, and amp(5) = jS,
.-.since ((A)y+'J' = B, we h&ve
x\oga — y.a + 2m7r = log h,
and yloga-\-x.a-{-2m7r = ^ + 2n'7r.
„ log g . log 6 + (g + 2m7r)(^ + 2'y^7r)
nence, x- (ioga)2+(« + 2m7r)2
and (/3+27i7r)loga-(a + 2m7r)log6
(logtt)2+(a + 2m7r)2
Thus we see that Log^B is a two-fold many-valued
function of A and B.
The value of Log^5 obtained by putting m = 0, ?i = 0 in
the above result is called the principal value of Log^5,
and is denoted by log^5.
Thus, we have
j^_\ogaAogb + a§.§\og analog b
^^^^^~ (loga)2 + a2 "^^ (loga)2 + a2' '
where a and 6 are the moduli, and a and /3 the principal
values of the amplitudes of A and B,
If a = 0 and ^ = 0, so that A=a and B = b, we obtain
the known arithmetical formula
\ogab = ,^^-
° los: a
COMPLEX INDICES. 471
300. In the following examples the method of the last
article is applied in special cases.
Example 1. — Find the general value of Logee.
Let 0=(r, 6).
Assume Loge2=^+3/^
then {{e)Y+y^=z={r, 6).
We have also ((e))*+^=exp{(.r+?/i)Loge}
= exp{(^ +3/0(1 + 2w7r . ^)}
= exp{.27 - y . '2,imT + i{y + x . ^tnir)},
or {{e)y^y'' is a number of modulus exp(^-y . 2m7r) and of amplitude
1/ + X . '2.17177.
Hence, ^-y - 2wi7r = log r,
and ?/ + ^. 2m7r=^+2w7r.
Solving for x and y we obtain the result
-r ^_logr+(^ + 2?i7r). 2w7r .6'+2^7r-2m7rlogr
^' l+(2m7r)2 "^^ l + (2m7r)2 "
If m=0, i.e., if the value of e be restricted to (e, 0),
we have Log(e, q)Z = log r + 1{ 0 + 2w7r),
and . '. Log(e, o)Z = Log z.
If s=l, thenr=l, ^=0,
and we have LogJ = ^^"- ^7"+^^
l+(2m7r)2
Example 2. — Find the general value of Log4( - 2), and shew that
one value is ^.
Let Log4( -2)=x+ yi,
then ((4))^+3'»= -2.
But ((4))*+2'» = exp{(ji' + yi) Log 4}
= exp{(.r + ?/0(log 4 + 2m7ri)}
= exp{x \og4-y. 2w7r + i{y log 4 + ^. 2w7r)}.
Also, -2 = (2, tt)
exp(.r log 4 - ?/ . 2m7r) = 2, or .^7 log 4 - ?/ , 2w7r = log 2,
and ?/log44-^. 2w7r = (2?i+l)7r.
Solving for x and y we obtain the result
T oo- r - 2^ -(log 2)^ + ^^7^(271 + l)7r .(2w + Dtt log 2 - mw log 2
^''^^ '^ 2(log2)2 + 2(wi7r)2 ^^' 2(log2)2 + 2(?n7r)2 '
If m = l, 71=0, we have the value ^.
472 COMl'LKX IM>l('KS,
ExAMPLR 3.— Find the general value of Lo^xl-^-¥i^'\ aiid
shew that one vahic is \.
Let L<)g,( - H»y )=.<''+.y*,
then ((l))x+r*„_j + 4vV3
But ((l))*+«"-exp{(^'+yOLog 1}
-exp{(^+yt).2wm}
= ex jX - y . 2m7r + ix . 2«i7r)
oxi)( -y . SmTr)"! or -y . 2mrP'=0
and X. 2w7r - ??^ + 2M7r.
3
Now m cannot equal zero, for then - J+*^- = exp(()) = l,anim-
]>o8Hible result ;
hence, .'/="0.
Weliavealso ' = -?-',
^A-^-^':yt^'
a real quantity in all cases.
If m=»l, 71 = 0, we obtain the particular value i.
Examples XXXL
1. Find tho values of ((l)y.
2. Express ((l+i))' in the form x + y% and shew that
mod(l 4- if = '46 approximately,
and amp(l+'i)^=slog2.
'.\. Find tho ni\i roots of a*^-\ and shew that their sum
is zero.
4. Prove that the real part of (V - ly*'^^'^"'"^^ is
e 8 cos(i7r log 2).
COMPLEX INDICES. 473
5. If {a - bs/^^iy"^^^ be exhibited in the form a + ps/^^,
where a and /3 are real quantities, find the values
of a and /3.
6. Shew that (a + biy+'i^ is entirely real if qlogm+2><l>
is a multiple of tt, where m = mod{a + hi) and
0 = amp(ct + 6i).
7. If -'^<0< J, shew that
(a+ia tan ^y''g(««ec0)-0f = exp{(log a sec 0)H^^}.
8. Reduce V- — r^!r^ — • to the form x-\-yi.
{a — hiy-'i' ^
9. Find the values ofLog^^{\/—l).
10. Find the general value of Log^2, and shew that one
value is i log 2/2^.
11. Prove that log2,._ 1^-1} = -.
12. Prove that Log^^^^{S/-l} = (2n+^y(2m+^,),
where m, 7i are any integers.
13. Prove that
where m, n, p, (/ are any integers.
14. If af^ =a(cosa + isina) find the general value
of X. If a = 2, a = 0, shew that the result gives
x=±^2.
CHAPTER XXII.
CIECULAR AND HYPEKBOLIC FUNCTIONS OF
COMPLEX NUMBEES.
301. The geometrical definitions of the circular and
hyperbolic functions imply that the variables are real
quantities. At the present stage sin(a + ^/3) and
co.sh(a + ^|8) are undefined and consequently meaningless.
Before extending the definitions, it will be well to
recapitulate briefly the properties of these functions so far
as real quantities are concerned.
Having defined the functions as the ratio of certain
lines connected with an angle, or with a hyperbolic sector,
we proved geometrically from a property of the circle
or rectangular hyperbola that
cos2i» + sin2a3=l,
and cosh^ic — sinh^aj = 1.
It was then shewn geometrically that certain addition
formulae were true, viz. : —
cos(a3 + y) = cos xcosy — sin x sin y,
cosh(ic + 2/) = cosh x cosh y + sinh cc sinh ?/,
sin(aj + y) = sin x cos y + cos x sin y,
sinh(i« + y) = sinh x cosh y + cosh aj sinh y.
From these formulae similar results for any number of
variables readily followed, and thence it was shewn that
474
FUNCTIONS OF COMPLEX NUMBERS. 475
the functions might be exhibited as convergent series in
powers of the variables, thus
cosa; = l-|+g-...,
coshaj = l+-2 + |^ + ...,
[3 [5 _
/y>3 /vi5
sinha; = aj + ^ + r5 + ....
It also followed from the definitions that each function
was one-valued, and that for real values of the variable
the circular functions were periodic, while the hyperbolic
functions were non-periodic, the hyperbolic sine, for
example, increasing continuously with the sector from
negative infinity through zero to positive infinity.
A division of the functions into the two classes of even
and odd functions was an obvious inference from the
geometrical definitions.
302. The series
[2 + 14 ••••
■ - + I+I+--
are convergent for complex as well as for real values of x,
and, when x is real, they are respectively equal to cos x,
cosh X, sin x, sinh x.
476 CIRCULAR AND HYPERBOLIC
We may therefore extend the meaoings of cos a;,
cosh X, sin x, sinh x as in the following definition : —
Def. — When x is any quantity, real or complex, the
series
aj2 . a;4
.-!+-...
'2 • 14
aj2 . x^
/yi3 /¥>5
are called the cosine of a?, the hyperbolic-cosine of a;, the
sine of a;, and the hyperbolic-sine of a?, respectively.
We further define sec x, sech a?, cosec a?, cosech a; as the
reciprocals of cos x, cosh x, sin a?, sinh x, respectively ; and
tan X, tanh x, cot a;, coth a; as the fractions
sin a; .sinh a; cos a; cosh a? ,. ,
, ) — -. — > -. — ) -V-, — , respectively.
cos a? cosh a; sinaj sinh a; ^ ''
Since each of the defining series is a one-valued func-
tion of Xj it follows that each of the circular and hyper-
bolic functions of a complex number is a one-valued
function of the variable.
The defining series for cos x and cosh x involve even
powers only of x, hence cos x and cosh x are even func-
tions of the complex variable. In like manner, sin x and
sinh X are odd functions of the variable. Thus, for a com-
plex, as for a real, variable we have
cos X = cos( — x), cosh X = cosh( — a;),
sina;= —sin ( — a;), sinh a; = —sin (—a;),
and so on for the other functions.
FUNCTIONS OF COMPLEX NUMBERS. ^^J'J
303. Exponential Values of the Circular and Hyper-
bolic Functions of a Complex Variable. — From the
definitions of art. 302 it follows, for all values of x, that
- , . , {xif , {xi)^ ,
GO^x + i^mx^l+xi+Y^-^^ ^ g+...
= exp(a;i),
(xi^ ('T/h ^
and that co^x—is>mx = l—xi+\ — o~r~oQ + "-
= ex-p{ — xi).
Hence, by addition and subtraction, we have
cosic = ^{exp(a;'^) + exp( — ^i)},
and 8mx = -^.{ex-p{xi) — exYi( — xi)}.
In like manner we have, for all values of x,
T^ X
cosh aj + sinh cc = 1 + aj + —-+——-— + .. .
1 . Z 1. . Z . o
= exp{x),
X X
and cosha; — sinha;=l—i»+— - — -——- + ...
1 . Z i. . Z , o
= exp{ — x),
cosh cc = ^{exp(i:c) + exp( — a?)},
and sinh a; = ^ { exp(ic) - exp( — x)}.
304. Periodicity . of the Circular and Hyperbolic
Functions of a Complex Variable. — If x be any
number, real or complex, and n any integer, positive
or negative, we have
exp(a:!i) = exp (xi + 2n7ri),
and exip( — xi) = exY)( — xi — 2n'7ri),
cosx = cos(x+2n7r), and sinfl? = sin(aj + 27?7r),
478 CIRCULAR AND HYPERBOLIC
i.e., cos a; and sin a; are periodic functions of the variables,
having for the period the number 27r.
Again, since
exp(7^^) = cos tt + i sin tt = — 1,
exp(iri) = — exp(a:;i -\-'ln-\-l . iri),
and exip{ — xi)=—ex'p{ — xi — 2n + l.7ri),
cos ic = — cos(a; +2n-\- Itt),
and sin a; = — sin(ic + 27i + Itt).
From the above results we see that tan x and cot x are
periodic functions of x, and that the period is the number
TT. So also secaj and cosecaj are periodic, the period
being 27r.
We have also
exp X = exp(fl3 + 2n7ri),
and exp( — fl3) = exp( — a? — 27i7ri),
cosh X = cosh (x + 2mri),
and sinh x = sinh(a; + ^niri),
i.e., the hyperbolic cosine and sine of x are periodic func-
tions of X having the imaginary period 2^1. It readily
follows that the hyperbolic tangent and cotangent have
the imaginary period iri.
Thus, we see that when complex variables are considered,
the analogy between the two classes of functions is com-
pleted by the establishment of the periodicity of the
hyperbolic as well as of the circular functions.
305. By art. 303, we have
cos a? -}- i sin a; = exp(a?i),
and cosa? — 'isina! = exp( — a;i),
therefore, by multiplication,
cos^a: + sin^a? = exp(O) = 1 .
FUNCTIONS OF COMPLEX NUMBERS. 479
In like manner,
cosh X + sinh x = exp(cc),
and coshfl? — sinha3 = exp( — cc),
cosh^i^c — sinh^^ = exp(O) = 1.
Again,
cos(ir + 2/)
= i{exp(aj + 2/-'^) + exp{-a) + 2/.^)}
= J {exp(fl3'i)exp(2/^) + exp( - a;7:)exp( - ?/i) } ;
and cos a; cos 2/ — sin i:c sin ;y
= i{exp(a;i) + exp(-a;i)}{exp(2/i) + exp(-2/^)}
-4p{expW-exp(-a;^)}{exp(2/i)-exp(-2/i)}
= i{exp(a3^)exp(2/^) + exp( - aji)exp( -yi)};
cos(i:c + 2/) = cos x cos 2/ — sin a; sin y.
In like manner, we have
sin(a; + 2/) = sin aj cos 2/ + cos a? sin 2/,
cosh(aj + 2/) = cosh x cosh 2/ + sinh x sinh 2/,
and sinh(a; + y) = sinh aj cosh y + cosh a? sinh y.
Thus, the fundamental formulae, and therefore also
all the consequences of these formulae established in
the preceding chapters, hold for complex values of the
variables.
306. Formulae of Interchange of Circular and Hyper-
bolic Functions.
We have cos(aji) = J{exp(a;i. ^) + exp( — a;^.^)}
= J{exp( — ic) + exp x]
= cosh X.
In like manner,
sin xi = i sinh x,
cosh xi = cos X,
sinh xi = i sin x.
480 CIRCULAR AND HYPERBOLIC
aud, consequently,
tan ict = i tanh x,
tanha;i = 'itanaj.
When ic is a real quantity we have defined gd ^^ as a
certain angle between — ^ and ^, and have shewn that if
0 = gd u, then tan - = tanh ^. We now extend the mean-
ing of gd u by the following definition.
Def. — If u be any number, real or complex, and ^ be a
number such that tan ^ = tanh-, then 0 is called the
gudermannian of u.
With this definition, we have for all values of u
l + tan^l l+tanh2|
sec 6 = ^ = = cosh u,
1— tan^^ 1— tanh^^
2 tan - 2 tanh ^
tan 6 = ^ = = sinh u,
l-tan2? l-tanh^l
and exp (u) = cosh t6 + sinh u = sec 0 + tan 0 = tanf J+^j,
or u = LogtSin(~+^.
The following example is introduced to shew the equi-
valence of the relations
tan ^ = tanh ^ and u = Log ^^^^f r +^)-
FUNCTIONS OF COMPLEX NUMBERS. 481
Example.— If w=Logtaii(^ + |) prove that tanli- = tan ^.
We have Log tan^| + |W log tan^| + 1) + ^mri,
| = ilogtan(J + |)+.^,^^•,
.-. tanh|=tanh{ilogtan(|+|)}
^exp{ilogtan(|+^)}-exp(-^logtan(|+|)}
~exp{ilogtan(| + |)}+exp{-ilogtan(^+|)}
exp(logtan(|+|)}-l
exp{logtan(^ + ^)} + ]
V4^2/ ^ ^
= 7i = tan -.
307. Inverse Circular Functions.
Def. — If cos(a? + m) = a + |8i, then x-\-yi is called the
inverse-cosine of the number a + /3i.
Since cos(2n'jr±x-\-yi) = cos(x-\-yi),
it follows that the inverse-cosine of a given number
a + fii has an infinite number of values. The notation
for this many- valued function is Cos-\a + pi).
For any given value of the real number x, we may
evidently determine a positive or negative integer n such
that either 2n7r + x or 2n7r — x shall lie between 0 and tt,
and this can be done in one way only. For example,
if ic lie between 0 and tt we make n = 0 in 2mr + x,
\i X „ TT „ Stt „ n = \m2nir—x,
iix
))
0 „ -TT ,.
71 = 0 in 2nir — x,
iix
5J
-TT „-27r ,.
71 = 1 in 2mr-\-x,
and so on.
2h
482 CIRCULAR AND HYPERBOLIC
The value of the general expression 2n'n'±{x-{-yi),
whose cosine is a + ^i, for which Inir + x or ^nir — x
lies between 0 and tt, is called the principal value of
Cos-^(a + /3i), and is denoted by QO^-\a + ^i), and we
'have Cos - \a + /3^) = 27i ± ttCOS - \a + /3i).
In like manner, if sec(a; + 2/'^) = a + /3i,
then iC + 2/'^ = Sec"^(a + ^i),
and the value of 2n'7r±(x-\-yi), for which 2mr-\-x, or
2mr—x, lies between 0 and tt, is called the principal
value of Sec-^(a + /3i), and is denoted by sec-i(a + ^i).
We have also ^QC~\a-\- fii) = 'tnir±^QQ,~\a-\- ^i).
Similar definitions may be given of the other inverse
circular functions, the principal values of the inverse
sine, cosecant, tangent and cotangent, having their real
parts between — ^ and ^. With this convention,
Sin-i(a+)Si) = '^7r + (-l)^sin-i(a + /3^),
Cosec-i(a + ;8i)=7i7r + (-ircosec-i(a + j8i),
Tan-l(a+iS^) = '^^'7^ + tan-l(a4-/3^),
Cot-i(a + j8i) = ?i7r + cot-i(a + /3i).
308. Inverse Hyperbolic Functions. — Definitions simi-
lar to those of the last article may be given for the
inverse hyperbolic functions of a complex variable.
The periods of the hyperbolic functions are imaginary,
and we take as the principal value of Cosh-^(a + /5^) or
Sech"^(a + /3'i) that value for which the imaginary part
lies between 0 and iri ; and for the principal value of each
of the remaining functions that value for which the
imaginary part lies between — ^ and + y*
FUNCTIONS OF COMPLEX NUMBERS. 483
With this convention,
Cosh - i(a + /3i) = 27i,7ri ± cosh - i(a + /3i),
Sinh-Xa + /3^)=7^7^^ + (-l)^sinh-l(a + /3^),
Tanh " \a + ^i) = niri + tanh -\a + /Si).
309. We add a few examples in illustration of the sub-
ject of this chapter.
Example 1. — Eeduce cos{a + (3i) to the form .v+i/i.
We have cos(a + /3t) = cos a cos f3t — sin a sin (Si
= cos a cosh ^ - sin a . i sinh /3
= cos a cosh j3-i sin a sinh /?.
Example 2. — Eeduce tanh(a+/5i) to the form x+yi.
We have tanh(a + ^*):='J"MM^^)^
cosh(a + ^^)
_ sinh g cos /? + 1 cosh a sin j8
cosh a cos /? + z sinh a sin ^
_(sinh g cos ^ + ^' cosh a sin ^)(cosh a cos ^ - 1 sinh a sin ^)
cosh'-^g cos^/^+sinh^g sin^/?
_ sinh g cosh g(cos2/5 + sin^^) + i sin /3 cos /^(cosh^a - sinh^g)
cosh^a cos^;8 + sinh% sin^jS
_ sinh g cosh g + 1 sin ^ cos ^
cosh^g cos'^/? + sinh^g sin^yS
_ sinh 2g + 1 sin 2/5
cosh 2g + cos 2j8 '
Example 3. — Express sin~Ya + /30 in the form x-\-yi.
We have g+/3^ = sin(^+^^) = sin;rcoshy + ^■cosa?sinhy,
sin ^ cosh y = g, (1)
and cos ^ sinh y=)S. (2)
From (2),
(l-sin2^)(coshV-l)=/3^
and . •. by ( 1 ), cosh^ + sin% ^d^+l + p^, (3) .
and consequently from (1) and (3), remembering that
cosh 3^ > 1 > sin ^, we have
cosh j/ + sin .r = */(g + 1 f + ^^
and coshy-sin^=\/(g-l)''^4-^^
cosh.v=i{v/(or+T)2+F+V(g- l)2 + iS-'},
and sin.r = i{V(g+l)2 + /?'-v'(g-l)2 + y82}.
484 CIRCULAR AND HYPERBOLIC
Now, since sin~Xa+/8i) is the principal value of the function, x
must lie between -J and J, therefore cos^ is positive, and there-
fore by (2) y and fS have the same sign.
Hence, finally,
±zcosh-^J{V(T+T7+^+ sl{a-\f-\-p%
the upper or lower sign being taken according as /? is positive or
negative.
Example 4. — To express ta.n~\a + fii) in the form x+yi.
Since tan~\a+^0 is the principal value of the function, its
real part lies between -^ and +^, i.e., -!^ < ;r<^'
We have a+^t=tan(^4-yi),
and, by an easy reduction, similar to that of Example 2,
. / , •v_sin 2a7+^sinh 2y .
cos 2.27+ cosh 2y
thus we have
^^ sin 2.g ,..
cos 2jp+ cosh 2y
B= ^^"^^3/ . (ii)
cos 2.r + cosh 2^^
From (i), a and x have the same sign, since -~ <cx<'^\ and from
(ii), yS and y have the same sign.
Affain g^+jg^— sin^2.y+8inh^2?/ cosh^2y - cos''^2^
^ ' '^ (cos 2.r+ cosh 2y)2~(cos2^+ cosh 2y)2
_cosh2y-cos2.r
~ cosh 2y + cos 2.r'
... l-a2-^2^ 2cos2^ (iji)
' cosh2y + cos2^
and l + a2 + /32= 2cosh2L^ ^i^^
cosh 2y + cos 'zx
From (ii) and (iv), tanh 2y = ^^^f^^,,
and therefore y=itanh~^ /— tso*
•^ 2 l+a2+^
FUNCTIONS OF COMPLEX NUMBERS. 485
2a
Also from (i) and (iii), tan ^x= ^ — m>
and therefore 2^=r7r + tan~-^- n — p^,
where r is to be taken so that _^<^<^, and that x may
have the same sign as a ; hence if a^ + /3'^ < 1,
and, if a^ + yS^ > 1,
the upper or lower sign being taken according as a is positive
or negative.
Hence, finally, if a'^+(3^<l,
andif a^ 4-/52 >1,
tan-Xa-f^z}= ±|+itan-l^_^, + ^itanh-^^J^^
the upper or lower sign being taken according as a is positive or
negative.
Example 5. — The roots of the cubic equation c(^ + ^IIx+G=0^
where H is positive, may be obtained in the following manner.
Let x=—. the equation becomes
m
Comparing this equation with the equation
sinh^w + 1 sinh u — ^ sinh 3w = 0,
and taking m and u such that m^ = —j^ ainh.Zu= —4m^Gj we see
that 2= sinh w.
From the equations for m and u, we have sinh 32^= -—7-3, a
211^'
real quantity ; hence u can be determined from a table of hyper-
bolic sines. Since sinh 3w=sinh(3w±27ri), the three roots are
2\fH8mhu, 2x/iS^sinh(w + i^) and 2v/Fsinh(w-t!li:).
486 CIRCULAR AND HYPERBOLIC
Example 6.— Given that
t*=logtanf^ + |j=a;+a3^ + a5^+a7;r^ + ...,
shew that x=u- a^u^ + a^u^ - a-jV? + . . . .
Silica ««==log tan(^+^j, we have x=gd. u,
and therefore tanh % = tan ^.
2 2
Now, tanh ^ = ^ tan ^*, and tan ^ = i tanh 5*,
' 2 i 2' 2 i 2'
tan ^ = tanh ^, or ui = gd(ari).
2 2
Hence, ^' is related to ui in the same manner as u is related to
X, and therefore if u=x + asx^ + a^s^ + . . . ,
we must have also
xi = ui + a^uif + a^{uiy + . . . ,
or :i7=w-a32i3+a5W^-....
Examples XXXII.
1. Shew that the difference of the squares of the moduix
of cos(a + j8i) and sin(a + ^i) is cos 2a.
2. If 2sin(a + )8V^T) = flj + 2/>v^^^, pi^ove that
aj2+2/^ = e2^+e-2^-2cos2a.
3. If sin(a + /3i)=a; + 2/^, prove that
X^ /Jj2 /jr.2 qj2
sin^a cos^a cosh^^ sinh^/?""
4. Ifsin(O4-0x/ — l)=p(cosa + /v/ — lsina),when p,a,0,(p
are real quantities, then will
tan a = -7 jcot^.
e'^ + e""^
r -n XT. i. X v/ . /D-\ tanhasec^/^ + isech^atan^
5. Prove that tanh(a + m)= , , . ■■ . . — oo ^.
^ ^ ^ l + tanh^atan^/S
6. Find the real and imaginary parts of sec{x-^ iy).
FUNCTIONS OF COMPLEX NUMBERS. 487
7. Express sin(a + ^x/^) sin(a -/g^/^) -^ ^^^^^-^^^^
^ sin(a-|8x/-l) sin(a + /3%/-l)
form.
o T> . 1 . i. / . • N sin2a; + isinh2'?/
S. rrove that tQ^mx-[-%y) = ^ — ; i-iv •
^ ^^ cos zee + cosh 2^/
9. If a; = (e^-{-e~^)cosa, 2/ = (e^ — e~^)sma, prove that
sec(a + 6 V - 1) + sec(a - 6/v/^) = ^alpT^a'
sec(a + 6V^-sec(a-6V^ = ^^^£2--
10. Prove that
log sin (ot + /3^) = J log J (cosh 2/3 — cos 2a) + cjii,
where 0 = amp (ein a cosh ^8 + i cos a sinh /5).
11. If a be a re^l number not numerically greater than
unity, prove that the equation ^\nx = a has no
imaginary root.
12. Eeduce cos"^(a + ^'i^) to the form x + yi.
13. If cos(0 + (pi) = cos a + i sin a, shew that sin^0 = sin^a.
14. If cos(a + /5i) = a + 6i, and sin(a + /3i) = c + <i'^, prove
that tan a = - = - - and hh " ^z? = (^2 4. d^)(b + c)2
15. Reduce tan-^(cos0 + isin 0) to the form A+Bi.
16. Assuming that sinu = ^. — and cosu = — ^ ,
shew that, li x-\-iy = cid,T^{^-\-iti), then
tan2f=— ^^^
c" — cc^ — 2/^
17. coshfcc + ^j = '^sinha;.
18. sinhf a:; + ^y=icosha?.
488 FUNCTIONS OF COMPLEX NUMBERS.
19. tanh(a;+^) = cotha;.
20. coth('a; + ^) = tanh£c.
21. ^Qch\x-^~\ = — i cosech x.
22. cosechraj + ^ j = — i sech x.
23. cosh('^— a;j= — isinhoj.
24. sinhf^ —xj=^icoshx.
25. ia.nh(l^-x) = -coth a;.
26. If 0 and u are connected by the equation tan ^ = tanh ^,
find the general value of 0 in terms of u, and, con-
versely, the general value of i6 in terms of 0.
27. If tan I = -^ tan |, then will log tan(| + 1) = ix.
28. If u = log tan(^^+|\ then will a;i = log tan('|+|^\
on I?- J i-u rj.T- • 1 COS 20 , cos 40
29. Find the sum of the seiies 1 — z — ^-f-
1.2 ■ 1.2.3.4
30. Shew that
— sin a + |T sin 2a+...acZ inf. = sinhf a; cos ^jsinf a; sin ^ j.
31. Shew that the roots of the cubic equation
x^-\-SHx-\-G = 0, when H is negative, may be
determined by the aid of a table of hyperbolic or
circular cosines, according as 6^^4-4^^ is positive
or negative.
MISCELLANEOUS EXAMPLES.
489
Miscellaneous Examples. IV,
1. Prove that
sin(^ — y) + sir«(y — a) + sin(a — ^)
+ 4 sin i(^ - y)sin i(y - a)sin i(a - ^) = 0.
2. If no two of the angles a, /3, y are equal, or differ by
a multiple of four right angles, then Esin(/3 — y)
cannot vanish.
3. Prove that
2sin(^ + y) X E sin(/3 — y) = 2 sin /3 sin y, cos ^ cos y, 1
sin y sin a, cos y cos a, 1
sin a sin ^, cos a cos |8, 1
4. If a, /3, y are unequal, and if E cosec a(sec ^ — sec y) = 0,
then will 2sin(^ + y) = 0.
5. If acos^cosy + 6sin/5siny
= a cos y cos a + 6 sin y sin a
= a cos a cos ^8 + 6 sin a sin /? = c,
then will g + « Bi"(«+./3)+|i"(^+y),
a 6 sin(a + y)
c c_sin(a — /3) — sin(/3 — y)
a 6~ sin(a + y)
2sin(/5 + y) = 0, and -+1 + ^=0.
6. If (XC0SaC0S|8 + fesin asin^ = c,
and a cos /3 cos y + 6 sin /3 sin y = c,
then will
(i+c^-i)^"'"'"^'^+S+a^-p)
11 1
sin a sin y= -5+10 — o-
' a^ Ir c^
490 MISCELLANEOUS EXAMPLES.
1. Having given
a^cos a cos /3 + a(sin a + sin j8) + 1 = 0,
a^cos a cos y + ci(sin a + sin y) + 1 = 0,
prove that
a^cos ^ cos y + a(sin ^ + sin y) + 1 = 0,
^, y being unequal and less than x.
2. Having given
a^cos a cos j8 + a(sin a + sin ^) + 1 = 0,
a^cos a cos y + a(sin a + sin y) + 1 = 0,
prove that
cos a + cos ^ 4- cos y = cos(a + i^ + y),
j8, y being unequal and less than x.
3. If acos(0+^) + &cos(e-^)+c = O,
a cos{(f) + V^) + 6 cos(^ — \lr) + c = 0,
a cos(i/r + 0) + 6 cos( V^ - 0) + c = 0,
and, if 6, 0, yp- are all unequal, then
j^ a sm^6 + bsm^(l) _ h sin^^ + c sin^o^ _ c sm^O -\- a sin^cp
Fco^O + cco^ "" c cos^O + a coii^<t> a cos^O + h cos-^^'
then will a^ + 6^ + c^ = Zahc.
5. If
a cos a cos ^ + 6 sin a sin /3 = a cos /3 cos y + 6 sin ^ sin y
= a cos y cos ^ + 6 sin y sin (5 = a cos ^ cos e + & sin S sin e
= a cos e cos a + 6 sin e sin a = c,
then will
sin(a + /3) + sin(/3+y)+sin(y + ^)+sin(54-e) + sin(e+a) = 0
and
sin(a + y) + sin(y + e) + sin(e+/3) + sin(^+^) + siii(^ + a) = 0.
MISCELLANEOUS EXAMPLES. 491
6. If
a cos a cos /3 + h sin a sin /5 = a cos /3 cos y + Z> sin p sin y
= (Xcosy cos(5 + ^siny sin ^ = acos ^cose+6sin ^sine
= a cos e cos a + 6 sin e sin a = c,
y-
1. Eliminate ^ and ^ from the equations
tan 0 + tan (p = a,
tan 0 tan ^(cosec 20 + cosec 2(p) = 6,
cos(0 + 0) = c cos(0 — 0).
J, j^ X cos 0^ , ysinO^_^ x cos 0^ y sin 0^ _
Z. XL 7 1. 7 i,
a 0 ah '
Oi — ^2 = 2«' ^^^^ will ""2 + 12= sec^a.
8. Prove that the elimination of 0, (p from the equations
r cos(20 —a) = 711. cos'^0,
r cos( 20 — a) = 7n cos^(p,
tan 0 = tan 0 + 2 sin /3,
m cos a
gives r = .j 2 — ^-g-^.
^ 1 — cos^a srn^p
4. Eliminate 0 from the equations
aj = asec(0 + a), y = htsiu((f) + /3).
5. Eliminate 0 from the equations
cos^O + acosO = 6,
sin^O + a sin 0 = c.
6. Eliminate 0 from the equations
4(cos a cos 6 + cos 0)(cos a sin 0 + sin 0)
= 4(cosacosO + cos'0-)(cosasin 0 + sin \/r)
= (cos 0 - cos ^)(sin 0 — sin xj/-).
492 MISCELLANEOUS EXAMPLES
s.
1. Shew that (l + sm0)(3siii^+4cos0+5)
is a perfect square.
2. Solve the equations
cosaj+cos2/ = a,
sin X + sin y = h.
3. If X sin^A cos B-y sm^B cos A + 0(cosM - cos^^) = 0,
and
z sin^Ccos A —xsin^A cos C+ y{cos^C— cos^J.) = 0,
where A, B, C are the angles of a triangle whose
sides are a, b, c, then ax -hy = cz.
4. Eliminate 6 from
sin 3(1+0) + 3 sin(|+ e) = 2a,
sin3g-O) + 3sing-0) = 26.
5. Eliminate 0 from the equations
-cos 0-1 sin 0 = cos 20, - sin 0+| cos 0=2 sin 20.
a o ah
6. Eliminate 0 from the equations
(a + 6)(a3 + 2/) = cos 0 + sin 0 sin 20,
and 2(a2 ~ b^)(x^ - 2/^) = 3 sin 20 + sin320.
1. The sides of a triangle are a(l— m), a, and a(l+W/),
where m is small, shew that the mean angle
TT ,- 3^3 ,_
= g -^ V 3 m^ — -^Y~ '^* — 3 V 3 m^. . . nearly.
2. Eliminate 0 and 0 from the equations
a cos(0 + a) sin 0 .,,. \,7«/,, \
r= — /A ; = ^-g, asm(0-a) + 6sm(0 + a) = c.
0 cos(0 — a). sm0 ^ _ ^^
MISCELLANEOUS EXAMPLES. 493
8. Eliminate Q and 0 from the equations
cos 6 + cos 0 = a, cot 6 + cot 0 = 6, cosec 6 + cosec 0 = c.
4. Establish the relation
tan((X + 6 + . . . + '^ + '>^)
_ 2 1 + cos 2a 2 l+cos2m
~~ sin 2a— tan 6— sin 26— "* tan-T^
5. Prove that if m be a positive integer,
cos(m + l)0
1 1
= cos 7YiO\ 2 cos Q •
2cos0- 2 cose
where 2 cos 6 is repeated m times.
6. Solve the equation
1111
— — cos 6)
2tane+
tane+ 4tan0+ tane4- 4tane+*"
= 2x/3 cosec 20,
6 being a positive acute angle.
i-
1. Shew that if 0 be a positive angle not greater than
a right angle, sin 0 will be less than 0 — -^.
2. Evaluate ^log(l+^) ^^^^^^ ^^^
1 — cos fl?
3. Find the limit, when x is indefinitely diminished, of
e^-l + log(l-a;)
sin^cc
4. Find the limit, when x is indefinitely diminished, of
sinic + sinhfl? — 2a;
x^ *
5. Shew that, as x continually diminishes,
Sx~^tsinx + Sx-^ia.n~'^x — 6x-^
continually approaches the value unity.
494 MISCELLANEOUS EXAMPLES
6. If regular polygons of the same number of sides be
inscribed and circumscribed to the same circle,
then, when the number of sides is very large, the
difference between the perimeter of the circum-
scribed polygon and the perimeter of the circle
is double of the difference of the perimeters of
the inscribed polygon and circle.
V'
1. If tan 0 = ic, tan nQ = a, we have
a = 7^ — 1^ » hence shew that
L-ms
tan 0 + tan(e + -) + tan('e-l-— ) + ... to n ten
z= ^n cot n(j + e\
2. Prove that
tan 6 tsLnfe + -\m (e-\-^\..ton factors
= (-1)1 or (-1)V tan 710,
according as n is even or odd.
3. Prove that
tan20 + tan2^0 + ^Vtan2(e +—) + ... to n terms
4. Find the value of
cot 0 + cot^O + -) + cot^0 +-^'\-\-... ton terms.
MISCELLANEOUS EXAMPLES. 495
5. Find the equation whose roots are
tan^—j -, tan^^, tan^—, tan^^j, tan^-^.
6. Find the value of
tan^^ + tan^l^+tan^^^ + tan^^ + tan^^.
a
1. Prove that ^-3!^+^^- J^-,...=^2.
2. Sum the series
1.22.3"^5.62.7^9.102.11"^13.142.15
to infinity.
3. Prove that, if J<^<7r, then
sin J. + J sin^ J. + i sinM + . . .
= 2(cot^ + icot3| + ...).
4. Prove that, if 0< ^ < J, then
\ sinM + \ sin* J. + \ sin^J. + . . .
= 2(tan2| + itan6^ + ...).
5. If _^<0<^, then will
2 sin^e + 1(2 sin20)2 + 1 (2 sin20)3 + . . .
= 2(tan20 + \ tan«0 + i tan^^^ +...).
6. Prove that
sin0 + isin3e + isin50+...
= 2(s;n0-isin3O + ism50-...).
496 MISCELLANEOUS EXAMPLES.
K.
1. Obtain the expansion oi QOsnO in terms of powers of
cos Q by comparing the expansions of the equival-
ent expressions
log(l + 2a;cos0+a;2) and log(l+aJ0)+Wl + -\
where 0 = cos 0 + i sin 0.
2. If cos(O-l-J) be positive, expand logcos(^+|^j in a
series of sines and cosines of multiples of 6.
3. Expand cos^"0 + sin*^0 in a series of cosines of multiples
of a
4. From the formula 2 sin 0 cos 0 = sin(0 + </>) + sin(0 — (fi),
deduce
[Tjlri |3 |2yi-2 |5 |27i-4 |2ti+ 1 [0
_(a+l)2n+l4.(^_l)2n+l
2[2;yH-l
•^- ^^l+^^+^2 = ^o+%'^ + «2^'+--- prove that
a^ + ctiflj cos Q + agflJ^cos 20 + . . . ad. inf.
_ m + 'yipa;^ + {n + mp + 'nga?^)a; cos 0 + mga;^cos 20
~ 1 +p V + g2^4 -f 22Jiz;(l + ga;2)cos 0 + 2g'a;2cos W '
6. Sum the series
0 sin 0 + 2c2sin 20 + Seisin 30 + . . . to ti terms.
X.
1. If Vn+\ = pv„ - gvn-i, and Vi =p, ^0=2^, shew by mathe-
matical induction that
, . -, y'yi(ti-r-l)...(?i-2r+l) n-2w 4.
■^^ ^ 1.2.3...r ^- ^-T"--
MISCELLANEOUS EXAMPLES. 497
2. Find the sum of n terms of the series
sin0 + 2sin(0 + a) + 3sin(0+2a)+.....
3. Sum the series
l^sin 0 + 22sin20+32sin3O+... to n terms.
4. If n be equal to 3m ± 1, prove that
{U^}-|— j2— + ^^ 1^
f7i(n-l)(n-2)(7i-3)_^(77.-l)(^-2)(7i-3)(7i-4)) _
+ 1 ^ + ^ |3 -...-0.
5. If r and s be unequal inte£:ers, prove that
Ssecf-^- + e)sec(?^+0) = ^ or 0.
l-(-l)2cOS7l0
according as n is even or odd ; the summation
extending for all integer values of r and s from 0
to n — 1 inclusive.
6. Shew that the series whose first term is
_^ a2^uia + a^^m'2^a-\- ...-\-an^m{n — '\)a
% + ttgCOs a 4- (XgCOS 2a + . . . + a„cos(?i — l)a'
and 7'*^ term
_^ar+isina + o^r+2sin2a+...+<Xr-isin('yi-l)a
tty + ttr+icos a + . . . + a^_iCOs('yi •- l)a
any term being formed from the preceding by a
cyclical interchange of letters, is an arithmetical
progression, whose common difference is a, pro-
vided that ua = ^ir.
2i
MATHEMATICAL TABLES.
LOGARITHMS OF NUMBERS.
No.
0
1
2
3
4
5
6
7
8
9
1004
0017337
7770
8202
8635
9067
9499
9932
0364
0796
1228
13
0056094
6523
6952
7380
7809
8238
8666
9094
9523
9951
1270
1038037
8379
8721
9063
9405
9747
0089
0430
0772
1114
1340
1271048
1372
1696
2020
2344
2668
2992
3316
3640
3964
1478
1696744
7038
7332
7626
7920
8213
8507
8801
9094
9388
1655
2187980
8242
8505
8767
9030
9293
9554
9816
0079
0341
1769
2477278
7524
7769
8015
8260
8506
8751
8997
9242
9487
1835
2636361
6597
6834
7071
7307
7544
7780
8017
8254
8490
2296
3609719
9908
0097
0286
0475
0664
0854
1043
1232
1421
2438
3870337
0515
0693
0871
1049
1228
1406
1584
1762
1940
39
2118
2296
2474
2652
2830
3008
3186
3364
3542
3720
2726
4355259
5418
5577
5736
5896
6055
6214
6374
6533
6692
30
4361626
1786
1945
2104
2263
2422
2581
2740
2899
3058
83
4445132
5288
5444
5600
5756
5912
6068
6224
6380
6536
2820
4502491
2645
2799
2953
3107
3261
3415
3569
3723
3877
3079
4884097
4238
4379
4520
4661
4802
4943
5084
5225
5366
3157
4992746
2883
3021
3158
3296
3434
3571
3709
3846
3984
58
4121
4259
4396
4534
4671
4809
4946
5084
5221
5359
3290
5171959
2091
2223
2355
2487
2619
2751
2883
3015
3147
3555
5508396
8518
8640
8763
8885
9007
9129
9251
9373
9495
61
5515720
5842
5964
6086
6208
6329
6451
6573
6695
6817
498
MATHEMATICAL TABLES.
499
No.
0
1
2
3
4
5
6
7
8
9
3570
5526682
6804
6925
7047
7169
7290
7412
7534
7655
7777
3667
5643109
3228
3346
3464
3583
3701
3820
3938
4056
4175
68
4293
4412
4530
4648
4767
4885
5004
5122
5240
5359
3701
5683191
3308
3426
3543
3660
3778
3895
4012
4130
4247
10
5693739
3856
3973
4090
4207
4324
4441
4558
4675
4793
40
5728716
8832
8948
9064
9180
9297
9413
9529
9645
9761
3920
5932861
2971
3082
3193
3304
3415
3525
3636
3747
3858
25
8397
8507
8618
8729
8839
8950
9060
9171
9282
9392
54
5970367
0476
0586
0696
0806
0916
1026
1135
1245
1355
4021
6043341
3449
3557
3665
3773
3881
3989
4097
4205
4313
4169
6200319
0423
0527
0631
0736
0840
0944
1048
1152
1256
4472
6505018
5115
5212
5309
5406
5503
5601
5698
5795
5892
4512
6543691
3787
3883
3980
4076
4172
4268
4365
4461
4557
4670
6693169
3262
3355
3448
3541
3634
3727
3820
3913
4006
4984
6975780
5867
5955
6042
6129
6216
6303
6390
6477
6565
5040
7024305
4392
4478
4564
4650
4736
4822
4909
4995
5081
5410
7331973
2053
2133
2213
2294
2374
2454
2535
2615
2695
5690
7551123
1199
1275
1352
1428
1504
1581
1657
1733
1810
6258
7964356
4425
4494
4564
4633
4703
4772
4841
4911
4980
7045
8478810
8872
8933
8995
9057
9118
9180
9241
9303
9365
8000
9030900
0954
1008
1063
1117
1171
1226
1280
1334
1388
500
MATHEMATICAL TABLES.
LOGARITHMS OF CIRCULAR FUNCTIONS.
Cosines. Sines. — Continued. Secants.
Angle.
LCOB.
23° 17'
9-9631082
18
9-9630538
53 14
9-7771060
. 15
9-7769369
67 10
9-7341572
Sines.
Angle.
X^sin.
1° 0'
8-2418553
16 23
9-4503452
24
9-4507747
35 14
9-7611063
37 14
9-7818002
40 8
9-8092691
9
9-8094189
41 13
9-8188250
14
9-8189692
27
9-8208358
47 19
9-8663534
48 26
9-8740085
46
9-8762361
47
9-8763468
60 19
9-9389076
Angle.
jLsin.
60° 20'
9-9389796
67 35
9-9658764
36
9-9659285
74 18
"9-9834872
76 7
9-9871236
8
9-9871549
41
9-9881628
Tangents.
Angle.
i^tan.
4° 48'
8-9241363
49
8-9256487
19 11
9-5414678
12
9-5418747
21 34
9-5968776
34 0
9-8289874
1
9-8292599
36 21
9-8668291
22
9-8670937
38 5
9-8941114
6
9-8943715
49 13
10-0641556
Angle.
i/sin.
21° 34'
53 38
39
100315215
10-2269815
10-2271532
Cotangent.
Angle.
iy cot.
24° 13'
10-3470119
ANSWERS TO THE EXAMPLES.
PART I.
La.
1. 206086", 33° 44' 35". 2. -7755, -64672.
6. 80°, 60°, 40°. 6. 150°, 25°, 5°.
3. 135°. 4. 16.
7. 30°, 66°, 108°, 156°
Lb.
1. 127127", 58° 28' 21". 3. -09176234567901, -9450661728395.
3. 156°. 4. 20. 6. 36°, 72°, 72°.
6. 45°, 75°, 105°, 135°. 7. 3, 6.
,- 15 15 17
**• TT> "8") T5-
13. JL, 7, V74.
V74 5 7
le ct2-62 2a&
23. 30°. 24. 0°
28. 0°, 90°.
31. 60°, 90'.
14.
ILa.
_2 3^ 3
Vl3' Vi3' 2'
^^ V(10 + 2V5)
4
25. 15°.
20. 30°, 60°.
33. a2(62_i)=l.
_^ 9 41 41
12. TIT) -9-> TTT-
15 V5 3 A
3 2 V5
22. V5-i
26. 15°. 27. 45°
30. 0°, 60°.
34. a2 + 62 = c2 + d2.
35. (cc' - aa'f = (ab' - hc'){a'h - b'c).
n.B.
,, 60 60 11
**• ¥TJ H' bit*
-J, 21 21 29
!*• ZTJ 2^» 2 1-
17.
V5-l
V(10 + 2v5)*
_« 12 12 13
12. TS") -5-) TT-
15.
22. I
4^5 19 4v5
21 ' 4^/5' 19 '
23. 0°.
501
_« 12 35 37
*«• ¥T> ^T> 3T-
16.
l-m2
-w3
l+m-*' 2m
24 15°. 25. 0°
502 ANSWERS TO THE EXAMPLES.
26. 0°.
27. 0°, 60°.
28.
30°.
20. 60°.
30. 90°.
31. tan-12.
32.
eos-t-
33. a2 + 62 = a=&2.
34.
«2 + «2 =
32 + r3.
36 . {mm' -pp'f = {mn' + np'){m'n + n'p).
III.
1 - cos^a sin^a
1. ; ""V o ■ a. 1. 3. tan6^ + 3tan4^ + 3tan2^ + l.
2 + cos''asm-a
6 45° 30° 7 ^v^(^-^^) \/(l-^'^)
' * * nsJ{m^-\)' ^/(m2-l)*
lO. a2 + 2c = l. 11. aM(a^ + 6^) = l. 12. 2a6.
IV. A.
_ 6 28 185 o 6 3 6« « 2 013 1318
!• UffTJ TTT' 2. -g-g-, ^-5. a. -JXTTS* TTOT-
2. cos a cos jS cos 7 - cos a sin /3 sin 7 - sin a cos ^ sin 7 — sin a sin § cos 7.
_g tan g + tan /3 + tan 7 - tan a tan j3 tan 7
1 - tan /3 tan 7 - tan 7 tan a — tan a tan ^'
__ 7 49 17 __240 720 10 3 3 8
ao. 3V5, 2 ?V5. 21; JL, 24 22 ^ 23. 0, JL, IZ.
7 7 15 V2 7 \/2' 52
24. W(2 + \/2), W(2-\/2), v^-1. 30. tana. 31. sec^a.
120
34. sin a. 35. cos 2a. 40. tan"^-— .
41. 5cosa-20cos3a + 16cos5a. 40. tan 2a. 50. -cot—.
66. 45°. 67. 15°. 68. 18°. 60. 0°, 60°.
60. 6°, 45°. 61. 0°, 30°. 62. 20°, 45°, 90°.
63. 0 = 30°, 0 = 0°. 64. 2a2 - 6 = 1. 66. a* + 62 = 2a2.
66. 2a%{h + c) = {c'^ + d?-a^-W){c'^ + (P-h^). 67. 1ahc = h'^-a\
IV. B.
, 989 1431 « 110 5 33 5 « 21 140
4. 1,|. 6. -mf. 7. tan 2a.
12. sin a cos j3 cos 7 + cos a sin /3 cos 7 + cos a cos /3 sin 7 - sin a sin /3 sin 7.
ANSWERS TO THE EXAMPLES.
503
13. |.
17.
36 4m(m^-l)
' 77' m4-6m2 + l' '
7.3 4 7. 7.
TOlTJ 7 2' 8"
4 3 3
"SJ T' 4'
18.
24 5^/11
18 *
21.
22. ft
23.
25'
JL 2
V2' 5'
-Q 4tana(l - tan^a)
1-6 tan-a + tan^a
49. sin3a/sin5o.
66. 75°.
L ^' a
V2' 128'
41. 5 sin a. - 20 sin^a +16 sin^a.
60. 2 cos22a/cos 3a.
57. tan~^J. (The root tan'^S is inadmissible in Part 1, being greater
than 45°. )
58. 0°, 45°. 59. 7° 30', 30°. 60. 0°, 9°.
61. 30°. 62. 0°, ir 15'. 63. 15°.
64. a2 + 6 = l. 65. {a^+¥-l){a" + b^-2) = 2{a-\-l).
66. (ah + cd)tsin{a-p) = bc-ad. 67. 2¥=3b-a.
3. sin(a + ^ + 7)+sin(a-/3-7) + sin(a-jS + 7) + sin(a + ^-7). 4. ^2,
7. c2=a2 + 62 + 2a6cos(a-^). 7=tan-i-^^^5^±^^H^.
a cos a + o cos /3
18. ^ + 20 = 90°.
VI. A.
1.
5-3967519.
2.
1 •8757579.
3.
7-5929813.
4.
3-9615188.
5.
2-7315929.
6.
176681.
7.
-00523945.
8.
16387 1.
9.
•0202245.
lO.
143-638.
11.
-00880535.
12.
26-7453.
13.
•0207381.
14.
1-4714612.
15.
1-5599642.
16.
1-3028511.
17.
1 9093311.
18.
1-5038295.
19.
1-9918678.
20.
19° 0' 10".
21.
16° 2' 34".
22.
15° 0' 4".
23.
42° 45' 48".
24.
80° 56' 34".
25.
28° 17' 51".
26.
•2088221.
27.
-3215359.
28.
•7620948.
29.
21° 48' 8".
30.
78° 27' 47".
VI.
B.
1.
5-8010626.
2.
2-7105548.
3.
1-8641900
4.
4-7251420.
5.
3 •8982874.
6.
1-65006.
7.
-235567
8.
20-8484.
9.
249525.
lO.
•000176042.
11.
-000235408.
12.
1 -19680.
13.
•191998.
14.
1-8980337.
15.
0-5478477.
16.
1 -9260104.
17.
1 -0087298.
18.
1-4072257.
19.
1-9892056.
20.
69° 4' 25".
21.
73° 6' 16".
22.
61° 7' 41".
23.
18° 29' 21".
24.
32° 56' 39".
25.
69° 56' 6".
26.
-3290360.
27.
-7283544.
28.
2-2213844.
29.
3.3° 23' 2".
30.
36° 37' 10".
504 ANSWERS TO THE EXAMPLES,
Vil. A.
1.
5 = 54° 33',
a = 99-759,
6=140-115.
a.
^=65U1',
a = 2771-21,
6 = 1252-22.
8.
^=48° 43',
6 = 793-72,
c= 1203-00.
4.
^=25° 57',
a = 19526-0,
c = 44622-0.
6.
^=34° 28' 21",
B = 55° 31' 39",
6 = 586-971.
6.
A = 74° 52' 54",
5=15° 7' 6",
a = 2946-37.
7.
^=39°19'48^
B = 50° 40' 12",
c = 1295 -39.
8.
A = 54° 38' 26",
5 = 35° 21' 34",
c = 971-138.
9.
5 = 27° 44' 18",
6=130-941,
c = 281 -330.
lO.
^=26° 36' 21",
5 = 63° 23' 39",
6=182521.
11.
A =1^° 45' 57",
a = 384-014,
6 = 104-580.
>3-
^=38° 37' 5",
5 = 51° 22' 55",
c = 642-499.
^
VII. B.
1.
P=18°45',
a = 3800-03,
6 = 1289-94.
a.
^=30° 12',
a = 2868 -22,
6 = 4928-10.
3.
^=79° 48',
6 = 8-6366,
c = 48-7708.
4.
^=64° 17',
a =1285-23,
c = 1426-53.
6.
A= 8° 29' 42",
5 = 81° 30' 18",
6 = 6864-90.
6.
.4=57° 4' 18",
5 = 32° 55' 42",
a = 47928-7.
7.
^=28° 57' 36",
5 = 61° 2' 24",
c = 1059-48.
8.
^=57° 18' 11",
5 = 32° 41' 49",
c = 9520-64.
9.
A=i 9° 45' 44",
6 = 4068-56,
c = 4128-34.
lO.
^=38° 56' 45",
5 = 51° 3' 15",
c = 93 -8615.
11.
A = 13° 18' 45",
5 = 76° 41' 15",
a =119-967.
12.
^=58° 45' 4",
a = 6159-71,
6 = 3737-64.
vni. A.
1. 449 ft. 8 ins. 2. 131 ft. 2 ins. 3. 34 ft. 2 ins.
4. 13-24 ins. 6. 7-899 miles ; 9-349 miles.
6. 9ft. 7 ins. ; 116ft. 5 ms. 7. 82 ft. 2 ins. 8. 501 ft.
9. 2 m. 342 yds. ; 42° 20' 52" E. of N.
lO. 485 yds. 1 ft. 11. 74 ft.
12. 7 m. 4413 ft. 13. 27° 36'. 14. 2° 9' 33".
16. 2 ft. 5-12 ins. 16. ils/2j\/5 - 1.
18. 351 ft.7 ins., 455 1ft. 7 ins. 19. acos^cosec(a + /3), 52 ft 11 ins.
ao. tB.n-^^^^-^L^^-, 2.A>-g!^-^^'Mmiles.
^(4 tan^a - tan^/S)' V \ 4 tan^a - tan2/3/
ANSWERS TO THE EXAMPLES. 505
VIII. B.
1. 101ft. 7 ins. 2. 12ft. Sins., 36ft. Sins. 3. 107 '07 ft.
4. 4° 32' 54". 5. r 37' 14". 6. 77 ft. 3 ins., 372 ft. 4 ins.
7. 110 ft. 3 ins. 8. 500 ft. 7 ins. 9. 124 ft. 7 ins.
lO. 190ft. Gins. 11. 38.39ft. 12. 5182ft.
13. 3° 13' 45". 14. 2M9'54". 15. 386 ft. 1 in.
17. a tan o sec ^. 19. 3978 -78 miles. 20. 103 ft. 9 ins.
Miscellaneous Examples. L
a.
I. 26° 35' 41" ; 150° 3. 45°, 90°.
• 1885J 18 8 5- ^' ^" 11 ^" •
1. 60. e. ^=43° 35' 42", 5=46° 24' 18", a=9762.
7. 85 ft. 11 ins.
7.
1. 33°, 60°, 87°. 3. 0°.
6. 64cos7a- 112cos^a + 56cos^a-7cosa. 6. sin5a/sin7a.
7. 11 ft. 6 ins.
5.
2. coBa = '2q{p + q)j{p^ + 2pq + q^), ta.na=p{p + 2q)/2q(p + q), etc.
3. 3, A «. -^
7. 459 ft. 9 ins., 2029 ft.
3. 3, -^ 6. J[=22°2'49", 5=67° 57' H", c=2730-67.
1. 30°, 25°, 125°. 3. 0°, 45°. 4. ^, f.
8 8
e. ab^=a + 2h, 7. 5-38 ft. per sec.
f.
1. cosa=(27H-l)/(27i2 + 2n+l), sina = 2w(7i+l)/(2n2 + 2n+l), etc.
2. c2(a + 2) = a62. 6. 5 = 70° 24', a = 335-452, & = 942-057.
7. 1255 ft. 6 ins.
6. 5=80° 6', 6=406-812, c=412-961. 7. 12° 46' 46".
3. m, 1. 5. 15°, 30°; 0°, 30°.
506 ANSWERS TO THE EXAMPLES.
e.
». Hii' fi. 0°, 90°. 6. a62=l. 7. 2.
*• r-~^^)^^^^^■ «• 2co83a. 6. 0', 7°30', 90'
2. Vt- 3- coseca. 5. 45°. 6. 6^^=40^^^ + 62-02).
IX. A.
1. ^ of a radian. 2. 6-981 feet. 3. 3957 miles.
4. 3-142. 6. 6 m. 150 yds. 8. 7r/180.
0. •7rr2. lO. -01745. 11. 50m = 27/i.
12. 11° 13' 52^-932; 0°0'33"-048; 78° 18'4"-212.
18. 32^ 48' 61" i ; 84" 21' 4" '§38271604 ; 48^ 13' 58"' -024691358.
- ITSlTT . 470035
• 9000 ' 200000'
IX. B.
1. 1/2880. 2. 2094-4 miles. 8. 7 miles.
4. 3-14. 6. 847849 miles nearly. 8. 7r/180.
9. 7rr2. 10. -052. 11. 250s = 81<T.
12. 48° 25' 46" -668; 4°36'31"-716; 0°0'0"-972.
13. 5^^ 61' 54" -320987654; 93^^ 57' 39' '209876543 ; O" 0' 15" '432098765.
14. 210297r/43200 ; 228137r/250000.
1. 38'. 2. 200/7r. 8. 50°, 27°. 4. 30n7r/(l+60?i).
6. 56if , 60°, 63xV. «. 85^°, 4^°. 7. 150°, 135°.
8. 49f4:°, y radians.
9. 45°, 75°, 105°, 135°; jr/i, 57r/12, 77r/12, 37r/4.
10. 47r/35, 97r/35, 27r/5, 197r/35, 247r/35. 11. (47r-3)/37r.
12. 65°27'16iS:". 14. 15°, 45°, 120°.
XL A.
8. -(2 + V2 + \/3 + /v/6). 4. 2/i7r±^J, 2nir±^.
5 5
6. 2nir+~. 6. nw, 7i7r + tan"^2, nir --.
~3 4
ANSWERS TO THE EXAMPLES. 507
3 "6 6' 4' 4 6
_Q TT TT TT 5t TtT 27r 5x IItT
* l2' 6' 3' 12' 12' y T' l2 *
XLb.
2. 0. 3. V3-2. 4. W7r±J, W7r±J. 6. mr±l.
4 3 6
6. nir + {-\r-^, n-rr-i-lT-^. 7. (2?i±l)5, 2(3n±l)7r
' 0 ^ ' 2 3 9
8. 2«7r±^, (2/i + l)7r+cos-il. 9. 60°, 120°, 240°, 300°, 420°, 480°
o 3
lO. J |, |:, 1^. 11. (2n + l):r + .|. 12. 2n:r + ^.
XII. A.
6. cos(a + /3 + 7 + 5) + cos{a + /3-7-5) + cos(a-/3 + 7-5) + cos(a-^-7 + 5).
34. tana. 35. Umr + ^, wt±J. 36. nx+—, nx-¥^.
3\ 2/' 3 -10 -10
37. O''"^)"' ^^" + ^"^^i 38. n^, {2n + lf-.
r~s ' r+s
44. (2«-i)7r±|. 45. (27i-l)7r-a±/3.
46. 71 . 360° + 69° 12' 13", n . 360° - 32° 20' 1".
50. iUs + s/5 + ^j5-^5), J(\/4 + N/6 + v/2+V4-V6 + Ay2),
i(V4-v/6 + x/2 + V4 + V6-N/2), -^^•
51. 4H7r + ?^ and 47^ + 5". 65. -^ + ^ = 1.
2 2 a'* 62
_Q +secasecj3
* l-7(sec2a-l)(sec2p-r)"
508 ANSWERS TO THE EXAMPLES.
XII. B.
86. nir + l, !^. 36. ""^^ ?^. 37. ^t + |, (6n±lW3(r-l).
38. (2n + l)5, ?If. 89. (4n-l)5, (471+1)^^. 40. !^.
41. «7r, 7i7r±^. 42. «7r + {-l)"^, n7r + (-ir^, n7r-(-l)«?^.
4 6 10 10
43. TO7r + (-l)"^-l 44. (w-i)7r + (-l)«sin-iA-
46. W7r + 5, TOT + ^. 46. Ji. 180° + ( - 1)«(12° 55' 15") -26°33'54".
4 2
«0. K\/4 + v/6-V2-V4-x/6-^/2), l-{V2(x/2 + l) + ^2(^2- 1)},
t(V4 - V6 - ^2 - 74 + ^/6 + ^/2), i(-N/4+N/6-v2+>/4-x/6 + V2).
61. 4n7r+?!rand4w7r + ^. 66. ^ + ^=1.
2 2 a^ 62
66. -1. 67. 45°.
xni.
14. _+a.
16. ^ = (2w+ l)7r, 0 = - (2w - J)7r ;
^=2n7r-2tan-i>^, <^= - (27i-|y + 2tan-i^.
2 \ 3/ 2
16. (2n+l)J ^ + (-I)n^. 17. (47i-l)|, (47i + 3)|.
18. "^ + {1)%. 18. 7i7r + tan-i$^if^±^.
2 12 sinasin/3
20. ±1±V2. 21. 1, ^-^^.
22. 0, 0; 00, I; 0, |; -<«, ^; 0, ,r; oo, ^; 0, ^; -oo, ^^j
0, 27r.
24. 4a2-8a + l = 0ora2=l. 25. coseca = 2.
26. (ac - 6/)2 + (a/- 6c)2=(a2_ 62)2^ where/=a6ci/(cd-a2-62).
27. ^y2.tana. 34. -7, --gT- 69. a;2=sin2y.
60. 2^=(m + 2n)7r + (-ina + /3)±(7 + a), etc.
62. ^ = m7r±^, 0 = 2w7r; d = 2mir, 0 = 2n7r + ^.
I
ANSWERS TO THE EXAMPLES. 509
63
64
(2.-i)„(2».,|)j
65. n7r + icos-^('sm'%osec^y 06. mr±a, W7r + 2a.
67. 2mr±a, 2w7r ±008-^-2 cos a). 68. !^ + ( - 1)'Y^- a\
60. (2n + l)^. 71.r!r + itan-^ msin2a
10 2 ?i 4- m cos 2a
72. ^ = 2?i7r±^ + a, 0=-2w7r + ^ + a. 73. 2mr, 2w7r-^.
74. (2m + 2wH-l)^, (?i-m)7r; {m-n)ir, (2rn + 2?i-l)5; (6m + 2H+l)J,
6 6 8
(2wH-6n-l)5; (3«-w + l)-, (3m-w-l)^.
8 4 4
76. 13. 83. Jab{ab-^)^{a + h)U\\c.
84. (6c-a)(ac + 6)-0. 85. a^ + fts^c^sec^l
87. aa' sin (j3 - j8') + 66' sin (a - a') = a'h sin (a - /3') + a6' sin (/3 - a').
88. a = 6(l+c). 92. 2cos(/3 — 7)cos(7-a)cos(a-/3).
93. cot a + e cos(a - j3)cosec a. 98, 3 + 2(cos5 + sin^).
99 1 1 102 ^' + ^'-2 «'-^'
3' V3' 2 ' ^M^'
111. a + ^=2mr + (t>+\p. 112. (a^ + c^- 8ac)(a2+c2)= ±4ac(a2-c2).
113. cos a- cos /3, -cosa-cos/3; -cosa + cos^, cos a + cos ^.
115. ±V3±1.
XIV. A.
1.
6.
FTj 2 6> "ST*
3^7 3 3
-%-' 4' 4
2.
7.
120°.
6 2 3
^/85' V85' 5'
&. 6 = 16, c = 8(x/3 + l).
8. h h I
9.
56 10864
65> 125455
168
19 3-
lO.
16 5 6 24
"• if-
12.
4=90°, i5 = 60=
', c = l.
510 ANSWERS TO THE EXAMPLES.
XIV. B.
1. 135^ 30% 15". a. 120°. 6. ^3-1, ^3 + 1.
j,12664 „2 14 «1313
«. T^» TS^ S' ■'• 5;^. ;^. 5' 8. t, % 1^.
-» 4 0 2 4 496 --. 84 4 _80 _, 7
12. ^=75°, 5=15°, c:a = 2V2:^/3 + l.
XV.
16. \{a + h + c).
23. ^=90°, 5 = 30°, C=60°; AB:BG=^:2; AB:AC=l:2.
25. ,^/{6c(a + & + c)(i + c-a)}^(6 + c). 27. 4, 5, 6.
38. The sum of the cosines of any two angles greater than the cosine of
the third.
XVI. A.
1.
A =
18° 56' 20",
6 =
7275-78,
c = 5043-04.
2.
A =
36° 18' 13",
b =
8484-69,
c = 7282-61.
3.
B =
48° 30' 40",
a =
9485-25,
c = 4686-30.
4.
B =
20° 16' 23",
G =
117° 27' 59",
a = 0932-81.
5.
A =
18° 7'54",
G=
41° 36' 26",
b = 3902-34.
6.
A =
32° 42' 21",
B =
68° 58' 55",
c = 551 -992.
7.
B =
22° 45' 46",
C =
107° 59' 44",
c = 6744-14.
8.
B =
29° 6' 37",
0 =
130° 34' 13",
c = 9530-82; or
B =
150° 53' 23",
c=
go 4^, 27",
c= 1917-53.
9.
Triangle impossibh
3.
lO.
A =
65° 18' 40",
c=
73° 3' 20",
a = 1104-07; or
A =
31° 25' 20",
0 =
106° 56' 40",
a = 633-508.
11.
A =
69° 10' 20",
B =
46° 37' 20",
a = 9531 -82.
12.
A =
78° 35' 5",
B =
60° 36' 39",
C= 40° 48' 16".
13.
A =
31° 4' 50",
B =
134° 56' 5",
(7= 13° 59' 5".
14.
A =
41° 41' 16",
B =
49° 41' 19",
a =88° 37' 25".
XVI. B.
1.
A =
28° 44' 25",
b =
8208-28,
c = 4670-13.
2.
B =
9° 42' 10",
a =
8756-13,
c= 13061 -7.
8.
0 =
87° 47' 50",
a =
5980-80,
6 = 2808-01.
4.
B =
65° 7' 15",
0 =
46° 9' 45",
a = 514-791.
5.
A =
50° 52' 17",
(7 =
34° 50' 43",
6 = 87-9286.
6.
A =
54° 40' 34",
B =
21° 5' 26",
c = 464 -759.
7.
B =
62° 12' 40",
G =
90° 35' 10",
c = 1188-37; or
B =
117° 47' 20",
C =
35° 0'30",
c = 681 -797.
ANSWERS TO THE EXAMPLES. 511
8.
B= 37° 17' 32",
C=
83° 1'48",
a = 11063-4.
9.
Triangle impossible
lO.
A= 44° 53' 17",
B =
48° 6' 39",
a = 677-158.
11.
5= 99° 33' 15",
G =
55° 25' 31",
6 = 4996-59; o
B= 30° 24' 17",
C =
124° 34' 29",
& = 2564-37.
12.
A= 38° 29' 2",
B =
66° 29' 21",
C = 75° 1'37".
13.
^ = 142° 6' 10",
B =
21° 43' 16",
(7= 16° 10' 34".
14. ^ = 153° 28' 37", 5= 16° 50' 52", G= 9° 40' 31".
XVII.
2 126° 18°- ^('^ + ^5) 8
' V(10 + 2>y5)' \/(10 + 2V5)'
3. a = 7306-72, & = 11545-35, c = 12545-35.
4. a =1192-64, 6 = 820-64. 7. 5 = 80° 46' 26", (7= 63° 48' 34".
XVIII.
1. 1960-95 yds. 2. 131 ft. 6 ins., nearly.
4. a sin(/3 - 7)sin a cosec(/3 - a) ; 86 ft. 2 ins.
5. 2072-49 yds. 6. 1352-07 yds. 7. 1113'92 yds., 1980-54 yds.
8. 272-14 yds. 9. 80 ft. 7 ins.
lO. asin(a + j8):(a + 6)sina. 11. 495-21 ft.
12. 699-95 ft.; 60° 19' 11". 16.264ft.
17. /isin(a + /3)cosec(a-i3); 780-28 ft. 18. 1574-18 ft.
19. tan-i (< + ^Osinasin^ ^ ^^ ig ^^^ g j^^^
^'cos a sin /3 - i sin a cos j3
21 ^^^''^ , "^^"'^ ; where a and /3 are the angles sub-
l-tan(a + j3)tani3' cos(a + 2/3)
tended by the flagstaff and the tower.
23. miles ; -^!-—-: miles. 26. cos"^(cosacos iS).
V(4-V3) 13
27. tan-i^ (a + 6)sin a sin^^ 3^^ 41° 2'in direction N. 33° 55' E.
6 cos a sin /3 - a sin a cos /3
34. tan 0 = (tan a - tan j3 cos ^)cot j8 cosec ^ and tan 5 = tan a cos <p, where
5 is the inclination of the common section to the horizon and 0
the angle between its direction and that of the dip of the first
plane.
35. tan-i(a - 6)cota/c S. of W. ; 27° 38' 20" S. of W.
36. hcotaco^d. 37. 29-72 ft.
512 ANSWERS TO THE EXAMPLES
39. Direction of wind is <t> E. of N. where
, , tan a sin 6
tan <b - , „ ;
tan a cos ^ - tan /3
velocity of wind = "^^^^"j"^ ft. per sec.
40. 5(V7-l)ft.
41. Length of path = 59 '55 miles, height at appearance = 93 '04 miles,
height at disappearance = 34 00 miles.
(p + g)7r
43.
acos^
4(7i-l)
2 sin P"" sin ^^
4(»-l) 4(7i-l)
46. I cos a cosec(a +/S), where tan /8 = tan 8 cos 6.
49. ha - 6)tan a cot^ j ha + 6)tan /3 tan X
2 2 2 2
XIX.
1. 270 sq. ft., 16296 sq. ft. 3. ^bhm2A, if A, a, b he given.
4. 6 sin J Va^ - fe^ siii'^!4, if ^, a, 6 be given.
6. ia2sinj5sinC/sin(^+(7). 16. 3. 17. 6. 18. 5 + ^^5:8.
19. 3(2 + ^3)a2. 28. 32^, 9, 14, 35, 90 inches.
29. 63J, 9^, 12, 60, 190 inches. 30. 2 + /^3 inches. 60. 2Jsq. ins.
l-sin:|
65. 8(^2- l)r2. 79. r ?, etc.
1 + sin^
82. TT, J. 107. rsec^^.
2 2
121
Ip-Pi-Ps, Ip-Pz-Pi^ lp-Pi~Pi- Ip + Pj + Ps
Miscellaneous Examples. II.
a.
1. 3-1416. 3. VE; (2n + l)^, ^ + (-ir^.
^ 31 5
/3.
3. a = (2/i + 1 )1 or iS - 7 = nw. 6. 86° 16' 47".
XSWEB.^ TO THE EXAMPLES. 513
7.
3. mr, iiT±^ I --, — . 6. 8.Mt. 1 111.
0 2 3
6. 396 ; 6O5 ; 4 ; 41, 44, 198.
5.
4, ^^^" ~ ^^l 5. 66° 29' 10", 4236-29.
2.-4 sin ^ sin B sin (7. 3. vir + ^ and ?/7r + ^.
4 4
5. 2a sin a, 2a cos 2a. 6.
^ - sm'
71
3. -^-. 5. j5 = 75°, (7=90°, r = 2v/2; or 5 =105°, (7 = 60*, c = ^/6.
I
'7-
3. ^J - ^^f cos (8 - a) + ^'I = sin2(|8 - a). 5. .4 = 106° 53' 54", /? = 32° 42' 6"
0" ab O"
^- 3' ¥' ^^^^' ^3' V3' ^^^' ^' ^- •'^' ^' *' ^^' "'•
4. ?i7r, »7r+tan-i^ . 5. 5926-61.
\.
1. cc^-px^+pqx-p = 0. 3. a+/S + 7 = ??-7r ; o + /3 + 7 = (2?? + 1)'
4. a sec A tan ^ tan (7 ; 230'966.
(I.
1. 6, S. 3. 'ij, (4»-l)J- ^
2 8 Va(26-a)
1. k, cc. 3. ^\^ ; 7?7r-l-tan-^2, )?7r + tan-MN/5-2), 7i7r - tan " Mx/5 + 2).
4. (/)- + (7'-)cot a +yi9(sin a - 2 cosec a).
2k
514 ANSWERS TO THE EXAMPLES.
XXII.
1. IT radians. 2. pfq. 3. ^. 4. 1. fi. 1.
8. 1, e 2 J or 0, according as m is <, = , or > 2. lO. 1.
XXIII.
1. cos ^(n + 1 )5 sin ^n^/sin i^. 2. sin^na/sina. 4. cos ^^ sin w^/sin i^.
^
6. |cos(a + 1^^ + ( - 1 )'»cos(a - ^^V) } / 2 cos ^
6. i(sin 2"+!^ - sin 2^). 7. coti^-cot2"-i<?.
8. cot^-2"cot2M. 9. tan 2"^- tan ^.
11. ^-cos(?i+l)^sinn&/2sin^. 13 ,-(2 cos 5 - cos 3^ - cos 5^).
2 lb
16. — — + — 17 ^^1^^^"^ sing 21. 0
* w' w 4}i' ' 2"cos2"g + l 2co8g+i"
23. sin 2na/2 sin a. 25. sin ^(3/i + 1 )a sin f wa/sin fa,
26. ^ + cos(n+l)asin7ia/2sina.
27
. lcos{3a + (n-I)|}sin^^/sin|^ + ^^os{a + (,.-l)^}sin|/sin|
28. ^-cos{2a + (?i-l)i3}sinw^/2sin/S. 33. l cot j^ - cot a ; --—!-.
^ Zi Jt n, tan ci
34. sin%S/sin^co8acos(a + w/3). 35. tan"^(l + 7i + »t2)_^.
4
36. cosec2a;--^(cot2^ + lY 87. sinwj8/4sini3-*icos{2a + (n- 1)^}.
38. w(r2 + \x-), r = rad. of incircle, x = dist. of point from the centre,
n = number of sides.
46. i(cosec g - cosec 3«g). 47. sin^g - 2"sin2^.
48. ^cosec-g-2"-icosec22«g. 49. ^t&n^n{n+l)e.
60. |(tan 3"a;/3" - tan ar). 61. —^{aeondaec{n+l)d-8&oe}.
2sin0
53. a sin na/2 sin^a 8in(7i + !)«, where a = dist. of 0 from AB.
64. (-l)'»-i?^. 66. 2«cosh«^sinh— .
i2n 2 2
lUin^^4lBu4-n]. 69. cot%-(lcot "V-f^f] _1).
2\ 2/ 2 J \2« 2"/ 3\ 4V
61. 2''-2cosec2«+ig-Jcosec2g.62. ^(tan 3"^ - tan g).
•-•2
ANSWERS TO THE EXAMPLES. 515
63. if cosec2^ - cosec2?^Y
SV 2 2 /
64. sin !| sin '^cos 'Lz\a + /3) /sin | sin |
XXIV.
14. w/2«-i. 18. 7r2/8; 7r796. 22. 7rV384. 23. tt^^.
25. ^(l-0 20. log(l-cos2^)-log(l-cos2^_^)-»log2.
43. (l)%even; (2) 7iodd.
XXV.
4. 2. 7. ^ = 14° 54' 13", B = 75° 5' 47", h = 5598-37.
8. 5 = 149° 53' 7" -5, C = 6'52"-5.
lO c = a + 6-_^^ 4- «^ ab(a-b)d'^ „_ &g _ah{a-b)d^
2(a + &)' a + 6 6(a + 6)3'' a + 6 6{a + 6)3 '
11. dab cos a, where a, b are the sides, a the included angle, and 6 the
change in the angle.
13. 6= -(f>- x//, y = {c(f> + b\f/ COS A +xsinB)cosec A.
z = {c(j) cos A+b\p-\-x sin C)cosec A ,
14. 0 = (?/ sin (7 - 2 sin 5 - Z;^ cos C),
a
^ = (z sin ^ - y sin G -cd cos ^), x = y cos C + s cos B + 6b sin (7.
a
15. 0 = f^-^ + ^cot^Van5, xly = (^-%- dcotA - d cotB)t3inB,
\b a I \a b I
z-{x-y cos G-Qb sin (7)sec B.
16. ^ = (a;-ycos(7-scos5)/6sinC, etc.
18. ^ = - W^ • ^, 0 = - iVL^f ;/, = ^v^Tj^ . therefore 5 can be deter-
70 70 ^ 70
mined with the greatest accuracy.
10. 3960 miles. 20. ^fl-^"\ % 23. '07 inch.
a \ 2 / a-
24. -0000952. 26. Decreased by cd sin A sec C.
Miscellaneous Examples. III.
7-
., - _i cos 5 + cos (7-
6. tan ^ . p — ^-p^
sin ZJ - sin G
oiG AXSW/'JRS TO THE EXAMPLES.
5.
a tun a ^ /.
2tan/3
tan /3 - tan a
6.
1. 5v/3, Ln^^. 4. ~^a.
a. 2:3:3. 6. 7r + 4a-2sm2a : 7r-4a + 28in2a
1. 0, - a ; or . ^^ = -^- = -^ , where a = ^^ or ^.
sin 2a sin3o sin a 7 7
a. 0, -a, ^(tVS-l); ±a, 0, +a;
or ^A^ = J^. = . ^^ =^ , where a = Z^T or i^^.
sin2a 8in3a sin 4a sma 9 9
8. 008 1:, 008 1?. 4. xiu^'. sinlO^.
4. 21ogsin^-21ogsm- -7ilog2. 6. I/tt.
X.
2. 2/m-^. 3. 3. 4. 7r3/16. 6. - ^.
Examples XXVI.
7. C08(4a; + 5y) + i8in(4a; + 52/). 8. cos(a + /3-7- 5), isin(a + j8-7-5).
9. cos(5a-4/3) + isin(5a-4^). lO. cos(10^ + 12a)-isin(10^ + 12a).
11.-2-'. 12.-2^0. 14.-1. 15. -^, i(±^3 + 0.
16. ±K^2(V3 + ^), ±i</2(-l + iV3).
17. ±V2(co8^ + isin-^-), ±V2(sin^2-**^°«F2)'
V2(co8^^ + i8inj^), V2(co8^ + mn3i^), -;/2(8in^2 + *'^°«r^)-
^2(co8^..8in^'^), ^2(cos^^.^8inl^-),
V2(cos^-Msin^).
18.
10.
ANSWERS TO THE EXAMPLES. 517
20. cosC2r + ^y + isin^2r + ^'\|, whei'er = 0, 1,2, 3or4.
21. cos(2r+ 1)T +ism(2r + 1)^, where r = 0, 1, 2, 3 ... 9.
22. ^^^f« - ^) {cos i(a + /?) + / sin i(a + /3)}.
cos^(a-/3)
25. 2-+i(-lfsin'»^-rJ'cos'^J^-\ weven;
2n+i( _ 1 )^'sin" t^ sin ^^+1), 71 odd.
26. /) = r"7^'", d = 7yia-na', where (r, a) = a+ <6, (r', a') = a' + e&'.
28. ^ cos 0 = r'^cos ma + r'"cos na', J? sin 0 = r"'sin ma + r'"sin na',
where (r, a) = a + i6, (r', a') = a' + i6'.
31. i-\ < cos??.a + wcos[ wa + M +w-cos( ?ia4--^ W.
33. ±i. 34. ±i.
35. sin(a + /3)cos(2^ + a + /3)
= sin(^ - ^)cos(^ + 2a + /3) - sin(<^ - a)cos(^ + a + 2/3).
4. 1,
Examples XXVII
7r
(^'l)--«- »-(i'I
6. -y2iisina-2sinlla + sinl0a)/(5-4cosa).
7. (cosa-a;)/(l -2a;cosa + a:^). 8- a;sina/(l +2a;co&a + a;-).
9. {cos^^ sin d - cos"+i^ sin(ri + 2)^ + cos'*+2^ sin( w +J_ )^}/sin-^.
10. {cosa-a;cos(a-/3)-a;"cos(a + n^) + a;"+^cos(a + ?<-- lj8)}/(l-2a:cosj3 + x%
{sina-a;sin(a-j8) -a;"sin(a + n^) + a;"+isin(a + n- l/3)}/(l -2a: cos /3 + x^).
14. sin ^(cos ^ - sin ^)/( 1 - sin 2^ + sin^^), sin'-^/( 1 - sin 26 + sin^^).
11 6 = ^ , the series oscillate.
2
15. V2sin^/(3-2;^/2cos^).
■«+l n
16. {sin?i0 - ^/2sin(n+ 1)^ + 2 ^ sin ^}/{2-'(3- 2^/2 cos ^)}.
18. 0, cot 6, if d=\=mr ; <x,0,ii d = nir.
Examples XXVIII.
1. M..., 18° 26'. 3. 2"cos""cos/^^* + iy, 2"cos""sin^^^ + l'\o.
6. 2"-'cos"^J"^cos"(<^ + 0) + 2»-icos»^±^cbs"(^-0). lO. -985.
2 ' 2 2 2
518 ANSWERS TO THE EXAMPLES
Examples XXIX
3. 2-7183, 7-3891. 4. 1-649, 4-482. 6. ecos(tan0), e8m(tan(').
7. exp(a:cos^)cos(a;sin^). 9. exp(ccos^)sin(a + csin/3).
lO. exp(co8Scot^)8m(^+sin5cot^). 11. 2cos^exp(^cot^).
13. exp(a cos a:)cos(a; + a sin cc).
Examples XXX.
1. -35 + tY-+2wirY 2. 3-00 + i(7r + 2n7r). 17. tan-V^^"^^
\4 / l-a;co8 6^
19. ilog(H-m)-ilog(l+2mcos2a + ?n=^). 24. tan-^(cot^).
26. taii-i{tan('|-|U. 28. ^ log(4 x/33i^).
29. Oor^, ifa;=|=(2n+l)^; ^ifa; = (27i + l)|.
30. -log2 + co82^-icos4^ + icos6^- ....
31. - 2{a cos 0 + ia2cos2e + Ja'co8 30 +...}.
Examples XXXI.
/rt X « 61oga + 2rr , ,. . 61oga + 2r7r
1. exp(2w7r). 3. cos — ^ +ism 5
5. exp( - c0)cos(c log r), exp( - c^)sin(c logr), where {r, <p) = a-hs/ -1.
8. cos 2(;?0 + g log r) + i sin 2(p0 + g' logr), where (r, 0) = a + 6t.
9. (47i + 1 )/(4m + 1 ). lO. — - * |5^, where 7n=l=0.
14. If a; = (r, 6) then
(a + 27i7r)sina + logacosa g, _ (« + 2M7r)cos a - log a sin a
~ a ' a
Examples XXXII.
6. 2cosa;coshy/(cos2« + cosh2y), 2isina;sinh?//(cos2a; + cosh2y).
gl- cos 2a cosh 2^ .
cosh 2/3 - cos 2a *
12. C08-H{ V(a + 1)2+^-' - s/(a - 1 )2 + j32}
+ icosh-H{\/(a + l)- + ia^+ V(a-l)=^ + j8--^}, the upper or lower
sign being taken according as /3 is positive or negative.
ANSWERS TO THE EXAMPLES 519
15. A - ± -> B = h tanh~^(sin ^), where A and cos 6 have the same sign.
4
26. ^ = 27i7r + 2tan-i(tanh^y ?i = 2?i7rj + 2tanh-i( tan ^V
29. cos(cos ^)cosh(sin 6).
3
31. If G^ + 4H^ is positive, let cosh 3m = numerical value of \G({- H)^^
then the roots are ±2sJ - H co^Yiu, +'2sJ - Hcosh.iu + i''-~\
+2 sj - jSTcoshf ?< - i — ], the upper or lower signs being taken
according as G is negative or positive.
If G^ + 4,H^ is negative, let cos 3^ = numerical value of
I G/{-Hf, then the roots are ± 2 \/^;^cos 6, ±2 V^cos ( ^ + ^) ,
±2\/- //cos( ^--;^ )i the upper or lower signs being taken
according as G is negative or positive.
Miscellaneous Examples IV.
7-
1. a = h{l + c). 4. ^-^-^^t^h-^=sinHa-p).
^ xslb^ + y^ b- + y^
6. Two equations in sin 26 may be deduced from the given equations.
6. cos^a = cos,m(f> -xp).
5.
2. a; = 2m7r + ^±cos"^-, y = 2mr + 6 + cos~'^^, where {r, d) = a + bi.
4. {a + b)^+{a-b)^ = 2. 5. (^ + UY+(^-KY = 2.
\a bj \a bj
6. {ax + by)^ + {ay + bxf = 2^.
2. tan2a = (a2-62)2/(2S6V-2a^). 3. a{ib^ + {b^ - c^)^} = Sbc. 6. -.
o
a. 2. 3. -J. 4. /^.
V-
4. ncotnd. 5. a;5_55a^4.330:e3_4(j2a:2 + l65x-Il = 0. 6. 2365.
.21) .L\'>yirA7Av TO TIJl-J i:xAMrLi::s.
2. ""i 1
M
1}
2. - log2 -- sin 2d + i cos 4^ + J sin 66 - J cos 8d - ^sin 10^+ ...
6. {csind-c3sind - (»+ l)c"+^sin(w+ l)d + 2(n+ l)c"+^sinnd
- (n + 1 )c'»+38in(„ -1)6 + nc"+hin{n + 2)d - 27ic''+^sin{n + 1 )d
+ nC+^sin n6}/{ 1 - 2c cos d + c2)2.
i{6 + h{n - 2)a}8in '-^ / 2 sin^^ - n cos{5 + ^(2??. - 1 )a}/2 sin "
Z ' z 2
3. (2m - 1 )sin n6l4. sin=^ - sin '^ sin ^^^^ A sin-^^-n2co8(2!'+i)^ /28in t
Z 2, 2t I Z 2/2
\
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