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University of California • Berkeley
From the Papers of
Prof. Edmund Pinne
THE ELEMENTS
OF
PLANE AND SPHERICAL
TRIGONOMETRY
DESIGNED FOR
THE USE OF STUDENTS
By JOHN HIND, MA.
FELLOW OF THE CAMBRIDGE PHILOSOPHICAL SOCIETY,
FELLOW OF THE ASTRONOMICAL SOCIETY OF LONDON, AND LATE FELLOW
AND TUTOR OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE.
SECOND EDITION.
CAMBRIDGE:
Printed by J. Smith, Printer to the University;
AND SOLD BY DEIGHTONS, STEVENSON, NEWBY,
BARRETT, HALL, JOHNSON, GRANT, AND HATT, CAMBRIDGE J
AND BY G. B. WHITTAKER, LONDON.
1828
Digitized by the Internet Archive
in 2008 with funding from
IVIicrosoft Corporation
http://www.archive.org/details/elementsofplanesOOhindrich
PREFACE.
The Work now laid before the Public consists
of two Treatises, the former on Plane and the latter
on Spherical Trigonometry, and in the execution of
them both the Author has adopted that Arrangement
which appeared to him the most natural, and at the
same time the most elementary. The whole of the
former Treatise with the exception of the last chapter,
has been made to depend solely upon the Propositions
usually read in Geometry, and the first Principles of
Algebra. For an account of the particular Articles
which may be found in the work the reader is referred
to the Table of Contents, but the following brief
outline will put him in possession of the general
plan upon which the performance has been con-
ducted.
The first Chapter contains the Definitions of the
subject and the Terms made use of in it, accompanied
by various Observations and Deductions of great im-
portance to the complete understanding of the subse-
quent parts of the work. With respect to the
explanation and elucidation of the geometrical Lines
which form the leading feature of this part of Mathe-
IV PREFACE.
matical Science, it may be observed that a Notation
has been adopted, in which small Figures are suffixed
to, or placed under, the Letters employed: this does
away with the necessity of introducing a greater
number of different letters, and has also the ad-
vantage of establishing and pointing out a Connection
between the geometrical Lines and their algebraical
Affections, which cannot fail greatly to assist the
progress of the student ; and it is of course always to
be understood, when neither of the above-mentioned
objects is to be attained, that the suffixes may be
altogether suppressed.
The second Chapter comprises what is generally
called the Arithmetic of Sines, and commences with
a geometrical Demonstration of a Proposition which
fonns the basis of the whole Doctrine: from this
proposition and the definitions of the preceding-
Chapter, all the other parts of it are either directly
or indirectly derived, various Examples of great utility,
or at least remarkable either for the frequency of
their occurrence or for the singularity of their results,
being occasionally introduced.
The third Chapter is a short treatise on the Con-
struction and Verification of sets of Tables adapted to
practical purposes, and commonly knovm by the name
of the Trigonometrical Canon, It is this part of
the subject which is most laborious, and renders it
PREFACE. V
available in the concerns of life. In this Chapter
some Approximations to the numerical value of the
Circumference, or to the Rectification, of the Circle,
have been made.
The fourth Chapter contains the Application of
Trigonometry to the determination of the Relations
between the different Parts and Properties of Triangles
and other rectilinear Figures, and it will be seen that
a variety of Problems has been expeditiously solved
by it, in which the operations of common Geometry
would have been long and tedious. The Properties
of regular Polygons have been here introduced, but
for the subject of Pohjgonometry in general, the reader
is referred to ' Polygonometrie, ou de la mesure des
figures rectiligiies i^ar Simon LliuiUer,' or to a
masterly extract from it, contained in the third
volume of Dr. Hutton's Course.
The fifth Chapter exhibits the Solutions of all
cases of triangles that usually occur, points out briefly
the methods to be preferred under different circum-
stances, and concludes with several examples of their
application in the Mensuration of Heights, Distances,
&c. This Chapter with the assistance of the tables of
whose formation a short account has been given in the third,
constitutes the general practical use of Trigonometry.
The sixth Chapter presents the subject in a more
general aiul analytical point of view, and treats of what
vi PREFACE.
was termed by Vieta and the Mathematicians of the old
School, Angular Sections: this is, in fact, a generaliza-
tion of the Arithmetic of Sines, and in a work not
designed for the purpose of affording elementary Infor-
mation, might have rendered unnecessary many of the
propositions demonstrated in the second Chapter of
the present.
The last Chapter on Plane Trigonometry is made up
of such Propositions as could not without a violation
of method be disposed of in any of the preceding
divisions of the work.
The Treatise on Spherical Trigonometry is also
comprised in seven Chapters, and in a great degree
a similar plan of arrangement has been adhered to.
Of course this Treatise has been made to depend almost
entirely upon the preceding one, and its division into
Chapters seemed so obvious that it is unnecessary here
to attempt to assign any reasons for it, the only novelty
in addition to the substance of the generality of works
on the same subject being a Chapter on Polyhedrons.
The reader will readily learn what he may expect to find
in it by casting his eye over the Table of Contents.
Throughout the whole of both Treatises it has
been the Author's object to present to his reader every
proposition proved in a plain concise form; and with
the view of forwarding the purposes of Academical
PREFACE. VU
Instruction, for which the work is principally intended,
the leading propositions are stated in Italics, though
it may be observed that the Corollaries and Deductions
sometimes involve results of no less importance than
the articles which have been so distinguished.
A collection of Theorems and Problems connected
with the substance of each Treatise has been annexed
in two Appendices at the end of the work, and they
have been partially allotted to the respective Chapters
in order to direct the student in some degree to the
knowledge necessary to enable him to attempt their
solution.
In the Table of Contents asterisks have been prefixed
to such articles as may be reserved for the student's
perusal after he has made a partial progress in some of
the other subjects of his Academical Education.
, -~«
Cambridge^
Nov. 25, 1828.
Lately published by the same Author,
The PRINCIPLES of the DIFFERENTIAL and
INTEGRAL CALCULUS, Vol. I.
Also in the Press,
The ELEMENTS of ALGEBRA.
CONTENTS.
On a first perusal, the articles marked with asterisks may be omitted.
PLANE TRIGONOMETRY.
Chap. I.
Article Page
1. Definition of Plane Trigonometry 1
2 — 6. Measures of Angles, &c. 1
7 — 11. Division of the Circle, &c ^ 4
12 — 14. Definitions of Complement and Supplement 7
15 — 16. Enumeration, &c. of Trigonometrical lines 8
17 — 20. Definition, &c. of Sine 9
21—25. Definition, &c. of Cosine 11
26—31. Definition, &c. of Versed Sine.' , 13
32—37. Definition, &c. of Chord 15
38 — 42. Definition, &c. of Tangent 17
43 — 47. Definition, &c. of Cotangent 19
48 — 52. Definition, &c. of Secant 21
53 — 57. Definition, &c. of Cosecant 23
58 Mode of finding the Algebraical Signs of Lines 25
59 — 62. Method of changing the Radius, &c 26
Chap. II.
63 — 69. Sines and Cosines of A±B, &c 33
70 — 72. Sines and Cosines of A and B in terms of A + B 39
73 — 74. Sine and Cosine of (72+1) A, &c 41
A
75 — 80. Sines and Cosines of 2^, — -, &c. Delambres Formula. . . 42
A
*81— 82. Sine and Cosine of — , &c 46
b
X CONTENTS.
Article ' Page
A
83 — 92. Sines and Cosines o£3A, — , &c. Euler's and Legendre's
Formulae of Verification 49
*93— 100. Versed Sines of ^ + B, 2A, SA, &c. ^, ^, ^^, &c. 5S
Ai O Ai
*101— 107. Chords of ^ + 5, 2.4, -,4^&c... 56
108—1 11. Tangents and Co-tangents of ^ + ^, &c. 60
1 12—113. Tangents and Co-tangents of 2 .4, — , &c QS
1 1 4—1 1 6. Tangents and Co-tangents of 3 ^ , — , &c , ^5
3
A A
*117— 123. SecantsandCo-secantspf J + jB,2^,-,&c.3^, -,&c. 67
*124— 130. Sine, Cosine, &c. of ^+J5 + C 71
*131— 136. Sine, Cosine, &c. of^+J5+ C + &c 75
Chap. III.
*137— 141. Sines and Cosines of Arcs less than 30<^ 79
*142. Sines and Cosines of Arcs between 30^ and 45^ ^S
*i43. Sines and Cosines of Arcs between 45^ and 90^ 84
*144. Sines and Cosines of Arcs greater than 90^ S5
*145 — 146. Tangents, Cotangents, &c. of Arcs S6
*147 — 151. Trigonometrical Canon, and Formulae of Verifica-
tion, &c. 87
*152 — 153. The Ratio of the Circumference of a Circle to the
Radius, &c 90
*154 — 156. The length of a circular Arc in terms of the Radius. . 90
Chap. IV.
157 — 191- Properties of Triangles 92
*192 — 200. Properties of certain Quadrilaterals II6
*201 — 209. Properties of regular Polygons 123
*210— 216. Properties of the Circle 128
CONTENTS. XI
Chap. V.
Article Page
2 17-— 219. Statement of the Cases of Triangles 132
220—228. Solution of right-angled Triangles 134
229 — 242. Solution of oblique-angled Triangles 140
*243— 256. Mensuration of Heights, Distances, &c 151
Chap. VI.
*257 — 262. Algebraical Expressions for the Sines, Cosines, &c.
of simple Arcs l64
*263 — 266. Algebraical Expressions for the Sines, Cosines, &c. of
multiple Arcs I68
■*267 — 269. The Formulae of Demoivre 172
*270— 271 . Sine, Cosine, &c. of a multiple Arc 174
*272 — 275. Powers of the Cosine and Sine of an Arc, &c 177
*276— 284. Sine, Cosine, &c. in terms of the Arc, &c 183
Chap. VII.
*285. Solution of Quadratic Equations I9I
*286— 289- Solution of Cubic Equations 193
*290. Demoivre's Theorem for the Solution of certain Equa-
tions 197
*291 — 293. Cotes's Theorems and Properties of the Circle 199
*294 — 296. Sine and Cosine of an Arc in continued Products, and
WalUss Theorem 203
*297 — 301. Expressions for the Length of a circular Arc, &c 205
*302— 306. Certain Solutions of Triangles, &c 208
*307 — 308. Expressions for the Cosines and Sines of multiple Arcs 212
*309 — 327. Summation of Trigonometrical Series, &c 214
SPHERICAL TRIGONOMETRY.
Chap. I.
1. Definition of Spherical Trigonometry 227
2—10. Section of a Sphere 227
1 1 — 12. Sides of a Spherical Triangle, &c 229
Xll CONTENTS.
Article Page
13 — 16. Polar Triangle and its Properties i 230
17—20. Angles of a Spherical Triangle, &c 231
Chap. TI.
21 — 26. Cosines, Sines, &c. of the Angles of a Spherical Triangle
in terms of the Sides, &c 233
27 — 30. Cosines, Sines, &c. of the Sides of a Spherical Triangle
in terms of the Angles, &c 238
31 — 36. Napier's Analogies, &c 241
Chap. III.
37. Statement of the Cases of Spherical Triangles 245
38 — 39. Solution of right-angled Triangles by Napier's Rules for
the Circular Parts 246
40. Solution of quadrantal Triangles by Napier's Rules for
the Circular Parts 251
41 — 48. Solution of oblique-angled Triangles 252
Chap. IV.
*49 — 61. Surface of a Spherical Lune, Triangle, &c. in terms of
the Angles, Sides, &c 260
*62_64. Measures of Solid Angles, &c 267
Chap. V.
^65 — 66. General Properties of Polyhedrons 271
*67 — 73. Properties of the regular Polyhedrons 272
*74 — 78. Properties of the Parallelepiped ^79
*79 — 81. Properties of the Triangular Pyramid 281
*82. Miscellaneous Observations on Polyhedrons 282
Chap. VI.
*83 — 84. General Remarks on the Variations of the Sides and
Angles of Spherical Triangles 286
*85 — 89. Variations of the Sides and Angles of right-angled
Spherical Triangles 288
*90 — 95. Variations of the Sides and Angles of oblique-angled
Spherical Triangles 29I
Chap. VII.
^■g() — 108. Miscellaneous Propositions 297
PLANE TRIGONOMETRY.
CHAP. I.
DEFINITIONS AND INTRODUCTORY OBSERVATIONS.
Article I. Definition I.
Plane Trigonometry in its original acceptation is that
part of Mathematical Science which treats of the mensuration
of the sides and angles of plane rectilinear triangles ; but it is
here used in a more comprehensive sense, and includes the
general doctrine of plane rectilinear angles, as well as their
relations to one another, and to the straight lines by which
they are formed, or with which they are in any manner con-
nected.
2. In a chxle of given radius, the arcs may he considered
as the measures of the angles which they subtend at the centre.
For, let C be the centre of the circle, of which FQ, P'Q\
are any two arcs subtending the angles PCQ, P'CQ! re-
spectively: draw the diameters ACD, BCE at right angles
to each other, which divide the circumference into four equal
parts called Quadrants : then (Euclid, 6. 33.) we have
arc PQ : arc P'Q' :: angle PCQ : angle PXQ' :
A
2
that is, the arcs are proportional to the angles which they
subtend, and may therefore be taken as the measures of them
to the given radius C^.
3, If the radii of circles he supposed to he of different
magnitudes, the angles at their centres will he directly pro-
portional to the arcs which suhtend them^ and inversely pro-
portional to the radii; and every angle may he measured hy
the fraction , ( — - — ) .
\raaius/
For, let PQ, P'Q! be two arcs of different circles subtending
at the common centre C, the angles PCQ, P'CQ! respectively :
draw the diameters ACD, BCE at right angles to each other
then
arc PQ : arc JB :: angle PCQ : angle JCB,
PQ
or angle PCQ = — — angle ACB; again,
A. D
arc P'Q! : arc A'B' :: angle PXQ' : angle AXB",
or angle P'CQ' = ~ angle AXB' :
\ angle PCQ : angle P'C& ::
PQ P'O'
~ angle ACB : ^. angle AXB:
s
but the angle ACB is the same as the angle A'CB\ and the
quadrants of circles are proportional to their radii, therefore
we have
angle PCQ : angle P'CQ' ■■ J^ ■ ~i^ >
that is, the angles are as the arcs directly and the radii inversely :
PQ P'Q
and the fractions, — ^ , , may therefore be taken as the
JxL/ j4 (_/
measures of the angles PCQ and P'CQ! respectively,
4. In Article 2. we have seen that an angle may be
measured by the corresponding arc of a circle, whose radius
is given ; and in Article 3. that an angle may be measured
(arc ^
: — I ; now it is manifest that
radmsx
these measures will not be upon the same scale, unless the given
radius in Art. 2. be supposed to be 1 : hence, therefore, adopting
this hypothesis, we conclude generally that any angle may be
measured by the corresponding arc of the circle whose radius is 1,
(Ixth
- ) part of the corresponding circular arc whose radius
is r; or generally by the fraction, ( : — )• that is, angle
Vradius/
arc
radius
5. Cor. 1. Hence, therefore, if a represent the length
of the arc which measures a given angle to the* radius 1, and
a' to the radius r, we shall have a= l~^ J } or a=ra.
6. Cor. 2. If the angles at the centres of different circles
be of the same given magnitude, the arcs by which they are
measured will be proportional to the radii ; and, if the arcs
be of the same given magnitude, the angles will be proportional
to the reciprocals of the radii.
4
Ex. 1. If the arc whose length is a measure a given angle
denoted by A to the radius 1^ then will an arc whose length is ra
measure the same angle when the radius is supposed to be r.
Ex. 2. If the arc a of given length measure an angle A to
the radius 1, then will an arc of the same length be the measure
(A\
— J when the radius used is r.
7. Def. 2. If the radius of the circle be supposed = 1,
the circumference (hereafter proved=6. 28318, &c.) is represented
by 27r, and is supposed to be divided into 360 equal parts
called degrees : each degree is again supposed to be divided
into 60 equal parts called minutes : each minute into 60 equal
parts called seconds, and so on. These are expressed by the
characters ^, ', '\ '\ &c. placed above the line to the right
of the numbers : thus 45° 35' 25" 14'" represent 45 degrees,
25 minutes, 25 seconds, 14 thirds.
In the following pages we shall always suppose the radius
to be 1, unless the contrary be expressed.
8. Cor. 1. It appears from the last definition that one
71"
right angle will be measured by - or 90^;
lit
Two right angles by tt or 180°;
Three by — or 270°;
2
Four by 27r or 360°.
9. Cor. 2. If we suppose the circumference of the circle
to be taken a second, third, &c. time, w^e may in the same man-
ner represent the sum of any number of right angles whatever :
thus,
Five right angles will be measured by — or 450° ;
5
Six right angles will be measured by 3 tt or 540® j
Seven or 630° ;
&c &c.;
UTT n
n — or w.go .
2
10. Most modern mathematicians, with the exception of
the English, divide the circumference of the circle into 400
equal parts, which they call degrees ; each degree into 100
equal parts, which they call minutes, and so on.
In this division of the circle
One right angle will be measured by - or 100°;
2
Two 7ror200°;
&c &c.;
nir
n —
2
or n . 100°.
11. Cor. Hence, we may easily investigate rules for re-
ducing degrees, &c. in either of these scales into degrees, &c. in
the other.
For, let IV=the number of degrees, &c. in the English
scale, then since 360 English degrees = 400 foreign,
we shall have 1 English degree = -— foreign
= -— foreign ;
.'. JV English degrees = foreign
= :\ -} loreign,
which gives the first Rule ;
To the number of English degrees, 8cc. add one-ninth
part^ and the sum will be the number of foreign degrees^ 8cc.
Ex. 1. Represent 31^ 45' 5l" English in the foreign
scale.
Here N = 31^ 45' 51'' = 3l\ 765833....
N
9
3^529537...
N
.'. IV 4- — J or the number of foreign degrees^ 8cc.
• =35\29537....=35^ 29'53" &c.
Ex.2. Hence also, 1° English =1^ ll' ll"&c. foreign;
l' English = r 85' 18" &c. foreign ;
l" English = 3" 08'" 64"" &c. foreign.
Agairij let w = the number of degrees, 8lc. in the foreign
scale, then
9
1 foreign degree = — English ;
.'. w foreign degrees = — English = ?z English,
which gives the second Rule :
From the number of foreign degrees, &c. subtract one-
tenth part,, and the remainder will be the number of English
degrees, 8ic.
Ex. 1. Express 25° 44' 89^' foreign, in the English scale.
In this case n = 25^.4489;
fi
.'. — = 2^ 54489;
ft
n J or the number of English degrees, 8cc.
= 22\ 9040 1=22*^ 54' 14" ^c.
Ex. 2. Similarly we find that
1° foreign = 0^ o4' English ;
l' foreign = O' 32". 4 English ;
\" foreign = 0".324 English.
We may here observe that in both cases minutes, seconds,
&c. must be expressed as decimals of a degree, and that it is
usual in practice to express all inferior denominations as deci-
mals of seconds.
12. Def. 3. If A represent any angle or arc, then 90^
-- A, or f — ^ ^ is called its Complement.
Ex. 1. The complement of 34^ Id' = 90^ -34^ 15' = 55° 45':
and the complement of 23° 2?' 53".67 = 90°- 23° 2?' 53".67
= u6° 32'6".33.
Ex. 2. The complement of
a±^)=i-(;±^)-a'=-')-
Ex. 3. The complement of
(i±^)-i-(ii-')="-
Ex. 4. If the angles of a plane rectilinear triangle be A,
B, C, whereof C is a right angle^ then since A+jB+C = 7r,
we have A-}-B = 7r — C = 7r—-=-, and .*. ^ = ^ — -D,
8
B= - — J, or each of the acute angles of a right-angled
triangle is the complement of the other.
13. Def. 4. If il be any angle or arc, 180^ — ^^ or
TT—A is called its Supplement.
Ex. 1. The supplement of 44° 16' = 180° - 44° 16'
= 135° 44', and the supplement of 173° 3' 13^81 = 180°-
173° 3' 13''.81 =6° 56' 46". 19.
Ex. 2. The supplement of
Ex. 3. The supplement of
(tT ± J[) = TT - (tT ± 1) = + A.
Ex. 4. If A, B, C be the angles of a triangle, and,
therpfnr^ A-\ B^C = Tr, we shall have A = tt — (B -f C),
J5 = TT - (A + C), and C = TT — (1 + £) : that is, each of
the angles of a triangle is the supplement of the sum of the
two others.
14. If in the last two articles, the foreign scale be used^
the complement of A will be = 100°-— A, and the supplement
of A = 0^00^-. A.
15. Since the magnitudes of angles or of the arcs by
which they are subtended, cannot without inconvenience be
determined by actual measurement, and since all measures found
by means of instruments are subject to error, both on account
of the unavoidable defects in their construction and the neces-
sary inaccuracy of their application, angular magnitudes as well
as the relations of angles to one another, are more easily, and
on that account more generally, found by means of certain
straight lines which are supposed to belong to all arcs and
angles, and are termed trigonometrical functions of the same :
9
these are the Sine, Co-sine, Versed sine, Chord, Tangent, Co-
tangent, Secant, and Co-secant : to' which are sometimes added,
though but little used, the Co-versed sine, Su-versed sine, Co-
chord, and Su'Chord. These lines are properly called the
Natural Sine, Natural Co-sine, &c. when the arc is supposed
to be the measure of the angle_, or the radius is supposed to
be 1.
16. If positive quantities be represented by lines measured
in any direction from a given point, it is easily shewn that
negative quantities will be represented by lines taken in the
opposite direction ; and it therefore follows conversely, that
if lines drawn in any direction be considered positive, those
lines which are drawn in the opposite direction must be con-
sidered negative. Again if lines drawn from 'the centre of
a circle through any point in its circumference be called
positive, those lines which are drawn from any point in the
circumference, through the centre,, must be supposed negative.
Also, if a point be taken in the circumference of a circle, and
through it a diameter be drawn, the positive arcs being mea-
sured in one direction from this point, the n^ative arcs will
be measured in the other.
The algebraical signs of lines are also sometimes determined
from the principle, that every quantity which admits of different
magnitudes has its sign changed* in passing through zero or in-
finity.
In the following pages, arcs of the circle which we begin
to measure upwards from the extremity of a diameter are
termed positive, and those which commence from the same
point downwards, negative. Also, all lines, in whatever man-
ner drawn, are termed positive for any arc not greater than a
quadrant.
I/* Def. 5. The sine (sin) of an arc is the straight line
drawn from the end of the arc, perpendicular to the diameter
passing through the beginning of it. Thus,
B
10
PiE
The sine of APi is N^Pij which is positive ;
^Pg is JV2JP2) positive;
-4P3 is N3P35 negative;
JP4 IS NiP^j negative.
It appears therefore, that the sine is positive in the first and
second quadrants, and negative in the third and fourth.
Hence it is also manifest^ that if the arc A Pi be called A,
NiPi is also the sine of the arcs denoted by 9,7r + J, 47r + ^,
67r-\-A, &c.(2w7r + ^).
18. Ex. "'From the definition, we have sin 0 = 0;
sin 90^ = sin (^^ ~ CP = 1 ;
sin 180^= sin (tt) = 0;
\= CE= - 1 ;
sin 360°= sin (Stt) = 0.
Hence, in the first quadrant, the magnitude of the sine
lies between 0 and 1 ; in the second, between 1 and 0 ; in the
third, between 0 and — 1 ; and in the fourth, between — 1
and 0.
19. Cor. 1. Sin (-A)= - Si?i A,
For, take APi = A, AP^ == — A ; then it is manifest, that
P4N4 = PiNi, or sin {- A)=—sm A, by (16), that is, the
11
algebraical sign of the sine of an arc changes with that of the
arc itself.
By means of this corollary, we have sin ^ = — sin { — A)
=— sin (Stt— ^'^)= — sin(47r— J.)=&c. = — sin(2w7r— ^).
Also^ from this corollary and (17) we have
sin (272 TT + ^)=— sin (Stztt — -^).
20. Cor. 2. Sin (x-A) = Sin A.
For, let AP^=:J, AF2 = {7r- A)\ therefore BCP^^^ir
-ACP2 = 7r-{7r-J) = A=ACP,, and .*. P^Nz = P,N„
or sill (tt — y4) = sin A; in other words, the sine of an angle
or arc is equal to the sine of its supplement.
From this corollary we have immediately, sin A = sin (tt — A)
= sin (Stt — ^)= &c. = sin { (2;z— 1) tt - ^}, and therefore
from (17) we conclude that
sin {2n7r-\-A) = sin {(Q.n— l) tt- A].
21. Def. 6. The cosine (cos) of an arc is the sine of its
complement, and is therefore equal to that part of the diameter
which is intercepted between the centre and the sine. Thus,
The co-sine of APj^ is P^M^ = CNi, which is positive;
AP2 is P2M2 = CiVg, negative;
AP^ is P3M3 = CN3, negative;
^4P4 is P4M4 = CIV4, positive.
IS
Hence, the cosine is positive in the first and fourth quadrants,
and negative in the second and third.
Here, as before, CA^i = cos ^ = cos (27r+ ^) = cos (47r4-^)
= &c. = cos (2?2 7r + A).
22. Ex. We have, therefore, cos 0 = CA = 1 ;
cos 90" = cos ^- J =0;
cos 180° = cos (tt) = CD= — 1;
cos 270° = cos {—) = 0;
cos 360° = cos (27r) = C^ = 1.
By these Examples it is seen, that in the first quadrant the
magnitude of the cosine lies between 1 and 0 ; in the second,
between 0 and — 1 ; in the third, between — 1 and 0 ; and in
the fourth, between 0 and 1.
23. Cor. 1. Cos ( - A) = Cos A.
For, take AP, = ^, AP^= -^; then cos(-^)=CiY4
= CNi = cos A ; that is^ whether an arc be considered positive
or negative, the algebraical sign of the cosine is the same.
From this and (2J) it folio wSj that cos ^ = cos ( — J.) =
cos ('27r± ^) = cos (47r± ^)=;&c. = cos (2rt7r + A),
24. Cor. 2. Cos(7r - A)= -- Cos A.
For, let APi = A, AP^r^^Tr— A) ; therefore, as before, we
shall have CN^^CNj,, or cos {7r—A)= -cos J, by (l6), that
is, the cosine of any arc is equal to the cosine of its supplement
with a different algebraical sign.
Hence also, cos A — —cos (tf — A)= —cos {o7r—A)=^
Ici
See. = — cos {(2/* -- 1) TT — ^} ; and by the last corollarj,
cos (2W7r±^) = — C0S{(2«— 1) TT — ^}.
25. Cor. 3. From the right-angled triangle CNiPi, we
have (Euclid, i. 47.)
P,N,' + CK = CPi\ or sin^ J + cos' A=l;
.*. also sin^ -^ = l—cos^ Az=z{l + cos -^)(1 —cos J)-,
and cos^ ^ = 1 - sin^ ^ = ( 1 + sin ^) ( 1 - sin J).
Ex. If7l=45", we shall have
1 = sin' 45' + cos' 45*^ = sin' 45^ + sin' 45°
= 2 sin^ 45^ = 2 cos^ 45^, by (21),
and .'. sin 45*^ = — ;;r-=cos45'.
V 2
26. Def. 7. The versed sine (vers) of an arc is the part
of the diameter, intercepted between the beginning of the arc
and the sine. It is sometimes called the Sagitta. Thus,
The versed sine of AP,^ is ANi, which is positive ;
^Pg is AN2, positive;
AP^ is AN3, positive;
AP^ is ANi, positive.
Hence, the versed sine is positive in every quadrant.
Also, it is clear that vers il = vers (2 7r+ A) = vers (47r + i4)
= &c. = vers (2 «7r + ^).
27. Ex. It follows therefore, that vers 0 = 0;
vers 90' = vers (-) =AC = \',
vers 180' = vers (tt) =AD=zQ,;
vers 270' = vers (-~) =^C = 1 ;
vers 360^ = vers (27r) =0.
14
From these Examples it appears that in the first quadrant,
the versed sine lies between 0 and 1 ; in the second, between
1 and 2 ; in the third, between 2 and 1 ; and in the fourth,
between 1 and 0.
28. Cor. 1. Vers{-'A)= Vers A.
For, let AP^=A, AP^=—A) then is AN^ = AlSi^y or
vers (—--4) = vers A,
From this we have vers A = vers ( — j4.) = vers (Stt — A)
= vers (47r — ^) = &c. = vers (2«7r— J.) : and therefore also,
vers (2w7r + ^) = vers (2;i7r — ^).
29. Cor. 2. Fers (tt — A) = 2— Fer5 A.
For, let J.Pi=^5 AP^ = {7r — A)y then vers {tt — A)
= AN^ = DN, ^AD- JN, = 2 - vers A,
This is called the Su-versed sine of A, because it is the
versed sine of its supplement.
Hence also, vers A =2 — vers (tt— ^) = 2 — vers (Stt — J.)
= &c. = 2- vers {(2;z— l)?:-^}.
30. Cor. 3. Since AN^ = AC — CN^, we have vers
A = l— cos A, and cos A= I —vers A, Also, vers {tt — A)
= 2 - (1 — cos 1) = 1 -f cos J.
31. Cor. 4. The versed sine of f^—Aj is BM^,
which = £0- CM^ = BC- N^P^ = 1 ~ sin A, and is called the
Co-versed sine of A, since it is the versed sine of its com-
plement.
Ex . Hence vers 45^ -= 1 — cos 45° =1 7=- ;
su-vers 45° = 1 + cos 45° = 1 H j^ ;
and CO" vers 45° = I — sin 45° =1 pr •
x/2
15
32. Def. 8. The Chord (chd) of an arc is the straight line
which joins the beginning and end of the arc. Thns,
B
^ J'/
The chord of ^Pj is AP^, which is positive;
J.P2 is AP2, positive;
J.P3 is AP3, positive;
J.P4 is iLP4, positive.
Hence, the cliord is positive in every quadrant.
And we likewise observe, that chd J. =chd (^tt-j- A) = chd
(47r + A) = &c. = chd {Qnir + A).
33. Ex. From the last Article we have chd 0 = 0;
chd 90' = chd (^) =AB = ^AC + BC = J2;
chdl80° = chd (tt) =AD = 2;
chd 270" = chd (^—^ =AE = ^AC'-{-EC'= ^ ;
chd 360^ = chd (^tt) =0.
Hence, then, in the first quadrant the chord lies between
0 and \/'2.; in the second, between ^2 and 2; in the thirds
between 2 and ^^2 ; and in the fourth, between .^/^"and 0.
34. Cor. 1. Chd {- A) = Chd A.
For, let AP^ = Af AP4= —A; then it is manifest, that
AP^=zAP,, or chd (- A) = chd A.
16
Therefore we have chd A = chd ( — .4) = chd (Stt - J)
=:chd (47r — ^) = 8cc. = chd (^wtt — A); and by (32) chd
(2}i7r + A) = chd {Inir-A).
35. Cor. 2. Chd (tt - A) = ^4 - CA^' A.
For^ let APi=A, AP2 = (7r — A); .•. the straight line
AP^=DP, =sJaD'' - ^P,", orchd(7r-J) = ^y4-chd'^ J.
This is called the Su-chord of ^.
36. Cor. 3. From the right-angled triangles, AN^P^,
AP,D, we have AP,^ -AN,^ + NiPi\ or chd" J = vers^ J
4-sin^ J, and .*. chd ^= ^ vers^ J + sin" A.
Also, JPi^= A-D.Ai\^i, orchdM = 2vers^ = 2-2cos74,
and chd A =■ >^2 — 2 cos A ; and therefore chd (tt — -^)
= V^2 + 2 cos 1, from {35).
37. Cor. 4. Chd (^^ - J.) = J5Pi=^PMi' + MiPf
= a/ci— sin ^)^ + cos''^ = ^^2-2 sin A, by (25).
This is called the Co'chord of j4.
Ex. If the arc IP be taken equal to 60', and CP, AP, PN
and PM be drawn, it is manifest that the triangle ACP is equi-
lateral, and that CA is bisected by PN: hence it follows that
^^^"^
p---
p
/
I c
N I
A
chd 60' = JP= ^C= radius = ]
vers 60' = ^1^= | JC = ^ radius r= f
17
Again, we have
cos 60' = CIV = I IC = i radius = | = sin 30" ;
, , \/3
sin 60" = PN = V CP'- CN''^ s/T^ = -^ = cos 30^
Hence also,
su-chd 60^ = >v/4~^^~ckF60^ = x/4-l= 1/3 = did 120^ ;
co-chd 60' = »y2 — 2 sin 6o' = V^^-~V^= chd 30^
38. Def. 9. The tangent (tan) of an arc is the straight
line touching the arc at the beginning, and terminated by the
radius through the end of it, produced. Thus,
The tangent of ^Pi is ATi, which is positive;
^IPg is J. To, negative;
APs is AT3, positive;
AP4 is A 2 4, negative.
We observe, therefore, that the tangent is positive in the
first and third quadrants, and negative in the second and fourth.
Hence likewise, tan A =tan (^tt + ^) = tan (47r + A)=&c.
= tan (2«7r + A).
39. Ex. From this definition, we have tan 0 = 0;
tan 90' = tan (J^J = 00;
C
18
tan 180" = tan (tt) =0;
tan 270^ = tan ( — } = ~ QO ;
tan 360^ = tan (Stt) =0.
These two articles prove that in the first quadrant, the mag-
nitude of the tangent lies between 0 and oo ; in the second,
between — oo and 0 ; in the third, between 0 and QO ; and in
the fourth, between — co and 0.
40. Cor. 1. ^«y^ (-A)= - Tan A.
For, take JPi = A, AP^ = — ^ ; then it is manifest that
AT^ = ATi, or tan ( — A) = -tan J, by (16); that is, the
algebraical sign of the tangent of an arc changes with that of the
arc itself.
Hence therefore, tan A =■ — tan { — A)=- — tan {^tt —A)
= ~ tan (47r— ^) = &c. = ~ tan (2/27r- A).
And, from (38) we have likewise tan (2/Z7r+i4)= —
tan (2;i7r — A).
41. Cor. 2. Tan {it ^ A) =^ — Tan A.
For, take APi = A, AP^ = w — A ; therefore Z ACT2
= zACTi, and AT^ = ATi, that is, tan (tt — ^)= — tan ^,
by (16); or the tangent of an arc is equal to the tangent of its
supplement with a different algebraical sign.
From this we have also, tan ^ = — tan (tt— A)= —
tan (37r— ^) = -tan (57r — J[)=&c.= — tan {{2n-~ l)7r — J.};
and also from (38), tan(2w7r + ^)= -tan {(2^i-l) tt-A],
42. Cor. 3. By the similar triangles CNiPi, CATi,
we have
CNi : NiPi :: CA : ATi,
or cos J. : sin :4 :: 1 : tan A,
and therefore tan A = —7.
cos A
that
19
Ex. From this corollary and the preceding pages^ it follows
tan 30^ =
tan 45" =
tan 60^ =
sin 30^
cos
30" ^3'
sin 45"
COS45"
sin 60"
1;
cos 60'
0 = x/3.
43. Def. 10. The co-tangent (cot) of an arc is the tangent
of its complement, and is therefore the straight line touching the
circle at the end of the first quadrant, and terminated by the
radius through the end of the arc, produced. Thus,
ts
U
B /;?
I
>x
C
-y^
y^
y
t
D
1
^
/x
.
V
/
^J
V
/
^x^d
/
The co-tangent of AY^ is -B^j, which is positive;
AV^ is ^t^y negative;
A^n, is 5^3, positive;
^P4 is Bt^, negative.
The co-tangent is therefore positive in the first and third
quadrants, and negative in the second and fourth.
For the same reason as before, we shall have cot A =
cot (Stt + A) = cot (47r + A)= &c. = cot (Q,tnr + A).
44. Ex. Hence therefore, cot 0= co;
cot go'' = cot (^^ =0;
cot 180° = cot (tt) = — CO ;
cot 270° = cot {^^} =0;
cot 360° = cot (Qtt) = CO .
From these two articles, it appears that the co-tangent
in the first quadrant is between co and 0 ; in the second,
between 0 and -co; in the third, between co and 0; and in
the fourth, between 0 and — c» .
45. CoR.l. Cot {'-A)= - Cot A.
For, let APi^A, AP^= —A; then, it is evident that
Bt^ = Bt^, and therefore cot (— ^)= -cot A, by (l6); or the
algebraical sign of the co-tangent of an arc changes with that of
the arc itself.
Therefore also, cot ^ = — cot ( — A) = — cot (Stt — ^4)
= — cot (47r — A) = 8cc. = — cot (Swtt — A)\ and thence
by (43) we have cot(2«7r + A)= - coi {Q.nir - A).
46. Cor. 2. Cot {tt - A)= — Cot A.
For, let APi — Ay JP2 = tt — A\ then, it is manifest that
Bt^ — Bti, or cot (tt — A)=— cot A, by (16); that is, the
co-tangent of an arc is of the same magnitude as the co-tangent
of its supplement, but with a different algebraical sign.
From this corollary, we have likewise cot A = — coi {tt — A)
= - cot (Stt— A) = &c. = - cot {{Q.n — 1) tt — A] ; and
therefore also by (43), cot {^n'Tr-\~A)= —cot {(2«— \)'7r—A}.
47. Cor. 3. By the similar triangles, CMiPi, CBti,
we have
CM, : M,P, :: CB : Bt,,
or sin A : cos ^ :: 1 : cot A,
cos A 1 . .
and .*. cot A = -: = r , by (4'2).
sm A tan A ' -^ ^
21
Ex. It follows from what has been already proved, that
cot 30^
cot 45'
cot 60°
cos 30
sin 30^
cos 45°
tan 30" -V3;
sm 45 tan 45^
cos 60° 1
"^iiTeo^ ~
= 1
tan
60° - V5 •
48. Def. 11. The secant (sec) of an arc is the straight
line drawn from the centre through the end of the arc, and
termmated by the tangent. Thus,
The secant of APi is C2\, which is positive;
AP2IS CTo, negative;
^Pg is CT3, negative;
' AP4 is Cl\j positive.
Therefore the secant is positive in the first and fourth
quadrants, and negative in the second and third.
Hence also as before,, secil = sec (^tt-}- -4)=^ sec (47r -j- j4)
= &c. =:sec i2n7r + A).
22
49. Ex. We shall therefore have sec 0=1;
sec 90^ = sec ( - ) = 00 ;
sec 180^ = sec (tt) =— 1;
sec 270^ = sec ^ — J = — go ;
sec 360^ = sec (27r) =1.
We conclude then^ that the magnitude of the secant in the
first quadrant lies between 1 and oo ; in the second, between
— 00 and — 1 ; in the thirds between — 1 and — co ; and in the
fourth^ between oo and 1 .
50. Cor. 1. Sec (- A)=^Sec A,
For, let APi = J, AP4 = - A; then CT4 = CT^, or
sec ( — ^) = sec J.; that is, the magnitude and algebraical sign
of the secant is the same whether the arc be positive or negative.
Hence also^ sec ^ = sec ( — A) = sec (Stt — A) = sec
(^TT— A) = 8cc. = sec (SwTT — A); and therefore by (48) we have
sec (2w'7r + ^) = sec {^utt — A),
51. Cor. 2. Sec i'7r — A)= - Sec A,
For, let lPl = ^ AP2^7r-A; then, CT^^^CT^, and
,'. sec {t7--A)= —sec A, by (16); or the secants of an arc
and of its supplement are of the same magnitude, but have
different algebraical signs.
And as before, sec j1 = — sec (tt — ^)= — sec (Stt — A)
= &c.= -sec {(2?i— 1) ir-A}: also by (48), sec (2/7 7r + A)
= —sec {(2yi— 1) tt— ^}.
52. Cor. 3. From the similar triangles CNiP^, CATi,
we get
CTi : CA :: CPi : CNi,
or sec A : 1 :: 1 : cos A,
and therefore sec A =
cos A
Also, from the triangle CATi, we have CTi' = CA^ + AJ?,
that isj sec^ A=-\'\- tan^ ^, and .*. tan^ A = sec" A — 1,
Ex, From either of these formulae, we shall have
2
sec 30'
,^, sec 45^ = ^2, and sec 60^ = 2.
x/3
53. Def. 12. The co-secant (cosec) of an arc is the secant
of its complement, and is therefore the straight line drawn from
the centre through the end of the arc, and terminated by the
Thus,
t^ t^.
B ts
f
\,K
<"
^>r
/
/y
X
/
J)
A
/x
\
\
/
\ "I
V
3*^
/
The co-secant of A Pi is Ct\, which is positive;
ilP2 is Ctc^i positive;
AT^ is C^3, negative;
J.P4 is C^4, negative.
Hence, the co-secant is positive in the first and second
quadrants, and negative in the third and fourth.
Alsoj cosec A = cosec (2 tt 4" ^) = cosec (4 tt -f- A) = &c. . . .
= cosec (2/Z7r -j- A).
54. Ex. This definition gives cosec 0 = co ;
cosec 90 = cosec
G) =
1;
24
eosec 180^ = cosec (tt) = oo ;
0 /^'^\
cosec 270 =cosec f — I = — 1;
cosec 360^ = cosec (Qtt) =00.
In the first quadrant therefore^, the co-secant is between co
and 1 ; in the second^ between 1 and co ; in the third,
between — co and — 1 ; and in the fourth, between — 1 and — co .
55. Cor. 1. Cosec { — A) = — Cosec A,
For, let APi = A, ^P4= —A; then Ct^ = Ctu or cosec
( — J.)= —cosec A, by (I6); that is, the algebraical sign of the
cosecant changes with thjit of the arc.
Hence also, cosec A= -- cosec (— A)= — cosec (2 tt — A)
= — cosec (47r — A) = &c. = — cosec (Q.mr — A).
And therefore by (53) j cosec (2;?7r + A) = — cosec
{2n7r — A).
56. Cor. 2. Co.sec (tt— A)= Cosec A.
For, let APi = A5 AP2 — 'n-- A; then Ct2=^Cti, as is
evident; that is, cosec (tt— A) = cosec A, by (I6); or the co-
secant of an arc is equal to that of its supplement.
So also, we have cosec A = cosec (tt - A) — cosec {Sir — A)
= &c.= cosec {(2«— 1) TT — A] ; and therefore by {o3)t cosec
(2«7r4- A)=:cosec {(2w— 1) TT— A}.
57. Cor. 3. From the triangle CBt^, Ch'^^ CB^ -^ Bti\
that is, cosec^ A = 1 + cot'^ A, and .*. cot" A = cosec'A — 1.
Also, cosec^ A =z\ -j- cot^ A
cos^ A sin^ A 4- cos^ J I
= 1 -f . . . = 7-2-1 = -:^T-r y by (25),
sin A sm A sm A
and therefore cosec A
sin A
25
Ex. These formulae _, with what has gone before, give
cosec 30^ = 2, cosec 45° = /n/2, and cosec 60^ =
58. In the preceding articles, we have determined the
algebraical signs of the tangent, co-tangent, secant^ and co-
secant, from an examination of the lines which represent them
in the figures, according to the principles assumed in (16);
but it may be observed that they are all easily deducible from
those of the sine and co-sine previously found.
IhuSj tan Ari= 7-— must be positive, since sin Ar^i
cos A Pi
and cos APi are both positive:
. _ cos APo . . . . „
and, cot AP2= ~ r-~ must be negative, since cos A-Tg
sin AP2
is negative, and sin AP^ positive :
also, sec AP3 = 7^=r- must be negative, since cos AP3
cos AP3
is negative :
and. cosec APa^^ . . .^ must be negative^ since sin AP4
sm AP4
The same method may be used to determine the magnitudes
of the same functions. Thus,
sin 0 0
tan 0 = = - = 0;
cos 0 1
i,, C)
sm
X7r\
tan
sec 0 =
cosQ
0
aj;
1
cos 0
1 __
" 1 ~
1 ;
D
26
sec
(f)
= ~ = CO :
cos
(I) °
and so of the rest, as alread}? found.
59. To transform trigonometrical formula constructed to
the radius 1, into others which shall be adapted to any radius r.
NA
N A
Let Ci4-I, CA'=r, and let ACP = AXP' be any
proposed angle, which is represented by A : then, to the radius 1 ,
we have
PN=s\n A,
CN=cosr A,
AN = vers A,
AP = chd A,
AT = tan A,
Bt = cot A^
CT= sec A,
Ct =cosecil;
and to the radius r, we have
P'iV' = sin A,
CN' =cos A,
A'N' = rers A,
.4'F=chd A,
A'r = t2iu A,
B' l' =cot A^
Cr =sec A,
Ct' =coseCil:
27
hence, denoting these lines on the latter scale by accents placed
over them, we shall have (Eucl. 6. 4.)
Sin^ : sin' A :: PN : FN' :: CP : CF :: 1 : r;
.*. sin A = " sin' A, and sin' A=r sin A
r
Cos A : cos' A :: CN : CN' .: CP : CP' ::
.*. cos -4 = - cos' Af and cos' A = r cos A
r
Vers A : vers' A :: AN : A'N' :: CP : CP' ::
.*. vers A — - vers' ^4, and vers' A=r vers il
r
did A : chd' A :: ^P : A'P' :; CP : CP' ::
.'. chd A = - chd' A, and chd' i4=r chd A
r
Tan^ : tan' A :: AT : AT' :: CA : CA' ::
•'. tan A = - tan' yl, and tan' A=r tan il
r
Cot vl : cot' A :: JB^ : 5'^' :: CB : CB' ::
.*. cot A = - cot' ii, and cot' A = 7- cot A
r
Sec A : sec' A :: CT : Cr :: CA : CA' ::
/. sec A = -- sec' J^, and sec' A = r sec A
r ;
Cosec A : cosec' A :: C^ : C^' :: CB : Ci3' :: I
.*. cosec A = -- cosec' A^ and cosec' tI = /• cosec A.
Hence, therefore, if we wish to make use of the radius r
instead of the radius 1, we have only to substitute in any proposed
28
formulae which are true on the supposition of the radius being 1,
sin A cos A ^ . . , r ■ A A
the quantities , , &c. ni the places ot sm A, cos A,
&c. respectively, and the results will be adapted to the radius r.
Ex. 1. We have seen in (25) that sin^ A+cos^ ^ = 1^ to
the radius 1 ;
(sin A\' , /cos A\^ . ..
/ "^ ( / ~ 1, to the raduis r,
or sin^ A + cos'^ A = r^, when the radius is r.
Ex. 2. By article (42)^ tun A = , to the radius 1 ;
cos A
sin A
tan A\ \ r /
(¥)
(tan A\ \ r ^
1 = — tQ tjje raduis r.
r y ycosA^ '
. /sm ^\ .
or tan A = r i I , it the radius be r.
vcosA/
Ex. 3. If the formula, cos m^=acos'" A-\-h cos*"~^^
+ c' cos*"~'" ^ + &c. were true on the supposition of the radius
being !_, then according to the article, we have
(QOsmA\ /cos^x*" ,/cosJ.\'"~^ /cos it\'"~''' „
or r*'*-^ cosmJ = « cos'" A+r/'; cos'"--^ ^4 +7''c cos""'"^ A+&c.
which would be true if the radius were represented by r.
6*0. Cor. 1. From the last example, which involves j
general expression, may be deduced the following Rule :
Render all the terms of any formula homogeneous, by
multiplying each by such a power of r as shall make its dimen-
sions equal to the highest involved in it, and the result will be
adapted to the radius r.
29
Ex. To the radius 1, we have seen in (52) that sec^ A
= 1 + tan A ; therefore to the radius r, we shall have sec^ A
= r^ + tan^ A, by making each of the terms a quantity of two
dimensions.
6l. Cor. 2. By similar substitutions, formulae deduced on
the supposition of the radius being R, may be transformed into
others which shall be true when the radius r is made use of.
For, let sin A represent the sine of A to the radius R,
sin' A ..r,
then, sin A : sin' ^ :: jR : r, as before,
and .*. sin A= — sin A ;
J-
similarly, cos A—— cos' A, &c. = &c.
;•
hence, for sin A, cos A, &c. we have only to substitute
R . , R ,
-^ sin A, — cos Ay &c. respectively,
and the results will be adapted to the radius r.
Ex. 1. The formula tan .4 = 2 -, is true for the radius
cos A
2, by (59); therefore if we wish to use the radius 3, we shall have
2 ^ ^ sin A , ^ sin A
- tan A = 2 , and .*. Ian A=^S .
S -t cos A cos A
Ex. 2. Sec" A= l6-f-tan' A, is true by (60) for the radius 4 ;
therefore if the radius be 10^ we shall have
(iysec^4=,6+(lyta.r4,
and .*. sec^ A = 100 + tai/ A, to the radius 10; and so on.
The same methods are applicable to transform any other
similar and similarly situated lines, from one radius to another.
30
62. By means of the relations between the trigonometrical
functions of an arc established in this chapter, we are enabled
to prove divers theorems, and to solve a variety of problems.
Ex. 1. It is required to prove that
in A = .y/2 vers A — vers^ A, to the radius 1 .
sni
By (25) we have sin^ ^=1 — cos^ J.= l — (1 — versil/ from (30),
= 1 — 1+2 vers A - vers^ A = 2 vers A — vers^ A ;
sni
sni
A = ^2 vers A — vers^ A, to the radius 1.
Also, to the radius r, by (59) we shall manifestly have
n j4 a/ yvers A\ ><vers ^\'
— = V "-K-j-)-KrT~) '
and .'. sin A =,^2r vers tI — vers" A,
^ _. , tan A + tan B . ,,
Ex. 2. lo prove that =- = tan A tan B. to
cot J + cot B
the radius 1.
From (42) we have
_. sin A sin B sin A cos B + cos A sin B
tan A -f tan B = H = :
cos A cos i) cos A cos B
also, from (47) we get
, . „ cos A cos B cos A sin B -f- sin ^1 cos B
cotA + cotJ5= - — - 4- - — ~= ^ — -i—. — 5 :
sm A sni B sm A sni i>
and the numerators of these fractions being the same, it follows
that
tan j4. 4- tan B sin A sin B /sin A\ /sin B
cot A + cot B cos ^4 cos B
tan il tan B, by (42), to the radius 1.
(sm A\ ysui n\
cosA^ \cosB/
31
tan A + tan B tan A tan B . ,.
Likewise — = 5 — — to the radius /', as
cot -4 -i- cot 15 r"
easily appears by means of the rule laid down in (60).
Ex. 3. Given m sin A=Ji cos A, to find the values of
sin A and cos A.
Here, m^ sin^ A =11' cos^ A = 71^ (1— siir A), by (25),
= n^ — ?t^ sin'^ -4 ;
/. (m^+w^) sin^ A^n^, whence sin A= ±
y
7?i^ + w^
also cos J. = 1 — sm ^1 = 1 —
m
and .*. cos -4= Hh / — r^=^, to the radius 1.
^m +71"
If the radius r be used, we have by what is proved in (59),
sin A n cos A m
^ i'^^m' + n r ^m^ + rr
whence sin ^ = ± — y -„ -„ , and cos A = ±
/— 2-^ 5 aim VV.D ^ — T. .— — — - .
Ex. 4. Given sin A = m sin JB, and tan A=w tan J3, to
find the values of sin A and sin B.
Since tan -4 = w tan B, we have
sin il sin -B , , ^ ^-
. = n -, by (42);
cos A cos .0
sin A cos A , . cos A
:. - — 5 = 71 5, that is, w = w -;
sm jD cos 75 cos i)
m
« ^^.2 A T _ ,-.,2 ^ 1 _ 2 ■'1
cos^ 1 _ 1 - siir A _ 1 - m^ siir B
cos' -B "" 1 - sin'' J? ~ 1 - sin^ B '
32
whence m^-^rri' sin^ B = ri^ - n^rt^ sin' By and
m^ ( 1 — w^) sin^ 3=^71^ — w', or sin^ 5 = — ^"^ ~^ ,
m (1 — w")
and /. sin jtf= + — V/ ;
ni ^ 1 - w^
wherefore sin il =#?2 sin jB= Hh 'y/ ^ ,
1 — IlL^
in both of which the radius is 1.
Adapting these values to the radius ? , since ni and n are
merely numerical magnitudes and therefore not considered of
any dimensions, we shall have by (60),
sm
A I 4 / W^ —if J • D I '' 4 /W^ — if
A = + r V , and sni 5 = + - \/
From these two equations all the other trigonometrical
functions of A and B are easily deduced.
CHAP. II.
On the relations between the Trigonometrical Functions of arcs
or angles, and those of their sums and differences, and also
of some of their multiples, sub-multiples and powers.
63, To express the sines and cosines of the sum and dif-
ference of tivo arcs, in terms of the sines and cosines of the arcs
themselves.
Let the arcs AP, PQ, be the measures of any two angles
ACP, PCQ denoted by A and B to the radius CA: draw QM
perpendicular to CP, and QR, MS, PN perpendicular to CA *
draw also MT parallel to CA. Then^
N A
PN =sm ACP, CIV = cos ACP ;
QM=sin PCQ, CM= cos PCQ;
QR = sin ACQ = sin {ACP ± PCQ) ;
CR = cos ACQ = cos (ACP ± PCQ) :
Now QR = RT ±QT= MS ± QT, the upper and lower
signs being used for the first and second figures respectively ;
E
34
but by similar triangles,
MS ; CM :: PN : CP,
and QT: QM :: CN i CP ',
PN CN
from which MS = CM— , and QT=QM-^;
PN CN
/. we have QE = CM— ± QM —
:=-^{PN.CM± CN,QM) = PN.CM ± CN.QM,
if the radius be supposed to be 1 ; that is,
sin iA± E) = sin A cos B ±cos A sin B (a).
Again, CR=CSTRS=zCS + MT;
but by similar triangles,
CS : CM :: CW : CP,
and Mr : QM :: PN : CP;
CN PN
whence CS= CM ;^, and Mr=QM-— ;
Ux Ox
CAT PN
.V we have CR=^CM~ + ^^ CP
^ JL(CN.CM+ PN.QM)= CN.CM+ PN.QM,
if CP= I, as before; that is,
cos {A±B) = cos il cos B ^ sin A sin B (/3)
Ex. 1. Let 5 = ^, or ^ + 5 = ^ + ^:
--^ -f ^ 1 = sni ~ cos ^ -I- cos - sm J
= cos A, by (18) nnd (22):
35
and cos I - + ^ I = cos - cos ^ — sin ^ sin A
= -sin A, by (18) and (22).
Ex.2. Lct^ = 7r, or A + B = 7r-\-A:
therefore sin (tt -\- A) = sin tt cos A 4- cos tt sin A
= -sin ^, by (18) and (22):
and cos (7r + 74) = cos x cos A— sin tt sin A
= — cos ^5 by (18) and (22).
64. Cor. 1. The construction and investigation above
given hold good whatever be the magnitudes of the angles
ACP, PCQy due regard being had to the algebraical signs of
the trigonometrical lines as determined in the preceding chapter :
and any three of the functions just mentioned may with great
facility be deduced from the remaining one. Thus,
mn(A-B) = sm {7r-(^-.B)}, by (20),
= sin {{7r-A) + B}
= sin (tt — A) cos B + cos (tt — A) sin B, by (a),
= sin A cos jB— cos A sin B, from (20) and (24):
again,
cos {A 4- B) = sin {3 - (^ + -B)| , by (21),
= sm |(^ +^) + i^j, by (22),
= sin f^ + Aj cos B + cos T- + A J sin B, by (a),
= 095 A cos B - sin A sin jB^ from (ti3) :
36
and, cos {A-B) = sm |^ -(^~ JB)j, by (21),
= ,n{(l-A)+B}
= Sin f^ —Aj cos B + cos C- - A J sin E, by (a),
= cos ^ cos £ -\- sill ^ sin i3, from (21).
The same values of cos (A ± B) are also easily deducible from
the equation cos (A± B) = -.^/l^--sir?(^±5).
65. Cor. 2. If the radius CP be supposed = ;•, we shall
have
sin {A± B)= - (sin A cos B ± cos A sin B) ;
and cos (^ + jB) = - (cos A cos B + sin y4 sin B),
r
which are the same as would have been derived from those just
found by means of the rule laid down in article (60).
66. Cor. 3. Let •S' and s be the sines of any two arcs^
C and c their cosines ; then by {QS) we have,,
The arc whose sine is 5 + the arc whose sine is s =
the arc whose sine is (*Sc + sC): also, the arc whose cosine
is C + the arc whose cosine is c = the arc whose cosine
is {Cc T Ss),
These are usually written abbreviatedly as follows :
Sin-' ,S ± sin"' s = sin"' {Sc ± sC) ;
Cos''C± cos-'c = cos-^' (CcT^5).
Ex. Sin-.(0+.i„-.©=™-.j2J + liS
37
125 25) ^ *
= (^), as appears from (18).
67. From (63)j we obtain by addition and subtraction
the following formulae :
sin (A + £)-{- sin (A ~- B) = 2 sin A cos B;
sin (A + B) — sin {A — B) = 2 cos A sin B ;
cos ( J[ — ^) + cos (^ + B) = 2 cos J cos ^ ;
cos (A — B)- cos(A 4- -B) = 2 sin ^ sin B.
These expressions furnish the following useful equations :
1. The sum of the sines of any two arcs is equal to twice
the rectangle of the sine of their semi-sum, and the cosine of
their semi-difference.
£. The difference of the sines of any two arcs is equal to
twice the rectangle of the cosine of their semi-sum, and the sine
of their semi-difference.
3. The sum of the cosines of any two arcs is equal to
twice the rectangle of the cosine of their semi-sum^ and the
cosine of their semi-difference.
4. The difference of the cosines of any two arcs is equal
to twice the rectangle of the sine of their semi-sum, and the
sine of their semi-difference.
68. From the same article we obtain by multiplication,
sin {A + B) sin {A - B)
= (sin A cos B + cos A sin B) (sin A cos B — cos ^ sin B)
= sin'^ A cos" B — cos^ A sin^ B
= sin" A (1 —sin' 5) — (1 —sin' A) sin^ B
= sin^ A — sin" A sin^ B — sin^ B + sin" A sin^ B
= sin' A — sin" B = (sin A -{- sin B) (sin A — sin B) :
or
= cos^ B — cos^ A = (cos B + cos A) (cos B — cos A).
Similarly,
cos (^4-B) cos(^-B)
= (cos A cos B — sin A sin JB) (cos A cos jB + sin A sin B)
= cos^yi cos^ B — sin^ ^ sin^ ^
= cos' A {I- sin' 5) - (1 -cos^ A) sin' B
= cos^ A — cos- ^ sin^ B — sin^ ^ + cos' A sin^ JB
= cos^ A - sin- B — (cos J + sin jB) (cos A ■— sin B) :
or
= cos^ B — sin' A = (cos B + sin A) (cos B — sin A).
69. By division, we have from the same article,
sin {A + jB) sin A cos B + cos A sin jB
sin (A — jB) sin A cos B — cos tI sin B
(sin J.X >^sin 5\
cos Ay Vcos JB/
(sin A\ >^sin Bx
cos tI/' Vcos B^^
(by dividing both numerator and denominator by cos A cos jB)
tan A 4- tan B
tan il - tan JB '
In like manner,
cos (A + B) _ cos A cos B — sin A sin B
cos {A — i^) cos A cos J3 + sin A sin B
_ 1 - tan A tan JB
~ 1 4- tan A tan B
above.
by proceeding as
39
The former of these expressions furnishes the following
useful proportion:
The sine of the sum of two arcs : the sine of their
difference :: the sum of their tangents : the difference of
their tangents,
70. To express the sines and cosines of tivo arcs in
terms of the sines and cosines of their semi-sum and semi-
difference,
therefore we have by means of (63),
.in ^ = sin {(i^) +(^^^)}
= sm (-^-; cos {^-^) + cos (-5-) sm (-^^ ;
and sin B = sin {{-~) - (^)}
Similarly, cos A = cos \ I — — / "^ ( ) i
= cos (— ^; cos {—^) -sm {—^) sm {—^)^
and cos B = cos | (^— ^J - i^—^) (
40
= eos (-^-) cos {-Y-) +s... {-^) «m {^-^-)
71. This article, by addition and subtraction, gives
sm id + sin B =2 sin f — - — 1 cos f — - — 1 ;
sin A - sin B = Q. cos I — 5°~~ I ^"^* I — 5 — J '
cos B + cos ^ = 2 cos I — - — I cos f I ;
, ^ . /A + B\ . /^ - J5\
cos J5 — cos A = 2 sin f I sm I 1 ;
in which expressions are comprised the Rules laid down in (67).
72. From the same article, we obtain by division,
'A-\-B\ /A-B^
. /A + B\ /A-B\
sin A 4- sin B
sin A ~ sin B /A + jB\ ^^-^
(A + ±>'\ . /A- ±S\
TJ^-gT' by (42) and (47);
tan
(^)
41
- cos jB + cos a
and
2 cos {—^) cos(^-^J
cosi)— costI . /<A -\- B\ . /A~B^
A+B\ /A-B^
cos
(^") - (^0
cot
From the former of these formula^, we have the following
useful Proportion.
The sum of the sines of any two arcs : the (hfference of
the sines :: the tanoent of their semi-sum : the tanc-ent of their
semi-difference.
By a similar process it is easily proved that
sin .i + sin E __ /A+ B\
cos J -{- cos B "^ ' \ 2 / *
and if i>=Oj we shall find by reduction that
J ] — tan ~
. ^ A / 1 - cos ^ , , 2
tan — = V -, and .". cos A = r .
2 ^ 1 + cos A ^ , . yl
1+tan- —
73. To express the sine and cosine of (n + 1) A, in terms
of the sines and cosines o/'nA, (n— 1) A, and A.
Here, attending to the formulce of (63) we have
sin (n 4- 1) A = sin 0^^ + A)
= sin 71 A cos A + cos ?iA sin Aj
and sin (/^— 1) A = sin (nA — A)
= sin nA cos A — cos n A sin A;
42
hence by addition, we have
sin (n-i- ]) A + sin (n — \) A = 2 sin ti A cos A,
and .'. sin {n -r I) A = Q sin nA cos A — sin (71 — \) A. .
Again, cos {12+ I) A = cos {nA-\'A)
= cos nA cos tI — sin nA sin ^,
and cos {n — \) A = cos (/i^ ~ A)
= cos 7f A cos A + sin //A sin A ;
whence as before, we get
cos (n -{- 1) A + cos (n — 1) A = Q cos 7iA cos il^
and .'. cos {n + l) A = 2 cos nA cos A — cos (?^ - 1) ^.
74. Ex. Let n be taken equal to 1, 2, 3, 4, &c.
successively, and we shall have
sin 2 J. = 2 sin A cos A ;
sin 3 A —9. sin 2 J. cos A - sin J.;
sin 4j1 = 2 sin 3 J. cos ^1 — sin 2^4 ;
&c = &c
cos 2 J. =2 cos A cos A — 1 ;
cos 3^ = 2 cos Q.A cos tI — cos A ;
cos 4^ = 2 cos 3 tI cos A — cos 2 J[ ;
&c = &c
75. To express the sine and cosine of twice an arc in terms
of the sine and cosine of the arc itself.
Here, by means of {QS), we have
sin 2 J. = sin ( A + -4) = sin A cos A + cos A sin A
= 2 sin A cos A ;
or = 2 sin A /s/ l—" s\n^ A ; or = 2 cos A ^ 1 -- cos'^ A
43
Also_, cos 2^ = cos (A -jr A) = cos A cos A — sin A sin A
= cos^ J[ — sin^ A ;
or = 2 cos^ tI — 1 ; or = 1 — 2 sin* A.
A
76. Cor. Putting A and — in the places of 2 A and A
respectively, we have from the last article^
sni ^ = 2 sni — cos —
2 2
and cos il = cos^ — — sin'' — = 2 cos'' 1 = 1—2 sin
A
2'
A . o A o A
— sni — = 2 cos^
2 2 2
also, from the latter of these we obtain
A
2 sin^ — = 1 - cos A = vers A = |- chd^ A, by (36),
A A
and .'. 2 sin — = chd A. or sin — = ^chd A ;
2 2 ^
that is, the chord of an arc is equal to twice the sine of
its half, or the sine of an arc is equal to half the chord
or its double.
77- By some writers, the properties just mentioned are
made the basis of many of the fundamental propositions of
Trigonometry, and they may be proved merely by a comparison
of the respective lines in the following figure : thus,
Let AP = A, AQ=^QP =^ — j then it appears from the
44
definitions laid down in the preceding chapter, that j4.P = chd A,
and QM=sin — - : it is also manifest that AL = LP. or AP
2
= 2AL, and AL=:QM: whence it follows that ^P = 2 QM,
j4 a I
or chd ^ = 2 sin — . and .*. sin — = - chd A.
2 2 2
78. To express the sine and cosine of half an arc in terms
of the sine of the arc itsef,
A A
Since cos^ \- sin' - = 1, by (25),
2 2
A A
and 2 sin — cos — = sin A^ by (76) ;
we have by addition,
o A .A A . ^ A . .
cos —4-2 sin — cos f- sni — = 1 -f sm J.
2 2 2 2
and by subtraction,
2 ^ .A A ^ , ^ A . ^
cos 2 sni — cos h sar — = 1 — sm A ;
2 2 2 2
whence, by extracting the square roots of both sides of these
equations, we obtain
j^ j^ ^ j^
cos~+sin -= V^J 4-sin A, cos ~ - sin — = + ^1 — sin A'-,
in the latter of which the positive or negative sign must be
used according as cos — is greater or less than sm — ;
2 2
.'. by addition and subtraction and division by 2,
A 1
cos — = -
2 2
and sm -
2
|x/(H-sin A) ± J\^\~ sin A)\ ,
-=- |x/(l+sin A) + V(l-sin 1)|
45
These two expressions are frequently used to examine the
accuracy of results deduced by other means, and on that account
are termed Formulte of Verification.
A 1
Ex. Let A=SO^, then - = 15^ and sin J= - , from (37),
wherefore sin lo^ = - {^1+sin 30^— ^1 —sin 30^}
and cos 15^ = - {^1+siu 30^ + >/l -sin 30^}
2
79- 2^0 express the sine and cosine of half an arc in terms
of the cosine of the arc itself
From (76)5 it appears that
cos — — sin — = cos A.
2 2 .
also, cos^ — + sin" — = ] , by (25) ;
A
.*. by addition_, 2 cos' — = 1 +cos A^
A
and by subtraction, 2 sin" — = 1 — cos A ;
whence,, dividing each of these equations by 2, and extracting
the square roots, we have
. A 4 /xl —cos ^X
sin — = + V I ) ,
2 ~ ^ V 2 y'
a„acosf=±v/(^).
46
A 1
Ex . Let A = 45^ then - = 22^ 30\ and cos A = — r- from
2 ^2
(25)_, whence we have
0 / A /l — cos 45^ A /l 1 ./Vq-I
sin 22° 30'= V = V 7- = V 7— '
^ 2 ^22l/2^2V2
eos 2.» 30'= x/i±^°^ = \/V^ = \/^?±^.
^ 2 ^22\/2 2 1/2
80. From articles {QS) and (75) may easily be deduced
what is called Delambre's formula.
For, sin {A-\- l^) = sin A cos l^ 4- cos A sin 1^^
and sin (^4. — 1^) = sin A cos 1^ — cos A sin l'' ;
whence by addition,
sin {A + 1^) + sin (J - 1^) = 2 sin A cos 1^
and .•. sin (A + 1^) = 2 sin A cos 1° - sin (A - 1^)
= 2 sin 1 (1 - 2 sin^ 30') - sin (A - 1°)
= 2 sin A - sin {A — 1^)— 4 sin A sin" 30'
= sin ^ + {sin J. - sin (A — 1^)} - 4 sin A sin^ 30\
8 1 . We have seen in the last article but one, that
• ^ \/^~~~^ 7
sin — = V - — - cos A
2 2 2
and from these are readily derived the following equations :
. A ./~i i A
sni -7T = V cos —
2- ^22 2
= \/i-iv/.-si.,»-^
47
4 = \/i^
1 A
sin — ^ = V — ;: cos -^
2 2 2 2
- V ^ - ^ V 1 - sin -, ;
2 2
(fee = &c
1 1
Sin — - = V ^ -- - cos
2" 2 2 2*""
Also, from the same article it appears that
2
whence we similarly obtain
cos - = V 5 + 5 cos 4
= V i + 5 \/l-siu^^
cos^.
i = V 2 + 5COS-
1 JL
cos -o = V " + ~ cos —
2^ 2 2 2-
- V - + - V 1 ~ sm -.
2^
ike = Sec r.
48
cos
COS
,n-l
82. From the last article, by substitution we shall have
sin
sni
cos ^ ;
Sec.
= 8cc.
sm — - =
v^^FI^
the radical sign being repeated ii times.
In the same manner, we get
cos
cos
A;
&c.
= &c.
COS — =
yl^Wl^lVl^lV^c \/i+icos^,
where the radical sign is repeated n times.
49
Ex. Let A= C-^ ; and /. cos A=0, and we have
{r4''^i-Wi*W^ n/^
COS
the radical sign in each case occurring n times.
83. To express the sine and cosine of thrice an arc in
terms of the sine and cosine of the arc itself.
Here we have from (63),
sin 3 ^ = sin (2 A-\-A) = sin 2 ^ cos A + cos 2 A sin A
= 2 sin A cos^ A + (1 — 2 sin^ A) sin A, by (75)
= 2 sin A - 2 sin^ il + sin A — 2 sin'^ A
= 3 sin A — 4 sin^ A ;
and from the same article,
cos 3 A = cos (2 A +-4)=: cos 2 A cos A — sin 2 A. sin A.
= (2 cos^ J - 1) cos A - 2 sin^ A cos A, by (75)
= 2 cos' ^ — cos ^ — 2 cos A + 2 cos^ A
= 4 cos^ A -- 3 cos A.
84. From the two formulae just proved, we have immediately
sin^ A=\{^ sin A — sin SA), and cos'' A=\{S cos A-\- cos 3 A).
85. By substituting in the formulae just investigated,
A and — in the places of 3 A and A respectively, we get
sm A = 3 sm — ' — 4 sm — . and cos A = 4 cos 3 cos — :
3 3 3 3
and thence the equations,
4 sin" (- } ■- 3 sin ("^ + sin A = 0,
G
50
and 4 cos^ ( ~~ ) "*" "^ ^^^ V "" ) — ^^^ il = 0 ;
by the solution of which, sin ( — ) and cos ( -7 ) may be found
in terms of sin A and cos A respectively.
Ex. 1. Let A= 180^ then sin ^1 = 0, and — = 60^;
therefore, 4 sin^ 60° — 3 sin 60^ = 0,
V^S 1
and sin 60^ = , and .*. cos 60^ = - .
Ex. 2. Suppose 1 = 90^ then cos ^ = 0, and -- = 30°;
hence, 4 cos^ 30^ — 3 cos 30^ = 0,
and cos 30" = , and .'. sin 30° = - .
2 ' 2
86. From the last examples, combined with some of the
preceding articles, we are enabled to find the sines and cosines
of 15^ 75', 105°, 165°, &c.
For, sin 15° = sin (45° — 30°) = sin 45° cos 30° - cos 45° sin 30°
I VS I \ 1
a/2 2 \/2 2 21/2
(/3- 1) = cos 75°;
similarly, cos 15° = — -r- ( ^^3+ 1) = sin 75°.
2 V 2
Again,
sin 105° = sin (60° +45°) = sin 60° cos 45°+ cos 60° sin 45°
= ^ -i- +i J- = -V(V^3 + l)=-cosl65°;
2 V^2 21/2 2/2
similarly, cos 105°= - TT/^ ( /3-l)= —sin l65°.
51
87. Articles (75) and (83) afford us the means of deter-
mining the sines and cosines of 18^ and 72^.
For, since sin 2^1 =2 sin A cos A,
and cos 3^=4 cos^ A — 3 cos A :
if ^ = 18^ we have 2^=36^ and 3^ = 54^
also, sin 2.1 = sin 36^ = cos (90° — 36^) = cos 54° = cos 3 yl :
hence 2 sin 18° cos 18° = 4cos'^ 18° — 3 cos 18°,
and 2 sin 18° = 4 cos" 18°— 3= 1 - 4 sin" 18°;
.'. 4 sin^ 18°+ 2 sin 18°= 1, which gives
V5- 1
4
and cos 18° = ^/l - sm^ 18° = — = sin 72°.
4
88. Hence the sines and cosines of 36^ and 54° are easily
found.
For, sin 36° = 2 sin 18° cos 18°
= P /^^-^\ x/lQ + 2^1 _ ( l/5~l) v^lO + 2 j/J
~ 4 "" 8
sin 18°= = cos 72°;
V 10 -2 1/5
= cos 54° :
4
and cos 36° = 2 cosM8°-l
2(10 + 2 1/5) 1/5 + 1 0
= 1 = = sm 54 .
16 4
89. From the last two articles is derived Eider's Formula
of Verification, which is
sin A = sin {36'' + il) + sin (72°- A)
- sin (36° -_4)- sin (72° + J):
62
For, sin (36^^ + A)- sin (36°- A) = 2 cos 36° sin A,
and sin (72^ -\- A)- sin (72° - ^) = 2 cos 72° sin A :
therefore by subtraction^ we have
sin (36° + ^) + sin (72° - Jl) - sin (36° -A)- sin (72° + A)
= 2 sin -A (cos 36°- cos 72°)
= 2 sin A \ \ = sin A,
A f ^^ + ^ _ Jl^zlII
90. Legendre's Formula of Verification is, in fact, the
same as Euler's, though different in form, that is,
cos A = sin (54° + A) -{- sin (54° - A)
-sin (18°-|-il)--sin (18°-^).
Here, sin (54° + ^) + sin (54° — ^) = 2 sin 54° cos A,
and sin(18° + ^)+sin (18°— 1) = 2 sin 18° cos A-,
therefore by subtraction, as before, we have
sin (54° + A) + sin (54°- A) - sin (18° + A)- sin (18°— .4)
= 2 cos yl {sin 54° -sin 18°}
= 2cos^|i:^i±i-J^^]=cos4.
14 4 J
91. By means of (78) the sine and cosine of 9^ and 81°
are easily obtained from the sine of 18°; from the sines and
cosines of 9^ and 15°, the sines and cosines of 6° and 24°,
and therefore of 84° and 66^ are very readily deduced ; from
the sines and cosines of 6° and 84° may be found those of
78° and 12°; and so on.
92. By a process in every respect similar to that used in
article (83), we readily obtain the following results :
sin 4 A ={4 sin J. — 8 sin^ A) cos A ;
sin 5A—0 sin A — 20 sin^ A + Id sin^ A;
sin 6 A =(0 sin 1—32 sin" JL + 32 sin^ A) cos A ;
&c =&c
53
cos 4^ = 8 cos* ^ — 8 cos^ A + I;
cos 5 ^ = ] 6 cos^ A -20 cos^ A + 5 cos A;
cos 6 A =32 cos^ ^-48 cosM + 18 cos" .4 - 1;
&c =&c
93 . To express the versed sines of the sum and difference of
two arcs in terms of the versed sines of the arcs themselves.
From (30) we have
vers {A±B)=zl- cos (A ± B)
= 1 — cos A cos -B + sin A sin B
which, by (30) and (62),
= 1 -(1- vers A) (1— vers B)
± ^J{2 vers A - vers^ A) (2 vers 5— vers^ B)
= vers A + vers B — vers A vers B
+ V (2 vers ^— vers° A) (2 vers £— vers" B).
94. From the expressions just proved we have immediately,
vers (A + B) + vers {A - B)
= 2 vers A + 2 vers J3 — 2 vers A vers B ;
or = 2(1 — cos j1 cos J5);
also_, vers (A + B) — vers (A — B)
= 2 /vy(2 vers A — vers^ A) (2 vers ^ — vers" B);
or = 2 sin A sin B ;
and vers {A 4- i>) vers (J — B) = (vers ^ — vers JB)^,
95. The vahie of vers (A + B) may however be exhibited
in a much neater form than that in which it is given in the
last article but one. Thus,
vers (A ± B) = vers A 4' vers B — vers A vers B
± x/(2 vers A — vers^ J) (2 vers B — vers" B)
54
= ^ {vers -4. (2 — vers B) -f vers B (2 — vers A)
± 2 ^(2 vers A - vers^ ^4) (2 vers B — vers^ B)}
= "I- {vers tI vers (tt — B) + vers 5 vers (tt — A)
+ 2 A^vers ^4 vers (tt — J) vers jB vers (tt — B) }
— 2 Is/ ^^^^ ^ v^^s (tt — B) ± M^ vers J5 vers (tt — A)}'^,
96. Cor. 1. In (93), suppose B = J, and we have im-
mediately
vers 2 A = 2 vers A — vers^ ^ + 2 vers -^ — vers'^ A
= 4 vers JL — 2 vers^ ^ = 2 vers A (2 — vers A)
= 2 vers J. vers (tt— A), by (29).
97. Cor. 2. In (95) for JB put 2 J, and we have vers 3 A
= "l^ {^vers ^ vers (tt — 9. A) + ^versQ,A vers(7r — ^)}^
= -^ {v vers A (2 — vers 2^) + /y^ vers 2 ^ (2 — vers ^)}'^
= ^ {/^2versy4(l — vers^)" + ^2 vers A (2 — vers ^4)^ }^
= f { (1 — vers ^) ^2 vers ^ + (2 ~ vers ^) ;y/2 vers j4 }^
= ~{(S — 2 vers^) ^2 vers 1}^ = vers A (3 — 2 vers ^)^
A similar process may be used to express the versed sines
of 4 A, 5 A, &c. in terms of vers A.
98. Since vers 2 A = 4 vers ^ — 2 vers" A, by substituting
A and — in the places of 2^ and A, we shall have
2 *^
vers A = 4 vers 2 vers — ,
2 2
, 3 ^ ^ vers A
and .'. vers — — 2 vers — = — ,
2 2 2 '
55
, c, A ^ A ^ 2 - vers A
whence vers 2 vers ■— + i =
2 2 2
and .*. vers — = 1 +
o —
v/(--^|:ii) .
Ex. 1. Let 1 =90*^; .'. vers ^ = 1, by (27), and — =45°,
whence vers 45^^ = 1 — v - = 1 7- = — 7— ( v^2— l).
^2 1/2 t/2^ ^
Ex. 2. Let A = 60^ .-. vers^= -, by (37), and - =30',
.-. vers 30° = 1 - V^= 1 - — = - {2 - ^3}.
4 2 2 ^
A A
Similarly the versed sines of — , — , &c. may be found in
terms of vers A, by the solutions of a cubic^ biquadratic, &c.
equation respectively.
99* By means of the last article we have the following
equations :
^ 4/1
vers — = 1— v 1— - vers A ;
2 2
A ./ \ A
vers ^ = ^ - V 1 - - vers - ;
A ./ \ A
vers^ = 1 - V 1- o^ers-,;
^c . — &r
and hence
vers
A ./ \ A
-5 = 1 — V 1 vers -
56
1 _ \/i 4-- \/l-iversA;
A
Q.
= 1 - \/7-
veis — , = 1 - V 1 — " ^ers -5
= - \/RvT^I7r:i^i;
2 2
&c =&c
and generally, vers — = 1 - V 1 — J vers ^^^j
= ' - \^\^WWz\/\ +&C.... \/l- iversl,
where the radical sign is repeated n times.
100. By substituting in the expressions for vers {A±B
(A + B\ , /A-B\
) and i — - — 1
in the places of A and B respectively^ the values of the
versed sines of J and B will be exhibited in terms of the
versed smes or I I and f 1 .
101. To express the chords of the sum and difference of two
arcs in terms of the chords of the arcs themselves-
From (36) we have
chd^ U ± -B) = 2 vers (1 ± 5) = 2 { 1 - cos(A ± ^)}
= 2 { 1 — cos A cos B + sin A sin B } :
now, from the same article it is easily proved that
, 2-chdM , . , V^4chd^^- did* A
cos A = — , and sm A = : ;
57
cos i5 = , and sin B = -^^
2 ' 2
therefore chd' (A ± B)
(2 - chd^ J) (2 - chd' B)
= 2 |l -
4
.y (4 did" A - chd' ^) (4 chd" B - chd"^)|
- 4 ^
= i. {4 - 4 + 2 chd^ A +2 chd' B - chd' A chd' B
± >v/(4 did' A - chd' A) (4 chd"- B - chd^ 5)}
= I- {2 chd' A + 2 did' 5 - did' A chd' jB
± >/(4 did' ^ - did' A) (4 did' ^ - chd' B)]
and chd (A ± B)
= -^ {2 did' 1 + 2 did' 5 - did' A did' 5
K 2
± V(4 chd' A - chd' 1)(4 did' ^- chd* ^)!^.
102. From the result above found^ we shall have
chd" (A + JB) + chd' {A - i^)
= 2 chd" A -b 2 chd- ^ - chd' A chd- 5 ;
or = 4 vers A + 4 vers B — 2 vers ^4 2 vers JB;
or = 4 ~ 4 cos 4+4 — 4 cos 7^
— 4 (1 — cos A — cos B + cos j4. cos B)
= 4—4 cos A cos J5 = 4 (1 — cos A cos 5);
and chd= {A+B)- chd' (1 - B)
= V(4 chd- A - chd'^ 1) (4 chd' B - chd^ jB) ;
or = chd A chd B ^/(4 — chd^ 1) (4 - chd" B) ;
or = 4 sin 4 sin B.
H
58
103. A more convenient expression for the chord of
{A + ^>) may easily be deduced from that found above.
Thus, chd' U ± B)
= I {did' A (4 - chd' B) + chd' i] (4 - chd- A)
± V^'tid' ^ (4 - chd' A) chd^ B (4 — chd' B)]
= ^ { chd- A chd- {TT- B) -h chd' B chd' {TT- A)
± 2 Vchd' A chd' (tt - il) chd' B did' (tt- J5)}
= |- {chd^ chd (tt-JB) + chd JB chd(7r-i4)}^
and .-.chd {A±B)
= |- {chd A chd (tt - 5) ± chd B chd (tt-A)},
104. Hence, by the common operations of arithmetic,
we have immediately,
chd (A+ B) + chd (A- B) = chd A chd (tt - jB) ;
chd (^ + jB) - chd {A - B) =z chd B chd (tt - ^);
and chd {A + B) chd (J - B)
= ^ {chd' A chd' (tt - 5) - chd' 5 chd' (tt - A)\ ;
1 { chd" ^ (4 - chd' B) - chd^ i? (4 - chd' 1) }
or
= chd' A -chd- 5= (chd A + chd jB) (chd A - chd B).
105. In (101) if we suppose B=^ A, we shall obtain
chd 2 A = >/4 chd' A — chd'^ ^ = chd A ^4 -chd' A\
or = chd A chd (tt— A).
Hence also, by the solution of a quadratic equation,
chd ~ = Y/s-^I-chdM or = ^S-cluKTr-il).
59
TT
Ex. Let A = 90^; then we have chd (tt — A) = chd -
= V2, by (33) ; and therefore chd 45^ = >/2- \/2.
By making jB successively equal to 2 A, 3 A, &c. the chords
of 3^1, 4<Aj &c. will be found in terms of chd A, and by the
. . . . A A
solutions of a cubic_, biquadratic, Sec. equation^ chd — ^ chd — ,
Sec. may be expressed in terms of the same line.
106. As in the last article, we obtain the followino: results:
chd ^ =
Va-
v.-
chd^-
2
= n/«-
- chd (tt
-!)■
1 1^
.x/...
V.-
■ M'i
= \/'i-chd(x-^);
&c = ik.c
chd| = \/'2- v/4-chd^^
and therefore by substitution, we have
4 . .. ..
chd ^ = a/ 2 - v^2 4- chd (tt - .1) ;
- = ^/ 2 - x/2 + x/2 4- chd (tt ~ A) ;
chd
2^
&c =&<
60
chd - = Y 2- V 2 -fVs +&c....Ay2 + chd(7r- J),
in which the radical sign occurs n times.
Ex. Let A=^TT\ therefore chd (tt— A) = chd 0 = 0, and
chd ~ = \/2 — n/^ + \I^ + Sec... /v/2.
107. By pursuing the method pointed out in (100),
the chords of A and i5 may be expressed in terms of the
chords of ( ) and ( ) ; but as the results possess
no elegance, and are at the same time, of little use, the
operations in this case as well as those in the Article just
alluded to, are omitted.
108. To express the tangents and co-tangents of the sum
and difference of two arcs in terms of the tangents and co-
tangents of the arcs themselves.
From (42) we have
sin {A ± B) sin A cos B + cos J sin B
tan ( J. ± IJ) =
cos {A + B) cos A cos B + sin A sin B
sin A sin B
cos A ~ cos B . „. . tan A + tan B
as m (69), =
_ sin A sin B 1 + tan J tan B
^ + -A B
cos^ cosi>
Again, from (47) we get
( A X ry\ __ ^os {A + B) __ cos A COS jB + sin A sin B
"" sin (^ + B) sin A cos B + cos A sin B
COS A cos B _
sin A sin B 1 ■ 1 ^^^ -^ ^'ot B + I
= — '■ — ~- 5 bv a similar process, = —
cos B cos A col i5 4- cot A
_ f- _ —
sin B sin A.
61
109- Either of the expressions given in the last article,
might have been deduced from the other.
Thus, cot (A±B) = -—-±-— by (47),
tan (A + B)
1 1 + tan A tan B
tan A 4- tan B \ tan A + tan B
(tan A ± tan p \
1 + tan A tan b)
_ 1 1
1 +
cot A cot B cot A cot B + 1
1 1 cot J5 + cot A
+
cot A ~~ cot 5
m . 0 A. t'*^" 45*' + tan A
Ex. 1. Tan (4o'' + .1) = — := r=o 7
- 1 + tan 45° tan .4
1 + tan A r ,.^K
J as appears irotn (42).
1 + tan A
tan 90° + tan A
Ex. 2. Tan (90° 4- 1) = _ -» ;
- 1 + tan 90 tan A
tan A
GO + tan A ~ CO _ 1
1+00 tan A 1 _ tan A
h tan it
00
= + cot Af as is manifest from (39) and (47).
110. Cor. Let T and t be the tangents of any two arcs^
T' and t' their co- tangents ; then, using the kind of notation
adopted in {QQ)y we shall have
Vi + rJ'
Ian' T ± tan-' t = tan"
1 (T't'+\
cof' I" + col"'(' = cot
/£_L±J\
62
1 1
3 1
Ex. 2. Cot~* - +cot~' ~ = cot
4 7
Ex. 1. Tan-^- +tan*^- = tan*' ^
2 3 i 1
=:tan-'(l) =45^from (42).
m
= cot-' ( _ 1) = - 45%r = 135^ from (47) and (46).
Ex.3. Tan~' - +tan-'- 4-tan-'-« ^tan"'-
3 5 7 8
=:{tan-i+tan-i} + {tan"' i + tan^ i}
= tan~' - +tan-'— = tan*' (l) = 45^
7 11
111. To express the tangents and co-tangents of two arcs
in terms of the tangents and co-tangents of their semi-sum and
semi-difference.
si„„..(i±^)+(i^-).
from article (108) we have
tan A —
/A -h B\ /A + 7^\ '
-t-'(-T-)'-("^)
63
/A + B\ (A- B\
tan ( J — tan ( I
tan B = — _ ;
/A + B\ /A + B\
/A-B\ , (A + B\ '
/A + B\ /A-B\
eot (-—) cot (-^) + i
/I - B\ /A + B\ '
cot A =
cot J5 =
112. To express the tangent and co-tangent of ttoice an arc
in terms of the tangent and co-tangent of the arc itself.
tan A + tan A
— tan .
2 tan A 2
Here, tan 2l =tan {A + A)= r- , from (108),
1 — tan A tan A
1 - Ian" A I
— tan A
tan A
2 cot J 2
cot A — 1 cot A — tan A
^ ^ .^ cot A cot A — I
Also, cot 2^ = cot {A-{-A)= , from (108),
cot ^ + cot J ^ '
cot^ A — I cot A 1
2 cot J 2 2 cot A
1 — tan^ ^ cot A — tan J
2 tan -4 2
64
113. To express the tangent and co-tangent of half an arc
in terms of the tangent and co-tangent of the arc itself
By substituting in the expressions found in the last
article, A and — in the places of 2^ and A respectively, we
shall have
2 tan -
2
A — , and cot A =
cy A
1 - tan^ -
2
cot^ 1
2
A
2 cot —
2
of which gives
oA , 2 A
tan^ 1 7 tan - =
2 tan A 2
1.
A ~ 1 ± V 1 + tan^ A
and ,*. tan — =
2 tan A
— 1 4- sec A
or
tan J
; or = 4- cosec A — cot A ;
and from the latter we get, cot^ 2 cot A cot — = 1,
2 2
whence, cot — = cot J. + x/ 1 + cot' A :
'2 ~ ^ '
or = cot A i cosec A.
Ex.1.' Let ^=90°; .-.tan ^ = oo , and cot ^ = 0;
and from the equations above given we get
tan 45° = 1 = cot 45^
Ex. 2. Let J = 45° ; /. tan ^ = 1 = cot ^ ;
whence, tan 22°30'= Vl±t^!l_A = ^g- 1 = cot 67° 30':
tan 45
and cot 22° 30' = cot 45° + x/H^'c^?45° = >v/2 -j-l = tan 67° 30'.
65
114. To express the tangent and co-tangent of thrice
an arc in terms of the tangent and cotangent of the arc
itself.
As before we have,
tan 3i4 = tan (2^ + A)
tan %A -f tan A
1 — tan 9.A tan A
2 tan A
, by (108),
3 tan A - tan^ A
/ 2 tan A \ .
(1:1^^7:1) + '='"^ _ ■ ■-■
tan'^ iL - 3 tan il 3 cot~ J— 1 1-3 cot^ A
3 tan^ ^ — 1 cot ^1 — 3 cot A 3 cot il — cot^ A
also, cot 3 ^ = cot (2 1 + A)
cot 2 A cot A
, by (108),
cot 2 A + cot ^
/cot^ A — 1\ .
I 7- I cot A - 1 3 ,
V 2 cot A / , , X cotM — 3cot4
= 2— i > bv (112), = — ^—
/cot^ A - 1\ . , - V /^ 3 cot^ A- 1
I — ) + cot A
\ 2 cot A /
3 cot A — cot^ A 1-3 tan^ A 3 tan^ A - 1
1 — 3 cot^ A 3 tan A — tan' A tan A - 3 tan A '
115. By substitutions similar to those used in some of the
preceding articles, we readily obtain,
I
66
A , A
3 tan tair — 3 cot cot" -
3 3 , , 3 3
tan A = — , and cot A = — -; — :
n A ^ A
1 - 3 tan^ — 1-3 cot^ --
3 3
and thence the equations,
tan^ 3 tan A tan^ 3 tan f- tan A = 0,
3 3 3
A A A
and cot^ 3 cot J cot^ 3 cot — + cot ^ = 0 ;
A A
by the solution of which, tan — and cot — will be expressed in
terms of tan A and cot A respectively.
A
Ex. 1. If ^1=90^ we have tan ^ = oo , and —=30°;
•J
therefore, 1 — 3 tan^ 30° = 0, and tan 30° = -^=^ cot 60°.
V3
Ex. 2. Let A = 180°; then tan il = 0, and - =60°;
therefore, tan^ 60° - 3 tan 60° = 0, and tan 60°= '/3 = cot 30°.
11 6. The method used in the last article to determine
the tangent and cotangent of 3 Aj may be applied to express
the tangents and cotangents of 4 Ay oA, 6 A, &c. in terms of
tan A and cot A ; and there will result
. 4 tan ^ — 4 tan^ A
tan 4 A=z
tan 5 ^ =
1—6 tan^' A + tan"* A '
5 tan ^ - 10 tan^ A + tan^ A
tan 6 ^ =
1 - 10 tan^ A + 5 tan^ A '
6 tan ^ — 20 tan^ A + 6 tan^ A
1 - 15 tan^ A-i-\5 tan^ A - tan^ A '
&c = &c
cot 4 A =
67
cQt^ A -6 cot' A + 1
4 cot^ ^ — 4 cot ^ '
. c A ^^^ J -10 cot^ A + 5 cot A
cot 5 J. = 5 i ;
5 cot' ^—10 cot^ ^+1
cot^ A -15 cot^ ^ 4- 15 cot^ ^ - 1
"" 6 cot^ ^ - 20 cot^ ^ + 6 cot ^ '
&c = &c
117- To express the secants and coseca?its of the sum and
difference of two arcs in terms of the secants and cosecants of
the ai'cs themselves.
From (52) we have
1 1
sec {A±B)=^
cos {A i B) cos A cos B + sin A sin B
1 I
cos A cos B . ^ V sec A sec B
,, as m (69), =7"=-
cos A cos B _ sin A sin B 1 + tan A tan jB
cos A cos i^ cos -^ cos B
sec il sec JB
. by (52);
1 + v^(sec' A- l)(sec' 5-1)
also, from (57)
cosec (^± -B) = -
sin (A + -B) sin il cos B ± cos il sin B
1 1
sin A sin 5 . ., . cosec A cosec £
, similarly _, =
sin A cos 5 , cos A sin 5 ' cot 5 + cot A
sin A sin 5 ^ sin A sin J5
cosec A cosec B
>/cosec^ 5— 1 ± >/ cosec" A — 1
, by (57).
68
118. The functions in the last article may be expressed
in terms somewhat different,
Thus^
1 1
sec {A ± B) =
cos {A ± B) cos A cos J5 -f sin J. sin B
1
_ _ _ J
sec A sec B cosec A cosec B
sec A sec B cosec A cosec B
cosec A cosec J5 + sec J. sec B
and
cosec {A±B) =
sin (A ± B) sin ^ cos B ± cos ^ sin B
1
_ -^ _ _ _
cosec A sec B sec A cosec B
__ cosec .4 cosec jB sec A sec B
cosec ^ sec J. + cosec A sec jB
119. By means of the substitutions used in articles (72)
and (107)^ the secants and cosecants of two arcs are expressed
in terms of the secants and cosecants of their semi-sum and
semi-difference.
120. To express the secant and cosecant of twice an arc
in terms of the secant and cosecant of the arc itself.
HerCj from (117), we have
sec A sec A sec^ A
sec 2 il = sec ( J + -A) = :; 24,, = ;; 2-7 5
1 — - sec ^ + 1 2 — sec tI
69
and
cosec A cosec A
cosec 2 4 = cosec (A + A) =
2 .y/cosec^ 4 — 1
cosec* J.
2 />/cosec^ ^ — 1
Or thus, by (118), we get
sec A sec A cosec A cosec A
sec 2 il =
cosec A cosec A — sec il sec A
sec^ A cosec^ A
cosec^ ^ — sec^ A '
and
cosec A cosec A sec A sec A
cosec 2 A =
cosec A sec A + cosec A. sec A
cosec^ A sec^ A 1
. = - cosec ^ sec A.
2 cosec A sec A 2
121. 2o express the secant and cosecant of half an arc in
terms of the secant and cosecant of the arc itself.
By the requisite substitutions in the last article we obtain
i.±K/-
2 sec A
sec
2 - ^ 1 + sec A'
and
A
cosec
— = + \/ ^ cosec" A ± 2 cosec A ^ycosec^ A — 1,
Ex. Let A = 90^ ; therefore since sec A = oo , cosec A = 1,
and — = 45^, we shall have
sec
45° = 1/2 = cosec 45^
70
122. lb express the secant and cosecant of thrice ati arc in
terms of the secant and cosecant of the arc itself.
From (117) we have
sec SA = sec (2A + A)
sec 9>A sec A
1 - >/(sec^ 21 — 1) (sec" .1-1)
sec^ A
4 — 3 sec^ A
also,
cosec 3il = cosec {2A -^^ A)
cosec ^A cosec A
y by substitution and reduction ;
^cosec'^ A — 1 + ^ cosec 2 A — 1
cosec"^ A
3 cosec^ 1—4
5 by the same process.
123. Let A and — be put for 3 A and A respectively in
o
the last article, and we get
3^
sec-
4
-3sec'-
3^
cosec —
;osec 4
3
3c
which give the following equations,
A „ A
sec^ h 3 sec 1 sec' 4 sec A = 0,-
3 3
and cosec'' — — 3 cosec 1 cosec^ —4-4 cosec 1=0;
3 3
71
by means of which the secant and cosecant of — are expressed
in terms of the secant and cosecant of A.
Ex. 1. Let A = 90°; then sec A =^ oo , cosec il = 1,
and - =30%
3 '
therefore 3 sec"^ 30^ - 4 = 0^
2
and sec 30 = —7- = cosec 60 .
1/3
Ex. 2. Let A = 180^; therefore sec A = — \, cosec
il = 00 , and — = uO :
3
hence sec^ 60^ - 3 sec^ 60^ + 4 = 0,
from which we obtain sec 60^ = 2 = cosec 30**.
It may be observed that the formulae in these two articles
might have been deduced from those in (83) and (84), by
means of (52) and (o7) ; and by continuing the process we
should in a similar manner obtain the values of
sec A A, cosec 4^. &c. sec — , cosec — , &c.
4 ' 4
124. To express the sine and cosine of the sum of three arcs
in terms of the si?ies and cosines of the arcs themselves.
By considering the sum of two of the proposed arcs as one,
we have
sin (^ + 5 4- C) =: sin {{A + B) ^ C]
= sin {A + B) cos C + cos (A + B) sin C
= (sin A cos B + cos A sin B) cos C
+ (cos A cos 5 — sin ^ sin jB) sin C, by {63),
= sin A cos B cos C + sin B cos A cos C
+ sin C cos A cos B ~ sin il sin jB sin C ;
72
and cos (A+ B + C) = cos {(A ■\- B) + C]
= cos {A + B) cos C — sin (^ + JB) sin C
= (cos -4 cos B — sin ^ sin B) cos C
— (sin A cos J5 + cos A sin B) sin C, by (6S),
= cos A cos B cos C — cos A sin ^ sin C
— cos 5 sin il sin C — cos C sin J. sin B.
Ex. 1, If we have (A + B -h C) = 2w - , or riTr, then will
2
sin A sin ^ sin C = sin A cos B cos C
4- sin B cos A cos C + sin C cos A cos ^.
Ex. 2. U A + B -\- C = (9.JI - 1) - , we shall have
2
cos A cos B cos C = cos A sin ^ sin C
+ cos B sin A sin C + cos C sin A sin B.
125. Cor. Let A = 5=C, then the formulae in the last
article become
sin 3 A = 3 sin A cos^ A — sin'^ A = 3 sin A — 4 sin^ A,
and cos 3 A =cos^ A — 3 cos A sin" A =4 cos"* A — 3 cos A ;
which have been already proved in (83).
126. By a process similar to that used in (124)^ we may
prove that
sin (A + B) sin (5 + C) = sin A sin C + sin B sm{A-\-B+ C).
For,
sin (A+J3H-C) = sin A cos (5 + C) + cos A sin (5 + C)
= sin A cos B cos C - sin A sin B sin C + cos yl sin (5+ C) ;
73
.*. sin B sin {A + B + C) + sin A sin C
= sin A cos B cos C sin B + sin A sin C (1 — sin^ B)
+ cos j1 sin B sin (5 + C)
= sin A cos -B (sin B cos C + cos B sin C)
+ cos A sin 5 sin (B 4- C)
= sin^ cosB sin (B + C) + cos A sin 5 sin(B + C)
= (sin ^ cos JB + cos A sin B) sin (-S + C)
= sin (A+B) sin (J5+C).
Similarly, sin {A — jB) sin (C — B) = sin il sin C
— sin jB sin {J~B+ C).
Ex.1. Let^ + 5+C = 7r; then sin (^ + 5 + C) = 0,
and it follows that
sin {A + B) sin (5 -}- C) = sin A sin C.
Ex. 2. Let A - JB + C = 0 ; then sin (1 — 5 + C) = 0,
and we have
sin (A — B) sin {C — B) — sin J. sin C.
127. From (93) and (101) the versed sine and chord
of (A + jB + C) are obtained after the same manner.
128. To express the tangent and cotangent of the sum
of three arcs in terms of the tangents and cotangents of the arcs
themselves.
Proceeding as in (124) we have
tan (^ + ^ + C) = tan {(A +\B) + C]
tan {A -\- B) + tan C
1 - tan {A 4- B) tan C
tan A + tan B
by (108),
("
^ , + tan C
tan A tan B/
, by (108),
/ tan A + tan i/ \ ^
( A ^ ) tan C
\1— tan A tan B/
74
tan A 4- tan B + tan C — tan A tan B tan C
"" 1 - (tan A tan 5 -f- tan A tan C + tan jB tan C)
and cot (1 + B -f C) = cot {(A + 5) + C)}
cot (A + 5) cot C - 1
cot (A 4- i5) + cot C
cot J. cot -B — 1
, by (108),
/cot A cot i5 — 1\
( ^ \ cot C - 1
V cot A + cot JB / ^ ^^^ .
= , by (108),
/cot A cot ^ — 1\ , r.
/ ^\ _|- cot C
\ cot J + cot ii /
_ cot A cot 5 cot C — (cot j1 + cot JB + cot C)
~" cot A cot 5 + cot A cot C + cot -B cot C - r
Ex. 1. Let (.4 + 5 + C) = (2w - 1) ~ ; therefore
£
tan (A+5 + C)= 00, and cot (J + 5+C) = 0; and hence
tan A tan B + tan A tan C + tan 5 tan C = 1 ;
also, cot A cot jB cot C = cot A + cot jB + cot C.
Ex. 2. Let (il + B + C) = 2w - = WTT ; then
tan (A + B + C) = 0, and cot (^ + jB + C) = oo ;
therefore tan A + tan 5 4- tan C = tan A tan B tan C ;
and cot ^ cot B + cot v^ cot C + cot B cot C= 1.
129. Cor. In the last article^ suppose ^ = JB = C; then
3 tan A — tan^ A , ;^ , 3 cot ^ - cot^ A
tan 3 A = — 2— — 5 and cot 3 A = r— — ;
1 - 3 tan^ A 1-3 cot^ A
which are the formulas proved in (1 14).
130. The same method leads to expressions for the secant
and cosecant of (A + ^ -|- C) in terms of the secants and
cosecants of A, B and C
75
131. By separating the arcs into two parts as has been
done in some of the preceding articles, we are enabled to
determine the sine, cosine, &c. of the sums of 4, o, &c.
n arcs : but as the methods are so very simple, notwithstanding
the prolixity of some of the results, we shall not pursue the
subject further in this place except to notice a curious property
of the tangent and cotangent of the sum of any number of arcs,
which shall be the subject of most of the remaining articles of this
chapter.
132. If ^i denote the sum of the tangents of n arcs,
Af B, C, D, &c. Kj L; So the sum of their products taken
two and two together ; S^ the sum of their products taken three
and three together ; and so on : then will
tan (^4--B + C + &c. + K + L) =
I- S^ + S^- &c.
T- 1 /,^^v .A ^v tan A -\- tan B Si
For, by (108), tan U + B)- - ^
1 — tan ^ tan jB 1 — .Sg '
again, by (128), tan {A -\- B + C)
_ tan A + tan B + tan C — tan A tan jB tan C _ •S'l — S^
l-(tan ^ tan jB + tan A tan C + tan B tan C) "" 1 — Sc^ '
and so on : and generally, if
tan(^+J5 + C+&c. + i^)
6'i-^3 + S5~&C.
1 - ^2 + ^4 - &C. '
we shall have, from (108),
tan (^ + i? + C + &c. + K + L)
- tan (J + B -{- C + ^c.-\- K) + tan L
~" 1 - tan (A + jB + C + &c. + K) tan L
Si — S^ -\- S5 "— &c.'
C' - :-' + ?^ - ^^-^ + tan L
VI — ^\ + S.^ - &c./
76
{S, - ^3 4- ^5 - &c.) + (1 - ^2 4- ^4 - &c.) tan L
(1-^2
+ s.
- &c.) -
•(S,
t - Ss + '% -
&c.) tan
L
(5,+ tar
iX)-
■iSs + S,
tan
L)+(^5+'S4
tan X) -
■&c.
1 - ("Sa + ^1 tan L) + (^^4 + S^ tan X) - &c.
which expression is manifestly formed after the same law as
the preceding one :
Therefore^ if the form be true for the tangent of the
sum of tt— I arcs, it will also be true for the tangent of the
sum of n arcs. Now it has been shewn that the law obtains
for the tangents of the sums of two and three arcs : hence it
obtains also for the tangents of the sums of 4, 3_, &c. arcs ;
that isj generally for the tangent of the sum of n arcs.
Ex. ]. If (^ + 5 + C + &c. + X + Jv) = 27i - , or
2
fiTT, we have
and therefore 5^ + S,^ + &c.... = *S3 + ^Sy + &c.
Ex. 2. If (A + 5 + C + &c. + X + X) = (2w - ]) - ,
we have
^S,- Ss + S,- &c.
and thence 1 + ^4 -}- &c. = ^Sg + Sq -f &c.
133. Cor. Supposing A = B=C = &ic. to n terms, we
shall manifestly have
n tan J — ?i i j ( j tan A + &c.
tan nA = ~
1 — n C- j tan^ A + &c.
77
134. In the last article but one, let il = - — A',
2
B = - - JB', C= - - C; &c. = &c., then will
A -\' B ^ C 4- &c. to w terms
= — - I A' + K + C + &c. to n terms} ;
and tan A = cot A! , tan ^ = cot B'^ tan C = cot C\ &c. = &c.
also, on this hypothesis, we have
S^ = the sum of the cotangents of A , B\ C\ &c.
aS'2 = the sum of their products taken two and two together ;
&c. = &c.
Hence if 7i be even,,
cot (^' -f ^ + Cr + &c.) = cot 1^ - (J + 5+ C + &c.)|
tan (1 + J3 + C + &c.) S,-S^-\-S-,-^c. '
and if n be odd,
cot(A' + 5^+C+&c.) = cot 1^ -U-f5 + C+&c.)|
•^i - 8.^ + 5^5 - &c.
= tan (J -f- B + C 4- &c.) =
1 - 6^2 + 'S'4 - &c.
135. Cor. If A' = i^' = C = &c. to n terms, we shall
have, when n is even,
n - 1
cot nA
1 - w (— ;7— ) cot" A' + &:c.
n coiA' - ;. (^) (^) cot^^ 1' + &c.
78
and when n is odd,
n cot A' - n {~-) C^—-\ cot^ A' + &c.
cot 71 A' ■= — — .
1 «- n (— — -) cot^ A' + &c.
136. In the various articles of this chapter, the trigo-
nometrical functions of {A + B) and (^1 — B) have each been
deduced by a separate process; but this is unnecessary, for in fact
the corresponding functions of both are contained in the same
expressions.
Thus, if we put — jB in the place of B, and — sin B, — tan B,
— cot B, — cosec B in the places of sin B, tan J5, cot B, and
cosec B respectively, the rest remaining unchanged agreeably to
what has been proved in Chap. I ; any trigonometrical function
of either {A-\-B) or {A — B) will be changed into the cor-
responding one of the other. Thus,
since sin (-4 -f- i^) = sin A cos B + cos A sin B ;
by changing B into — B, and sin B into — sin B^ we have
sin (^ — J5) = sin A cos B — cos -4 sin jB :
_,^ tan A — tan B
agam, because tan {A — B) =
1 4- tan J tan J5 '
.*. by putting - B for JB, and — tan B for tan JB, we get
, . „, tan j1 + tan i?
tan (A + jB) = ;: : and so on.
1 — tan A tan B
79
CHAP. Ill
On the computation of the sines, cosines , 8fc. of one, two, three,
Sfc, minutes, and succeeding arcs, and on the construction of
the Trigonometrical Canon. On the uses of Formulcz of
Verification- On the Logarithmic sines, cosines, Sfc. of arcs.
On the ratio of the circumference of a circle to its diameter,
Sic.
137. To express the sine and cosine of one minute in
terms of the radius 1.
In the last chapter at (81), it has been proved that
. A
sin
2
i=\/i-i\/;r~rA,
^ • ^ \/^ 1 . / . 2 ^
and sm ~ = V - - - V 1 - sni -^^
A = 30^, we have sin tI = - , from (37), and thence
sin 15^= - V 2 - VS = .2588190 8cc.
sin 7' 30'= i \/- — ^= .1305262 &c.
2 2
&c....= &c = &c
80
and it is manifest that by this process we shall obtain succes-
sively the sines of 3^ 45/ 1^ 59! 30'\ &c.
Now, at the end of the tenth division from 3(f, the arc be-
comes l' 45'' 28'" 1"" 30'"" and its sine .0005113269 &c.; also
at the end of the eleventh, the arc becomes 52" 44'" 3"" 45""'
and its sine .0002556634 &c. : from which it appears, that
when the operation above-mentioned has been repeated so many
timesj the sine of the arc is halved at the same time that the arc
itself is bisected ; that is, the sines become then proportional to
the arcs : hence
sm 52 44 3 45 : sm 1 :: 52 4 3 45 : 1
:: .0002556634 &c. : .0002908882 &c.
and therefore sin l' = .0002908882 &c.
also, cos l' = »/ 1 - sin^ l' = .999999957 &c.
138. Cor. From what has been proved in the preceding-
article, it is clear that the sine of any number n of seconds may
be obtained simply by a proportion. Thus,
sm n : sm 1 :: n : 1 :: n : DO,
and therefore sin it" = -rr- sin l'; and the cosine may be deter-
60
mined by means of the equation, cos A = \/ I — sin^ •^, as
before.
139. To express the sine and cosine of 2, 3, 4, 5, <5)T.
minutes in terms of the radius 1.
From (73), we have the equation
sin (;i + 1 ) A=Q. cos A sin n A — sin (w — 1 ) ^ ;
and if in this we suppose A to be l', and n to be taken equal
to the numbers 1, 2, 3, 4 &c. successively, we get
sin 2=2 cos l' sin l'
= .0005817764 &c. = cos 89^ 58';
81 ^
sin 3' = 2 cos l' sin 2' -sin l'
= .0008726645 &c. = cos 89^ ol' \
sin 4' = 2 cos l' sin 3' - sin 2'
= .0011635526 &c.=cos 89** 56^
sin 5' = 2 cos 1' sin 4' — sin 3'
= . 0014544406 &c. = cos 89^ 5o ;
&c.... = &c =&c
Again, from the other forntiula proved in the same article,
cos (w + 1) il = 2 cos ^ cos ?^ J. — cos {n — \) A,
we shall have by the same substitutions,
cos 2' = 2 cos 1' cos 1' — 1
= . 9999998308 &c. = sin 89° 58' ;
cos 3' = 2 cos 1' cos 2' —cos 1'
= . 9999996192 &c. = sin 89° 57' ;
cos 4' = 2 cos 1' cos 3' — cos 2
= . 9999993231 &c. = sin 89° 5Q' ;
cos 5' = 2 cos 1' cos 4' — cos 3'
= . 9999989423 &c. = sin 89° 6o ;
&c....= &c = &lc
We may observe from the latter of these sets of equations,
that when an arc becomes very nearly equal to 90^ and 0°, the
changes which the sine and cosine respectively undergo, are of
no value as far as five or six places of decimals.
140. The process adopted in the last article being con-
tinued would enable us to determine the sines and cosines of all
L
82
arcs whatsoever ; but by reason of the long and tedious numeri-
cal operations that are required, expedients of various kinds have
been had recourse to, to facilitate the computation : thus, by
means of the formula,
sin {A + B) = 2 sin A cos B — sin {A — B)
(R\
l-2sin^— )-sin {J — B)
— 2 sin A — sin (A ~ B) — 4 sin A sin^ — ;
if we suppose B~ \', and A to take the values l', 2', 3', &c. in
succession, we shall have,
sin 2' =2 sin l' — sin O' — 4 sin l' sin" SO"
sin 3' = 2 sin 2' — sin l' - 4 sin 2' sin^ 30"
sin 4' = 2 sin S' - sin 2^—4 sin S' sin" 30"
&c.= &c
Again, if ^=l', and A assume the values 1°, I*' l', 1° %\
&c. we get
sin 1^ l' = 2 sin 1^— sin 59' — 4 sin 1^ sin^ SO" \
sin 1^ 2' = 2 sin I*' l' - sin 1^-4 sin 1^ l' sin^ 30" ;
sin 1° 3' = 2 sin 1^2' — sin 1° l' — 4 sin 1° 2' sin^30'';
&c. =&c
and so on ;
and these operations are somewhat less laborious than those
which would be necessary by the former method.
141. The formulae,
sin {A + B) sin {A — B) — (sin A + sin B) (sin A — sin B),
and
cos {A -f B) cos {A — JB) = (cos A + sin B) (cos A — sin B),
83
proved in (68)^ will also enable us to deduce the sines and
cosines of arcs from the sines and cosines of others pre-
viously determined, and may likewise be the means of verifying
the results found by the preceding methods : thus by making
J5= 1^5 and A = QP, 3^, &c. in order, we obtain
_:„ oO_ (sin 2^ + sin 1°) (sin 2° -sin 1^)^
sin 1«
sin 4*^ =
(sin 3° + sin 1°) (sin 3° - sin l*")
sin 2^
&c.=
&c
cos 3° =
(cos 2^ -f sin 1°) (cos 2' -sin f)
cos 1
cos 4 =
(cos 3^ + sin 1^) (cos 3^ -sin 1°)
cos 2^
&c.=
&c
and the values thus found may be checked by assigning different
values to A and B, so that their sum may still remain the
same.
By one or other of these methods we may proceed to de-
termine the values of the sines and cosines of all arcs as far as
30 J after which the tediousness of the numerical operations may
in a great degree be avoided, by means of certain formulae which
have already been investigated in the second chapter.
142. To express the sines and cosines of arcs greater than
30^ and less than 45^ in terms of the radius 1.
■»
It has been shewn in (67), that
sin (^ + jB) + sin (A - i?) = 2 sin ^ cos 5 ;
therefore sin {A + B) = 2 sin A cos B — sin {A — B),
84
and if A be made equal to 30^ and B be assumed equal to l',
2', 3', ficc... successively, we get, since sin 30^= - , from (37),
sin 30^ 1' = cos 1' - sin 29^ 59';
sin 30^ 2' = cos 2'— sin 29^ 58' ;
sin 30"^ 3' = cos 3' - sin 29"" 5?' ;
&c.... = &c
and thus the sines of all arcs as far as 45^ may be derived from
the sines and cosines of those previously found :
cos (A - 5) - cos (1 + 5) = 2 sin A sin B, by (67),
we have
cos (^ 4- jB) = cos ( Jl — B) — 2 sin A sin S :
and if A be supposed = 30° as above, and jB equal to 1 , 2', 3 ,
&c. in succession,
cos 30° 1' = cos 29° 59' - sin 1' ;
cos 30° 2' = cos 29° 58' - sin 2' ;
cos 30° 3' = cos 29^ 5/ - sin 3' ;
&c..-.= &c
and hence the cosines of all arcs up to 45° may be determined
by means of the sines and cosines of those which are less than
30°.
143. To express the sines and cosines of arcs greater than
45° and less than 90° in terms of the radius 1 .
Since by (12) and (21), sin (45°+l)=cos (45°— i.), we have
sin 45° l' = cos 44° 59'
sin 45° 2' = cos 44° 58'
sin 45° 3' = cos 44° 57'
<&c.... = &c
85
and since cos (45° + jI) = sin {4i5^ — A), by the same articles, we
get
cos 46^ l' = sin44^59';
cos 45^ 2' = sin 44^ 58';
cos 45^ 3' = sin 44^ 5?' ;
&c....= &c,
and thus the sines and cosines of all arcs as far as 90° may be
found.
From this it is manifest that if the sines and cosines of all
arcs up to 45^ were formed into a table, such a table would
serve for the sines and cosines of all arcs as far as 90°.
144. jf'o express the sines and cosines of arcs greater than
90^ in terms of the radius 1.
From (63), we have
sin (90° + A) = sin 90° cos A + cos 90° sin A = cos A ;
cos (90° + A) = cos 90° cos A — sin 90° sin ^ = — sin A :
again,
sin (180°+^) = sin 180° cos ^-j-cos 180° sin A = -sin A;
cos (180° + ^) = cos 180° cos ^ — sin 180° sin A = - cos A :
and,
sin (270° + J) = sin 270° cos A + cos 270° sin A~ - cos A;
cos (270° + ^) = cos 270° cos A — sin 270° sin A = sin J :
&c....=&c
therefore the values of the sines and cosines of all arcs greater
than 90°, will be the same as the sines and cosines of corres-
ponding arcs less than 90° : and if a table be formed to contain
the sines and cosines of all arcs less than a quadrant_, such table
will contain the sines and cosines of all arcs greater than a
86
quadrant, proper regard being paid to the algebraical signs of the
quantities according to the principles laid down in (16) and applied
in the subsequent articles of the first chapter.
145. The sines and cosines of all arcs being determined
by the methods just explained, the tangents, cotangents, secants,
and cosecants are immediately deduced from the following equa-
tions :
sin J. . cos A I
tan A = , cot A = -: , sec ^1 =
cos A sin A cos A
1
and cosec A =
sin A
and the versed sines and chords, if necessary, from the equations,
^
vers j1 = 1 — cos A, and chd A = ^ 2 — Q cos A, or = 2 sin — .
2
146. The tangents of arcs greater than 45^ may however
be easily found from the tangents of those that are less, by
simple addition only.
. . sin A cos A
For. smce tan A — cot A = :
cos A sm A
sin^ A — cos^ A cos 9>A
= : — - = -2-: = — 2 cot 2^ :
sin A cos A sm 2^1
if we suppose, A=45^-^B, and .*. 2J = 90^ + 2JB,
we shall have
tan (45°+ B) - tan (45° - ^) = - 2 cot (90° + 2 J5) = 2 tan 2 B ;
.-. tan (45° + 5) = 2 tan 2^ + tan (45° - B) :
hence, assuming B to be equal to 1°, 2°, 3°, &c. successively,,
we have
tan 46° = 2 tan 2° + tan 44° ;
tan 47° = 2 tan 4° + tan 43° ;
87
tan 48° = 2 tan 6^ + tan 42° ;
&c =&c
These expressions may also be used to try the correctness
of the values of the tangents deduced by the other method.
147. The sines, cosines^ &c. of all arcs being thus cal-
culated and tabulated, form what is called the Trigonometrical
Canon; and it is easily seen that the sole difficulty in con-
structing such tables arises from the application of the
fundamental rules of arithmetic to numbers consisting of many
places of figures ; and some of the expedients generally resorted
to, to remove this difficulty ^ have already been explained. As
a check upon such computations, Formulcz of Verification have
been introduced, which involving the dependance of the trigono-
metrical functions of arcs upon one another, may be applied to
ascertain the correctness of a numerical calculation from the
known accuracy of one or more others.
Formulae of Verification might be multiplied indefinitely,
but the most useful and those most generally used, have been
proved in (78), (89)? and (90), and their utility will be manifest
from the two following articles.
148. In article (78) it has been proved that
sin— = - {x/(l + sin A) + ^(1 - sin ^1)},
2 2
A 1
and cos - = - { ^/(l + sin A) ± s/ {^ - sin A)\ :
2 2
now if we assign any value as 25^ to A, we shall have
sin 12^ 30' = - { v^(l+sin25') - ^ {\ - sm Q.o'')} ,
2
and cos 12^ 30' = ^ { ^(H-sin 25") + ^(1 - sin 25")} ;
88
hence if the results of these equations be the same as the sine
and cosine of 12^ 30' calculated by the method before given,
we may conclude with a considerable degree of certainty that all
the operations concerned are correct.
The formulae just mentioned might manifestly have been
likewise employed to deduce the sine and cosine of — im-
mediately from the sine of J,
149. In Elder's formula proved in (89) we have seen that
sin ^=sin (36^ ^) + sin (72°-^)-sin (36^-^)~sin (7^''+^):
and if A be taken equal to 5^, we shall have
sin 5^ = sin 41 V sin 67^ — sin 31°- sin 77°:
therefore if the values of the sines of these arcs already com-
puted satisfy this equation, they may each be reasonably
presumed to be correct, and the contrary.
Again, in Legendre's formula,
cosil=sin(54° + ^) + sin(54°-^)-sin(18°4-A)-sin(l8°- A),
which is proved in (90), if we suppose A = 7°, we get
cos 7° = sin 61° + sin 47° -sin 25° -sin 11°,
from which the same inferences may be drawn as before.
Similarly, of the sines and cosines of other arcs.
150. In the Trigonometrical Canon, constructed and
verified by these methods, the radius has been supposed to be
equal to 1 ; but as the logarithms of quantities afford great
facilities in the multiplication, division, involution and evolution
of large numbers, it is desirable that the logarithms of the sines,
cosines, &c. of arcs should also be tabulated, many of which
from their nature would to this radius be negative. On this
account the Tabular Radius has been assumed equal to ten
89
thousand millions, and consequently each of the sines, cosines,
&c. thus computed, must be increased in the same proportion,
and their logarithms will then become positive quantities. Thus,
since to the radius 1, we have sin l' = .0002908882 8cc.
.*. to the radius 10^^ we shall have sin l' = 2908882. &c.
and hence, log sin j' =log 2908882. &c.
=6.4637261 Scc.andsoon:
and a table constructed on this principle, is called a table of
logarithmic sines, cosines, &c. by the use of which most of the
practical applications of trigonometry are greatly facilitated and
generally performed.
151. If the logarithmic sines and cosines of all arcs be
found as in the last article, the logarithmic tangents, cotangents,
secants and cosecants, as also the versed sines and chords, may
be deduced from them by the operations of addition and sub-
traction only. Thus,
log tan A=]oglr j) =logr 4 log sin ^ — log cos tI
= 10 + log sin A — log cos A ;
(cos A \
r - — — ) = log r + log cos A — log sin A
sm A. /
= 10 + log cos A — log sin A ;
log sec A = log ( 1=2 log r — log cos A
\ cos A/
= 20 — log cos A ;
log cosec A = loo- ( I = 2 losf r — log sin A
^ Vsin A/ ^ °
— 20 — log sin A ;
A
= iog\ — - — y = log 2 +
log vers A = logV^ -J = log 2 + 2 log sin log ?•
M
90
= log 2 + 2 log sin — ■— 10 ;
2
and log chd A = log ( 2 sin — ) = log 2 + log sin — .
\ 2/ 2
152. To Jind the numerical ratio of the circumference of
a circle to its radius and diameter.
In (137) where the radius is supposed to be 1, the sine
of l' has been shewn to be .0002908882 &c. and ithas been
proved also that the sines of arcs so small as l' are very nearly
equal to the arcs themselves : hence_, since the number of
minutes in the whole circumference is 360 X 60 = 6 X 60 X 60^
we shall have
the whole circumference = . 0002908882 8cc. x 6 X 60 X 60
= . 0017453292 &c. x 60 x 60
= . 1047197520 &c. x 60
= 6.28318512 &c.
which was assumed in (7) to be represented by 27r; therefore
the circumference of a circle : the radius
:: 6.28318512 &c. : 1 ;
and the circumference of a circle : the diameter
:: 6.28318512 &:c. : 2 :: 3 . 14159256 &c. : 1.
153. By the method of converging fractions, approxima-
tions to the ratio just found are 3 : 1 ; 22 : 7 ; SSS : 106, &c.
which are alternately less and greater than_, but more and more
nearly equal to, the true ratio, and may be adopted in most cases
of practice without sensible error.
154. To find the magnitude of the angle which is subtended
by an arc of the circle equal to the radius.
Since 6 .28318512 &c. or the whole circumference subtends
four right angles^ or is equivalent to 360^ on the same scale
on which 1 represents the radius^ we shall have
360^ : the required Z :: 6.28318512 &c. : 1,
91
and therefore the required angle will be
360^
6.28318512 8vc.
= 57^2957795 &c.=57^ 17' 44" 48''' &c.
156. 3b express the length of the arc zohich Jiieasures
a given angle, in terms of the radius.
Since the arc subtending an angle of 57° 17' 44" 48'" Sec.
is in every circle equal to the radius, because the arcs are pro-
portional to the radii, when the angles which they subtend at the
centres are equal ; if A^ be the magnitude of any angle^ and a the
arc subtending it, we have
r : a :: 57^2957795 &c. : A"",
and .'. the arc a expressed in terms of the radius r
A'
5 f .0.951195 &c,
E
(1 \^ / 1 \^
— ) , (tit) , succes-
sively, then we have
the lenoth of one degree = r ( — „ — ) ,
V57^2957795 &c./
= ?• (.017453292 8cc.),
the length of one minute = r (.0002908882 &c.),
the length of one second = r (.00000484813 &c.).
156. Cor. if 57^2957795 &c. be represented by r^,
we shall have a = r —(T , and thence - = -tt , which agrees
r r r
A^
with what is assumed in (4) : also if r=z\, then will a = —q-.
92
CHAP. IV.
On the relations between the sides, angles, areas, circumscribed
and inscribed circles, ^c. of plane triangles. On the relations
between the sides, angles, diagonals, areas and circumscribed
circles, Sfc, of certain quadrilaterals. On the perimeters,
areas, Sfc, of regular polygons. On the periphery, area, S^c.
of a circle,
157. 2^HE sides of a plane triangle are proportional
to the sines of the angles which they respectively subtend.
Let ABC be a plane triangle, of which the angles
A /^'
D ' V B
are A, B and C; with centres A, B, and radius \, describe
circular arcs cutting CA and CB, or these lines produced
in the points a and /3 ; draw afx, jiv and CD perpendicular to
AB (produced if necessary) ; then by similar triangles
AC : CD :: Ja : afx :: 1 : sin J, by (17);
and CD : BC :: f^v : Bfi :: s'm B : 1, by (17) or (20) ;
.'. by compounding these proportions^ we have
AC : BC :: sin B : sin A;
93
similarly, AC : AB :: sin jB : sin C;
and AB : BC :: sin C : sin il ;
and therefore generally
BC : AC : AB :: sin A : sin B : sin C.
158. This fundamental property of plane triangles may
hkewise be proved as follows :
Suppose a circle to be described about the triangle ABC,
and let its centre be O, and its radius equal to R ; then it is
manifest that the sides of the triangle are the chords of the
arcs they respectively cutoff, to the radius R ; join AO^ BO, CO,
and by (oQ) we have
CB
R
chd BOC = 2 sin ("V") ~ '^ ^"^ ^ ' ^^ ^'^^'
-— = chd AOC = 2 sin ( ) = 2 sm B ;
R V 2 / '
J 7?
and -— = chd AOB
R
2 sin (
iOB\
2 )
= 2 sin C;
whence
CB CA AB
:: 2 sin ^ : 2 sin B : 2 sin C
R R R
that is, CB : C^ : AB :: sin ^ ; sin B : sin C
94
If the sides which subtend the angles A, Bj C be called
a^ b, c respectively, we have
a : b : c :: sin A : sin jB : sin C;
a __ sin A a s'm A b sin jB
b sin B ' c sin C ' c sin C '
159. Cor. 1. By either of the last two articles, we have
a : b :: sin A : s'm B,
.*. a + b : a— b :: sin ^1 + sin J3 : sin ^ — sin B
/A + B\ /A-Bx , ^ ■
'"''" (-^)^tan(-^-),by(72);
similarly
rA-\-C\ /A-C\
« + c : a — c : : tan I | : tan | I :
and o-jrc : 0^ c : : tan I I : tan I ) .
V 2 / V 2 /
Hence^ in a plane triangle, the sum of any two sides : the
difference :: the tangent of the semi-sum of their opposite angles :
the tangent of the semi-difference,
160. Cor. 2. Let CE drawn to bisect the angle ACB
c
A E B
meet the base AB in £, then by (157) we have
AE : AC :: sin ACE : sin AEC
:: sin BCE : sin BEC :: BE : BC',
.\ AE : BE ;: AC : BCi
95
that is, the segments of the base have the same ratio which the
other sides of the triangle have to one another.
Also, if AE = a'j BE = b', we have a : h' w b : a,
whence we find a = — — r , and h =
a + /; ' a+b'
and .• . a b' =^ ab I ) :
\a + b)
or a b' : ab :: c~ : {a-^bf.
A similar process may be used if the exterior angle be bisected^
and it will appear that
a'b' : ab :: c^ : {a^bf.
161. Cor. 3. If CF be supposed to bisect the side AB,
we have from (157),
A F B
sin ACF : sin CAF :: AF : FC
:: BF : FC :: sin BCF : sin CBF;
.-. sin ACF : sin BCF :: sin CAF : sin CBF:
or the sines of the segments of the vertical angle are proportional
to the sines of the corresponding angles at the base.
Also, if A' and B' represent the segments of the angle C,
we have
sin A' : sin B' :: sin A : sin B,
and .*. sin A' + sin B' : sin A' — sin B' ;:
sin A + sin B : sin A — sin B,
96
o.. tan {~:r-) •■ t"" (-^-j ■■■■ '=»> {-^) ■■ t-» (-i- j ;
tan
(^)
V 2 7 . C /A - X>\ O
— , or = tan I I tan —
2 ' V 2 / 2
- tan
from which, and the equation A' -{-B' ■= C, the values of A and
J3' become known.
162. To find the relations betiveen the sides and angles of
right-angled triangles.
Let ACB be a triangle having its sides represented by «, h, c,
as before, and the angle at C a right angle, then
A
BC : AB :: sin BAG : sin ACB
:: sm A : sni ~
2
:: sin A : 1, by (18);
whence BC-AB sm A - AB sin (^ ~ b\ =: AB cos B;
similarly, AC = AB sin J5 = AB sin N - J j = JJB cos A;
97
again, BC : AC
;in BAC : sin ABC
sin J : cos A
tan A : 1, by (42),
/. 5C = ^Ctan A = ^C tan ^- - Jb") =AC cotE;
similarly, ^C = 5C tan B=:BC tan /^- - a) = JBC cot A :
whence we have
AB =
BC
sin ^4
5C
COS
B
= i?C cosec A = BC sec 5;
7iC TiC
also, sin A = — ^ = cos £, and tan A = —— ; = cot B.
AH AC
163. To find the relations between the sides and angles of
oblique-angled triangles.
Let ACB be an oblique-angled triangle, draw CD perpen-
dicular to AB, and let the sides subtending the angles Ay B, C
be called a, b, c respectively :
c
A D B
then, c=1jD + ED
= AC cos A + BC cos B, by the last article,
= b cos A -T a cos B ;
similarly _, b = a cos C -{- c cos A ;
and a = b cos C-^c cos B:
and from these equations any one of the quantities involved
may be found in terms of the rest.
164. Cor. The last article combined with the property
proved in (157), is sometimes applied to express the sine of the
sum of two angles in terms of the sines and cosines of the
angles themselves. Thus,
since c =b cos A +a cos B, we have ~ = - cos A -|- cos B ;
a a
N
98
c sin C sin (tt -- C) sin (A -{- B) b sin B
but - = -^ — 7 = : ;; — = : — -: , and - = -. — j ;
a sin A sni A sni A a sm A
., „,e shall have ^^liA^ = fjlL^ eos ^ +cos B,
sui A sin iL
and thence, sin {A •{■ B) =■ sin jB cos A + sin ^ cos B
= sin j1 cos B -\- cos il sin By
as has been already proved in (63).
Since ^-f- jB is less than tt, the proof just given may at first
sight seem partial ; but by means of the relations established in
the first chapter, it is easily extended to the sine of the sum of
any two arcs whatever.
165. To express the cosines of the angles of a plane triangle
in terms of the sides.
If ^, I^, C be the angles of any plane triangle, a, bj c the
corresponding sides which subtend them, we have seen that
a =^ b cos C + c cos J5,
b = a cos C + c cos A,
c = a cos B -{■ h cos A ;
and multiplying both sides of these equations by a, b, c re-
spectively, we obtain
d^ =zab cos C-\- ac cos B,
h^ =.ab cos C -{-be cos A,
-2
C
ac cos B-\- be cos A ;
therefore,, by addition,
a" +5' -f-c-^ = Q.ab cos C +Q.ac cos B'{-2bc cos A :
from this equation, subtract successively 2a^, 2Z>", 2c^ and their
equals, and we have
b'^ -h c^ — a'^ ~2bc cos A ;
J. a^ -\- c" — b'^ = 2ac cos B ;
a--^b'~ c'':=:Qab cosC:
h'
+ c'-
a'
9.hc
2
a
+ c-°- -
b'
2ac
a'
+ 6^-
■c'
99
from which equations immediately result
cos A =
cos B =
cos C= ^ ,
l66. The values above found are frequently deduced by
means of the twelfth or thirteenth Propositions of the Second
Book of EudicVs Elements.
For, BC^ = AC--{ AK T 9.AB . AD,
but AD = AC cos A, by (l62),
or = AC cos (tt - A) by (l62), = - IC cos ^, by (24) ;
.-. BC' = AC' -h AB'-^AB. AC cos A,
or a' = ^'^ + c^ — 2 6f cos ^,
and .*. cos A = ;^ , as before.
2bc
It may here be observed, that the Propositions of Euclid above
referred to, are in reality proved in the last article.
For_, since a^ = b~ + c^ — 2b c cos J, we have
BC' = JC + AB'-qAB. AC cos A
= AC''^AB^^'2AB. ADy
as appears from (l62).
Ex. 1. Let a = b, or the triangle be isosceles: then
c' r c _
cos A = - — = r" — 771 — cos ^ j
also, cos C = 5 — = 1 - 77-3 , .and vers C = -;:--y
2a 2a 2a
100
Ex. 2. Let a — b^c, or the triangle be equilateral: then
cos il = ^ = cos 60^ = cos B = cos C.
167. Cor. 1. Since cos C= ; , we have
2ab
c^ = a^ — 2ah cos C -H b^, and thencec = ^a^~2(2 6 cos C-i- b'^,
which is the value of one side expressed in terms of the two
others and their included angle,
Ex. ]. Let C = 90", then cos C =0, and /. c^ = a^ + ^>^
which is the 47th Proposition of the first book of Euclid^ s
Elements established by the Principles of Trigonometry.
Ex. 2. If C = 60^ we have cos C = X^ from (37),
md .'. c"^ = a" — ab -}- b'
2
a' + ¥
a -{- b
Ex. 3. If C = 120^ we have cos C = - ^, by (24),
" 2 7 , 12 a —b^
and .•.£' = « -h cib -\- b = .
a— b
168. Cor. 2. From (165), we have immediately
2 6c- cos A = b^' -h c" - a^
or a' — b" = c^ — ^bc cos A =Q.c (- — b cos A \ :
V2 ;
that is, the difference of the squares of the sides is equal to
twice the rectangle contained by the base, and the distance of its
middle point from the perpendicular.
Again, {a + b) {a — b) = c (c — 2b cos J),
ore : a + b :: a — b : c — %b cos A :
that is, the base : the sum of the sides :: the difference of
the sides : the difference or sum of the segments of the base
made by a perpendicular let fall upon it from the opposite
angle, according as it falls within or without the triangle.
101
169. Cor. 3. From the preceding articles it is seen, that
^ =• c^ •\' h' — %ab cos C
= a — ah cos C + b^ — ab cos C
= a {a^ b cos C) + b {b — a cos C) :
suppose now AP and BG to be drawn from the angles A and B
respectively perpendicular to the subtending sides ;
then « - 6 cos C = i?C - CF= BF,
and b — a cos C = AC — CG =^ AG;
hence replacing a, b, c by 5C, AC and ^J5 respectively,
we have
AB^=:BC.BF+ AC, AG:
or the square described upon any side of a triangle is equal
to the sum of the rectangles contained by the two others
and their segments respectively cut off by perpendiculars let fall
upon them (produced if necessary) from the opposite angles.
170. Cor. 4. If the angle C be bisected by the straight
line CE meeting the opposite side in E, the value of this line
may be found; for by (I66) we have
A E B
CE' = AC'' + AE'-2AC.AE COS J
b'c- 2 6
c
(a + by a + b
b'c" {a^-b^)b bc^
= b'' + , . ... +
{a-\-br a\h a-^h
b'c' ^ be'
''^'^{a-\'bf a + 6
abc^
=.ab —
(a + bf \a -\-bJ \a-]r b)
or AC. BC = AE,EB-\-CE^:
that is, if any angle of a triangle be bisected by a straight line
which cuts the opposite side, the rectangle of the two other sides
is equal to the rectangle of the segments of the divided side
together with the square of the dividing line.
171 • Cor. 5. Supposing CF to bisect the side AB in F,
we shall have by (I66),
CF'==CA' + AF'-'2AC . .IF cos A =6'+ {^^ - be cos A
■2 j2i2 2 2i7 2
c b '\-c — a a -\- 0
= 6^ + -
4 2 2 4
therefore a' + 6' = 2 (- j + 2CF^
which shews that the sum of the squares of any two sides
of a triangle is equal to twice the square of half the other side,
and twice the square of the straight line which is drawn from the
opposite angle to bisect it.
103
172, Cor. 6. If we suppose CF = h, and the coi-
responding lines drawn from the angles A and B to bisect the
opposite sides, equal to k and / respectively, we shall have from
the last article,
a-^ + i' = 2 g)+2/i^
6^ + c= = 2g)'+2ft^
and therefore by addition,
and thence
that is, three times the sum of the squares of the sides of a plane
triangle is equal to four times the sum of the squares of the lines
drawn from the angles to bisect the opposite sides.
173. To express the sines of the atigies of a plane triangle
in terms of the sides.
From {25)y we have sin" A-= I — cos" A
■b^ + e- a\' {2bcf - (b- + c' - a-f
4b'c'
(Qbc + 6" + c^ - fl^) (^bc - b--c^+ a')
4 /re'
{(b + cf--a'}{a'^{b-cf]
4b'c^
104
_ ia'{-b-^c) {b-\-c - a) (a -j- c - b) (a-\-b''c)
4>b'c'
and therefore
1
sin A = —r- />7(a+6 + c) (6 + c- a) (a4-c-6)(a+6-c);
Qbc
similarly,
sin 5 = ^J{a + b + c){b-\'C-a){a^i^c-b){a + b-c)',
and
sin C = — r i^{a-\- b + c) {b + c— a) (a+c- /;) (a + 6 — c):
assume now, 2AS' = fl -\- b +c = the sum of the sides ;
then Q.{S-a) = b ■\- c- a,
2{S - b)=a + c — b,
2(S- c)=a -{- b -c;
whence by substitution we obtain
sin ^ = ~ \/s (S~a){S-~b)(S-c);
be ^
sin5 = — \/s (S-a)(S-b)(S-c)',
ac ^
sin C = — \/ S {S — a){S-b)(S-c\
ab ^
Ex. 1. Let a = b, or the triangle be isosceles; then
in^=-^ \/ S{S-a)iS'-a){S'-c)
sm
ac
^^'-^\/7^,
ac
-y(-3("-i)
105
4 ?/ — <? — sin B ;
and sin C = '-i^ \/s(S-c) = J \/(« + fJ („-i)
Ex. 2. Leta = Z> = c, or the triangle be equilateral, then
sm
..iv/.„.-.>-=iN/(fy©^
= =sin 60° = sin B = sin C.
2
Ex. 3. If the sides of the triangle a, h^ c be respectively
equal to 3, 4, 5, we shall have
2 ^ = 3 + 4+5=12, and *S = 6,
.-. S — a==S, S—b = 2, and 5-c= 1 :
2 A / 12 3
and sin 1 = — V 6 . 3 . 2 . 1 = — = - ;
20 ^ 20 5
. T. 2 4 /^ 12 4
sm jB=— V 6.3.2.1=—=-;
sin C = -^ V 6.3.2. 1 = ~= 1 =sin^:
12 12 2
TT
hence C = - , or the triangle is right-angled at C.
til
174. Cor. From the last article we have
V 6' (5 - a) {S -b){S-c)=: ^' sill C, wliich, if C = -
gives 2 ,yT(^^~«)(5-6)(.S~c)^«^:
O
106
and by substituting for a, b, c, the quantities 71^ — Ij 2;/ and
n^ 4- 1 respectively, it will be found that this equation is verified ;
and therefore the sides of any rational right-angled triangle may
be represented by these quantities^ n being assumed at plea-
sure equal to any quantity greater than unity.
175. To express the sine, co-sine, tangent , Sfc. of half
an angle of a triangle in terms of the sides.
From (79) we have
^A , b~'\-c"'-a~ %hc'-J)'-c^+a^
2 snr — = 1 — cos A= \
a^-{b-cf __ {a-\-b'-c){a^-C'-b) ^ 2 (S-b) 2 (S -c)
Qbc ~ 2bc Q,bc
_ . A _ /(S-b)(S-^)
sin
similarly,
$m — = V
2 ac
. C /(S-a)(S-'b)
and sm — = V "■ ; — ~~~ =
2 ab
Again, from the same- article, we have
A'
cosec —
2
1
cosec
2
]
cosec
c
2
,A . , Z»' + c^~«"- Qbc-i-b' + c^-,
£ cos'— = 1-f cos 7i= 1 +
2 2bc 9.bc
{b-\-cf-a' ___ {a + b-^c){b + c-a) _ 2S 2(8 -a)
2b~c 2bc "" oOc
COS
107
^ A/SiS-a) 1
= n/
2 ^ be A'
sec —
2
, B /s{S-b) 1
ally, cos - = V = 7t;
similar ,, _ ,
ac B
sec —
2
, C /SiS-c) 1
and cos — = y/^
2 ^ ab C
sec —
o
Hence .*. we shall have by (42)
^ A ./iS-b){S-c) 1
tau — = V — tttt: ^ — = •,
cot —
o
cot —
o.
. C /iS-a)(S
and tan — = y -— —
b)
2 ^ S{S-c) C
cot —
2
176. Cor. 1. If the angle C be a right angle^ we shall
have
. C . , } . C
sm — = sin 45 = — 7- = cos 45 = cos — ;
2 V 2 2
1 ./{S-a)(S--b) /S(S--c)
hence — 7- = v ; = V ; — ;
and therefore
a6 = 2(S-a)(6"-6) = 2 6' (5-c),
and {S-a){S'--b)=:S (S-c),
108
177- Cor. 2. From (175) we may easily deduce what is
proved in (173).
A A
For, sin A = 2 sin — cos — , by (76),
2 2
=.\/^
-b) {S - c) S{S- a)
be be
= ^s/S{S-a){S'~b){S~-el
he ^
as before : similarly of the others.
178. To express the area of a plane triangle in terms of
the sides.
D B
The area of the triangle ABC ^- AB . CD
= - AB , AC sin A, by (l62),
= ^^ -^, s/S{S--a){S-b){S'-e\ by (173),
2 b
= ^/^^(5-«)(S-6)(S-c).
179* The area above found might easily have been deter-
mined without assuming the expression for the sine of an angle
of the triangle. Thus,
since «^ = 6^ + c'-2r AD, by (l66), .-. AD = ' "^^ "
2<:
CD"- = AC^-AD^- = ,^~(^^±^)
109
- (Q-t-^-^g) {h-hc- a) (a + c-6) ja + b-c)
4c^
Q.S 2{S-a)<^{S-b) 2 (S-c)
hence the area = = yj S {S - a) (S — b) (S — c),
From either of these articles, we obtain the following
Rule :
From the semi-sum of the sides, subtract each side sepa-
rately; multiply the semi-sum and the three remainders together,
and the square root of the product will be the area.
Ex. 1. Let « = ^, then the area of an isosceles triangle
whose base is c = {S^a)sJ S (S — c)
=i^/(-n)("j)_
4 4
Ex. 2. If a = 6 = c, the area of an equilateral triangle
whose side is a = S^ (S" — a)^ = ( — ) ( "" ) ~ " •
Ex. 3. If a, b, c be equal to 18, 24, and 30 respectively,
we shall have
5: = i (18 4-24 + 30) = ^ (72) = 36;
2 2
.-. S-a = 36- 18 = 18,
.S- A = 36-24=12,
S ^ c = 36 - 30 = 6 ;
110
and .'. the area = ^36.18.12.6 = ^36.36.36
= 6,6.6 = 216.
180. Cor. 1. Hence the perpendicular drawn from any
angle to the opposite side is easily expressed in terms of the sides
of the triangle : for
4S{S-a) {S-'b){S-c)
and . . LIJ =
181. Cor. 2. The area of the triangle may very easily be
expressed in different terms.
Thus, by (178) the area =z^bc sin A;
or = V — — — j^ S {S- a)bc
A .
= sin — as/ S {S — a)bc ;
or = \/^Z^ ^(S-b)iS-c)bc
be
A ,
= tan — S (S - a): &c.
182. Cor. 3. If the triangle be right-angled at C, we
shall have
Ill
c
tan — or 1
^ ^/(S-a)(S-b)
and .*. the area = S (6' — c), or = {S — a) (S — b) :
that is, the area of a right-angled triangle is equal to the
rectangle contained by the semi-perimeter and its excess above
the hypothenuse ; or to the rectangle contained by the excesses
of the semi-perimeter above each of the sides containing the
right angle.
183. Cor, 4. From the values of the area above deter-
mined, it may be demonstrated that the areas of similar triangles
are in the duplicate ratio of their homologous sides.
Let A, B, C ; a, by c, and A' , B\ C ; a, b\ c' be the cor-
responding angles and sides of two similar triangles ;
then if 2 *S = a + 6 + f, and 2 5^ = a ■\- b' -\- c\
. ^ ../{S-b){S-0 . A ./{S'-U){S'-c)
wehavesm— = V 1 =sni— -= y rn ;
A /S{S-a) A' /S'iS'-a)
and cos — = V ; = ^'os — = \/ — — -^-7 ;
2 ^ he 2 ^ be
.'. area of the triangle ABC : area of the triangle A'B'C
:: ^S{S~a){S~b){S-c) : ^ S\S'-a){S'-b')iS'-e)
,, , e shi B , c sin B 2 ,q
:: he : be' :: c . ,, : c . ^, , by (157), :: c : c .
sm C sm C
184. Cor. o. Smce ABC : AlB'C ::^ he : Ve, if we
suppose 1BC = A'jB'C', we shall have be^b' e\
and /. h : h' \: c : c :
or, if the areas of two triangles which have one angle of the one
112
equal to one angle of the other, be equal ^ the sides about the
equal angles are reciprocally proportional : and conversely.
185. To express the radius of the circle inscribed in a plane
triangle in terms of the sides.
Let ABC be the triangle, its angles and corresponding
opposite sides being denoted by A, B, C; a, b, c as before:
bisect the angles A and B by the straight lines Ao, Bo
meeting in o, draw o«, ob, oc perpendicular to BC, ^ICand
AB respectively; then o is the centre, and o« = o^ = oc the
radius of the inscribed circle ; let this be called r :
now by (157), we have
Ac sin ^oc cos oAc
oc sin ou4c sin oAc tan oAc
1
tan
A'
', ilc =
oc
; snnilarly. Be =
tan
and
A
tan —
2
Ac -\- Be = 7' \- —
B
tan —
2
B'
tan —
o
(./ S(S-a)
'' {^ (S-b)(S-c)
— tan — S
^2 2>'
. / ^ (S - b) 1
V,., \,o J> by (175),
(S-a) iS-c))
( S(S-a) + S(S-b) )
\s/S{S-a)(S-b){S-c))
113
26' - (a + b)
i^S{S-a) {S-b){S-'c)}
rcS
whence we obtain
^S(S-a)(S- b)(S -c)
w
{S -a){S- b) (S - c)
S
Ex. 1. Let a = by or the triangle be isosceles, then
/ o \ \ /^ "" ^ ^ \ /^^ -^
o 2 2a + c
Ex. 2. If a = b = c, or the triangle be equilateral, we
shall have
== x/^^ - of ^ sj~i_
S 12 SV'S
186. Cor. 1. Since r5 = ^^^ (5 - a) (5 - ^>) (5 - c),
we have
/a + 6 + c\ , ^ , . ,
J. I I — the area or tlie triangle.
This is also manifest from the consideration that the triangle
^JBC = the sum of the triangles AoB, AoC, BoCj
AB.oc AC.ob BC.oa /a-^b + r
= -\- -j-
2
/a-j-0-\-c\
and from this property the value of r is very easily obtained ;
thus
2 area js/ S {S - a) {S - b) [S - c) , .
r = — = -^^ — , as before.
a + b -\- c S
P
114
187- Cor. 2. We may hence find the segments of the
sides of the triangle made by the points of contact with the
inscribed circle.
From (185) and (175) we have
B
Ac 2 S—a Ac S — a S — a
Be A S—h' " AB 2S-{a + b) c '
tan —
2
wherefore Ac = S — a = ^{b -\-c~-a) :
In the same manner Be = S — 6=^(a-|-c— 6):
and similarly of the rest.
Hence also, Ac. Bc = (S'-a) (S — b); and -jr- = ^ ■ . ;
DC o — 6
and so of the rest.
188^ To express the radius of the circle circumscribed about
a plajie triangle in terms of the sides.
Let ABC be the triangle, the angles and sides being
Af B, C; a, b, c as before: bisect the sides AC and BC
in the points 5 and a; draw bo, ao at right angles to AC
and BC respectively, meeting in a ; then is o the centre, and
115
Ao = Bo = Co the radius of the circumscribed circle : call
this jR. Then
^0 1 1 1 ^ , V
"7T = "- — T~l = -A — = "■: — ;; > (Eucl. S. 20.);
Ab sm Job . Aoc sm B
sin
2
^ Ab b 1
ac
abc
4^S{S-'a)(S'-'b){S-c)'
Ex. 1. Let a ■= b, then in an isosceles triangle we have
2 9.
n __ _^ ^_f _ a
Ex.2. If a = b=^c^ we shall have for an equilateral
triangle,
3 3
I89. Cor. 1. By means of the last article, we have
abc
Q,R =
^^S{S-a)(S'-b)(S-c)
ab
"" 2 ^S{S-a){S-b){S^)
c
= -T7f:y as appears from (180);
and .-. Q,R,CD = ab ^ AC . BC i
116
or the rectangle contained by any two sides of a plane triangle
is equal to the rectangle contained by the diameter of the cir-
cumscribed circle, and the perpendicular let fall upon the
remaining side from its opposite angle.
190. Cor. 2. The property just mentioned which may be
proved by means of similar triangles, is frequently made use
of to determine the radius of the circumscribed circle.
^ . ^,, «6 , ^ AB.CD abc
ror, smce CD = — ~ , therefore = — ^;
^ AB,CD ^ . . • 1
but = the area of the triangle
= ^S{S-a){S-'b){S''c\
hence ^ = ^ S {S - a) {S - b) (S-c\
4 it
and R =
4 V*S(*S'-«)(6^-6)(5-c)'
as before proved.
191. Cor. 3. The segments of the angles A, B, C made
by the radii of the circumscribed circle may easily be foluid.
TT TT
For, /L oAb = - — Aob = - — B = / oCb:
TT TT
similarly, /. oAc ='^ — ^oc = - — C= Z oBc,
• 2 2
and Z oBa = - — Boa = - — A = Z oCa.
2 2
192. To express the cosines of the angles of a quadrilateral
in terms of the sides^ two opposite angles being supplemental
to each other.
Let the angles of the quadrilateral be denoted by the letters
at its angular points, j4, Bj C, D: also join AD, and suppose
in
AB = a, BC = b, CD = c, and DA=d: draw the diagonals
AC, BD and let BD = a, AC = f^ : then by (1 65) we have
2 ad cos A = rt* + d'' - a^, from the triangle ABD ;
also,
26c cos C = b' -{- c^ — a^y from the triangle BCD
now cos C = cos {tt — A)= — cos J.^
therefore —2bc cos A = />>^ + c^ — a :
hencCj by the elimination of a, we obtain
2 (a J 4- be) cos ^ = «- 4- ^"^ - /r -^6-' ;
a^-\-d'~b'-c'
and cos ^ =
2 (ad + be)
cos
C:
sHTiilarly, cos ii = ; — = — cos JJ.
^ 2{ab-]-cd)
193. To express the sines of the angles in terms of the
sides of the quadrilateral.
As in (173) we have sin^ A
•a'^d^-b'^-c-^
= 1 —cos'
1
~^~ V 2{ad-^bc) J
4(af/+ bcY
2{ad + bc)
{4{ad-{-bcf-{a' + d'-b'--cy]
1
4{ad + bcY
118
4(ari + />c)
4(ac? + ^>c)-
{(a+^' + ^-cXa+c + rf-^Xfl + ^ + c-^iX^+c + cif-fl)}:
now let cf + 5 + c + c? = 2 6' ;
« + 6 + 6;— c=2(6'-c),
4(a^ + 6c)
(flc? + 6c)
and sin A=—- — J (^S ~- a) {S - b) {S - c) {S - d) :
ad + be
and similarly of the rest.
194. By a process very similar we may express the sine,
cosine, tangent, &c. of half an angle of the quadrilateral in
terms of the sides.
Thus, 2 sm^ — = 1 — cos ^ = 1
2 Q. (ad + bc)
ad-i-^hc-a^-d'^ + b' + c^ (b + cy-{a-~dy
2(ad-i-bc) 2(ad + bc)
(a + b + c-d) (b + c+d-a) 2 (S - a) 2(S-d)
2(ad-{-bc) Q.{ad + bc)
. . ^ ./(S'^a){S-d) 1
and .*. sHi — = \/ ; = ^ ;
2 ^ ad+bc A
cosec —
2
119
similarly, cos
A^ ^(S-h)iS^) 1
ad-\-bc A
sec —
2
A ./{S-a){S-d) 1
and .*. tan — = \/
o ^ ^S'-'b){S-c) A'
cot —
2
similarly of the others.
195. On the same supposition to express the diagonals of
the quadrilateral in terms of the sides.
The construction and notation remaining the same as in
(192), we have
2 I j2 2
a +0 —a
cos A —
and cos C = — cos A
therefore
^ad
Q.bc '
2ad 2bc
hence a' (ad + be) = (a^ + d') b c -{- (b' -}- c') ad,
. /(a^ + d') bc + (b^ + c^)ad
and a = V JTT
ad + be
_ ./a^c + d'bc-hb'^ad + c'^ ^ ./{ac + bd){ab+c
" ^ ad + bc ^ ad + bc
similarly, since cos B= — cos D, we shall have
_ y/{a''+b')cd + (c^ + d')ab
_ y/ a^c d + b\d-^c^ab-\- d^b _ ./{ac-^-bd) {ad-\- be)
ab-\-cd ab + cd
120
190. Cor. 1. From the last article, we have immediately
a/^ = ac + bd, or AC.BD = AB,CD'\' BCAD:
that is, the rectangle of the diagonals is equal to the sum of the
rectangles of the opposite sides.
a _ ab + cd AC _ AB.BC + CD. DA
' f3 " ad + bc' ""'bd" AB.AD + BC.CD'
that is, the diagonals are to each other as the sums of the
rectangles of the conterminous sides respectively meeting their
extremities.
197* Cor. 2. The former property deduced in the last
article, which may be proved geometrically, is sometimes made
use of to express the sine of the sum of two arcs in terms of
the sines and cosines of the arcs themselves.
For, if A and B be the proposed arcs, take yJP = 2A, PQ
= 2l?; draw the diameter PR and join AP, JQ, AR, PQ,
QR:
then AP = chd 2A = 2 sin J, PQ = chd 2B = 2 sin B,
^l^ = chd (tt- Q.A) = 2 sin (^ — Aj =2 cos A,
QR = chd (7r-2.B) = 2sin (^-b\ = 2 cos B,
and ^Q = chd (2^ + 2 6) = 2 sin (A-f-B):
121
now PR . JQ = AP . QR + AR . PQ, by the last article;
.*. 4 sin (A-hB) = <2. sin A 2 cos 1^ + 2 cos ^ 2 sin B,
or sin (A + B) ~ sin A cos B + cos A sin i)j as before.
198. On the same hypothesis, to express the area of the
quadrilateral in terms of the sides.
The area of A BCD = the area of the triangle ABC + the
area of the triangle ACD
= — sin B H sin D, by(l78),
2 2 f J ^ y^
ah . ^ cd ah + cd .
= — sin B-i sm (tt - /5) = sm h
2 2 2
ab + cd 2
^{S-a){S-h){S''c){S-d)
2 06 + <:c?
= ^(6:- a) (S-b) {S - c) {S^d).
199» Cor. From the last article, it appears that
A ABC = ""^ , VC'^ - «) ("^ - ^) (-^ - f) (-^ - A
ab + cd
and
ab + cd
Also, if 0 denote the angle in which the diagonals intersect
each other, and AG, CH be drawn perpendicular to the diagonal
BDj we manifestly have
the area of ABCD =aABD-^A CBD = ^^^ + ^^^T^
= ^ {AG+CH}=^ [AE sin 0+CE sin 0}
= ^ sm 0 ;
Q
122
whence
sin 0
2 ABCD 2 ^/{S - a) {S-b) {S -c){S- d)
AC.BD ac + bd
200. A circle may be circumscribed about the above-men-
tioned quadrilateral, and its radius may be determined.
Let Jf^ be the radius of the circle circumscribed about the
triangle ABD, R that of the circle circumscribed about the
triangle BCD: then by (188) we have
aad aad a
Jx =
4 A ABD ad . , 2 sin A '
4 — sm A
2
similarly^
tbc abc
R' =
4 A BCD ^c . ^, 2 sin C
4 — sm C
now sin C = sin (tt — i4) = sin A, and therefore R' = R:
or the circle which can be circumscribed about the triangle ABD,
will also be circumscribed about the triangle ACD, and there-
fore about the quadrilateral ABCD.
Also, R =
A A ABD
a {ad-{- be)
4 ^(S-a) (S- b) iS-c) (S-d)
, by (198),
-W%
b + cd)(ac + bd)(ad + bc)
, from (193).
{S-a){S'-b){S-c){S-d)
By making any one of the sides of the quadrilateral equal to
nothing, all the formulae just proved will manifestly be true of a
triangle.
123
201. To express the area of a regular polygon in terms of
the side.
Let ABy BC be two adjacent sides of the polygon each = a :
bisect the angles at A and B, by the straight lines AO and BO
meeting in O; then if n be the number of sides of the polygon,
and therefore the number of angles, we have (Euc. 1.32) the
sum of the angles
= (271 — 4) -=(«-2)7r;
(n— 2\
■ I TT :
Draw Oa perpendicular to the side AB, then Aa = Ba, and
the area of the polygon
,^^ AB.Oa a a ^^ nc? /;* — 2\ tt
— ni^AOB^n = /j tan U^ tt = --- tan
2 2 2 4
/n — 2\ TT
\ n ) i
na /TT 7r\ na tt na
= — tan I I = — cot - = — .
4 \2 7^/ 4 n ^ IT
4 tan -
n
Ex. Let n be taken equal to 3, 4, 5, &c. successively,
and we shall have
. , 3a- 1 TT ?:>a^ ^0 a'Vs
area of a triangle = tan - - = — - tan 30 = :
^4324 4
4 a~ 1 TT
area of a square = — tan - - = a^ tan 45^ = a^ :
^ 4 2 2
5 a" 3 TT 5q' 0
area of a pentagon = tan - - = tan 54 : and so on.
4 5 2 4
124
202. To express the radius of the circle inscribed in a
regular polygon, in terms of the side.
Let AB and BC as before be two adjacent sides of the
polygon, n the number of sides : bisect the angles at A and B by
the straight lines AO and BO meeting in 0; draw Oa, Ob per-
pendicular to AB and BC respectively, and therefore bisecting
them ; then will O be the centre and Oa = Ob the radius of the
inscribed circle ; let this = r :
now
Oa
m OB a sin OBa
Ba sin a OB cos OBa
(TT 7r\ TT
- — - ) = cot - ;
2 n/ n
— tan OBa
a TT
/. 7' = - cot -' =
2 n
a
TT
2 tan -
n
203. Cor. Hence from article (201), we have
.1
the area =
na
4 tan
TT
2
TT
2 tan -
nar r ■ .
= " pernneter.
204. To express the radius of the circle circumscribed
about a regular polygon, in terms of the sides.
Bisect the angles A and B by the straight lines AO and BO
meeting in 0 ; then it is manifest from the fourth book of
125
Euclid's Elements, that O is the centre of the circumscribed
circle, and OA = OB the radius^ which call R ; then
OA sin AaO
A a sm AO a
(i - «■•«)
SHI
COS OAa
(~); -G-d
Sin -
n
R =
a
2sm -
71
205. Cor. 1. Hence also the area of the polygon
na cos
11 a K cos -
na
4 tan
TT
21 sni -
TT
it cos -
2
perimeter.
206. Cor. 2. From articles (202) and (204) we have
directly
R
a
TT
tan -
a n
r
. TT
TT . TT
2 sin — 2 tan - sin — cos —
n n 2 n
126
Ex. Let n be taken successively equal to 3, 4, 5, Scc, then
. , ^ 1
m a triangle — = — -q = 2 :
r cos 60
in a pentagon - = ^„ = V^- 1 :
and so on.
207. To express the perimeter and area of a regular
polygon inscribed in a circle^ in terms of the radius.
Let r be the radius of the circle, n the number of sides of
the polygon ; then the angle at the centre of the circle subtended
by each side AB \s — ;
a:
now by article (59), we have
— = did AOB = chd — = 2 sin -, by (76),
r n n
.*. Ai) = 2r sui - :
and the perimeter of the polygon = nAJB = 2«r sin - ;
Again,
AO.BO r' 27r
A AOB = — sin AOB, by (178), = — sin — ;
127
iir^ . Qtt
and the area of the polygon = n A AOB = — sin
^ Tl
r TT . TT r TT .
= " COS - 2/zr sm - = - cos -« perimeter.
2 w n 2 n
208. 7^0 express the perimeter and area of a regular
polygon circumscribed about a circle, in terms of the radius.
Let r be the radius, n the number of sides of the polygon,
then the angle at the centre subtended by each side AB\s — ;
n
draw Oa perpendicular to AB which is therefore bisected in a:
then;, AB = 2lrt = 2 0« tan AOa — Q.r tan -;
n
therefore the perimeter :=. Q^nr tan - .
w
.^i> AB.Oa . ^ 2 TT
Also, aAOB== =Aa.Oa = r tan -",
.'. the area of the pologon = nr^ tan -
r TT r .
= - 9,?ir tan - = - perimeter.
209. Cor. From the last two articles, if P, P' and ^,
A represent respectively the perimeters and areas of the inscribed
and circumscribed regular polygons of the same number n of
sides, we perceive that
128
„ Sin - . sm —
jP 7i__ TTjI ^*_ 9 '^ ,
P' TT A< ' A' TT n'
tan - 2 tan -
n n
210. 3o express the periphery of a circle in terms of the
radius.
Let p and p represent the perimeters of two regular
polygons of n sides, the former inscribed in, the latter cir-
cumscribed about, a circle whose radius is 1 ;
then p = 2n sin - , by {9.01),
and p = 'In tan - , by (208) ;
n
sm -
therefore — = = cos — :
p IT n
^ tan -
n
and if we suppose the value of n to be increased indefinitely, the
value of -- will be indefinitely diminished,
n
and .*. cos - = 1, or p = p :
now the periphery of the circle evidently lies between p and p\
and therefore in this case is equal to either of them; hence
on this supposition — th part of the perimeter of the polygon is
equal to — - th part of the periphery of the circle ;
. TT 27r TT , TT TT TT
that is, 2 sm «- = — = 2 tan - , or sm -«='-= tan -- ;
n n n n n n
therefore the perimeter of a polygon described about the circle
whose radius is r
TT
= ^Zur tan - ,
129
TT
and the circumference of the circle = 9,nr tan -« when n is
n
Hihnite^ = 2wr - = 27rr.
211. Cor. 1. Let a be an arc of a circle whose radius
is r, A^ the angle subtended by it at the centre ; then by
(Eucl. 6. SS) we have
27rr : a :: 27r : A^
and thence A = = - :
2 TT r 7'
or an angle is equal to the corresponding arc divided by
the radius.
212. Cor. 2. From what has been proved in (210), if r
and r be the radii of two circles, D and D' their diameters,
C and C their circumferences, it appears that
C = 27r?', and C' = Stt/;
C 27rr r 2/' J)
and
' C 27r/ r' 2/-' jy*
that is^ the circumferences of circles are proportional to their
radii or diameters.
The properties proved in this and the preceding article were
assumed in articles (3) and (4) ; but it may be observed that no
conclusion was drawn from them, upon which any of the
propositions on the trigonometrical functions of arcs or angles
in any way depend.
213. Cor. 3. From the demonstration of (210) it appears
that if a circular arc be continually diminished, it ap-
proaches continually to a ratio of equality with its sine or
A A
tangent : also, since chd A = 2 sin — by (76) = 2 — (if A be
indefinitely diminished) = A, we conclude generally that the sine,
R
180
chord, and tangent of a circular arc are all ultimately equal
to one another^ and to the arc itself.
214. To express the area of a circle in terms of the
radius.
If A and A' denote the areas of two regular polygons of
the same number n of sides,, described, in and about the
circle whose radius is r, we have seen by (207) and (208)
that
iir" . 9>7r , „ TT
A = sni — , and A =iir' tan - ;
2 n n
sm
A 1 n
hence — -. = -
^ 2 TT
tan ~
n
•=■ cos^ - = 1^ if n be indefinitely increased ;
n
.*. A — Al\ and on this supposition the area of the circle is
equal to either of them, that is,
• 1 wr^ . 27r . . ^ .
the area of the circle = sm when n is infinite,
nr 9,1? ^ ^ 9
- — by (210) = Ttr^ :
2 n
T r
the area also =- 27rr = - the circumference, from (210).
2 2
If the radius = 1, we have the area = tt : that is, ir which
represents the semi-circumference of a circle whose radius is 1,
will also represent the area.
215. Cor. 1. By means of the last article, the area
of a circular sector is easily found.
131
FoFj let A^ be the angle of the sector, a the arc ; then
(Eucl. 6. 33) the area of the sector : the area of the quadrant
:: a : :: - : - :: A° : -, from (211);
4 7-2 2
.'. the area of the sector = — area of the quadrant
TT
2
2A TT?'" ^ ^ A?'^ r ^ r
= — from (214) = = - A r = - the arc, from (21 1).
TT 4 2 2 2 ^ V /
2l6. Cor. 2. Hence it is easily shewn that the areas of
circles are proportional to the squares of their radii, diameters^ or
circumferences.
For, let r, / be the radii of two circles^
D, ly the diameters, C, C- the circumferences ;
then by (214) we have A = 7rr^ and A' =.7^'^ ;
A irr" 7-^ 4/^ D"- 0
and .*. — = -^^, = I^ = ZZr- =•■ Tvl' = 7^^ *^>' (-^^^'
132
CHAP. V.
On the Solution of Triangles and the Jpplication of Trigono-
metry to the Mensuration of Heights, Distances, &,-c.
217. In every triangle there are six parts, the three
sides and the three angles ; and if a, b, c represent the former^
and A, B, C the angles subtended by them respectively, it
has been proved in (l63) that their mutual dependence upon one
another is expressed by the equations
a = b cos C + c cos B,
b = a cos C + c cos A,
c = a cos B -jr b cos A :
now, since n independent equations are in general necessary and
sufficient for the determination of n unknown quantities, it is
manifest that if three of the above-mentioned quantities be
given, the other three may generaUij be found.
On further examination however, it will appear that wlien
the three parts given are the angles, the magnitudes of the
sides will be indeterminate, though their ratios to one another
may be found ; for, in addition to the dependance expressed
in the equations just mentioned, the sides and angles are
further connected by the equations proved in (158), namely
a sin A a sin A b sin B
1 = - — ^, - = —7;, a"(^ - = ~^~7^-
o sm i5 c sm L c sm C
hence, by division and substitution we have immediately
- = cos C + - cos B = cos C -I — ^ — =r COS B ;
b b sm i>
133
. , « „ . sin A ^
= cos A -\ — cos C = cos A -{ — : — - cos C;
c sin C
= cos i> -f- - cos A = cos i) + -: 7 cos A :
« r^ sin A
therefore if A, B, C be given, and the latter sides of either of
these two sets of equations be called w, n, p, respectively,
we have
a h c
-. = m, - = ;^ and - = p^
be a
from which it is evident that the magnitudes of a, b, c cannot be
determined, though their ratios to each other are found.
From what has been said, it follows that in every triangle, if
any three parts not all angles be given, the remaining parts can
be found ; and the reason of the exception above stated is still
further apparent from the circumstance that the lengths of the
sides of triangles may be increased or diminished, while the
magnitudes of the angles remain the same.
218. If one of the angles of the triangle be equal to 90^,
or the triangle be right-angled, it follows that this angle may
in all cases be considered as one of the parts which are given,
and therefore that only two other parts will be necessary
and sufficient for the determination of all the rest ; the same
exception being made, and for the same reason as in oblique-
angled triangles.
219. From the considerations of the last two articles,
it is manifest that the solutions of all right-angled triangles
are comprised in those of the two following cases :
I. When one side and one angle are given :
II. When two sides are given :
and the solutions of all oblique-angled triangles in the following
four :
134
I. When one side and two angles are given:
II. When two sides and the angle opposite one of them,
are given :
III. When two sides and the angle included by them,
are given :
IV. When the three sides are given :
and the investigations of the solutions of these cases in order,
will be contained in the following articles.
Solution of right-angled triangles.
Case I, in which one side and one angle are given.
' 220. Let cf, h, c be the sides of the triangle. A, B, C the
angles subtended by them respectively, C being the right angle ;
then since A -\- B = - , (Euc. 1. 32), if one of these angles be
2
known, the other is likewise found :
a sin A sin A
= tan A — cot B ;
= sin A = cos B ;
= sm B — cos A :
and from these equations, if il or jB and any one of the
quantities a, by c, be given, all the rest may be determined.
221. Ex. 1. Given the side a and tlie opposite angle A,
to find the rest.
Here i> = - —A. and is therefore found :
o '
.so, -
sin B
cos A
a
sin A
sin A
^
—
—
c
sin C
. IT
Sin -
2
, b
sin B
sin JB
and -
=
=
c
sin C
sin -
2
135
also,
a
sm
A
sni
A
=
tan A J
or
b
a
=
- a
cot
A
b
sin
B
cos
A
tan^
and
c
a
sin
sin
c
A "
i
1
C :
=
a
sin A
a
cosec
A:
iin
A' "'
hence b and c are also found.
These values of b and c being adapted to the radius r by
means of (60), become
ra ra
tan A^ sin J. '
and taking the logarithms of both sides of each, we have
log b = log r -f- log a — log tan A ;
log c = log 7' -f log a — log sin A :
from which, by means of logarithmic tables, the logarithms
of b and c, and therefore b and c themselves, may be found.
222. In the expressions for the logarithms of b and c
just found, the radius r has been introduced, because the natural
sines, cosines, &c. being all calculated to the radius 1, the
logarithms of many of them would of course be negative or
decimals ; and to avoid this, the radius used in the tables of
logarithms as has been observed in (loO) is generally supposed
to be ten thousand millions, and consequently its logarithm
to be 10.
Hence, therefore the equations above given become
log 6=10+ log a — log tan A ;
log c = 10 + log a — log sin A .
To illustrate what has just been said, let us suppose
a = 4, and A = 53^ l' 54'';
.-. B = 90^-53' f 54" = 36^ 52' 6":
136
also Jog ^» = 10 + log 4 - log tan 53^ 7' o4"
= 10 + 0.60206 - 10.12494
= 0.47712
= log 3 ;
therefore 6 = 3:
and log c = 10 + log 4 - log sin 53^ 7' 54''
= 10 + 0.60206 - 9.90309
= 0.69897
= log 5 ;
therefore c = 5.
223. The quantities, 10 — log sin tI, 10 — log tan J, &c.
are called the Arithmetic Complements of log sin A, log tan j4j
&c. and it is manifest that if we denote these complements
by colog sin Ay colog tan A, &c. we shall from (222) have
log 6= log a + colog tan A;
log c = log a -f- colog sin A :
and it may here be further observed that to obtain the arithmetic
complement of a logarithm, it is necessary merely to subtract
the first digit to the right-hand from 10, and all the rest from 9
in succession.
224. Ex. 2. Given the side c and the adjacent angle A,
to find the rest. ^
Here, we have B = 90^ — A, which is therefore known ;
c sin C 1 . .
and - = -: — - = —, , or a = c sm A ;
a sm A sm A
c sin C 1 1 .
also 7 = -: — ^ = ~ — ^ = ~ 7, or 6 = c cos ^;
b sm i> sm Ji cos A
whence a and b are known.
137
Adapting the expressions for a and b to the radius r, and
taking the logarithms of both sides as before, we have
sin A cos A
a = c , 6 = c ;
r r
.', log a = log c + log sin A ^ log r = log c + log sin A — 10
= log c — ( 10 — log sin A)= log c — colog sin A ;
log b = log c + log cos A — log r = log c -{- log cos A— \0
= log c — ( 10 ~ log cos A) = log c — colog cos A :
and from these equations, the logarithms of a and b, and thence
a and 6 themselves, are found.
Case II, in which two sides are given.
225. Using the same notation as before, we have (Euc. I.
47. )j c^ = a^ + b^y whence if any two of the quantities a, b^ c
be given, the remaining one is found :
a sin A sin A
also, - = - — — = ; = tan A = cot B ;
b sm B cos A
a sin A sin A .
- = ~ = = sm A = cos B ;
c sin C . TT
= sin B = cos A ;
from which, if any two of the quantities a, 6, c be given, the
angles of the triangle will be found.
226. Ex. 1. Given the sides a and 6, to find the rest.
sin ^
2
b sin jB
c sin C
sin B
sm ^
2
Here, c = t^a^ + b^ is found ;
a sin J. . 1
also, 7 = - — =: = tan A
b sin B tan B
138
,\ tan A — T 3 and tan ij = - ;
b a
and hence A and B are found.
Adapting the expressions above deduced to the radius r, and
taking the logarithms of both sides of the equations, we shall
have
A ^ T^ ^
tan A = r - J tan i5 = r - ,
0 a
log tan A = log r + log a — log b
= (10 — log b) + log a = colog b -\-\og a;
and log tan B = log r + log 6 — log a
= ( 10 — log a) -j- log b = colog a + log b ;
and thus, by means of the tables, the logarithms of tan A and
tan B, and therefore A and B themselves, are found.
227. If the values of a and b be expressed in numbers, we
have only to add together their squares, and by extracting the
square root, to obtain the value of c ; but if these quantities
involve trigonometrical functions of angles, the value of c may
be adapted to logarithmic computation by the following pro-
cess :
smce c
= Va^ + 6"- = a \/l + ^;
assume a subsidiary angle 6 such that tan 0 = — ;
a
therefore c = a ^Z 1 + tan^ 6 = a sec 0;
jp b CL seo'v
and to the radius r we have tan ^= — , c = ■ — ;
a r
whence log tan ^ = logr + log /; — log a = 10+ log 6 — log a,
from which 0 is found :
139
also log c = log a + log sec 0 — log r — log a + log sec 0 — 10,
from which the value of c is obtained ;
or if the logarithmic secants be not found in the tables,
log c=log a + 20 — log cos ^—10, by (lol)
= 10 + log a — log cos Of
from which the value of c is readily determined.
228. Ex. 2. Given the sides a and c, to find the rest.
In this we have, b = \J (? — c^ ^ which is found;
a sin A .
and by (lo8), - = -; — — = sni A = cos B ;
c sni C
whence A and JB are determined.
As in (226), we have to the radius r, sin A = r - = cos B,
c
.*. log sin ^=log ?' + log a — log c= 10 + log a— logc = log cos B,
from which, by the tables, A and B are found :
also, as in (227), since b = ^ c — (^ — c y 1 — — ;
assume
cos ^ = - , then h — c sj \ — cos'"^ 0 = c sin d ;
^ ?vz c sm 0
and to the radius r we have cos \3 = — , and b=- :
c r
hence log cos d — log v + log a — log c = 10 + log a — log c,
which gives the value of Q ;
and log 6 = log c -f log sin 0 — log r = log c + log sin 0—10,
from which b is determined.
140
Solution of oblique-angled Triangles.
Case I, in which one side and two angles are given.
229. In the triangle, let a, b, c; A, B, C be the sides and
opposite angles respectively; then since A-}- B-\'C = ir, (Euc. 1 .
32), if any two of the angles A, B, C be given, the remaining
one is found :
a sin A a sin A b sin B
also, - = - — ~ , - = - — ~, - = - — - ;
o sm B c sm L c sin C
from which equations it is manifest that if any one of the quanti-
ties a, bj c, and any two of the quantities A, B, C be given, the
rest may be found.
230. Ex. Given the side a and the angles A^ B, to find
the rest.
Since A -jr B -Jr C = w, we have C = tt - ( A + jB), which is
found ;
a sin A sin B
also, - = -: — -, .'. h = a ~ -;
o sm i) sm A
a _ sm A __ sin A ^ _^ sin (^ + B)
c sin C sin (A -f J5) ' sin A '
and thence b and c are also determined.
The values of b and c just found being already adapted to
any radius, we have immediately,
log b = log a + log sin B — log sin A ;
log f = log a + log sin ( J. -|- J5) — log sin ^ ;
therefore, by means of the tables, the logarithms of b and c, and
thence b and c themselves are obtained.
Case II, in which two sides and the angle opposite one of
them J are given.
23 1 . Retaining the notation of (229), we have
« _ sin ^ a sin A b sin B
b sin B' c ""sin C ' c ~ sin C '
141
and from these equations, if two of the quantities a, b, c and a
corresponding one of the quantities A, B, C be given, all the rest
may be found.
232. Ex. Given the sides a, b and the angle Aj to find
the rest.
a sin A . . „ b
Here, - = - — 7, , . . sm b = - sm A,
o sm Jb a
from which B is found^ and hence C = tt — (A + jB) is determined :
a sin A sin C
also, - = -: — -:, , .'. c = a - — - ,
c sm C sm A
which is therefore found.
The side c as here found involves the value of C which is
not one of the quantities given, though it has been determined in
the previous part of the solution. The same side may however
be found in terms of a, b and A only :
for, cos A = , by (105),
206'
.*. 26c cos A = &^ + c^ — a^, and c^ — 9.bc cos A=a^ — b'.
from which c = b cos A ± sj c? — Jf sin^ A.
The equations sm o = - sm A. and c ■= a — — - ,
^ a sm A
are already adapted to an v i adius ;
.*. log sin B = log b — log a + log sin A ;
and log c = log a -f log sin C — log sin Ay
which_, by the tables, give the values of B and c.
The equation c = ^ cos A + ^ cT — b^ sin^ A, is not much
used, owing to the difficulty of adapting it to logarithmic compu-
tation, and is indeed rendered almost unnecessary by the facility
with which the angles B and C are determined.
142
233. Cor. In the last example we have seen that sin B = -
a
sin A ; and because the sine of an angle and the sine of its sup-
plement (20) are equal, we are left in doubt whether the angle B
should be acute or obtuse. If however the side b adjacent to
the given angle A be less than the side a which is opposite to it,
it follows (Euc. 1. 18) that the angle B is less than the angle Aj
and therefore the ambiguity is in this case removed. But if h be
greater than a, the case remains ambiguous, as is also easily
shewn by geometrical construction.
For, at the point A in the indefinite straight line AD make
the angle CAD equal to the given angle A; take AC equal to
the side b, and with centre C and radius equal to the side a,
describe a circular arc, which, since a is less than b, will cut AD
in two points B, B on the same side of A : therefore each of the
triangles ABC possesses the same data, and consequently each of
the required parts admits of two different values.
The same construction shews that if a be greater than b^
there is no ambiguity, the intersections B, B being then on
opposite sides of A.
Case IIIj in zohicJi two sides and the angle included by them,
234. The same notation remaining, we have from (159)
fA + B\ C
tan I ) cot —
a-\-b _^ \ 2 J £
rT^ " /A-Bx" /T - B\ '
143
/A+C\ B
tan ( I cot —
a -\- c \ 2 / 2
a — c /A — C\ /A — C
tan
J^4-C\ A
tan I I cot —
b + c
h - c /B-C\ /B^ C\'
tan
(^) -r-i^)
from which, if any one of the angles A, Bj C and the two
sides containing it be given, the difference of the two remaining
angles is found : also the sum of the same angles being the
supplement of the given one is known, whence the angles them-
selves are found : and the remaining side is then determined
from (157).
235. Ex. Given the sides a, b and the included angle C,
to find the rest.
c
, cot --
XT U "^ + ^ ^
Here we have
a-b /A-B
tan I
/A-B\ /a-h\ C D
■'■ '"" {--J-) = to) ""'i =''"¥' ^"''P"^^'
from which ( ) is found = — ; then we have
\ 2 / 2
A-\-B TT C ,A-B D
= — , and = — :
2 2 2 2 2
whence, by addition and subtraction, are obtained
2 V 2 / 2 V 2 /'
which therefore both become known :
144
- c sin C sin C sin C
also, - = — — -r =
a sin ^ . /TT C-Dx /C-D\'
sill I \ cos I I
a sin C
(^")
, which becomes known.
cos
The value of c may also be found directly, without
previously determining the angles A and B, for in (167) it has
been proved that
c = ^a^-{-b^-C :'^ cos C.
The equation, tan [—^) = (^^:p^j cot -^ ,
is already adapted to any radius, and if a and b be numerical
magnitudes, it is also adapted to logarithmic computation ; thus,
log tan ( — —j = log {a - b) — log (a + b) -{- log cot — ,
from which by the tables, / J becomes known.
If however a and b be not numbers, but involve trigono-
metrical functions of angles, assume j = tan 0,
V 2 / \a + bj 2 }a
,'. tan
/tan 0- 1\ C ^_ 0^ C
= I TTTT I cot — = tan (t^~ 45 ) cot -- to the radius 1 ;
Vtan^+1/ 2 2
, ,. . a tan 0
.*. to the radius r we have - = ,
b r
/A-B\ tan (0-45^) C
and tan I I = cot — :
{^)-
145
hence log tan 0 = log r -f- log « — log ^=10 + log a — log 6,
from which 0 is found ;
and log tan ( — — ) -= log tan (0— 45^) + log cot log r
C
= log tan {0 ~ 45^) + log cot - - 10,
from which ( ] is determined:
V 2 /
again, log c = log a + log sin C — log cos ( J ,
from which, by the tables, c becomes known.
The other expression for c is also easily adapted to lo-
garithmic computation, for
c = ^a"-\-b^'-Q.ab cos C = >>/ (a - 6)" + 2 « 6 ( 1 — cos C)
= \/(a -hY + 4.ab sin^ ^ = {a-h) \/ x + -^^~^^ :
let .*. the subsidiary angle 0 be such that tan^ 0 = ^ -r> sin^ — ,
{a—oy 2
whence c = {a-' h) ^J 1 + tan" 0 = (« — h) sec 0 ;
and to the radius r, we have
o ^ 4 a ^ . o C , ■, , ^ sec 0
tan' 9 = —7 sm' — , and c — (a—o) ,
{a — by 2 ?'
therefore taking the logarithms of both sides, we get
log tan 6 = -^ [log 4a b -^ 2 log sin — — 2 log (a — ^)},
which determines the value of 0 ;
and log c = log (« — b) + log sec ^ — log r
= log {a — b) 4- log sec ^ — 10,
= log (fl — 6) + 10 •— log cos Oy
from which c is found.
T
146
236. In practice, when two sides AC^ AB oi a triangle
and the included angle A are given, a perpendicular CD is
generally let fall upon one of the given sides AB from the
opposite angle C ; the triangle is thus divided into two right-
angled triangles ACD^ BCD, of which the sides and angles are
easily found by the methods already laid down ; and hence the
remaining side and angles of the proposed triangle may be
determined.
Case IV, in which the three sides are given.
237. The notation of the preceding articles remaining, we
have seen by (173), that i( %S = a + b + c,
sin JL = -^ ^S(S-a)(S-- b) {S - c);
DC
sin B = — JS{S-'a){S-h)(S -c);
ac ^
sin C=— J^S{S-a)(S-b){S''c)
also, in (175), it has been shewn that
. A /(S-b)(S^)
^"? = V ^^ ;
sm
2 ac
. C ./(S-a)(S'-b)
sm - = V — -T
2 ab
and cos
2
147
2 ^ ac
C ^/S(S-c)
s — = V ; :
Q ^ ab
A ^ ^(S-b)(S~c)
and .*. tan „ ^ ^ .
2 ^ S(S-a)
^"^2 ^ SiS^b)
C . /{S ~a){S- b)
tan = V ^-Tq T" •
2 S (S -- c)
and from any of these sets of equations the values of the angles
A, B, C may be determined.
238. Ex. Given the sides a, b, c, to find the angle A.
From the first set of equations given in the last article,
we have
sin A = ^ ^S{S-a)iS^b)(S-^c),
be
which being adapted to the radius r by means of (59) becomes
sin A = ^ ^S{S-a)(S''b)iS'^c),
be
and in logarithms gives
log sin 4 = J 0 4- log 2 — log 6 — log c
+ f {logS + log(5-«)+log(S-6) + log(5-c)},
from which the value of A is found.
This solution might at first sight seem to be all that
is necessary, and sufficient for the determination of an angle
148
in all cases ; but upon examination of the tables of logarithmic
sines, and from (139), it appears that when an angle becomes
nearly equal to 90^, its logarithmic sine does not differ by any
significant figure from those of several other angles nearly equal
to it, though this does not happen in other cases. If therefore,
from the relations of the sides of the triangle, we perceive that
no one of its angles is nearly equal to 90^, tliis method of solution
will not be liable to objection^ owing to any imperfection in
the construction of the trigonometrical and logarithmic tables,
and may therefore be applied without apprehension of great
inaccuracy in the result.
239. The second set of equations mentioned in (237), gives
^ y/'{S--b)(S-c)
sin — = Y , to the radius 1,
... A ./(S-J^HS-c)
and . . sm — = r y , to the radms r ;
2 he
.\ log sin- = 10+ - {log(6'-6) + Iog(^-c)-log^»-~logcl,
from which — , and therefore A, may be found.
To determine in what cases it may be expedient to make
use of this solution.
Since sin a' — sin a — 2 cos ( — ^ \ sin / \ , by (()7),
sin a — sin a (a + a\ V ^ /
we have -. == cos I 1 ', 3
a - a V 2 / /g - o-\
which, if a and a be very nearly equal to one another, becomes
sin a — sin a c (ox c^\ .
-, = cos a, nearly, as appears irom (2io;.
149
that is, corresponding to a given change in the angle, the change
of the sine will be nearly proportional to the cosine, or the greater
as the angle is the less ; and hence it follows that this method
of solution is to be preferred when the angle required is acute.
From this it is also manifest why the changes in the sines of
angles nearly equal to 90° are very small, the cosines being then
nearly equal to 0; and also, that this second method of solution
may be employed when the angle of the triangle which is
required is nearly equal to a right angle.
240. From the third set of equations enumerated in (237),
we have
- = v-^'^' ~ "*
cos — = v ; -, to the radms 1,
2 be
A /s{S-a) ^ . ,.
and .*. cos — = ?• V/ : > to the radius r :
£ ^ be
A 1
hence log cos — = 10 + - { log «S -(- log {S — a) — log ^ — lo
from which — , and therefore A. becomes known.
To determine the eligibility of this solution in anj case.
a -T a
\
6 "- ]
o" c
m\
Since cos a "^ cos a = 2 sin y—^ ) si" ( — ^) > ^y (^
cos d '^ cos a . (a -r o.\ \ ^ J
we have ^_^ = s.n (^-^-j , _ ,
cos a '^ cos a . ,
which becomes -, =sm a, nearly,
a — a
when a and a are nearly equal to one another: hence the
change in the cosine corresponding to a given change m the angle
varies nearly as the sine of the angle, and will therefore be the
150
greatest when that angle is nearly equal to 90°. This solution
will therefore be best adapted to those cases in which the angle
considerably exceeds a right angle, and consequently when its
half differs not greatly from a right angle.
Hence also^ this solution and the next preceding one will
respectively have the advantage, according as the angle sought
is obtuse or acute.
241. The fourth set of equations given in (237)j has
A . /{S ~b){S- c) . ^. ^
2 o (o — a)
/ . . ^ K/(S-h)(S-c)
and . . tan — = r V — ~—^ r — , to the radius r ;
2 o {o — a)
A
.*. log tan ^
= 10+ I {log {S - ^»)4-log (S-c)-log 6^-log (^~«)},
and hence — and A are found.
2
As in the two preceding methods of solution,
sin a sin a
snice tan a - tan a =
cos a cos a
sin a cos a — cos a sin a sin (a — a)
cos a cos a cos a cos a
we have
tan a — tan a 1 sin (a' — a)
a — a cos a cos a (a' -- a)
1
cos a
sec^ Uj nearly, if a be very nearly equal to a :
151
hence, if the change in the angle be given, the change in its
tangent will be proportional to the square of its secant^ and
therefore very great when that angle is nearly equal to 90 . This
method of solution therefore,, owing to the want of accuracy in
the proportional parts, will be least eligible when the angle
required considerably exceeds a right angle, but in other cases
may generally be used with advantage.
242. This case, like the preceding, is in practice generally
solved by drawing a perpendicular from one of the angles C
upon the side which subtends it, and thus dividing the triangle
into two others ACD, BCD having each a right angle at D:
then, since by (l68), the base AB : the sum of the sides
(AC+ BC) :: the difference of the sides {AC - BC) : the dif-
ference or sum of the segments of the base {AD^ DB) made
by a perpendicular let fall upon it from the opposite angle,
according as it falls within or without the triangle: therefore
the difference or sum of the segments of the base may be found,
and their sum or difference AB being given, the segments AD
and DB become known, and consequently two parts in each of
the right-angled triangles are determined, from which all the rest
are immediately derived.
Meiisuration of Heights, Distances, 8sf.
243. The solutions of triangles given in this chapter, will
in all cases enable us to determine the relations between their
different parts, and if the number of quantities which are given
be sufficient, to find the rest ; and the mensuration of heights,
distances, &c. is merely the application of these solutions to par-
ticular instances, together with the use of certain instruments, by
which the lengths of lines and the measures of angles are ascer-
152
tained. Gunter's chain of 4 poles or ^2 yards, or common
tape of 50 or 100 feet is used to measure distances ; a graduated
quadrant furnished with a plumb-line to measure angles of eleva-
tion or depression ; a theodolite, to measure horizontal angles,
and a sextant or quadrant, such as are oblique. This ap-
plication of trigonometry involves no principles but what have
been aheady explained, and no general rules can be laid down,
except that such lines must be measured and such angles observed,
as may most easily and conveniently lead to the determination of
those which are required. The following examples will be suf-
ficient to make this part of the subject understood.
244. Ex. 1 . To find the height of an accessible object stand-
iiig upon a horizontal plane.
Let AB be the object standing upon the horizontal plane
JBD; measure off BC = a feet, and at C let the angle ACB be
observed : then
AB sin C sin C ^
= -: — T = p, = tan C,
BL sm A cos G
.*. AB = BC tan C = a tan C, which is the height of the
object •
, ^C sin i^ 1
a»« --- = -, — - = = sec C,
i)C sm A cos C
.*. AC = BC sec C = a sec C, which is the distance of its
summit from the place of observation.
If the observed angle were C instead of C, and h, K the
heights deduced from these two angles, we have
K = a tan C', and h = a tan C,
w ; , ^,, .,, (sin O sin C)
/. h — h = a (tan C —tan C) = a < — , — -J.
(cos C cos C)
{sin C cos C -- cos C sin C) sin (C — C)
77, 7^ ( =« w, n,
cos C cos C } cos C cosC
153
h'—h a sin (C — C) a
and .*. —, — 7, = 7^7 t; — jTr 7^ — = 2-7,9 by (187),
C — C cos C cos C C — C cos C
(if C' be very nearly equal to C)
h h Qh
tan C cos'"^ C sin C cos C sin 2 C
hence, if a small error of given magnitude be made in the ob-
servation of the angle, the error in the computed height will be
inversely as the sine of double that angle^ and therefore the least
when that angle is 45° ; which consequently points out the place
in which it is desirable that the observation should be made.
245. Ex. 2. To Jind the height of an inaccessible object
above a hoi'izontal plane.
Let the point A denote the place of the object : draw AB
perpendicular to the horizontal line jBD, and in this line take
two positions C, D distant a feet from each other, at which ob-
serve the angles ACB, ADB equal to C and JD respectively;
then zDAC= L BCA - z.BDA = C-D:
, AC sin ADC sin D
and
CD sin CAD sin (C-D)'
CD sin D a sin D
AC =
sin (C-jD) sin (C~ D)'
AB sin ACB . ^
also, —-7 = — ^ = sm C,
AC sm ABC
.-. AB = AC sm C = . .^ ^r- ,
sm (C — D)
which is the height required :
Likewise
. ^ a sin D . ^ r* a sin C
AC 5= — k;? and AD =
sin (C-D)' sin(C-D)'
154
which are the distances of the object from the places of observ-
ation.
This example determines the distances of an inaccessible
object from two stations in the same plane with it.
246. Ex. 3. To Jifid the height of an accessible object
standing upon an inclined plane.
Let AB he the object upon the inclined plane, in which
take any two positions C, D in a line with it; suppose BC==a,
CD = b, and let the angles ACB, ADB be observed, and called
C and D : then
AC sin D ^ ,^ CD sin D b sin D
and ,\ AC
CD sin (C-D)' sin(C-D) sin(C~D)'
also, in the triangle ACB, we have by (234),
C
cot —
AC + CB 2
AC - CB
tan
(^)
/^-^\ AC-CB C
6 sin D-a sin (C-D) C E
= , — : — 7^: : — 77; 7^ cot — = tan — , suppose,
6 sm jD+fl sm (C — jD) 2 2 ' ^'
B-A E , B + A IT C
,'. = — , also = ^ -_ —
2 2 2 2 2
whence B = ( | and A = ^ — ( |
2\2/ 2V2/
AB __ sin C sin C
" IC" sin B "" /C~£\'
cos
im
2indAB= AC
155
sin C h sin C sin D
cos ( ) sin(C— jD)cos( \
which is the height required.
247. Ex. 4. Tojind the height of an object standing upon
a hill contiguous to a horizontal plane.
Let AB be the object^ C, D two stations on the plane in a
line with it ; produce AB to meet the plane in jE ; at C observe
the angles ACE, BCE, and let them be C and C respectively;
suppose CD = a, and let the angles ADE, BDE be called D
and Z)' : then
AC sin D , ^/n C^ sii^ ^ a sin D
, and ,'. AC =
CD sin (C~D)' ' sin (C-D) sin (C-D)'
BC sin D' CD sin D' a sin D'
also, 7^-7:: = -: — -p;-^ — Tv", and.*.i5C=
Ci) sin{C-D')' sin{C-D') sin {C-DY
hence in the triangle ACB we have found the two sides AC,
BC and the included angle ACB= C— C^ from which AB the
height of the object may be determined as in the last example.
Precisely in the same manner the distance between two
objects at A and B which are inaccessible to the observer at C
and D, and to each other, may be ascertained.
248. Ex. 5. From the top of an eminence of given height
the angles of depression of two objects on the horizon in the
same plane ivith it are observed, to find the distance between
them.
Let AB be the given eminence and = fl^ C, D the ob-
jects in the horizon: draw AE parallel to the horizon and
156
let the angles of depression EAC, EAD of the objects be C
and D ;
, AC sin ABC
then -^—r =
AB sin JCD sm E AC sin C'
AB a
•. AC
sin C sin C
CD _ sin CAD _ sin {C-D)
^" C^ ~ sin ^ DC"" sin D '
. ^7^ _ 6t sin (C - D)
sm C sni D
which is the distance between the objects.
Hence also the distances of the objects from the point of
observation are — — - and -: — - .
sm C sm D
249. Ex. 6. To determine the height of an object standing
upon a horizontal plane, by means of observations made at the
top of another object of given height on the same plane.
Let the height of the given object AB be a; observe the
angles of depression or elevation of the bottom and top of the
other CD, and let these be /3, a respectively ; then AE being
drawn horizontal to meet CD or CD produced in E, we have
aADB == DAE = /3,
= ~ 7c, whence AIJ =
AB sm A DB sin /3 sin /3
CD sin CJD sin (/3 ± a)
and
AD sin ACD
sm
i„(| + a)
157
^^ _ AD sin (/3±a) _ a sin (/3±a)
/TT _ \ COS a sni p
sm (j + «)
which is the height required.
Hence also the distances of the bottom and top of the
object CD from the place of observation A may be found :
for AD = — — - , as above,
sm p
, AC cos B ,„ AD cos B a cos B
and — — - = , /. AC = ~
AD cos a cos a cos a sin /3
250. Ex. 7. To determine the height and distance of an
object standing on the horizon, from two observations of its
altitude, one made on the horizon, and the other at a given
height above it.
Let A and B be the two points of observation in the
same vertical line, at which the angles of elevation are a,
)3 respectively; AB = a, CD the object whose height and
distance are required; draw A IE, parallel to the horizon, meeting
it in Ey then
CD = BD tan /3, and CE = AE tan a,
,'. AB=^ED=:CD- CE=:BD (tan /3-tan a),
, _,_ BA a cos a cos B
and BD = ^ = — : — -^ -~,
tan p — tan a sni (p — a)
which is the required distance :
^x^ T^T^ n « cos a cos /3 « cos a sin /3
also, CD = BD tan /3 = —7—73 ~ tan /3 = — r-— 5 ^,
sm (p — a) sm (p — a)
which is the height required.
251. Ex. 8. Given the distances between three stations in
a straight line with an object standing upon a horizontal plane,
158
TT
and the angles of its elevation at these points 0, -^ — 0, and 0,6
in order, 0 being unknown^ to find its height.
Let ^ J5 be the object, C, D, E the stations in a straight
line with it, CD = a, DE = b : then
zAEB = e, /.ADB:='^ -G, and^ACB^^O;
.-. ^EAC= ^ACB- zAEB = 2e-e^0== a AEC,
and .-. AC=CE', also, zE^i)= /. ADB- /. AEB
2 2 '
hence, in the triangle ADE, we hr^ve
DA sin AED sin 0
DE sm DAE
SHl
G-")
s'ln 0 _ . b sin 0
DA =
cos 20 ' ' * cos %0
again_, in the triangle ACD, we have
sin (- -e)
AC _ sin ADC _ ' \2 J _ cos 0
AD ^ sin ACD "" sin 20 ~ sin 2 0
, ^ . _^ cos 0 6 sin 0 cos 0 b
/. AC=^D . ^^ = -r— -7^ -r = — r = EC=a-i-b,
sin 2 0 sin 20 cos 20 2 cos 20 '
whence cos 20 = — ; — rrrj ^nd is .'. found;
2 (a -f- o)
and AB = AC sin 20 = (« + /;) V 1 ^——,
4 (a + 6)"
= — , Uic required height.
159
252. Ex. 9* Given the distances between three objects,
and the angles contained by the lines drawn from each of them
to a certain station in the same plane with them, to find its
distance from each.
Let Ay B, C be the objects, D the station, at which let the
angular distances of ^ and C^ C and JB be a, ^ respectively ;
then calling the sides of the triangle a, b, c as before, if the
angles DAC, DEC be 0 and 6',
sin a b sin 0' DC
we shall have -: — ;: = -p^-p^, and -: — -pr = ,
sin 6 DC sm p a
. s"^ cc sin 0' b . • /3' 7 • /3 • ^
• • ~ — 7r~- — 7^ = •" * or a sm a sni t^ = 6 sin p sin ^ :
sm 6 sin p a
but the angles of the quadrilateral A DEC being together equal
to four right-anglesj we have
0'=27r - (a + /3+C) — ^ = 0-^, suppose,
.'. b sin /3 sin 9 = a sin a sin {(p — 6)
■= a sin a (sin (p cos 6 — cos (p sin ^),
and b sin j3 tan 0 = c sin a (sin 0 — cos 0 tan 0)^
a sin a sin d>
whence tan d — — : ~7~7~i — '• — 3 *
a sin a cos <p-rb sin p
and therefore the angles CAD, CED become known, and
consequently the distances ADj ED and CD.
This problem may be solved geometrically, by describing
upon AC and EC two segments of circles containing angles
equal to a, /3 respectively, and intersecting each other in the
point D.
253. Ex. 10. To determine the height and distance of an
object from its observed equal elevations at two points whose
distance from each other is given, and its elevation at the
middle point between them.
Let AE hQ the object, C, D the two points whose distance
from each other is a, and E the middle point between them :
160
a the elevations at C and D, and /3 the elevation at E: join
BC, BD, BE, AC, AD, AE; then
tana BE rtrv
since BCE is manifestly right angled at E, because CBD is
isosceles ;
a sin BCE a tan a
BE=:CEt^n BCE =
2 cos BCE 2 ^tan-/3-tan^ a
whence AB = BE tan /3
a tan a tan /3 a sin a sin /3
"" 5 ^tan^/B-tan'"^ "" 2 ^sin(/3-a) sin (p + a) *
which is the height of the object :
^ fl cos a sin B ^^
also, BC = AB cot a = . ^ ^ -x = ^I> >
2 ^sin(j3 - a) sin (/3 + a)
^ „ . ^. ^ ^ sin a cos i8
and BE = AB cot /3 = ^
2 ^sin (j3 - a) sin (/3 + a) '
which are the horizontal distances of the object from the places
of observation.
254. Ex. 11. Given the elevations of an object above
a horizontal plane at three points at given distances from one
another in the same straight line, to find its height.
Let AB he the object standing at the point B on the
horizontal plane ; and at the three stations C, D, E in a straight
line, let the observed elevations be a, /3, y^ and suppose the
height AB to be represented by h: then
BC = h cot a, BD = h cot /3 and BE = h cot y :
draw BE perpendicular to CE, then from (l66), if CD and DE
be called a and b, we shall have
h^ coV a = a- + h:' cot^ (3 + 2aDF,
and h^ cot" 7 = // + /r cot^ ^— 2/; DF;
161
hence bh^ cof a = a b-{-bh^ cot^ fi + 2a b DF,
and ah^ cofy — ab^ + ah^ cor/3 — 2a6 DF,
.'.by addition_, we get
{a cot^ y + b cot^ a) h^ = a'b-\-ab' + (a + b) cot' fth\
and consequently,
{a cot^ y — {a-\'b) coV fi -{■ b cot" a] h^ = ab {a-\-b);
a/ ab (a + b)
whence h = y 5 — 5-75 — ; 2~
a cot 7 — (a + 0) cot p + b cot a
the required height :
I „.^ ; 4 / « 6 (a + 6)
also, xJC =:/iCOta = cot a v 5 — q-?^ — ; 5~ >
« cot 7 - (a + 6) cot^ ^ + 6 cot^ a
which is the horizontal distance of the object from one of the
stations.
If we suppose b = a, and 7 = a, the perpendicular height
and horizontal distance of the object will be the same as
determined in the last example.
255. Ex. 12. Four objects situated at unequal but given
distances in the same straight line, appear to a spectator in the
same plane zcith them to be at equal distances from each other,
it is required to determine his position.
Let Ay B, C, D be the objects, E the place of the eye;
draw EF perpendicular to DA produced if necessary, and
suppose AB = a, BC = b, CD = c;
also let lAEB= ^BEC= ^CED = (j), and zEAF=e:
sin 3(p _ sin AED __ AD
sin EAD ~ sin EAD "" ED'
sin EAD _ sin EAD __ EB
""""^ sin 0 "" sin AEB " ~AB'
X
162
sin 3 0 . 2 , AD EB
whence — : — p- or 3 — 4 sni 0 = ■— — — -—
sin (f) ^ AB ED
AB CD ac
and /. sin 0 = A/ ,
consequently 0 is determined :
again,
EF EF
sin 0 = sin EJF = -— , and sin (0 -■ 20) = sin ECF = -~ ,
AJb EC
sin 0 CE BC ^ ^ ^, ^ b
sin(y— -20) -4il/ 7li5 a
and a sin 0 = 6 sin (^ — 20) = 6 (sin ^ cos 20 — cos ^ sin 20),
6 sin 20
whence tan 0 = ; , and .*. 6 is found:
o cos 2(p — a
AE sin EBA sin (0-0) , , ^ « sin (0 - 0)
•'• ITB = "^ — TWB = • ^ ^ and ^E = . , ^ ;
il5 sm ^Ej« sin 0 sm 0
. 1 4 77 ^ I? /3 « sin (0 - 0) cos 0
whence also At = AE cos U = : — '- ,
sm 0
^ rj? AT? ■ D ^ ^"^ (0-0) sin 0
and EF=AE sm 0 — -. — f ,
sm 0
and thus the position of E is completely determined.
It is obvious that if (a-\-b-^c)b be greater than Sac, the
problem is impossible: also, if we suppose a = b = c, we shall
have sin 0 = 0, and AF and EF indefinitely great ; that is_,
equidistant objects in the same straight line appear to be so to
a spectator indefinitely distant.
256. It would be no difficult matter to extend the number
of examples on this subject, but from what has been already
163
done, the method of proceeding in other cases cannot but be
manifest, though few or no general rules have been given.
The Trigonometrical Survey of a Country or large tract of
land is conducted in a similar manner^ by selecting conspicuous
places which may be seen from one another as stations at which
angles are observed, for instance, the mutual bearings and
directions of such objects as it is intended to include ; and the
distances between two or more of these stations being found by
actual measurement, each of the other parts of the triangles
employed may be determined : this may also be verified by
taking the measure of a different line and proceeding with it as
with the first, so that if the whole be correct, the conclusions
will necessaiily be the same by each process.
In Navigation, the computations used in what is called
Plane Sailing are nothing more than the solutions of right-
angled triangles, the hypothenuse being termed the Distance, the
other two sides the Difference of Latitude and Departure, and
one of the acute angles the Course: and if any two of these
quantities be given, the remaining two may be found by the
methods already explained.
164
CHAP. VI.
On Algebraical Expressions fo?- the sines j cosines, Sac of arcs,
and their sums, differences, multiples, Sfc. On the general
relations between the sine, cosine, S^c. of an arc, and those
of its multiples and submultiples. On the general relations
between the poivers of the sine, cosine, <S)C. of an arc, and
the sines, cosines, S^c. of its multiples. On general ex-
pressions for the sine, cosine, S^c. of an arc in terms of
the arc, and impossible exponential quantities,
257. 2o express the sine and cosine of an, arc by means of
algebraical binomials.
Since, by article (25),
l=cosM+sin^J.=(cosA+ ^ — 1 sin -4) {cosA — ^ — lsmA),
, . 1
we have cos A— j^ — \ sin A = / : 7 •
^ cos ^ + ^ — 1 sm A
let then cos A + ^ — 1 sin A=^x, .', cos A— ^ - \smA= - ,
and by addition and subtraction we get
2 cos A = x -\r - , or cos A — - (:r + -|: and
X Q\x/
2 x/ - 1 sin A=x— - , or sin ^ = 7=- ( x | .
X 2^-1 V xJ
By means of the relations established in the first chapter,
the versed sine, chord, &c. of A might be expressed in terms
of X, if it were necessary.
165
1258. To express the sines and cosines of the sum and
difference of tzvo arcs by means of algebraical binomials.
Let cos A -\- ^ —• \ s'm A=^x, .'. cos A — a^ — I sin A= - ;
^ X
and cos B -\- ^ — 1 sin B =y, .*. cos B — yj — \ sin £ = - :
hence cos A = -- \ x -\ — \ . sin ^ = 7-=- ( x | ;
2 V x)' '^J-\ V x)'
and cos B = - ( y + - | , sin jB = - — 7== \y ) :
^K y) 2^-1 V^ y)
.• . cos (jI + 5) = cos A COS B— %\w A sin B
COS ( A — 5) = cos A COS B + sin A sin B
2 V 0^/ 2 V'^ y) 2 V^^ ^ ^'/ 2 ^ - 1 V-^ y)
4 1 y X j 2 (j/ a j '
sin (^ + B) = sin J. cos B + cos ^ sin B
7^H-41 =
7= U-y r;
sin (^1 — i>) = sill A cos B — cos ^4 sin B
166
259. Cor. 1. If C be another arc, and cos C+ >^ — l sin C
be assumed equal to z, we shall, by a process exactly similar,
have
and sin (^ + JB + C) = > \xyz v;
2 V — 1 I ^ j/ 2 )
and generally, if there be any number of arcs whatever,
cos (A + -B + C -{- &c.) = -i^xyz &c. +
xyz
&c.j'
in(iH-i^ + C + &c.) =
sm
_L_|
lj/2 &C.
xyz
&cj
260. Cor. 2. By multiphcation and addition, we obtain
from the expressions just deduced,
cos(il + jB+C+&c.)-i-A/^sin(^ + ii4-C + &c.) = j?3/2&c.
= (cos A H- ^ — 1 sin ^) (cos B + >^/ - 1 sin B)
(cos C + >/°^ sin C) &c.
and by multiplication and subtraction,
cos (1 + B + C + &C.) - V^ sin (1 4- jB 4- C + &c J
;ry2:&c.
(cos il + /^ - 1 sin 1) (cos ^+ ^— 1 sin B)
1
(cos C 'f V — 1 .''in C)
, &c.
167
= (cos A — ^— 1 sin A) (cos B ~ ^ — 1 sin £)
(cos C — v~ 1 sin C) &c.
20l. The properties proved in the last article are frequently
deduced by a direct process ; thus^ by actual multiplication
we have
(cos A ± ^/ - 1 sin A) (cos B ± ^ - l sin 7^)
= (cos A cos B — sin A sin B) ± ^/ — 1 (sin A cos B + cos A sin J5)
= cos {A-hB)± >/^ sin (A + B):
again,
(cos^l + ^y — lsin^)(cos B ± ^^ — 1 sinB)(cosC± /y/ — 1 sin C)
= {cos(7i + jB) ± v^^n'sin(^ + 5)} {cosC± v^^ sin C}
= cos(^ + -S+C)± ^~^ sm{A + B+C), as before;
and by the principle that if the formula be true for the sum
of n — I arcs, it will also be true for the sum of n arcs,
we have generally
(cos A ± v"--~l sin A) (cos B ± sj — 1 sin B) &c.
= cos(A + B + &c.) ± \f-^\ sin(^ + jB + &c.)-
262. Cor. From the last article is easily deduced what
was proved in (132).
For, cos {1 + JB + C + &C.} -f >y/^ sin {1 + B + C + &C.}
= (cos A + s/ — ^ sin A) (cos JB + >>/ -— 1 sin J5)
(cos C+ ^/^ sin C)&c.
= cos7lcosJ5cosC&c. (1 + ^/— 1 tan J){\ + ^/-l tan B)
(1 + ^y - 1 tan C) &:c. = cos A cos B cos C &:c.
{14- V^ ^1 - ^2 - >s/^-~l '^3 + '^^+ V^-^o-^c.}
the notation used in article (132) being retained :
168
hence, equating the possible and impossible parts respectively
on both sides of this equation, we get
cos{J + B + C+&ic.} =cosAcosJBcosC&c. { 1 — ^Sa + S4 — &c.} ;
and
sin {^ + 5 + C + &:c.}=cos^ cos B cos C &c. {^'i-^Sa + ^s-Scc}
.-. tan (A + £+C + &c.) = p , /^ , Q — N
cos (A -{-3 + 0 + Sec.)
as before.
263. To express the sine and cosine of the multiple of an
arc hy means of algebraical binomials involving the same quau-
tities as the sine and cosine of the arc itself
Let cos A-\- ^ —l sin id = :r, .'. cos yi — ,>/ — 1 sin ^4 = - ,
X
and 2 cos A — x -\-- , 2^ — 1 sin A=x :
X ^ X
hence, cos 2 ^ = 2 cos' ^—1=2- ( x ■{ ) — 1
4 V x/
= - (x^-\--A. or 2 cos2 ^=x^ + -^;
2 \, x^J /'
J = 2 sin A cos A= 2 7-=^^- (x ] - ( x -{- - |
^^ - i \ x/ 2 \ ^ xj
= ;; — /- I x^ — -^ ) , or 2 /v/ — 1 sin 2 ^ = .r^ 5 : and so on :
2 V — 1 V ^' / ^ x^
and
sin 2
169
if then,
2 cos(//- 1) A = .r^^*' +—_-,,
and .-. 2^—1 sin (n — \) A — r"~ ^ ~ "X^rr >
we shall have
2 cos /iy4=:- {2 COS (;i - 1) J. 2 cos A — 2 sin {n — \) A 2 sin^}
= - (2 a^^^ + --) = .r^' + — , and .*. 2 ^/"^ sin //^ = ^" - - ;
2 \ X / X X
hence therefore it appears that if these formulae be true for any
one value of n, they are necessarily true for the next superior
value : also, it has been just shewn that they are true when
?i = 2, therefore they will likewise be true when ?^ is equal to
3, 4, 5, &c, ; that is^ they will be generally true.
It may here be remarked, that these two formulie are im-
mediately derivable from (259) by making A = B= C = &.c. and
therefore x=iy = z=- &c.
264. If n be an odd number, n-\-\ will be an even one,
and therefore, as we have seen by the last article, we shall
have
. cos ('^)l = . -+-!,,
X ^
and 2^ — 1 sni I 1 ^ = .r " -^^^ :
\ 2 / — 2~
170
but the formulae are equally true when n is even, and therefore
an improper traction.
Since 2cos^— = l-fcos ^4=1+- (i-f-W- ( Vx-{ y- \ ,
A y \ , /— .Ay 1
.*. 2 COS — = Va' + — /-, andSv^ - i sm — = Vx --r- :
2 Vx ^ 2 Vx
hence^ if 2 cos ( i A ^ x + ^_^ ,
^- ^
1 V >» — 1 1
. //<— 1\ , -IT 1
and .'. 2 >/ -1 sin (-~— ) ^ =
we shall, as in the last article^ have
X —
n — l )
2 COS (-^j ^=^ "^"^
. //i+l\ , ^^ 1
and 2 ^/ — I sin (—7-) ^ = :i "" -
and therefore^ as before^ the formulag will be generally true.
265. Cor. From the preceding articles of this chapter, it
may easily be proved, on the suppositions there made^ that we
shall have
2 cos {nA^mB)^xy + -^^-^; 2 cos {iiA - 7?2J5) = — +^ ;
2 J~^\ sin ill A + mB) = x^'f - ~^ ;
X y
171
these formulae are however more curious than useful.
266. The following singular property of the chords of a
circle, the discovery of which has been attributed to Vieta,
Waring, &c. may be immediately deduced from (263).
Let PA be the diameter of a circle, and let there be taken
any number of contiguous equal arcs ABf BC, CD, &c.
then, if the chords PB, PC, PD, &c. be drawn, and PB be
assumed equal to .r +- ; we shall have
1 ^^ , 1
PC = .r" + ^, PD = .r" + ~, &c. = &
.c.
tor, P^ = chd(7r — iljB) = 2sm ( )=2 cos = 2' + - ;
\2 2 / '2. X
PC = chd(7r-lC) = 2 sin f- -• ~\ = <z cos —
\2 2 / 2
= 2 cos 2 ( \ = X' + - , by (263) ;
PD = chd (tt- i4D) = 2 sm I ) = 2 cos —
\2 2 / 2
/AB\ - 1
= 2 cos 3 ( "IT ) = '^"+ — J ^"^ so on.
172
267. Demoivre's fornuilit, which are
and
(cos A ± v^ — 1 sin AT = con uA ± \/ — 1 sin n A^
(cos A ± ^ — It sin A)" = cos — yi ± \/ — 1 sin — A,
m
m
may also be proved by means of the expressions investigated in
Let X — cos A-^- sj — 1 sin A^ /. '", = cos ^4 — ^ — 1 sin ^
then we shall have
x'' - (cos A -^ ^ — 1 sin A)^ and - = (cos A - sj - 1 sin J)" :
but on the assumption above made we have shewn in (263)
that
x^ 4 =21 cos n
A, and x'' —% ^J -\ sin nA ;
therefore, by addition and subtraction, we get
1
a'"=cos 7f^-l- \J — 1 m\nA, and ~ = cos ?iJ. — \f — 1 sin?/ A;
I*
whence
(cos A + \/ — I sin ^ly = cos ?iA+ \/ - 1 sin ^iid,
and
(cos A — ^/ — 1 sin A)" = cos ^^4 — >>/ — 1 sin tiA,
Again, let nA z= mB, then (cos A ± s/ — \ sm A)"'
= cos 71 A ±^ /v/ — 1 sin ?iA = cos mB ± \/ — \ sin ^/zi^
17^
= (cos B± J - 1 sin 5)'" = (cos - A± x/~^l si" " -^V".
m
and .'. (cos A± v — 1 sm A) =cos - ^ + ^ — 1 sin — ^
m m
26s. Cor. 1. If the indices n and - be negative^ the
m
formulae will still hold good, by changing the algebraical sign
of the arc in the latter sides of the equations.
For, (cos A ± V--1 sin A)-" = . ^ j .-— -,
^ (cos ^ ± >/ — 1 sui Ay
, by (257),
Vcos J. + V — 1 sin A/
= (cos A + ^ - I sin A)" = cos «^ + V — 1 si" ^^A
= cos { — iiA) ± v—l sin { — iiA)',
and similarly of the other case.
269. Cor. 2. We may here observe that the formula? of
Demoivre contained in the last two articles are in reality only
particular cases of those which were demonstrated in (260) and
(261), and may be immediately derived from them by supposing
l = B = C = &c.
«
Also, from the truth of the formulae (cos A± \/ — I sin Af'
= cos — A + ^ — \ sin — A, if we put 2 cos A = x -r ^ ,
m ~~ ^ m X
we may prove conversely, that 2 cos — A = x H
174
n , m 1
and 2 ^ — 1 sin — ^=.r ^: and thence conclude that
m —
these formulae are general
270. 2'o express the sine and cosine of the multiple of an
arc in terms of the powers of the sine and cosine of the arc
itself.
By Demoivre's formulae proved in (267), we have
cos nA-^ ^ - 1 sin nA = (cos A -{- a^ — 1 sin A)''
= cos" 1 + w cos"-' A x/-^ sin A -71 C^^-—) cos"'^ A sin' A
- n C^^) C—A) cos"-' A J'~^\ sin^" 1 +&c. +( - 1)^ sin^' A,
by the binomial theorem :
hence, equating the impossible and possible parts of these ex-
pressions respectively, we obtain the follovt'ing equations ;
f>J -—A sin n A
n cos"-' A v^- 1 sin A-n C^-^-) (~T" ) cos"-'yl >/^ sin'^ + &c
.*. dividing both sides by ^ — ]^ we get
n— 1 \ /n — 2^
SHl
nA=n cos"-' ^ sin ^ - n C^—\ C- -\ cos""' J[ sin" J.
11 -^ \
+ &C. which will be continued to terms if // be odd
2
n
and to - terms if n be even :
2
175
n— 1
and cos jiA = cos" A — ?t ( ) cos" ^ A sin^ A -{- Sec.
n-\- 1
which will be continued to terms if n be odd, and to
- + 1 terms li // be even.
Ex. Making n equal to the numbers 2, 3, 4, 8cc. succes-
sively, we shall have
sin 2 A = 2 cos A sin A ;
cos 2 A = cos" A — sin' A :
sin 3 A = 3 cos^ A sin A - sin^ A ;
cos 3 il = cos^ A — 3 cos A sin^ A :
sin 4 A = 4 cos"^ ^ sin ^ — 4 cos J. sin^ ^ ;
cos 4 A =^ cos"* A — 6 cos^ A shr J. + sin'^ A :
8cc. = &c
The versed sines and chords of the same multiples may be
expressed in terms of the sines and cosines, and therefore in
terms of the versed sines and chords of the same arcs, by means
of the observations made in Chap. i.
271. Cor. From the last article, the tangent of the mul-
tiple of an arc may be expressed in terms of the powers of the
tanoent of the arc itself.
'&'
sin n A
row tan tiA =
cos
?iA
n cos"~^ A sin A-n C- \ C- — ^\ cos"~^ A sin^ A+&c.
- A-n (^) cos'^-- A sin" ^ + Scc.
n
176
sin A /n— \\ /n — 2\ sin'' A
cos A
fn— \\ /n - 2\ sill A
/n— 1\ sin A
' \ 2 / cos^ A
(by dividing both numerator and denominator by cos" A)
\ tan-* ^ + &c.
as has
n tan ^ — /* (—^ — j (~^) ^^"^ ^ "^ *^^*
1 ~ n (—^ — j tair A + &c.
been proved in (133), and in which if ?i be odd, the numerator
n + 1
and denominator will each be continued to terms, and
2
.. , n n
if n oe even, to - and - + 1 terms respectively.
2 2
This expression for tan nA might however have been found
without taking for granted those for sin nA and cos nA.
Ihus. tan nA =
COSH A
1 ((cos ti A + jsj — 1 sin ?i^) — (cos uA — /^ — \ sin?/^)>
^ — 1 ((cos iiA+ /v/ ~ 1 sin nA)-\-{cos nA — \/ — \ smnA)J
_ 1 |(cos A + aJ - 1 sin Ay - (cos A — sj — 1 sin AY\
V"-^ Kcos A + x/~^ sin ^r - (cos A - ^/^ sin A)"]
= __L_ (0 + \/^ tan ■ir~-(l - ^"^ tan ^r)
n/"^ kT+'y^^tan Ar + {\ - ^ITtan iy'i
?i tan A--n {j—^^ (~~~) ^''"^ ^ "'" ^^'
= — ~— — , as before.
1 — w ( — ^ j tan" ^ + ^c.
177
The expressions just found for the sine, cosine and
tangent of nA, have been deduced on the supposition that
ti is a whole number ; they are however true whether ii be
whole or fractional, but in the latter case the number of terms
will be indefinitely great, and consequently they are approxima-
tions to the true value only when the series converge ; and by
substitutions similar to those made above, we can find expres-
sions for the co-tangent, secant, and co-secant of nA, whether
71 be whole or fractional.
272. 3b express the poivers of the cosine of an arc in terms
of the cosines of its multiples.
Assume cos A + x/"--~l sin A^x, then 2 cos A = jt + - by (257) :
X
.', by the binomial theorem we have 2" cos" A= {x-\ — \
('■+i)-'('"--?^.)-('^)('-' + ^)--
First let n be odd, therefore the number of terms in
the expanded binomial will be n + 1 which is even, and there
will be an exact number of pairs of terms ; hence by
2
{2,^^) we shall have
2" cos'' 1=2 cos »il +27?cos(«-2)A4-2«^— — j cos{n-4)A
;z + 1
■\- Sec. to — terms ;
178
and .*. cos" A
_ -rlcosnA + n cos{n—Q.) Ai-n I | cos(w— 4)^+&c.to terms> :
Next let 71 be even, then the number of terms of the
expanded binomial being w + I will be odd_, so that in addition
to the pairs of terms above, there will be an additional one
which is the middle or ( - + 1 j th term of the expansion, and is
equal to
w(w-l)(?i-2)&c.{- + 1 )
^ ^ \2 ^ I 1.3.5 &c. (;2 -1)
^ — 2 ;
n n
1 . 2 . 3 &c. - 1 . 2 . 3 &c. -
2 2
hence 2" cos" ^ = (x''+ -A +?i{x'"^+ -7rr^?i -f &c. to ~ terms
, r 1 .3.5&C. (/i-l)
+ 2
n
1 . 2 . 3 &c. -
2
= 2 COS nA-\r^n cos (n — 2) A-\-&c. to - terms
I 1 .3 .5 &:c. Oi~l)
+ 2 ■;
n
1 . 2 . 3 &c. -
2
If w
and .*. cos" A = — — r ^cos w J. +w cos(7Z — 2)^ + &c. to - terms
2"- ( 2
- 1-1 1.3.5 &c. (w-1)")
n r
1 . 2 . 3 &c. -
2 ^
179
273. Ex. Let 71 be taken equal to the numbers 2, S,
4, 5, &c. successively, and the formulae just demonstrated
alternately give
cos'^ A = - {cos 2A-{-\} ;
2
cos^ A = " {cos 3il + 3cos ^} ;
cos'* A = ^ ^ cos 4^ + 4 cos 2 A +3} ;
cos^ A = — {cos 5A +5cos3A + 10 cos A} ;
&c. = &c
274. 2o express the powers of the sine of an arc in terms
of the sines or cosines of its multiples.
If cos A + ^ — ] sin A =^ X,
we have 2 ^/-- 1 sin A=x — , by (257) ;
therefore 2" (x/'^)" sin^'A= (x- -X
now n must necessarily be of one or other of the forms A-m,
Am + ], 4m + 2, 4m + 3> and therefore (v — i)" must admit
of four different values :
180
In the first place^ let Ji be equal to 4m, and therefore
( x/"^! )" = ( x/*^)'''* = (( s/~^)Y = l^'* = 1 ; then
2" sin" A
■('■ + f)-(--+p^)-'(=^)('-'+?^0
n 11.3.5 &c. (n - 1)
— <?fec. to - terms +2
2 n
1 .2. 3 &c. -
2
= 2 cos nA — 2n cos (w — 2) A + 2w ( --) cos (w — 4) J.
71 f 1 .3.5 &c. (n- 1)
— • &c. to z terms + 2 ;
1 . 2 . 3 &c. -
2
.'. sm
''^A
= ^_y I cos w A — w COS {n — 2) A + w ( ) cos {n — 4) A
n , 1-11.3.5 &c.(;z- l)"!
— oCc. to - terms + 2 f
2 ^^ > :
1 . 2 . 3 &c. -
2 -^
Secondly, let 7i = 4m+ 1,
hence
2" 7 - 1 sin" A = (x» - i;) - n (y- - ^ - ^)
+ „('l^)(."---i,)-Scc.to!i±ite™,
181
= 2 ^Z - 1 sin w^ - 2n ^ - 1 sin {n—Q) A
- . w -f- 1
1 sin (n — 4) A — &c. to terms ;
2
/. sin" A
= ^^_i < sin JiA — n sin (/« — 2) il + w ( ) sin (?z— 4) A
7Z+ 1
— &c. to terms
Thirdly, if n = 4m + 2, we have
• •. -2" sin'' A
w 4 1 .3. 5 &c. (?z- 1)
— &c. to - terms — 2
2 ^ ?i
1 . 2.3&C.-
2
= 2 cos nA — 2« cos (/z — 2) A + 2;z ( j cos {n- 4) A
^ 1 .3 .5 &.C. (72- 1)
— &c. to - terms — 2
2
1 . 2 . 3 Sec.
.-. sin" A
= ^^j I cos nA^n cos (?t— 2) A+/2 ( — ^ — jcos(/i— 4)A
T-i 1 .3 .5 &c. (w— 1)'
—Sec. to - terms — 2
o
.5 &c. (w— lA
2 . 3 &c. - \
2 J
182
Lastly, if // = 4m + 3, we have
=('-?)-"('■'■-.-) ^"(=ii)('"--p^)
w+ 1
-&c. to terms
4-27i
2 .>y — 1 sin nA-2nA>y — I sin (« - 2) ^
w— 1
1 sin (« - 4) yl — &c. to terms ;
sm
~ __ I sin nA — n sin {n — 2) A+n (— ^ ) 9in(/i— 4) A
n+l \
-- &c. to — terms f .
275, CoPt. The first and third cases of the proposition
proved in the last article are both comprehended in the formula,
sm
"A
= ± ~;;;:^ \ cos n A — n cos {n — 2) A -\rn ( jcos(/i— 4)J.—
1-1 1.3.5 8cc. {?i- lU
± 2 ^ -\
1.2.3&c.i' '
2 J
&c.
to - terms
2
and the second and fourth in the formula,
sin^ J.
= ± 1 |sinM7l-?/sin (n—9)A + n( — ^ — j sin(w — 4)^ ~ &c.
]
M+1
to terms
183
and attention to the algebraical signs might have made it
sufficient to divide tlie proposition into the two cases only,
in which 7i is even and odd.
Ex. If we suppose w to assume the particular values
2, 3, 4, 5, &c. successively, we shall, by attending to the different
forms, immediately obtain from these formulae,
sin^ A= (cos 2 A. — 1) = - (1 — cos 2^) ;
2
- (sin 3A'-3smA)= -
4 ^ 4
sin^ J. = — - (sin 3A'-3smA)= ^ (3 sin y^— sin 3 A);
sin* J.= -(cos4l - 4cos2yl -f 3)= -(3-4cosG^ + cos4J.);
8 8
sin^ A = — : (sin 5 ^ — 5 sin 3 A + 1 0 sin 1)= -~ ( 10 sin ^ — 5 sin 3 A + sin 5 A) ;
&c. = 8cc,
276. To express the sine of an arc in terms of the arc
itself
In article (270) we have proved that
sin72A = wcos"-^A ^\\\A-nC- j C- j cos"-^^sin^A+&c.
, ( sin A /w— ]\ /n-2\ s'm^ A ^ )
I cos .4 \ 2 / \ 3 / cosM )
= cos'^ A L tan A - 71 (~-) (~J~) ^^"^ ^ + ^c.j ;
Q
assume now 7iA=0, ox A =■ - ,
71
e { 0 /n—\\rn — Q.\ ^0 )
.. sin 0 = cos" ^^ [n tan ;^ - . (-^) (— ) tan^ - ,. &c. [ ;
0 . .
then if n become indefinitely great, - will be indefinitely small,
n
and
184
.°, we have cos -- = 1, tan - = - _,
fi 71 n
71 {71- 1) (W, — 2) = n^ (l \ (l \ = 71% &c.
. ^ e li' 0' 71' 0' . . ^ .
iience sin 6 = 71-- r- + ? — &c. in iniimtum
71 1 .2.3 rf ] .2.3.4.5 ?^^ ^
0' 0'
=:: ^ _j ^Q^ iji infinitum,
1.2.3 1.2.3.4.5 ^
277* Since in the expression for the sine of 7iA, the
quantities involved are trigonometrical functions of the arc
and therefore expressed in terms of the radius 1, it follows
that before the expression just investigated can be applied
the value of 0 must be found in terms of the same radius :
hence therefore if r^ = 57^.2957795 &c. the number of degrees
contained in an arc equal to the radius, and 0^ be the number in
any proposed arc, we shall have
r^ : 0^ :: 1 : -n = the value of 0^ in terms of the radius,
r
id.\sm^^= (-^) -(ii) + (-t) -&c.
VrV 1.2.3 VrV 1 , 2. 3.4. 5 V/'V
Ex. 1. Suppose d^ =■ 1°, theUj in (155) we have seen that
-0 = .0174532 Sec,
r
,'. sin 1^=. 0174532 &c. (.0174532 &c.)^+&c.
1.2.3^
= .0174524 &c.
(1.0 ^0
— ) = l'; then -5- = .00029088 &c.
00/ r
from (155),
and sin l' = .00029088 &c. (.00029088 SiccY-r&c.
= . 0002909 &x.
185
(1 \ 0 fi^
— j = \'\ we have -^ = .000004848 &c.
and .*. sin l" = .000004848 &c. (.000004848)^ + &c.
1 .2.3
= .000004848 &c.
and so on.
278. Cor. From the last of the examples above given,
it appears that the arc of one second and its sine do not differ
by any significant figure, and therefore in all practical applica-
tions of trigonometry, we may without sensible error assume
sin 1'' = 1'': similarly, sin 2" = 2 sin l"=2", sin S" = 3 sin l" = S"
&c. very nearly, by (74).
Also, if (j) (A) denote any trigonometrical function of A
expressed in terms of the radius, we shall have
sin T' : (b (A) :: l'' :
the value of the same function expressed in seconds, which
I'XD 0(A)
. . = — / , = ■^. — -jy seconds,
sin 1 sm 1
279- To express the cosine of an arc in terms of the arc
itself
It has been proved in (270) that
cos nA = cos" A-\- n ( \ cos"~^ A sin^ A + &c.
= cos" A \\-n (r—^ tan^ A + &c. |
.'. as before cos 0 = cos**- \\—n ( ) tan^ - +&c.(
w I \ Q. J n 5
n' e' n' 0' ^ . . ^. .
= 1 -^ H — -- &c. tn rnhmtnm,
1 . 2 «^ 1 . 2 . 3 . 4 n* -^
by reasoning as in (276),
A A
186
— 1 — \~ Sec. in mfimtum.
1.2 1 .2.3.4 ^
and pursuing the steps taken in article (277), we get
1.2 VrV 1.2.3.4 \r/
Ex. If the same values of 0^ be assumed as in the last
examples, we shall find
cos 1^ = .9998477 &c.
cos l' = .9999998 &c.
cos l"= .9999999 &c.
280. By means of the expressions for sin 6 and cos 6>,
found in articles (276) and (279), formulae for the other trigo-
nometrical functions are easily obtained. Thus,
(gi r\\
1 1 &c. I
1.2 1.2.3.4 /
= j- &,c. in infinitum:
1.2 1.2.3.4 ^ '
chd 0 = 2 sm - = 2 ^ -^ -{ -^ - &cA
2 (2 1.2.3 2' 1.2.3.4.0 2^ j
"" ^ "^ 773 ? "^ 1.3.4.5 ¥ " ^''* ''' '''^"'^'''^ '
Q ^
sinO 1.2.31.2.3.4.5
tan ^ = 2f ~ B^^ m —
cos d 6 6^
1 + &c.
1.2 1.2.3.4
^ . 20' ....
r= 0 H H- - — - — - + &c. m tnjinitum ;
J 87
0' 0*
^n COS 0 1.2 1.2.3.4
cot0 =
sin e 0^ e^
e + . — &c.
1.2.3 1.2.3.4.5
r: — &c. in infimtum;
^1.31.5.9
1 1
sec 9
1 + &c.
1.2 1.2.3.4
S^ 5 9*
= 1 H H h &c. m infinitum ;
^ 1 .2 1.2.3.4 -"
1 1
COSec 9 = -:— ;r = ^
Sin
sin ^ ^ 9 9""
9 &c.
1.2.3 1.2.3.4.5
1 . 9 14^' ^ . . ^ .
= 7; H i ^ + &c. ?w infinitum^
9 1 .2.3 1.2.3.4.5.6 -^
281. Cor. If the arc be small,, approximate values of
some of the preceding functions are readily obtained. Thus, if
9 be very small, we shall have
^ = ^-r:l:3 = H'-r:?:i) = K'-i^/ = ^<-^)*'
hence, log sin 9 = log 9 -{- ^ log cos 9,
and log 9 = log sin 0 — ^ log cos 9 :
9 9
again, tan 6> = 0 + — = ^(l+—) =
9\^
(1--)- (co.#
.*. log tan 9 = log ^ — 7; log cos 9,
2
and log 9 — log tan ^ + - log cos 0:
188 .
also, 2 mi e + tan 6 = ^0 h 0 -\ = 3 6:
1.2.3 ' 1 .3
and 8chd - - chd^^S ( 5)- (^ ; "Ts)
2 V2 1 . 3 4V V 1 . 3 2V
10^^ \ 0'
„ — 0 + =
1.32^ 1.32'
= 46-^^,-6+^^=36,
1 / f? \
whence 0 = -^ ( 8 did chd 6),
3 \ 2 /'
which is an useful approximate formula for practice; and in all
these instances, the arc is of course supposed to be expressed in
terms of the radius.
282. To express the sine and cosine of an arc in terms of
impossible exponential functions of the arc itself
1 3
XX X
In the expression e^ = 1 H h 1 1- &c.
^ 11.21.2.3
where e = 2.71828 &c. the base of hyperbolic logarithms, let
0 \/ — l and ■— d j^ — 1 be successively substituted in the place
of X, and we shall have
' ^1 1.2 1.2.3 ^1.2.3.4^
1 1.2 1.2.3 1.2.3.4
hence by exddition and from (279), we get
+ ,-e-.=.,{l_Jl+_i__Scc.S = 2cos6,
1.2 1.2.3.4
and . . cos v =
2
and by subtraction and from (276), we obtain
189
~=2yirr|0 — ^ + — ^ &c.
^ I 1.2.3 1.2.3.4.5
= 2 ^ - 1 sin 0,
hence .'. sin 0
eev/^i_^-9v/— 1
2x/-l
283* These two formulae, as before, enable us to find
expressions of the same kind for all the other trigonometrical
functions. Thus,
vers
e=l-cos0=!-
chd 0 = 2 sm -- =2
2 2V^=T x/- 1
sin ^
tan 0 =
sni ^ 1 e ' — e 1
cos^ 7-le^^-^+e-^^-^ 7-le^^^-i + l'
cos^ ,— ;e^-^+e-e-^ e^^^^+1
— e~~ ^ '
cosec
1 _ 2^-1
^^-1
sine e^^-i-.-^^-i ^ g2e^/^ .9v-:ri-
284. If the equations investigated in the last two articles
were established by any independent method, they might be used
to determine the relations between other different trigonometrical
expressions. Thus,
2 sin 0 cos 0 = 2 [^ 1 — \ (
-i_.-^^^. ^e^^^ + e-'^=^-
2 V^^
190
{e — e J (e' +e )
= i=. (/^ -^1 -,e-2e v^)=:si„ 2^, as in (76) ;
2 V -1
also, cos u = ( I
8 ^
_ 1 je +^ . Q <? +g I
~ 4 I 2 2 -f
= - {cos S0 + 3COS 0},
4
as has been before proved in (273) :
again, (cos 0 + \/ — I sin BT^
=(
2 ~
) =*■
2 -
= cos mO ■¥ fk.
2
/ — 1 sin w^.
which is Demoivre's formula already established by a different
process.
191
CHAP. VII.
On the application of Trigonometrical FormuliZ to the solution
of quadratic and cubic equations. On Theorems for the
decomposition of certain Functions into their simple and
quadratic Factors, On Expressions for the sine and cosine
of an arc by means of continued products, and their use.
On Expressions for an arc in terms of its tangent, sines of
its multiples, S^c. On the Solution of certain cases of triangles
by means of series^ and without the aid of Tables, S)C. On
Expressions for the cosine and sine of the multiple of an arc
in terms of the cosine and sine of the arc itself. On the
Summation of the Trigonometrical Functions of certain series
of arcs, S^c,
285. To f?id the roots of a quadratic equation by means
of a table of sines, cosines, Sfc.
First, let the proposed equation be x^±px-\-q = 0,
let sin" 0 = —q- ,
P
-rP^.-r
and .', x= + - {l +^l-sm^O\ = + - ! 1 +cqs^!:
9 0
now, ] •— cos 0=2 sin^ - , and 1 + cos 0 = 2 cos" - ;
2 2
6 9
,', the values of x will be + » sin^ - , and + p cos^ - .
2 2
192
These values may be exhibited in a different form : for
snice p = —. — - , we have
sni V
_ . ,e _ CL\^q . 2 ^ _ / 6
+ p snr - = + -: — T snr - = + v ^ tan ;- ,
^e „ 2A/r/ ,0 _ . e
and + » cos - = + -: — ~ cos - = + v 7 cot - .
^ 2 sni ^2 ^2
If these solutions be applied to practice^ the formulae must
be adapted to the radius r by (60), and then we get
log sin ^ = |- {20+2 log 2 +log g' - 2 log p},
Q
log J7 = + { log 2? + 2 log sin 20 } ,
Q
and log j; = + { log p + 2 log cos - — 20} :
similarly of the other solutions.
Next, let the equation proposed be a*+ px — q=:0,
from which we have x = + -^ ] 1 + y 1 -i 1 [ ;
let tan^ 0 = -f ,
P
and .-. j: = + ? { 1 + ^l+tm^^O} = + ? { 1 + sec ^} :
2 2
0 fi
now, I — sec 0= — tan ^ tan - , and 1 -hsec 0 = tan 0 cot - ;
2 2
.'. the values of x are + - tan 0 tan - , and T - tan 0 cot - :
~ 2 2 2 2
. . 2 \/^
and these, by substituting for p its value ^-^ become
tan 0
± V </ tan - , and + v </ cot - , respectively.
193
Each of these sets of formula? must be adapted to practice
as before ; and it may be observed that all these soUitions can
be advantageously employed in those cases only, in which p and q
are very large or very complicated quantities.
286. To find the roots of a cubic equation by means of
a table of sines, cosines, 8^c.
Let the equation be reduced to' x" — qx + r = 0, by taking
away the second term, and assume sin 0 = x,
.•. sin 30 = 3 sin ^ — 4 sin'^^, to the radius 1:
hence sin'^ 0 ~- sin 0 ■{ sin 3^ = 0, to the radius R,
4 4
, • 3 3R- R' . ^
that iSj x^ a H sin 30 = 0;
4 4
therefore equating the coefficients of the corresponding terms
of this and the proposed equation, we have
3R^ R^ . ^
= 7, and — sm 30= 4- r ;
4 4 ~
/o 3t
whence jR = 2 v - - and sin 30= H :
.*. to the radius 2 y - find an angle 3 0 whose sine is + — ,
and thus sin 0 or x will be determined :
3 7'
also, since + — =sin 30 = sin (Stt + 30) = — sin(27r— 30)
= sin (47r+30)= — sin (47r-30) = &c. by (17) and (20), the
values of .r will be
SHI
9,si„(^f + e).-si„(^=J-e), si„(ir+^)
(47r \
— — 0 1 , 8cc.
Bb
194
but sin (^ 4- eV - sin (stt - -y - e^ = - sin (-—- -~ e\ ;
-sin(ij^ - e) =sin(2.- ^ ^- o) = sin (^ +0);
&c =&c
(27r \
~^^ — h ^ ) 5
and ^ sin ( ^ J to the radius 2 S/ - ^ since after these
three, the same values continually recur.
To the radius 1, the values of x will manifestly be
2 \/i" sin 9, 2 \/^s,n (^ + O) , and -2 x/^in (^ - o) ,
because sin 0 to the radius R= R sin 9 to the radius 1.
If we had assumed cos 6 = x, the roots might have been
obtained in a similar manner.
Ex. Let it be required to determine the roots of the
equation .r"^ — 3 .r — - 1 = 0.
In this case ^ = 3, r=:l;,
.-. JR = £ \/ - = 2, and sin 30— = — 1, to the radius 2 :
3 q
hence to the radius 1, sin S6= - i = sin 210^, and 9 = 70^^ ;
.'. the values of a are 2 sin 70^ 2 sin IQO^, and - 2 sin 50^.
287. In the solution above given^, we have assumed
JJ = 2 V ^ , and therefore if q be negative, R will be
o
3r
impossible : again, if q be positive and ± — be greater than
195
Vi-
e shall have sin 30 greater than R, which is also
impossible : hence therefore in both these cases this solution by
the trisection of an arc fails^ but Cardati's solution does not.
two roots being then impossible, and it is observable that both
solutions succeed when + — =2 \/ -- , or two roots are equal.
- q ^ 3
288. Though the solution just given fails in the instances
above enumerated, trigonometrical formulae may still be applied
in finding the only root which is possible: thus, if the proposed
cubic be x^-\-qx-\- r = 0, we have by Cardan's rule.
assume tan' kj — _, or — = ^r-r-'
27 r' 4 27 tan- e
.-. a= V -\ [V l-sec^+ Vl+sec^l
= V - ^ |V sec^+1- V sec^- l|
— \/ ^ [\/^^^^ +^ _ 4 '/sec 0 -T|
~ ^ 3 (^ tan^ ^ tan^ )
let V cot- =cot<p, .*. V tan- = tan ^;
and .r = V ? {cot 0 — tan 0} = 2 y ^ cot 20,
3
wliiqh is the possible root.
196
If the equation proposed be x^ — qx + r =0, and
— be greater than 2 y 7 , we have as before^
■an ^^^ !!l _ 9
assume sin 0 =■ o ? o*' —
Q7r' 4 27sin^e'
.-. a = V - ^ I VI - c«se -r S/i+cos^l
* /7f\/l-cosa ^ y 1 4- cos e|
— V ? {tan 0 + cot 0} = 2 V
the possible root
{tan (^ + cot 0} = 2 \/ -^ cosec 2^,
289. The latter solution alkided to in article (286) might
without much difficulty have been deduced from that of Cardan,
but the process will be less simple than the one which would
result immediately from the assumption there pointed out.
For, since :.= \7 -i' + \/IIC+ \/ZIZ\/ZZl ,
let - l^a, and V ~ - '^ =/3./^, and .-. «' + ^^ = |r'
hence x = V a +/3>/^^ + V a - /3 ^ - 1 :
now
197
|3
and if we assume cos S6= , , and .*. sin S0= ,— — — ,
we shall havea±/3.y^= V^^ {cos 30+ V~^l sin 30};
.-. V a+/3.y^= Vl {cos0-{-.y~^ sin0},
o
and \/a-/3V~^= V- {cos 0 - V"^! sin 0} ;
whence a = 2 \/ -- cos 0 :
also since cos 30 = cos (Stt + 30) = cos (Stt - 30),
the two remaining values of .r will be
2\/Fcos(^;+e), and2sAcos(^-e),
as would have been found from (286).
This is the solution of what is called the Irreducible Case
of Cardan s Rule.
By an assumption similar to the one just made, that is,
if tan 0 = — , it is easily proved that
a
7ri?7^ = 7^"^' fcos -^ + ^~l sin ^-\ .
^ 7f n)
290. To decompose x*" — 2 cos 0 x^ -[- 1 i)(to its simple and
quadratic factors.
Assuming ar^" — 2 cos 0 x" 4-1=0, we obtain
a" = cos 0 ± ^y - 1 sin 0, and also by (1?) and (21),
= cos(27r + 0)± >y — 1 sin (27r + 0)
= cos (47r-f0)± s/ - 1 sin(47r4-0)
= &c
198
= cos {2 in- \)7r + 0) ± ^y - \ sin (2 (/^ ~ l)7r + ^):
but, by (267), we shall have from these equations,
0 i — . e
a- = cos — I- /s/ - I sin - ;
n n
27r-hO
sni
X = COS + n/ ~" ^
47r-i-^ / . 47r-f0
X = cos — + lU — I sni
11 n
ac. = &c,
2(7/-l)7r + 0 y . 2O/-l)7r + 0
x = cos ± V "~ ^ ^'"
hence, by the nature of equations, we have
x'"-2cos0x"+l
. 0\\
= [r - (cos ; + 7 - 1 sin ^)} {. - (cos ^^ - Vri sin ^J
f / 27r + 0 . /- . 27r + ^\l
f- (^"^-^; — ^ -^ ^ ^"^-T~)i ^^^-
Jx — ( cos
2(72-l)7r + ^ y— . 2(w-])7r4-^
— ^ h >^/ — 1 sni —
2(«— 1)77 + ^
■! 1 — ( cos — ^^ '■ ^y — 1 sin
. )1
2(;/,~l)7r + ^\|^
and combining each of these pairs of simple factors, we get
t"* - 2 cos (9 .r" 4- 1
= Tr" — 2cos - X + \\ ^1
27r + ^
2 cos X H- 1
199
/x^ — 2cos — - — T4- 1] ^c.lx^ ~ 2cos— ^^ 2+1 K
the number of factors being ;/.
This theorem of Demoivre contains the solution of the
equation a'" — 2 cos 0 .r" + 1 =0, all the quadratic factors of
which appear to be possible, and all the roots impossible,
unless some extreme value be assigned to cos 0.
291- Cor. 1. In the formula just investigated suppose ^=0,
or cos 0=1; then we have
^^2«_2.r"+l=(2^'-e2 + 1) /^x--2cos -^ X + \\
/o 47r \ /o 2(7i— Ott \
( x — 2 cos .r -j- 1 j 8zc. ( jt" — 2 cos x + 1 j ,
to 71 factors :
2(«- Dtt
now cos
cos
- l)7r / 27r\ 27r
= cos ( 2 TT I = COS — :
n \ n / n
2(w--2)7r / 47r\ 47r
= COS ( 27r ) = COS :
n \ n J n
&c =&:c =&c.
,'. :r'" -2.r" + l=(x'~2a^+l) /'/--2 cos — .r + l)
(rr - 2 cos x -{- \\ &c.
First let n be odd, then
:r^«-2A^"+l=(x^--2.r+l) Tr' - 2 cos — .r + A
200
(47r \" /a C^*""' 1) TT , \^
/_2cos .r + \j &c. (a^--2cos a^ + ij ,
n + 1
the number of factors being manifestly — - — ; and extracting
the square roots of both sides, we have
x" - 1 = (a: - 1) ( X- '- 2 cos — X -{■ \j
Next let n be even,
a A.
(27r \ / 47r \
2^" -2 cos .r+1) r/-2cos- — x -\- \ \ &c.
./-2C0S^^ — r + Ij {X^ — 2 cos IT X + ])
^(x^ — Qx-hl) (x"' -2C0S—X -{- l\ &c. (j:' + 2^:4-1),
the number of factors being ~ ; and extracting the square roots
as before, we get
a?"— l=(x- 1) r/ — 2cos— X + l\ &c. (a:-f l)
^^- 2 COS x-\-\\ &c. rr^-2cos ^ + 1 )
In the same manner, if we make 0 = 7r, ^^^^+1 may be
decomposed into its simple and quadratic factors, and the roots
of the equation x" + 1 = 0, will be determined.
201
These formulae which are from their inventor called Cotes^s
Theorems, include the solution of the equations j;" + 1 = 0,
the roots of which are the n roots of + ], and it is evident that
only one of them is possible when n is odd, and two or none
when n is even.
292. Cor. 2. By means of the formula of (290), we are
enabled to prove also Cotes^s Properties of the Circle.
For, since a;^*' —2 cos 0 x" + 1
= (x^ — 2 cos -' X + 1 j ( T^— 2 cos X + 1 ) &c. to n factors,
if we suppose ^ = 27rj we shall have
0 __2'7r 27r + 0__47r 47r + 0 __67r _ __
' n n ' n n n n ^
and .-. :i'"-2i" + l
27r \ / c 47r
(M 71" \ / 4 TT \
/ — 2cos ^^ j: + 1 ) ( / - 2 cos — x + \\ &c. to n
factors
now if P be any point in the diameter (produced if necessary)
of a circle whose radius is 1, and the whole circumference
be divided into 71 equal parts, AB, BC, CD, &c. we have,
if OP = x,
^^n „ 22"-M = 0P^"~20P''-i- 1 =(0P"-^0^'^)';
Cc
202
^'-2 cos —:r+l = 0P'-20P cos A0J5+1 = PP';
n
oc^^Q, COS -^:r + l = 0P'-2 OP cos ^0C + 1=PC';
n
x'-Q cos — :r+l= OP'-£OP coslOD+l = PD';
n
&c =&c.
.'. (op^'-o^^f =pJ3^pc^pJ)^&c.
and OP"'^OA''=:PB . PC . PD . &ic.
Again, if the arcs AB, BC, CD, &c. be bisected in the
points ttj b, Cj &c. and P«, Pb, Pc, 8cc. ; Oa, Ob, Oc, &c.
be joined, we shall have as before
OP^''^ OA^'' = Pa.PB,Pb.PC. &c.
^Pa.Pb.d^c, PB,PC. &c. = Pa. Pb . &c. {OP"'^OA''\
and /.P..P6.Pc.&c.= gpn^oA'^ =OP'^+OA\
293. Cor. If the point P be supposed to coincide
with J[, Pa, Pb, &c. will become the chords of -, , 8lc.
n n
and /. chd - chd chd &c. to n factors =2,
21 n n
TT . Sir . OTT
or 2 sm 2 sin 2 sm &c. = 2,
2/i 2?i 2^^
, . TT . Stt . 57r „ 1
and .*. sin — - sin — sm — &c, = 1 .
2^ 2w 2w 2"~
203
294. To express the sine and cositie of an arc by meajis of
continued products,
1 + 8cc. [ , if we
1.2.3 1.2.3.4.5 J'
assume sin 0 = 0, the corresponding values of 9 will be
0, ±'Jr, ±2 7r, ±37r, &c.
.*. by the nature of equations, we have
sin e=Ce {tt-B) (7r + ^) i^TT-O) (27r+0) &c.
= C7r^2V3V"-&€
e\ / G"
■»(-5)(-^)-
but when 0 is indefinitely small, we have seen in (213) that
-^— = 1, and .*. we have C tt^ 2" tt^ 3^ tt^ &c. = 1 ;
u
hence, sin 0 = 0 (\ -^ (\ ^-^ &c.
Again, smce cos t7 = 1 + — &c. if we
^ 1.2 1.2.3.4
suppose cos 0 = 0, the values of 0 will be
TT Sir OTT „
+ - , + — , ± , &c.
- 2 - 2 2
= C -7; jj- &c. '
2' 2
and if 0 be indefinitely small, we shall have
2 o2 2
cos ^= 1, or C— 2 — 2"" ^^' = 1 »
2 2~
hence, cos 0= {\ ^j ( 1 — 7^—2) occ.
2104
295. Cor. 1. It is manifest that if we suppose C=C\ there
will be the same number of factors in the expressions for
both sin 0 and cos 0 : omitting therefore one of the first of
the equal factors in the latter to make the number correspond
with that of the former, we have
w fSV'^ 5'w' 7V= 1 ,„,,„,, „ ^
2 ( 2" 2^ 2" J
^ TT 2^ 4' 6' &c. in inf. ^. ^ . ^^^ ^^. ,
and . . - = <, c, — TT— : — 7—7 3 which is Wains s expression
2 3" 5- 7~ &c. in mj. ^
for the circumference of a circle whose diameter is I.
Hence also from (214) it appears that the area of a circle :
the square of its diameter
2^ 4^ 6" &c. in inf, 8 24 48
:: ^g ^^ ^2 . . . r ' 1 :: - X — X — X &c. : 1.
3^ 5^ 7 &c. in znf. Q 25 49
296. Cor. 2. From the two theorems above proved, the
logarithmic sines and cosines of arcs are easily derived without
previously computing their natural sines and cosines.
m IT '
For, let ^ = — — ; therefore we have
' n 2
sin — — = — — ( 1 ~- ) ( 1 — -— — J Sec.
w TT / m^\ / m^ \
cos = I 1 ^ I I I ^-5 I &C.
n 2 \ ir) \ 3^n'}
and thence
log s.n _ _ = log ^ + log (-j +log(^l - ^,j +&C.
,„g,os- - =log(. - -,) +Iog (1 - -,-^) +&C, .
^05
297* To express the length of mi arc in terms of its
tangent-
From (282) we have
cos ^ 4- \A^ sin 0 = 6^"^', and cos 0- ^'^l smO = e'^"^^;
e^"""^ c:Q^m__ cos0-\-^ —\sin9 _ \ + ,y -ItanO
^''F^^^'' ~ cos^-^^sin^" l-V'^ltan^'
hence, taking the logarithms of both sides, we get
2^>y/"^=log(l + ^/^ tane)-log(l- x/-~i ^an^)
= >/^ tan^+;; tan'e- - ^'^ tan^ 0 - - tan'^ + &c.
+ ^/~^l tan^ tan^a- - ^7 - 1 tan^^+ - tan^0— &c.
2 3 4
and .*. 6 = t3in0 tan^ 0 + - tan^^ — &c.
3 5
Ex. 1. Let 6 = 45^, or tan 6= \,
.-. the arc of 45^= 1 4- + \-^c. in iff.
3 5 7 9 11
= (l I + f- - -I + (- 1 +&C. in infinitum
\ 3/ \5 7/ V9 11/ ^
2 2 2 . . . ^. .
H 1 h&c. 2w infinitum:
1.3 5.7 9.11
and the whole circumference of the circle whose radius is r
(1 1 1 .... 1
= I6r \ ( 1 h dc. ?7f infimtum\ .
ll ..3 5.7 9.11 -^
206
Ex. 2. Let tan 0 and tan 0' be taken respectively equal
to - and " , then
2 3
2 3 Q,^ 5 9.^ ' 3 3 3^ 5 3^
^il , . „-il _..o
but by (110), e + 0' = tan''- +tan-^- =45^;
.•.thearcof45°= f- +-^- - (-k + ~^^+- f -5 + -r") - &c.
\2 3/ 3 \2^ 3V 5 W 3V
298. Cor. 1. Since log u = (w - m"') - - (w^ - m"')
4- - (w^— w~^) — 8CC.5 if we suppose ti= \/ — 1, we shall have
log V~=^l
/ 1 If > 1 f / 1 1 o
= 2V~I {' - 5 + 5 - :) "^ &c.} = 2^^ ^, by Ex. 1;
.*. Stt or the circumference of the circle whose radius is 1
x/^i •
299. Cor. 2. Since w'^ = e"'^^^ if u = V^, we have
TT 1 TT^ 1 Tt''
= 1 " :: + T-;: :: — + &c. = .207879 &c.
2 1.2 4 1.2.38
The discovery of these singular formulie is due to the
celebrated John Bernoulli,
207
300. To express the length of an arc in terms of its sine,
and the sines of its successive multiples.
Since log .r = Cr-a?-') - - (x'^-jr"^) + - {x^ ■- x'^) --^c.
= sill 0 sin 20 + - sin SG - &c.
2 3
301. To express the length of an arc in terms of its sine,
and the secants of successive suhmultiples.
By article (76) we have immediately,
. e Q
sni y = 2 sm - cos - ,
2 2
. ^ .0 0
sm^ = '^«'"5i^«V^'
^ 0 6
sill-. = 2sin^cos^,
&c. = &c.
. e . 0 9
sin j;7T-r = 2 sm ^ cos - ;
therefore by multiplication, we get
.6000, 0
sm 0 = 2" sm — cos - cos — cos -^ ccc. cos —7 .
208
. 0 . ^ 0 0 0 ^ 0
and 2" sin — = sm u sec -- sec — sec — ^ cCc. sec - — :
2" 2 2 2 2"
suppose now n to be infinite, in which case sin ~ = — by (213),
9 6
,'. 0 =z s\n 0 sec - sec -^ &c. in infinitum;
2 2^ ^
which is generally known by the name of Euler^s formula.
302. Given two sides a, h of a triangle, and the included
angle C, to find the remaining side and angles hy means of
infinite series.
By (167) we have
c' = a^ -^b^ — 2ab cos C
= a' + b' - ab (e^''^' ^e'^^~) = {a^b e^''^)(a'- b e'^'^'^'\
.'. as before, 2 log c = log (« — ft e^' ~^) -r log (a — & e"^ ~^)
= 2 log « + log ( 1 - - e^^~^j + log ( 1 e~^^^~^ J
&c.
b b^ / ^
and loar c = loo- a cos C — — ^ cos 2 C — — it cos SC— &c.
^ a 2a^ Sa"^
the logarithms being taken in the system whose base is e or
2.71828 &c.
, . a sin A sin(5+C) „ , sin C
also, snice -r = -: — - = : — - — = cos C + — ,
b Hm B sin B tan B
, _, 6 sin C
we nave tan B
a — b cos C '
and .*. - -, / — ^—F= = T= —
or e^^"^^
209
and taking the logarithms of both sides, we get
2l^-v/^ = log (a- he-^''^~^^)-\o^ {a- hc^'^^)
__ ^ /^Cv'— ^-Cv^TZY.
^r/^V--T_^-2C^^>
a 2a
3a
and .'. B = - sin C + — 5 sin 2C -1 -, sin SC + &c.
a Q,ar Sa-^
which is expressed in terms of the radius 1 : and hy (278)
6 sin C h^ sin 2 C , h^ sin 3 C ^
^ -3 — ^ — ^ .J __ ^ ^ __ ^ ^ g^^,^
« sin 1 2 a sm J 3 a sm 1
b sin C 6^ sin 2 C ^^ sin 3 C
a sm 1 a . sni 2 a"^ sm 3
which is the value of B in seconds ; also, if b be much less
than a, a few terms of these series will be sufficient.
303. By processes similar to those pursued in the last
article, we shall find that the equation
n sin cb
tan u = -, cives
1+W COS0 "^
^ = n sin <p sm Q.(p ■\- — sin 3 0-*&c.
Dd
210
tan O^n tan (p gives
tan ^ = cos a tan ^ gives
0 = d) — tan* - sin 2d) 4- - tan"* - sin 4d) — &c.
and sin 0 =: sin a sin (/3 + ^) gives
^ = sin a sin /3 + - sin'^ a sin 2/3 + - sin^a sin 3/3 + &C.
2 3
304. To solve triangles without the aid of tables.
In (224) we have seen that « = c sin A, .*. by (277) we have
(/A\ 1 /A\' 1 /A\' )
also ta»*l^^ = -t , and /. by (297) we obtain
^•-'"{(-:)-3©'+j(-:)'— 1^
2 I f 2 2
again, since cos C = ^ — 7 , we shall have from (279)
and so on for all other cases.
305. By a similar process if r be the radius of the circle
circmnscribed about a triangle whose sides are 2^, 2&, 2c, it will
readily appear that
211
'a+b+c\ . 1 /a^^b' + c\ . 1.3 /a' + b'+c'
1.3.5 /a^ -\- b'^ + c\
H : — :: — - ( :; i "f &c. = the sean-circumference or
a circle whose radius is 1.
306. To solve triangles, two of whose angles are very
acute.
First, let the two very acute angles A, B and the side c be
given, then will the remaining angle C be found to be very
obtuse : also by (276) we have
A^ . ^ ,. B^
sin A = A , sin B=B-~
1.2.3' 1 .2.3'
and sin C = sin {A -{- B)-{A^ B) - ^ "^ ^ nearly :
csinil cA f 2.154-^^)
hence a = -: rr- = — { 1 + > ,
smU + jB) A-\-B\ 1.2.3 r
cs'mB cB ( , A^-{-2AB)
^"^ ^= sinUH-^) = TT^ r "^ Tr^rrl'
which are the true values after neglecting such terms as contain
four or more dimensions of A and B.
Hence the excess of the sum of the two other sides above
c A B
that which subtends the greatest angle = a'{'b — c = — , and
2
it may be observed that A and B are both understood to be
expressed in terms of the radius.
Next, let the two sides a, b and the included angle C which is
very obtuse be given : assume C = 7r--a, then by (l67) we have
c^ = «^ + 6^—- 2at cos C = a^ '^b'^-\-Q,ab cos a
= a~-^b' + 2ab (\ - -^) ^(a^-bf- aba\
1 aha
2 a + b
3
■ A a . ^ a . a \ a \
a2;ain, sin A= - sin C= - sin a = - la — ; ^ A
° c c c V 1 .2 .3;
__ rt r \ aba *^ \
"" rt + 6 I" 5 {a + 6)' "" 1.2. si
a a j («" + b") — a b a 1
^7+7^ r (a-f-W' 1.2.3)'
sin"^ J. a a , nb{a — b) a^
whence A = sin ilH ^ — ~ = 4-
sim
1.2.3 a+6 ' {a + bf 1.2.3
_ gg \ bja-b) g' I
which would have been found to be of the same value from the
equation B = a ~ A.
307. To express the cosine of the multiple of an arc in
terms of descending powers of the cosine of the arc itself
Since
(1 -ax) (\ - -\ =1- fa+ '-\ .r+/=l - x (a -\ x\,
if we assume rt 4- - = p, and take the logarithms of both sides
of this equation, we have
log ( 1 ~ « j:) + log M — '-< j = log \i—x(p- x)}, and
a „ a ^ a
« r + -r x^ + - a ' -f &c. + — i" + cScC. ■
2 3 n
213
2 3 n
therefore equating the coefficients of a'^ in both sides,, we get
«. ^ n « 2 n(fi—3) „^. n{n—4)(n—5) „ ^
^ y^(,^-77^--l)8cc. (7^-2;y^-^- l)^.-2.»^
1 . 2 . 3 &c. 7?^ ^ "^ '
in which the last term will be 2 or rip according as w is
even or odd :
nowif/9 = a + - = 2 cos il, we have 2 cos 72^ = a'* + — :
a a"
and if n be even^ we shall have
2 cos JiA
= (2cos^)"-n(2cos^r-'+ ^^ (2 cos l)"-*-8cc. ± 2;
1 . 2
also if n be odd, we shall tind
2 cos ?« J.
= (2 cos AT - n (2 cos 1)"" ^ + n{n-S) ^^ ^^^ AY'^-'&c.
1 .2
+ 7Z (2 cos A).
If we differentiate both sides of the equations just investigated,
we shall obtain similar expressions for sin nA,
308. To express the cosine of the multiple of an arc in terms
of ascending powers of the cosine of the arc itself.
Reversing the order of the terms of the expressions found in
the last article, wc get immediately, if n be even.
COS
214
- I 1.2 1.2.3.4 j
and if fi be odd,
cos7iil= +7/cos^ J I COS A-\ COS A—&cA,
- I 1.2.3 1.2.3.4.5 J
in which the upper signs must be used when 7i is of the forms
4m and 4m + 1^ and the lower when of the forms 4!m+2 and
4;72 + 3.
Differentiating both sides of these equations, we shall
immediately obtain expressions of the same kind for sin nA.
309. To find the sum of the sines of a series of arcs in
arithmetical progression.
Let sin^ + sin(^-}-^) + sin(^ + 2^) + &c. + sin(^+(«-l)^)
be the proposed series, then by (67), we have
2 sin - sin 0 = cos ( 0 | — cos ( 0 4- -) 9
2 \ 2/ V 2/
2 sin- sin {0-^^) = cos (o ^ ''^-cos/^^^- —V
&c =&c
2sin^sin(^ + (;z-l)^)=cos(^ + ?^^)-cos(^ + ?^^^
and denoting the sum of the proposed series by Sj
we shall have by addition^
215
sin ( ^ H 0 I 8111 —
A . <;? V ^ / ^
and . . o = ' --^ .
sin -
2
310. Ex. If I be taken equal to d, 2^, 3^, &c. suc-
cessively, we shall have
sin^ -I- sin 2^ + sin 30 + &c. + sin nQ
. (n^\\^. 71$
sin I I 6 sm --—
. 0
sm -
2
sin0 -f- sin 30 + sin o0 + &c. + sin (2/2- 1)0= --. - :
sm C7
sin 0 + sin 40 + sin 70 + &c. + sin (3w - 2)0
3n- 1\ ^ . 3n0
sin
in ( I 0sin
\ 2 / 2
. 30
sin —
2
&c = &c,
311. Coil. Hence also,
sin 0 - sin (0 + ^) + sin (0 + 2^) ~ &c.
^{sin0 + sin(0+2S)+&c.}-{sin(0 + ^) + sin(0 + 3^) + &c.}
may be found ; and if — ^ be substituted in the place of S,
the sum of the series sin 0 + sin (0 — ^) + sin (0— 2^) + &c.
will be obtained.
312 By proper substitutions in the formula above deduced,
the summation of the sines of any series of arcs in arithmetical
progression may be effected : also, if
2«-i ^ ^''K^-i)
^_| — ^ = (2wi — 1) - , the sum will be k— ,
2 2 .0
2 sm -
2
^16
and this has been erroneously called the sum of the series con*
tinued in i?ifinitum, but which in fact cannot be determined.
313. To find the sum of the cosines of a series of arcs
in arithmetical progression.
Let cos0 + cos(f)+^) + cos(^-h2S)-|-&c. + cos(O + (?i-l)^)
be the proposed series, then as before,
2 sin - cos ^ = sin (04-'-] — sin (O | ,
2sin -- cos(0 + ^) = sin {$-{ \ -sin (Oi- -V
&c =&c
2 sin ^ cos{e + (n- l)^) = sin (O-^—-! ^^ - sin (^0 + ?^^);
whence by addition, we have
2 sin ^ S= sin (o + ?^^) -sin (o - ^)
= 2 cos (^ + '^ ^^) sin ~ , by (67).
cos ( a H 0 ) sni —
V 2 / 2
w— 1
cos(6^ +
and /. S =
sui --
2
314. Ex. Let ^ be taken equal to 0, 20, 30, &c. in
succession, and
cos 0 -f cos 20 + cos 30 -f (fee. + cos nO
nO
C^)
cos I — — ] 0 sin
, 0
sni --
2
2117
cos 0 + COS 3 0 -h COS 50 + &c. -I- cos {Qn — 1)0
_ cos nO sin ti9
sin 9
cos 6 + cos 40 + cos 70 + &c. + cos {Sn- 2)0
/3n— 1\ _ . 3ft9
cos ( I 0sin
= V 2 / 2
. 30 '
sin —
&c =&c.
315. Cor. The sums of the cosines of the series men-
tioned in article (311) may be found by the formula above
deduced, as indeed may the sums of the cosines of any series
of arcs whatsoever in arithmetical progression.
31 6. As in the preceding articles^ the sums of the squares,
cubes, &c. of the sines and cosines of the same arcs may be
obtained by means of (272) and (274). Thus
sin^ 0 + sin' (0 + ^) + ^c, + sin' (0 + (« - 1 ) ^) =
1
icos 20 +
1 {cos (2 0-f(»-l)^) sin ?rS]
--- {cos 20 + cos 2(0 + ^) -f&c. -f cos2(0 + (/i- 1)^)}
2 2
_7i ^ I fcos {0 9-\-in — \)d) sm y?d|
*" 5 "" 2 I sin ^ J '
by (313), and so on.
Also, by successive diflferentiations of the examples given
in (310) and (314), the sums of such series as
sin 0 + 2'" sin 2 0 + 3'" sin 30+ &c.
and cos 0 + 2"' cos 2 0 + 3"^ cos 3 0 + &c.
will be obtained.
Similarly, by multiplying by dO, and integrating succes-
sively, the sums of such series as
. , 1 . ^ 1 . ^
sm 9-^ — sm 2 0+— sm 30+ <S:c.
Ee
218
and cos 0 -\ — — cos Q. 0 -] — cos 30 + &Ca
2 3
may be determined.
317. The series menlioned in the preceding articles might
have been summed by means of the expressions for the sine and
cosine of an arc investigated in (263) and (282). Thus, if we
suppose sin 6 = ^= ( oc — - ) , we shall have
sin ^ + sin 2 0 + sin 3 ^ -}- &c. to n terms
?.r -4- r'^ 4- r^ -I- &c. to ?i terms V
, .X + X -\- X' + &c. to ?i terms,
1 fl 1 1 }
. - <- + -17 H — 5 -|- &c. to n terms?
2 ^ _ I U x' x^ )
1 p (:r"- 1) l-x" )
2^^ 1 ^r-1 "^ .r'^(a:-- l)j
2^"=^ i x''{x-]) J
._((''-;;)C' -ip)
5 5
sm — sm I — I y
2 V 2 y
;, as berore.
I
. 0
sin -
2
Again, since cos 0 = we have
2
cos 0 -f cos 2^ -f cos 30 + &c. to ?i terms
219
= i {e^^-^ + e'S^-' + ^^^^^4. &e. to n terms}
2
+ i {e~^^-^ + e-'^^^-f^-''^-' + &c. to^^ terms}
5 I i -i i
. nO
sin — cos
— J as before.
. 0
sin -
2
It moreover appears from (73) that these are recurring
series^ and may therefore be summed by the rules laid down for
that purpose ; but there still remains to be explained another
method not inferior to any that have yet been given.
Let 5' = sin ^H-sin 30 + sin 50 + &c. + sin (2n— \)0,
.'. 5 sin 20 = sin 0 sin 20 + sin 20 sin 30 + sin 2 0 sin50+&c.
+ sin 20 sin (2w — 1) 0, and as appears from (6?) =
- J cos 0 — cos 30 + cos 0— cos 50 + cos 30 — cos 70 + 8cc.
2
+ cos (271 - 5) 0— cos (2?i— 1)0 + cos (2 n - 3) 0— cos (2??+ 1)0 [
= - {2cos0-cos(2«~ 1)0- cos(2?i+ 1)0}
= - {2cos0 — 2cos2?i0 COS0}, by (67),
2
_ cos 0 — cos 2;i0 COS 0 1 — cos 2 7^0 _ siir;/0
* ' " 2 sin 0 cos 0 ~~ 2 sin 0 "~ sin 0
220
318. To Jin d the sum of the series^ cosec 0 -i- cosec 0.8
+ cosec 2' 6 + S:c. to n terms,
9
Here, cosec 0 = cot cot 9.
cosec 9.9 =^ cot 9 -cot 2 9,
&c. = . . . . &.C.. ...
cosec 2^*-'a = cot 2"-'e-cot2"-'^;
.*. by addition, we shall have
cosec 9 4- cosec 20 + 8cc. + cosec 2''~^9;
9
= cot cot Q'^~^9,
2
319. To find the sum of the series^ tan 9 -^ 9, tan 9.9 +
2"^ tan 2' 9 + <S)C. to n terms.
It is easily proved that
tan 0 = cot ^ - 2 cot 2 9,
2 tan 2 0 = 2 cot 2 0 — 2* cot QT 9,
2^ tan 2^9 = 2^ cot 2'^ - 2^ cot 2^9,
&c. . . . = &c
2"-' tan 2"-'^ = 2""' cot 2"-'^ -2'' cot 2"^;
.*. by addition, we get
tan (9 + 2 tan 2 ^ -f &c. +2"-' tan 2"*^^ = cot ^ - 2" cot 2'*^.
320. To find the sum of the series
10 1 \ 9 9
^tan- + -, tan -, + ~, tan -
tan ;: + :::2 ^^'^ ~2 "i — 3 ^^^^ ~r &c. to n terms.
9 9
We have seen in (112) that tan - = cot - — 2cot0;
2 2
221
\ e \ e
,'. - tan - = - cot coty,
2 2 2 2
1 ^1 e \ e
-7 tan -^ = --2 cot -z — -^ cot - ,
<2^ 2^ 2^ 2-2 2
Szc = &c.
10 10 1 0
— - tan -r = — cot —- — 7 cot
2» £/j 2'* 2" 2""^ 2'^~^ '
1 9
hence by addition, the required sum = — cot ~ — cot 0 ;
and if « be infinite^ this becomes ^ — cot 0.
u
321. To find the sum of the series, (tan 0 + cot 0)
+ {tan 26 + cot 2 0) + (tan 2'0 + ^ot 2^0) -f ^c. to u terms.
In this case, we have
tan 6 + cot 0 = 2 cot 0 — 2 cot 2 9,
tan 20 + cot 2 0 = 2 cot 20 — 2 cot 2^0^
&c = &c
tan 2"-'0 + cot 2''-'0 = 2 cot 2"-^0 - 2 cot 2"0;
.'. the required sum = 2 cot 0 — 2 cot 2"0.
322. To find the sum of the series x sin 0 + x* 5/;/ 2 0
+ x^ sin 39 + S^c. to n terms.
By means of the formulae investigated in (282), if S be the
required sum, we shall have
.■..(•:^7::-^)^..c-^---"-'")
-j- &c. to ;/ terms
^22
2^-1
=^|^.e*^^^ + ^V'^-"^ + ^r^e'^^+&c. + ^"e"^^-')}
2 V -1 ^
■ ~ £^^ 1 x'-{e'^' + e-'^^^) :r + 1 I
■ 1 ( x{e^^-'^e-'^-^) )
.t'*+^ sin nO- ^" + ^ sin (/i + l) ^ + :r sin^
2~ ~ 2 cos 0 X -\- \
If :i^ be a proper fraction and n indefinitely great^ we
shall have the sum of the series continued in infinitum
X sin S
x^ •— 9, cos 0 X -^ \
323. To find the sum of the series x cos 0 -^ x^ cos 2 0
+ x^ C05 3 0 + Sfc to n terms.
As before,
2
~2
+ :r ( — ) + &c. to n terms
= - {^ e^^-^ + / e-^^^ + a^^^ ^'^'^ 4- &c. + a-" e"^^-^:
2
223
2
^^e^^-i cr^J^rzrr^n
« + 2 /gnQvCrr _|_ ^-«6
,6v _ i , ^-^vzn
11 ^ u
.r^ + ^ cos ne- 0?" + ' cos (w + l) ^ + x cos 0 - x"
x" — 2 cos 0 X -{- \
and as in the last article, the snni of the series indefinitely
continued
X cos 0 — x"^
"~ x'^ ~ 2 cos 9 X ■\- \ '
324. By means of the operations of differentiation and
integration as pointed out in article (v3l6), the sums of various
other trigonometrical series may easily be determined ; but the
almost entire absence of utility renders it unnecessary now to
devote more time to the subject. We shall however conclude
this Chapter with two or three instances in which some of
the preceding series are made available to the solution of
more important Problems.
325. From (322) it appears that the sum of the series
sin d-]rx sin 20+.r^ sin 30-|~Slc. continued in injinitum is
sin 9 sin 9
x^ — 2 cos 9x + 1 1 — .r (2 cos ^ — .r) '
and therefore bv actual division we shall have
224
sin e + x sill '^e+x' sm SO + Sic.+x'*~' siiiw^ + Scc.
= sin 0 {\-\-x{9, cos 0-x) + x^ {2 cos 0 - xY-{-^c.
+ ar"-2 (2 cos 0 - xT-^-hx''^' (2 cos 0 -^rr"' + 8cc.} :
whence by expanding the binomials and equating the coefficients
of x^^"^ on both sides, we shall obtain
sin nO = sin G ((2 cos OT''' - ^^^—^{2 cos Of-''
(.-3)(.-4) ^^^ 1^
Similarly by means of the series summed in (323), it may be
shewn that
cos;i0 = l.|(2cos^r--w(2cos^r-^^+^^^^i^(2cos^f--'-8cc.|,
which might have also been readily derived from the preceding
by the operation of differentiation.
27r
326. If we suppose 0 = — or 72*^, the second example
5
of (314) gives
^ cos2^sin2^ , sin40 sin(27r-^)
cost^ + cos3y= 7—7^ = ^ ■ ^ = ^ : — =— i;
sin a 2 sm^ 2 s,n^ 2'
also from (6?) we have sin 0 cos 30= ^ ^cos 4^ + cos 20 \
= §{cos(27r-^) + cos(27r-3^}= | {cos 0 + cos 30} = - ^ :
and from these two equations are immediately deduced
cos 0 = cos 72^= "^^"^ andcos30=cos2l6'= " "^ ^ " \
4 ' 4 ^
which are the same as would have been found by the methods
pointed out in the second chapter*
225
327. Let 0 = — or 170 = 7r, then from the example re-
ferred to in the last article we have
cos 80 sin 8 0
cos 04-cos 30-f COS50 + &C. + cos 150 =
;in0
I
sin 160 __ J sin (ti— 0) ^
sin0 ~2 ^iJT^ - 2'
assume now x = cos 0 + cos 90 + cos 130 + cos 1 5 0,
and y = cos 3 0 + cos 5 0+ cos 70 + cos 110;
then if these two quantities be multiplied together, and their
product be reduced by (67)j we shall obtain
X7/ = Q. {cos20 + cos40 + cos60 + &c. + COS 160}
= -2 {cos 150 + COS 130 + cos 1 10 +&c. + cos0}= - 1,
by what has just been proved :
whence the equations x-\-i/=—, and jry= — 1, give
i+x/Tt" , i-^'y?
X = 5: and y = ^^^ :
4 -^ 4
Again, let s = cos 0+cos 130, and ^=cos 9^4-cos 150,
also M = cos 30 + cos 5 0, and v=qos 7 0 + cos 11 0,
.K . ^. 1+n/^ ^ l-x/T7
so that s + ^ = ^^^ , and u +v — — ;
4 4
whence proceeding as before we shall obtain
St ■= — J: , and uv = — |:;
and thus the four quantities 5, t, u, v, may be determined :
hence since cos0+cos 130 = 5, and by (67), cos 0 cos 130
= i {cos 120 + cos 140} = - ^ {cos 30 + COS50} = - ^ .
u
5
the values of 0 and 130 are readily obtained.
Ff
226
This article enables us to determine the side of a regular
polygon of 17 sides inscribed in a circle whose radius is 1^
which is manifestly = chd 2^ = 2 sin ^ = 2>/l— cos^ 0,
In what we have just been doing, no reason has been
assigned for the assumptions there made : and in fact no reasons
can be given without entering upon a theory much too difficult
for a place in an elementary Treatise like the present. The
invention of such a theory is due to M, Gauss, Professor of
Mathematics at Strasburgh, and it may be seen fully developed
in his work entitled Disquisitiones Arithmetic cz^ which has been
translated into French by M. PouUet-Delisle under the title of
Kecheixhes Arithmetiques. On this subject the reader is referred
also to the last chapter of Barlow'^s Elementary Investigation
of the Theory of Numbers.
SPHERICAL TRIGONOMETRY.
CHAP. I.
DEFINITIONS AND PRELIMINARY PROPOSITIONS.
Article I. Definition I.
Spherical Trigonometry treats of the relations between
the sides and angles, &c. of figures formed by the intersections
of three or more planes with the surface of a sphere.
2. Everi/ section of the surface of a sphere made by a plane
cutting it, is the arc of a circle.
Let O be the centre of the sphere, ABC the section made
by a plane passing through it ; draw OD perpendicular to this
plane and produce it both ways to meet the surface in E and F,
join ADf BDj CD, and draw the radii of the sphere OA,
OB, OC: then by Euclid xi. Def. 3, ODA, ODE, ODC
are right angles ;
228
that is DA^ = DB' :=DC' = &c. or DA = DB=DC = &c.
and therefore the section ABC is a circle whose centre is D,
and radius - DJ = DB= DC = &c.
3. Cor. If the distance of the cutting plane from the
centre of the sphere be called d, and the radius of the sphere r,
we shall have the radius DA of the section = s^ OA^ — OD^
= A^r^ - d'^: and if d = 0, or the cutting plane pass through
the centre of the sphere, the radius of the section is equal to
the radius of the sphere, and its centre coincides with the
centre of the sphere.
4. Def. 2. The pole of a circle of the sphere is a point
in the surface of the sphere from which all straight lines drawn
to the circumference of the circle are equal.
6. Cor. Hence if the line OD be produced both ways
to meet the surface of the sphere in E and F, these points a e
called the poles of the circle ABC, the former the near, the
latter the remote pole.
6. Def. 3. When the cutting plane passes through the
centre of the sphere, the radius of the section being equal
to the radius of the sphere is the greatest possible, and the
circle is called a Great Circle of the sphere : in all other cases
the section is termed a Small Circle.
7. Cor. 1. If the section pass through the centre of the
sphere, the points O and D coincide, and the poles of a great
circle are the points of intersection with the surface of the sphere
made by a perpendicular to the circle passing through its centre ;
and it is manifest that the arc of the sphere intercepted between
the circumference of a great circle and either of its poles is
a quadrant.
8. Cor. 2. Hence two great circles of the sphere bisect
one another^ because they have a common centre^ and iheir
229
common section being a diameter of each therefore bisects
them.
Q. Def. 4. The arcs on the surface of a sphere are
always understood to be portions of great circles unless the
contrary be expressed : a figure formed by three such arcs is
a spherical triangle : by four a spherical quadrilateral, &c : and
by n such arcs a spherical polygon of 7i sides.
10. Def. 5. The angles of spherical triangles, &c. are
those on the surface of the sphere contained by the arcs of
the great circles which form the sides, and are the same with
the inclinations of the planes of those great circles to one
another.
11. Ani/ two sides of a spherical triangle are together
greater than the third, and the three sides are together less than
the circumference of a great circle.
Let ABC be a spherical triangle on the surface of a sphere
whose centre is O ; draw the radii of the sphere OA, OB,
B
OC to the angular points : then since the solid angle at 0 is
contained by the three plane angles AOB, AOC, BOC, any
two of which are by Euclid xi. 20. together greater than the
third, it follows that any two of the arcs which measure these
angles are together greater than the third: that is, AB + AC
is greater than BC, AB + BC greater than AC, and AC + BC
greater than AB.
Also, since the solid angle at 0 is contained by the three plane
angles JOB, AOC, BOC, which by Emc/?V7 xi. 21. are together
rso
less than four right angles, it is manifest that the three arcs AB,
AC and BC are together less than the circumference of a great
circle.
Hence if the sides be denoted by a, b, c, and the radius of
the sphere be 1, then a + b > c, a + c > b, b + c > a, and
a + b + c < 27r.
12. CoR. For the same reason, since {Euclid x\. 21.) every
solid angle is contained by plane angles which are together
less than four right angles, it follows that all the sides of a
spherical polygon are together less than the circumference of
a great circle.
13. Def. 6. If with the angular points of a spherical
triangle as poles, great circles of the sphere be described,
the figure formed by the intersections of these circles is called
the Polar Triangle, in contradistinction to which, the proposed
one is styled the Primitive Triangle,
14. The angular points of the polar triangle are the poles
of the sides of the primitive triangle.
Let A BC be the primitive triangle, DFE the polar triangle
D
described according to the definition, and let the great circles
be produced as in the figure : then
since A is the pole of DE, AD is a quadrant,
and since B is the pole of DF, BD is a quadrant:
231
.•. the distances of the points A and B from D being quadrants
are equal to one another,, and consequently D is the pole
of AB: for the same reason the angular points E and F oH the
polar triangle are the poles of the sides j^C and BC of the
primitive triangle.
15. The sides and angles of the polar triangle are the
supplements of the angles and sides respectively of the primitive
triangle.
The same construction remaining, and the radius of the
sphere being supposed =1, so that an angle may be equal
to the arc which measures it, we have
zA=HM=DH-DM=:DH+ME'-DE='n—DE',
similarly A B = nr — BF, and zC = 7r--EF:
again A D = GH= GB -h BH= GB + AH- AB^ir - AB:
similarly z£ = 7r-lC, and Z.F=7r—BC.
Hence if «, hj c. A, By C be the sides and angles respectively
of the primitive triangle, and a\ b', c, A\ B' , C' those of the
polar triangle, we shall have
a'=7r — u4, b' = TT — B, c'=7r— C,
and A'= TT — a, B' = tt — b, C' = tt — c;
and from these properties the polar triangle is frequently styled
the supplemental triangle.
16. Cor. 1. If one or more of the sides or angles of the
primitive triangle be quadrants or right angles, the corresponding
angles or sides of the polar triangle will be right angles or
quadrants.
17. Cor. 2. Hence the sum of the three angles of a
spherical triangle lies between two and six right angles.
For, since by (11) a''\-b'-\-c is less than Stt, it follows
that A + B + C = Stt — {a + b' -{- c) is greater than w or two
S32
right angles: and we manifestly \m\e A-\-B+C=37r—{a-}-b' -^c)
less than Stt or six right angles.
A spherical triangle may therefore have two or three right
angles, or two or three obtuse angles.
18. Cor. 3. Hence also the sum of any two angles of
a spherical triangle exceeds the third by less than two right
angles.
For, since by (11) a -{-b' is greater than c , we have
TT — A-\-7r — B greater than tt — C, or tt greater than A + B— C,
.*. ^ + -B — C is less than tt : similarly J + C—B, und B + C- A
are each less than tt.
19. Cor. 4. In the same manner if the sides of a sphe-
rical polygon be each less than a semicircle, and with its
angular points as poles great circles be described, another
spherical polygon will be formed which will be supplemental
to the former.
20. Def. 7. If one of the angles of a spherical triangle
be a right angle, it is called a right-angled triangle; if one of
the sides be a quadrant, it is called a quadrantal triangle, and all
others are called oblique-angled triangles.
S33
CHAP. II
On the relations between the sides and angles, S)X. of spherical
triangles.
21. 7^0 express the cosines of the angles of a spherical
triangle in term^ of the sides.
Let ABC be a triangle on the surface of a sphere whose
centre is O and radius = 1, the angles being A, J5, C and the
corresponding opposite sides a, b, c : let ADj AE touching
the arcs AB, AC nt the point A^ meet the radii OB, OC pro-
duced in D, E, and join DE : then we have
AE = tan AC = tan b, AD = tan AB = tan c,
OE = sec AC = sec b, OD = sec AB = sec c :
now in the triangle DOE, we have from (l65) PL Trig,
DE' = OjE' +0D'-20E. OD cos DOE
= sec^ b + sec^ c — 2 sec b sec c cos a
= 2 -f tan^ 6 + tan^ c — 2 sec 6 sec c cos a ;
Go
234
also, in the triangle EAD, we have by the same article
DE' = AE^ +AD^'-QAE.AD cos DAE
= tan^ b 4" tan^ c — 2 tan 6 tan c cos A :
whence equating and transposing we get
2 tan b tan c cos A = 2sec b sec c cos a — 2,
2 sec b sec c cos a — 2 cos « — cos b cos c
and .*. cos A =
2 tan b tan c sin b sin c
cos /; — cos a cos c ^ cos c — cos a cos 6
similarly cos JD= ^ : , and cos C= : : — 7
•^ sin a sm c sm a sni 0
Ex. 1. If a = 6, we have
cos a — - cos a cos c cos a (1 — cos c)
cos il =
2 sin^
sm « sm c sm a sm c
c
cos a 2 ^ 7 ^ r>
= -: : = cot a tan - = cot b tan - = cos ±> i
sm a sm c 2 2
cosc-cos^a 2 0
and cos C = r-^ = cos c cosec a — cot a.
sm a
Hence we have A ^^ B, or the angles at the base of an
isosceles spherical triangle are equal to one another.
Ex. 2. Let a=^b = c, then we shall have
cosa — cos^a cos«(l— cos«)
cos A =
sin a yy/ 1 — cos^ a
a
tan
4/1 —cos a « 2 _, ^
= cot a y — ; = cot a tan - = = cos B = cos C
1 + cos a 2 tan a
Hence every equilateral spherical triangle is also equiangular.
235
22. Cor. If the angle at C be a riglit angle^ we have
0 = cos c — cos a cos b^ and therefore according as cos a and
cos b have the same or different signs, cos c will be positive or
negative ; that is, according as the sides are of the same or
different affections^ the hypothenuse is less or greater than a
quadrant.
23. To express the sines of the angles of a spherical
triangle in terms of the sides.
Since sin ^ = ^ 1 — cos^ A = ^(1 — cos A) (1 + cos A),
cos a — cos b cos c
and cos A =
we have 1 — cos J. = 1 —
sin b sin c
cos a — cos b cos c
sin b sin c
sin b sin c — cos a + cos b cos c cos (6 — c) — cos a
sin 6 sin c sin 6 sin c
/a-^-b-cx . /a + c-^»\
^«^«(— i— )-(-^— )
r-f-. ^ ^, by (67), PI. Trig.
sm 6 sm c ^ n » o
, , . , cos a " cos b cos c
and 1 + cos A = 1 ^ : — ; — :
sm b sm c
sin b sin c + cos a — cos b cos c cos a — cos (b + c)
sin 6 sin c sin b sin c
= r— ^— : , by (67);
sm 0 sm c
\eta + b +c=Q.S, .-. 6 + c - a = 2 (5 - a);
a + c — 6 = 2 (5 — 6), and « + 6 — c = 2 (5 - c) ;
2 sin (5-6) sin (*S - c)
1 — cos A =
^nd 1 + cos A =
sin 6 sin c '
2 sin S sin (5 — a)
sin 6 sin c
2S6
2
whence sin -4= -: — - — : — /^ sin S sin {S — a) sin {S — b) sin (6* — c) -
sin 6 sin c
2 / : :
similarly sin B= — : — aV sin S sin (S — a) sin (S — b) sin (S — c),
sin a sin c
and sin C = -: r— r ^/ sin 5 sin (5 — a) sin (S — b) sin (S - c).
sin a sin 6
Ex. 1. Let a = b, or the triangle be isosceles, then
. . 2sin(^ — «) f- — -—.
sin A = — : : /v/ sin S sin (S — c)
sin a sin c
. c
% sin -
2
sm a sm c
\/sin(a+^)sin(a-0
V sin ( a H — \ sin ( a -
2/
c
sm « cos -
2
V sin(/>+ ''-\^mih -
-D
= sin ^ :
sm 6 cos -^
2
. 2sin(5-«) 1 . .. • .o V
and sin C = r-^ \J ?A\\ o sin (o— c/
2 sin -
• 2
sm
I ^ y ^ ^
\/ sin ( a -) — ) sin (a )
a ^ V 2/ V 2/
Ex. 2. Let a = ^=6', then in an equilateral triangle
we have
jin A
__ 2 /i^ sin *S' sin^ {S — a)
237
A / . 3a . ~ a
2 V sin — sni --
sin a . o a ofl
4 sm" -- cos
. 3a
sm
.2«
3—4 sin*^ - = sin J3 = sin C.
2 cos - sin
24. Cor. 1. Hence rejecting the common factors, we have
sin j4 : sin B : sin C = sin a : sin b : sin c ;
or the sines of the sides of a spherical triangle are to one
another as the sines of the opposite angles.
A
25. Cor. 2. Since 1 — cos -4 =2 sin"^ — , and 1 + cos A
2
= 2 cos' — , we have from (23) by reduction,
2
sm — = v
sin (5 — h) sin {S — c)
2 sin h sin c
^ ^ /sin 6' sin (8 - a)
and cos — = v ^ — t — :^ >
2 sin 0 sin c
A A / s'm (S — b) sm (S — c)
and thence tan ~ = y -. — ^—. — — ; —
2 sin o sin (o — a)
sin —
,. ^ „ , 2 ^ / sm a sm (S — b)
20, Cor. 3. Hence also 5 = V "^ — ; — • — 77^ ^ >
. B sin b sin (o — a)
sin —
2
and therefore according as a is greater or less than b, sin --
238
will be greater or less than sin — , and theiice A greater or
less than B : that is^ the greater side of every spherical triangle
is opposite the greater angle, and the contrary.
27. To express the cosines of the sides of a spherical triangle
in terms of the angles.
Let a, 65 c, Aj B, C be the sides and angles of the pro-
posed triangle, a, b', c , A\ B/ C those of the polar triangle,
then by (21), we have
. , cos a — cos h' cos c'
cos A = . ,. . ;; ;
sni o sm c
but cos ^' = cos (tt — a)= — cos a, cos a' = cos (tt ^ A) = — cos A,
cos b' = cos (tt — jB) = — cos B, cos c = cos (tt — C) = — cos C,
sin b'= sin (tt — JB) = sin B, sin c' = sin (tt — C) = sin C;
cos A + cos B cos C
/. the formula just given becomes cos a = : — j^ — : — 7; -
^ ° sm B sm C
cos B -f cos A cos C
similarly, cos 6 = : — -. — : — 7; ,
sm A sm C
cos C + cos A cos jB
and cos c = : — -. — : — fi •
sm A sm i>
cos A I + cos C
Ex. 1. Let A = B, .'. cos a = -: — ;; : — 7^—
sm A sm C
c c
= cot ^ cot — = cot B cot — = cos 6 ;
cos C + cos A „ ^ i . ..2 ^
and cos c = r-r—, = cos C cosec" ^ + cot A .
sm A
Hence if two angles of a spherical triangle be equal to one
another, the sides which subtend them are also equal.
239
Ex. 2. hetA = B=C, .'. cos a=cotyl cot — =cos 6=cosc:
2
wherefore equiangular spherical triangles are also equilateral.
cos A
28. Cor. If C =90", we have cos a = -: — —\ .'. cos a
sin i5
will be positive or negative according as cos A is positive
or negative : that is, a will be less or greater than a quadrant
according as A is less or greater than a right angle: or the sides
of right-angled triangles are of the same affections as their opposite
angles.
29. To express the sines of the sides of a spherical triangle
in terms of the angles.
From the last article but one we have
cos A + cos B cos C
1 — cos a = \
sin B sin C
sin 5 sin C — cos A — cos B cos C _^ cos A + cos (B -{■ C)
sin B sin C sin B sin C
A + B-{-C\ /B-\-C-A
2 cos
sin B sin C
cos yl + COS B COS C
and 1 + COS a = ] + • ti • ^^
sni ±> sin C
sin B sin C + cos A -^ cos i^ cos C __ cos A + cos (B - C)
sin £ sin C sin J3 sin C
2 cos
(l±|^)cos(^i±^)
sin B sin C
.'. assume 2.S' = A -f 5 + C, whence we shall have
2 cos S' cos{S' -A)
1 — cos a =
sin B sin C
and 1 -f cos a =
240
2 cos (5' - B) cos {S' - C)
sin ^ sin C
whence sin a
= ■ ^- ^^ -cosS'cosiS'-A)cos{S'-B)cos{S''-'C);
sin ±> sm C
similarly sin b
_ 9,
sin A sin C
and sin c
2
sin J. sin B
^-cosS' cos (5'-^) cos (S'-B) cos (S-C);
V - cos y cos (6"-^) cos (S' - jB) cos (6"- C).
It may here be remarked that since by (17) the sum of the
three angles of a spherical triangle is greater than two right
angles and less than six, -S' is manifestly greater than one right
angle and less than three, and consequently cos 5' is a negative
quantity; also since by (18) the excess of the sum of any two
angles of a spherical triangle above the remaining one is less
than two right angles, it follows that .S'— ^, S'— B and 5'— C
are all less than one right angle, and therefore that cos {S'— A),
cos {S' — B) and cos (S^ — C) are all positive, from which it
results that all the expressions just investigated are possible,
though they appear in an imaginary form.
30. Cor. 1. From the demonstration of the last article
we have
. 2 «
2 sin - = 1 — cos a= —
^ /A + B+C\ /B-{-C-A\
2 cos I 1 cos ( I
V 2 / V 2 /
sin jB sin C
/^4-B~C\ /A + C-B\
2 cos I ) cos ( I
V 2 7 V 2 /_
sin B sin C
/ /A+B + C\ /B + C-A\
/ — cos ( ) COS I I
I ■ ^^ \/ V 2 / V 2 /
whence sin - = y
2 sin B sin C
- 2 ^
and 2 cos -- = 1 -f cos a
s/
241
cos .S' (cos S' — A)
sin B sin C
/ /A + B-C\ /A + C-B^
^ /eos(~^-^— )cos(— ^ )
and cos ~ = V
sin B sin C
-V
cos (*S'- jB)cos (S'~ C)
and .*. tan
sin B sin C
5 ^ ^ cos(y-ii) cos(6"-C)
31. To express the tangents of the semi-sum and semi-
difference of two angles of a spherical triangle in terms of their
opposite sides and the remaining angle*
cos a — cos 6 cose cos c — cos fl cos 6
Snice cos A— : — ; — : . and cos C = : : —
sm h sni c sin a sin b
cos a — cos 6 cos c
.'. cos A sin c = : — ;
sm 0
cos a cos b
sm
b sin b
(sin a sin b cos C + cos a cos /;)
cos a . 1 r^ ^^^' ^^
—. —'sin a cos b cos L — cos a — — r-
sin b sin b
cos a . 1 r^ cos a .
= ~ — r — Sin a cos b cos C : — r -f- cos fl sm f/
sm b sin 6
= cos a sin & — sin a cos /; cos C :
similarly cos B sin c = cos ft sin a — sin ft cos a cos C;
.*. (cos A -}- cos jB) sin c = sin (rt -|- ft) ( 1 — cos C) :
sin j1 sin B sin C , ,,
but since from (24) — : = —: — r = — r —
sm a sin ft sm c
Hh
242
(sin A ± sin B) sin c = (sin a ± sin b) sin C J
hence using the upper sign we obtain
sin A + sin B sin a + sin b sin C
cosil+cosjB sin (a + 6) 1 -- cos C '
a — b^
/a- d\
cos I I
A + B V £ /
and by means of the lower we get
or tan = r— cot — ;
a-\-b\ 2
cos
sin A — sin B __ sin « — sin b sin C
cos A + cos 5 sin {a-\-b) 1 — cos C
iin I 1
( ) = — cot —
sin
These equations converted into proportions constitute what
are from their inventor called Napier^s first and second Analo-
gies.
Ex. If C = 90^, we shall have for a right-angled triangle,
a - ^A . /a — b'
/a - ^;\ . /a — 0\
\ 9. J /a + b\ \ 2 / . /a-h
cos
/A+B\
tan
\ 2 /
cos
m ^ ^ ' -c-f^)
32. Cor. From the two equations just investigated we
have
'a-\-b\ . /a + b'
. Q ) /A-^-Bx _ ^^"\ 2 .
/a- b\ ^ \ o )" ^a — bx
C V 2 / /A-\'B\ V 2 / /A-B\
cot — = 7— tan I I = — tan ( ).
2 /a-~b\ \ 2 / . /a'-b\ \ 2- /
cos
243
33. To express the tangents of the semi-sum and semi-
difference of two sides of a spherical triangle in terms of their
opposite angles and the remaining side.
Retaining the notation before used, we shall by the last
article but one have in the supplemental triangle
cos
/A' + B\ ^^^ V 2 ) C
tan I I = , . w cot — ,
cos
a-'b'
sin '
and tan I ^ — ) = ; — -7— cot — :
V 2 / . /a +h\ 2 '
sin I I
V 2 /
and by effecting the proper substitutions as in (27) we shall
obtain
a + h^
/a-i- o\
tan ( I = . . ^ tan r->
V 2 / /A-\'B\ 2'
cos
sin (^~^\
and tan (^)=-—^ tan-;
sm
which converted into proportions as before are Napier's third
and fourth Analogies.
34. Cor. Hence in the same manner as in (32), we have
cos I 1 , , sin ( — - — ) ,
cos(-^) s,n(-^-j
244
Ex. Let c = ~ , then in a quadrantal triangle we have
tan
\ 2 /
cos
(£zj?) ^ 3... (i^)
— ^ -— , and tan ( ) = =7-
35. To express the co-tangent of an angle of a spherical
triangle in terms of another ajigle, and the sides zohich include it,
cos a — cos 6 cos c ^ cos c— cos a cos 6
Suice cos A — : — - — -. J and cos C= -. -. — ;
sin 0 sin c sm a sm 0
.*. sin b sin c cos A = cos a — cos b cos c ;
sin C . • • 7 ri , L
but sin c = -: — - sin (?, and cos c = sin « sin b cos C + cosrz cos 6;
sin A
.*. cot A sin « sin 6 sin C=cos a — sin a sin b cos 6 cos C
— cos a cos^ 6 = cos a sin^ 6 — sin a sin 6 cos b cos C,
cos <^ sin 6 cos C
.'. cot A — —. -: — — — cos o — — —
sm a sin C sin C
= cot a sin /> cosec C — cos 6 cot C.
36. 2 a express the co-tangent of the side of a spherical
triangle in terms of another side, and the angles which are
adjacent to it,
cos J.+COS i3 cos C cos C + cos il cos ii
Since cos « = : — r—-. — — , and cos c = -. — - — : — ;
sm h sm C sm A sm h
we have sin B sin C cos a = cos ^-|-cos B cos C;
but sin C = - — sin A, and cos C=sin A sin B cos c— cos A cos B ;
sin«
.*. cot a sin A sin jB sin c = cos A + sin ^4 sin B cos JB cos c
~ cos A cos" jB = cos A. sin^ i^ + sin A sin 5 cos B cos c,
cos A sin jB cos c
whence cot a = -: -: 1- cos B —
sin A sm c sm c
= cot A sin B cosec c + cos B cot c.
CHAP. Ill
On the Solution of Spherical Triangles.
37. r*ROM the pieceding chapter it appears that the
following relations between the sides and angles of spherical
triangles have been established; namely, from (21)
cos a — cos b cos c _ cos b — cos a cos c
cos A = : — -—. , cos B = : : '
sni o sni c sm a sm c
cos c — cos a cos ^ , , .
cos C = : : — ; and from (27),
sm a sm o
cos A+cos B cos C , cos ^ + cos A cos C
cos a = : ^— : — p:; , COS 6 = : -— : ,
sin B sin C sm ^ sm C
COS C + cos A COS JB
cos c =
sin tI sin B
and since each of these sets contains three independent equations,
of the six quantities involved in them any three being given^ the
remaining three may be found ; but if one of the parts of the
triangle be a right angle or a quadrantal arc^ it is manifest that
only two other parts will be necessary for the discovery of all
the rest.
On this account therefore the solutions of spherical triangles
are distributed under the three following heads :
I. Solution of right-angled triangles.
II. Solution of quadrantal triangles.
III. Solution of oblique-angled triangles.
246
and a proper application of the propositions contained in the
last chapter, will enable us to effect the solutions of all their
particular cases.
I. Solution of Right-Angled Triangles.
38. From what has been said, it appears that all the
cases of right-angled spherical triangles may be solved by
means of either of the sets of formulae above given ; but it
may also be observed that the substitutions and eliminations
necessary to effect these solutions would in many cases be
too tedious for practice, and their results burdensome to the
memory. To remedy this inconvenience, Baron Napier the
celebrated inventor of logarithms devised two rules easy to
be remembered, which are sufficient for the solutions of all
cases of right-angled spherical triangles, and of which the
following explanation may be given.
If C be supposed to be the right angle, there remain five
other parts belonging to every triangle, namely, the two sides
or legs a, b, the hypothenuse c, and the two angles A, B : now
TT
the two legs a, ft, the complement of the hypothenuse - — c,
2
and the complements of the two angles - — J^, - — B are
by Napier termed Circular Parts, the right angle being left
entirely out of the consideration, and any one of these parts
may be assumed to be what he calls a Middle Part : then the
two parts which lie close on each side of it are called Adjacent
Extremes, and the two remaining parts which are farthest off
from it and separated from it by an adjacent part are termed
Opposite Extremes. This being premised, the two following
equations are found universally to obtain, and are called Napier's
Rules :
(1) Radius x the sine of the middle part = the rectangle
of the tangents of the adjacent extremes.
M7
(2) Radius x the sine of the middle part = the rectangle
of the cosines of the opposite extremes :
and from these equations if any two of the quantities involved be
given, the remaining parts may be immediately derived.
39. To prove Napier's Rules.
TT
First, let one of the legs a be the middle part, then - — JB
and h are the adjacent^ and -^ — A and - — c the opposite
extremes :
, . , cos JB + cos^ cosC cos £ . 0
now by (27) cos 0 = ^ — - — : — — = - — - , smce C=90 ,
sHi A sm C sHi A
. sin B sin a
.'. cos B =: sin A cos o = : — ;; cos 0, by (24);
sm 0
sin b cos B » t^
whence sm a = r -. — — = tan 0 cot B,
cos 0 sm B
or r sin a = tan 6 tan ( - -- Bj (1)
similarly r sm b = tan a cot A = tan a tan ( A\ (2)
^ sin a sin j1 . . . , .
agam by (24) —. = -: — — = sm ^, .'. sma = sm c sm A,
sm c sm C
or r sm a = cos (- - cj cos (- — A\ (3)
similarly r sin 6=sinc sin B=cos(- — c\ cos (- — B\ (4)
TV
Secondly, let the complement of the hypothenuse - — c be the
die part, then the adjacent extrei
and the opposite extremes a and b :
o
middle part, then the adjacent extremes are A and - — 5^
now by (27) cos c
248
cos C + COS A cos B
sin A sin B
cos J. cos B
sin J. sin B
cot A cot jB,
••• »• ^i" (i - <') ='"" (i - ^) '''" (i - ^) (^>
^ ^ ^ cose— cos a cos Z> ,
and by (21) cos C = : — ; — : = 0, .'. cos c=cos a cos by
sin 0 sm c
in (- — c) =cosa cos b (6).
or r sm
TT
Lastly, let the complement of one of the angles -- — ^ be
TT
the middle part, then the adjacent extremes are h and ^ — c,
TT
and the opposite extremes a and jB :
cos c
— cos b cos c
. . cos a — cos o cos c cos 6
now by (21) cos A = -
sin b sin c sin b sin c
cose (1 — cos^6) cose sin b sin 6 cos c
: — = tan b cote;
sin b cos 6 sin c sin 6 cos b sin e cos 6 sin e
or r sin / - — A ) = tan b tan ( - - e ) (7)
similarly r cos B — tan « cot c,
or r sin( ^)~ ^^"^ ^^'^ { " "" ^ ) (S)
. , ,^ , cos J. + cos S cos C cos tI . ^, -
again by (27) cos a = : — -— ^ — = ~r~~ , since C=90^
sin i5 sm C sm B
.'. cos tI = cos a sin B,
or r sin ( Aj = cos a cos f - — B | (Q)
249
similarly cos B = cos b sin A,
or r sin ( jB ) = cos b cos f - — ^ j (10),
and since out of five things taken two and two together, there
5.4
can be formed - — or 10 combinations, it follows that the ten
1 .2
equations above deduced include all the cases that can possibly
occur : moreover they are all adapted to logarithmic com-
putation.
We will now illustrate the use of these rules by the following
examples in which the radius is supposed to be 1.
Ex. 1. Given a and b, to find the rest.
From (l), sin a=tan b cot B, .'. cot B = = sin acotb:
tan b
(2), sin b=tan a cot A, .', cot A = = sin b cot a :
tan a
(6), cosc=cos a cos ^ ;
whence B, A and c may be found, and it is manifest that there
is no ambiguity.
Ex. 2. Given a and c, to find the rest.
sin a
From (3), sin a = sin c sin A, .*. sin A =
(6), cos c = cos a cos ^, .*. cos ^ =
sni c
cos c
cos a
(8), cos B = tan a cot c :
whence Ay c, B may be determined, and there can be no
ambiguity except in the value of A, and this is removed by
means of the circumstance stated in (28).
Ii
250
Ex. S. Given a and A, to find the rest.
_ . . • . i • sin a
From (3). sin a = sm c sm A, .'. sm c = -: — 7 ^
sin A
(2), sin Z> = tan a cot A :
cos A
{9)3 cos A = cos « sin B, ,\ sin jB =
cos a
hence the sines of c, h, and B, and therefore the parts them-
selves may be found : but it may be observed that there is
nothing to decide whether c, h, and B should be greater or less
than - or QO^^ and therefore the solution is ambiguous; and as
in Plane Trigonometry (233), it is readily shewn that there
may be two right-angled spherical triangles, which possess the
proposed data, and in which the required parts are supplemental
to each other.
Ex. 4. Given a and B, to find the rest.
From (1), sin a = tan 6 cot 13, .'.tan 6= =sinatani5:
cotB
/^\ Ti cos_B _
(8)5 cos jd = tan a cot c, .*. cote = = cosi5cota:
tan a
(9), cos A = cos a sin B ;
therefore h, c and A may be determined, and there is no
ambiguity in the solution.
Ex. 5. Given c and J, to find the rest.
From (3), sin a = sin c sm A:
cos c
(5), cos c = cot A cot ^, .*. coti5= r=cosctanJ^:
^ cot A
(7), cos -A = tan 6 cot c, .'.tan 6 = = cosAtanc;
cot c
whence a^ B and b may be found, and there can be no ambiguity
except in the first, which may be removed by means of the
considerations noticed in (28).
251
Ex. 6. Given A and B, to find the rest.
From {5), cos c = cot A cot 5 :
(9), cos j1 = cos a sin Bj .*. cos a =
cos ji
sin ii
cos
B
(10), cos 5 = cos & sin A, .*. cos 6 = — .
^ ^' ' sin ^
therefore c, a, and Z> may be found without ambiguity.
II. Solution of Quadrantal Triangles,
40. Let ABC be a spherical triangle whose sides and
angles are denoted by o, h, c, A, B, C respectively, whereof
c is a quadrantal arc : construct the polar triangle, and let
its sides and angles be expressed by a, h , c\ A , B', C respec-
tively as in (27), then it is manifest that C will be a right angle.
Now by the last article we have
r sin a' = tan h' cot B', and r sin a = sin c sin A :
whence by substitution we get
r sin {it — A)-=- tan (tt — B) cot (tt — b), or ;• sin A = tan B cot b ;
and
r sin (tt— A) = sin (tt — C) sin (tt — a), or ?' sin A = sin C sin a :
In the same manner all the ten cases of the polar triangle as
enumerated in the last article being resolved, those of the primi-
tive triangle will be immediately deduced from them, and it will
readily be observed that the two Rules of Napier above ex-
plained will be applicable to the solution of all the cases of
quadrantal triangles, if the two angles adjacent to the quadrantal
side, the complements of the two other sides, and the com-
plement of the hypothenusal angle, or angle subtended by the
quadrant be considered as the circular parts ; and to all
ambiguity of solution whether real or apparent, the remarks
S52
made in the different examples at the end of the preceding
article may be applied.
These solutions like the preceding are already adapted to
logarithmic computation.
III. Solution of Oblique- Angled Triangles,
41. Since every oblique-angled triangle has six distinct
parts, the three sides and the three angles, it follows that
the number of solutions in which from three parts given a fourth
may be found, will be equal to the number of combinations
that can be formed out of six quantities taken four at a time,
that is, = 15; but from a little consideration it will appear that
all the solutions essentially different will be comprised in the
six following cases :
I. When two sides and the angle opposite one of them are
given.
II. When two angles and the sides subtending one of them
are given.
III. When two sides and the included angle are given.
IV. When two angles and the adjacent side are given.
V. When the three sides are given.
VI. When the three angles are given;
and the resolution of oblique-angled spherical triangles will
be complete, if we can show that the expressions already
investigated can be applied to effect the solution of each
particular case.
42. Case I, in which two sides a, b, and the angle A
opposite one of them are given, to find the rest,
sin B sin ^ . ^ , . sm b .
Smce —. = — — by ('24), we have sm B = -: sni A,
sm A sm a sm a
which is therefore found :
253
again from (32), cot — = — tan ( )
cos I I
V 2 /
which is also determined;
c V 2 / /a + h^
c \ '^ J /« "^ ^\ • , .
and from (34), tan — = -. 77— tan ( — -— ) is obtained,
2 /A-B\ \ 2 /
cos
sin C . sin C .
or sin c = — : — r sm a = -: — =: sin o becomes known,
sin A sin B
All these formulas are adapted to logarithmic computation ;
but it must be observed that C and c are here expressed in
terms involving B which was not originally given_, but has
been determined in the previous part of the solution. This
is however by no means necessary, for in (35) we have seen
that cot A sin C = cot a sin b — cos b cos C, and to adapt it
to logarithms^ assume the subsidiary angle 0 such that
tan 0 = cos b tan A ; then we shall have
cos 5 sin C .
— = cot a sin 0 — cos 0 cos C,
tan 6
.*. cos b sin C cos 0 = cot a sin b sin 9 — cos b cos C sin 0^
whence
cot a sin 6 sin 6=cos 6 (sin C cos ^ + cos C sin O)=cos 5 sin (0+^),
and .'. sin (C-l-^) = cot rt tan ^ sin 0_, from which C + ^_, and
therefore C may be determined.
Again, from (21), sin b sin c cos A = cos « — cos b cos c ;
.*. if we assume tan ^ = cos A tan b, we shall have
cos b sin c tan 0 = cos a — cos 6 cos r,
and cos & sin c sin 0 = cos a cos 0 — cos b cos c cos 6,
254
whence
cos h cos {c — 6) = cos a cos 0, or cos (c — ^) = cos 0,
cos o
which is adapted to logarithms^ and gives the value of c — 0 and
therefore of c.
43. Case II, in zohich two angles A, B and the side a
subtending one of them are given, to find the rest.
sin 6 sin J5 . sin 5 .
Since -. — = -: — r by (24), we get sin b = -: — r sin a,
sin a sm A sin A
which is found :
A^- B
c \ 9. y /a + b
cos
\ 9. y ^n A- h\
becomes known:
/rA-- B\ \ 2 /
cos
and from (34) tan - = .A-B^ '"" (^)
(a + b\
~ir) z^+Bx.
also trom (32) cot — = tan ( ) is determined,
2 xg- b\ V 2 /
cos
. sin c . sin c .
or sm C = -. — sin A = — — - sin B is found,
sin a sm b
These solutions are all adapted to logarithmic computation,
and to these methods of finding c and C the same observations
may be applied as in the last case ; but articles {SQ) and (27)
by the introduction of subsidiary angles may as above be the
means of expressing in logarithmic forms the values of these
two parts without the previous determination of the side b,
44. Case III_, in which two sides a_, b^ and the included
angle C are given ^ to find the rest.
In (31) we have seen that
b^
A-\-B\ ™(_^ C
COS
a + b\ 2'
COS
255
'a — h^
sin
and tan { — - — ) = cot— ,
V 2 / ^ /^ + '
sni ( I
V 2 y
whence and — become known and therefore A and B:
2 2
sin C . sin C . , . ,
also sni c =*■ -^ — - sni a = ^ — — sni o is thence round,
sin A sin B
These forms are all adapted to logarithms^ but the side c may
likewise be expressed in terms of a, /;, C immediately and in
a form fitted for practice.
. cos c — cos « cos 6
For, since cos C = : -. — ; , we have
sin a sm o
cos c = cos a cos h + sin a sin h cos C
= cos a cos 6 + sin « sin ^ — sin a sin 6 vers C
= cos {a — h) -— sin a sin h vers C
.*. vers c = vers (« — h) + sin a sin h vers C
{sin a sin 5 vers C)
vers {a— o) )
, . , . sin rt sin h vers C 2 /^ i
which, by assuming ; = tan t/, becomes
vers ia — b)
vers (.fi ~~" o^ sec yj
vers c = 2 t^ ^^6 radius r, and is adapted as
before.
45. Case IV, in which two angles A, B and the adjacent
side c are given, to find the rest.
In (SS) it has been proved that
A-B
a f h\ \ Q J c
¥¥)
cos
/a •+- ft\ ,. ^ ,
tan I I = : ^^r-tan
\ 2 /
cos
256
sin (^-^\
,ndtan(-^)= ^ ., jg tan ^^ ,
from which and , and therefore a and b are known :
2 2
. sin c . . sin c . ^- . , , . .
and sm C = —. — sin A = -: — - sm i> is thus determined,
sin a sin 6
These formulie are all ready for logarithmic computation, but
as in the last case, the value of C may be expressed in terms
of the given parts A, B and c alone, and adapted to practice.
_ . , , , cos C + cos A cos B
tor. since by (27) cos c = -. — - — -. — =; , we have
sin A sin B
cos C = sin A sin B cos c — cos A cos B ;
.*. vers C = 1 — sin it sin jB (1 — vers c) + cos A cos B
= 1 — sin A sin JB + sin A sin jB vers c -j- cos A cos B
= 1 + cos ( J. + -B) + sin A sin B vers c,
^ . ,C ^ /A4-Bx . , . ^
.*. 2 sin — =2 cos I ■ I + sm A sin B vers c.
2 V 2 /
A , T) C , sin ^ sin B vers c'X
A 4- B\ <, - . sin A sin jB vers c
be
2/A + i^\ 2/1 f
= 2 cos ( — I sec t7, ir
V 2 /
. 2/-4 + jB>
2 cos
{^)
assumed = tan"^ 0, and thus the value of C may be found from
the tables.
46. Case V, in which the three sides a, b, c are giveriy
to find the rest.
In (23) and (25) it has been demonstrated that
2 ,
sin A = . , . — ^sin S sin (S - a) sin {S — h) sin {S - c\
sm 6 sin c ^ ^ / v / \ /?
257
. A . / sm{S — h) sin (S—c) A ^ /sin S sin (S—a)
sm--=y : — - — : , cos — = V ■ — ; — : ^
2 sin /; sin c 2 sm 6 sin c
A 4 / s'm (S — b) sm (S — c)
and tan — = \/ ^ ^ ^ : ;
2 sin S sin {S —a)
all of which when adapted to the radius r will be prepared for
logarithmic computation : and to which of these the preference
ought to be given above the rest, must be decided by means of
the remarks made in (238)^ (239)^ (240) and (241)/ of the
Plane Trigonometry.
47. Case VI_, in ivhich the three angles A, B, C are
given, to find the rest.
In (29) and (30), we have seen that
sin a = . ^ . ^ ^-cos^ cos(.S'-^)cos(S'-^)cos(.S'-C),
sin i5 sm C
. a A / — cosaS cos(6 — 7I) a a / cos{S — B) cos(S —C)
m - = V : — T, — ■ — 7; ,COS- = V : — ^ — : — 7; 5
2 ^ sm 5 sin C 2 ^ sm B sm C
a A / —COS S' cos{S' — A)
""•^ '^" 5 = ^ cos(S'-B)cos(S'-Cy
from any of which when adapted to the radius r, the value of a
or - may be logarithmically determined, the preference as to
method being given according to the remarks referred to in the
last article.
48. Though there has just been given a solution of every
one of the six cases above enumerated, and which, it was
•observed, are all that are essentially different from each other,
it may still be added that by supposing an arc of a great circle
to be drawn from one of the angles perpendicular to the opposite
side, Napier s Rules for the solution of right-angled triangles
are sufficient for the solution of all spherical triangles whatsoever.
For brevity's sake, we will exemplify their application in the last
two cases only.
Kk
S58
First, let the three sides a, b, c be given, to find the three
angles, and suppose the arc CD of a great circle to be drawn
from the angle C perpendicular to the side AB; then by
Napier's Rules we have
/' cos a = cos BD cos CD, and r cos ^ = cos AD cos CD,
cos a cos BD
whence
5 or the cosines of the segments of
cos b cos AD
the base are proportional to the cosines of the adjacent sides ;
cos
a — cos 6 cos 51)— cos j4D
cos a + cos b cos BD + cos AD
. AD + BD . AD-BD
sm sm — .
AD+BD AD^BD
cos — cos
2
c /AD—BDx
= tan - tan i )
2 \ 2 /
AD-BD . ^ , , AD + BD ^ . .
whence is lound, and being given^ two
parts in each of the right-angled triangles ACD, BCD become
known, and consequently all the angles of the triangle may be
determined.
Hence also the perpendicular CD may be found from
.. ' , . _,_ rcosa ^,^ rcosb
either of the equations, cos C jL)= 777^, or cos CD =
cos BD
cos AD
259
Nextj let the three angles A, B, C be given, to find the rest;
then the same construction remaining, we have by Napier s
Rules,
r cos A = cos CD sin A CD, and r cos B = cos CD sin BCD ;
, , ^ cos ^ sin ACD . . ^ ,
and therefore ~ = ^ — ttftf: j o*" the smes of the segments
cos B sm BCD
of the vertical angle are proportional to the sines of the cor-
responding angles at the base ;
hence if these segments be called a, /3 respectively, we have
cos A — cos B sin a — sin /3
cos A + cos B sin a + sin /3 '
tan ( — z — I tan ~
from which may be found, and or — bemg given,
the values of the segments a, /3 may easily be determined ; and
thus in each of the right-angled triangles ACD. BCD two angles
being known, the sides of the proposed triangle may be found.
As before, the magnitude of CD may be found from either
of the equations "^
r cos A ^^ r cos B
cos
sm ACD sm IJCD
260
CHAP. IV.
Ofi the Areas of Spherical Triangles, 8^c. and the Spherical Excess,
On the Measures of solid Angles, S^c,
49. The surface of a spherical Liine is proportional to the
angle contained betioeen the planes of the two semi-circles by
which it is formed.
Let APNQ be the lune formed by the two great semi-circles
of the sphere PAQ, PNQ: then it is obvious that if the arc
P
AN which measures the angle APNhe doubled, or increased in
any other ratio, the surface APNQ will be doubled, or increased
in the same ratio, because equal portions of surface will mani-
festly correspond to equal parts of the arc : that is, the surface
APNQ is proportional to the arc AN, or to the angle APN.
50. Cor. 1. Hence, if 5 represent the whole surface of
the sphere, the surface APBQ = — , and we shall have
2
the area of the lune JPNQ : - :: Z APN : 180°;
2
.% the area of the lune APNQ =
180° 2
261
and if the radius = lj we have seen in (£14) PL Trig, that the
<S
area of the circle = tt, and .*. — = 27r= 180^,
2
z APN
whence the area of the lune APNQ = „- 27r= Z APN.
51. Cor. 2. The area of the lune may be expressed
in other terms.
For since the spherical angle APN = the plane angle
AON= ^ao7i= — = -: — — — , we have the area of the lune
ao sjn Pa
APNQ= ""
sin Pa
52. To express the area of a spherical triangle iti terms of
its angles.
Let ABC be a spherical triangle on the surface of a sphere
whose radius is 1, and produce the sides AC, BC till they meet
in c on the opposite hemisphere ; then it is manifest that the arcs
Cac, Cbc are semi-circles; whence, since -4 Ca, BCb are also
semi-circles^ it follows that AC = ac and BC = bc: therefore
the angles at C and c which measure the inclination of the
same planes being also equal, we have the triangles ACB,
atb in every respect^ equal to one another:
26s
Now if the area of the triangle ACB be called x, and BCa,
ACb, CBa be assumed equal to a, (i, y respectively, we
shall have
A. S B S C S
^+"=Ti5^¥' ^+^=T85-«-5' ^+^=Ti3-»-i'
.*. by addition, observing that x + a 4-/3-|-7 = —, we get
S A + B-^-C S
1 , r .u . • 1 ^ + ^ + C--180" S
whence x. or the area of the triangle = ^
^ 180^ 4
= ^+E + C-180^ as appears from (214).
53. Cor. 1. Hence the area of the triangle is equal to
the excess of the sum of its three angles above the two right
angles, which is called the Spherical Excess.
54. Cor. 2. It follows therefore by (17), that the area of
a spherical triangle may be represented by any number of
degrees between 0 and 360, and also that if two of the angles
be right angles, the area varies as the third.
55. Cor. 3. If the radius of the sphere be supposed = r,
we shall have S = 47rr^, Diff, Cal.j and therefore the area of the
. , ^4-5 + C-180" 2 ^ . ,.
triansle = n ttv : and to the radius 1, ir re-
presents 180^ expressed in terms of that radius, therefore the
area of the triangle expressed hi seconds
= (A + i^H-C~180')/sinl".
56. Cor. 4. By means of this article the area of a spherical
polygon may likewise be expressed in terms of its angles.
For let A BCD Sec. be a polygon of n sides whose angles
are Aj B, C, D &c. : take any point F in its surface, and from
263
it to the angular points draw arcs of great circles of the sphere:
then the area of the polygon ABCD &c. = the sum of the areas
of the triangles AFB, BFC, CFD, DFE, &c. = the sum of
the angles of the polygon, together with the angles at P— w 180°
= ^ + 5 + C + D-f-&c. + 2.180''-wl80°
= A + 5 + C + D + &c. — (;* — 2) 1 80^ the radius of the sphere
being supposed = 1.
57. To express the area of a spherical triangle in terms of
two sides and the included angle.
Let a, b be the proposed sides and C their included angle,
and suppose the area of the triangle or the spherical excess
A + ^ + C — 180^ to be represented by E : then in (31)
we have seen that
b^
cos
{"¥)
A+B\ V 2 / C
/A + DA
tan I I = r-7- cot — ,
\ 2 / /a-^b-
cos
whence cot
_2
1 — tan ( " ' " ) tan —
V 2 y -
/a-b\ C /a + b\ C
I I cot 1- cos I I tan —
V2/ 2 \2/ 2
/a-b\ /ai-b\
cos I — I ~ COS I 1
\ 2 / \ 2 /
264
/a-b\ gC , /« + 6\ . 2 C
cos I 1 cos hcos I I sin —
. a . b . C C
2 sm ~ sin - sin — cos —
2 2 2 2
a b a . b / C . oC\
cos - cos - -f-sin- sin - | cos sin'— I
2 2 2 2 \ 2 2/
. a . b
sin - sm -" sm C
2 2
cot - cot f- COS C
2 2
sm c
58. Tb express the area of a spherical triangle in terms of
the sides.
It has been shewn in the last article that
^ cot - cot - -f- cos C
E 2 2
cot — =
2 sin C
a b (1 4-cos «)(1 -|-cos 6)
now cot - cot - = : : — ;
2 2 sin a sm b
1 + cos a + cos Z/ -|- cos a cos 6 ^ cos c — cos a cos Z>
: r-T , COsC= : r-j
sm a sm o sin a sin o
and sin C = -: : — - v sin S sin (*S — a) sin (5 — b) sin (S — c),
sin a sino
according to the notation adopted in (23);
E 1 4- cos a + cos b + cos c
/. cot — =
2 2 v^ sin 5^ sin (5 - a) sin (5 - b) sin (6' - c)
This formula for the spherical excess was discovered by
De Gua, but it is not adapted to logarithmic computation.
a b „
„ cot - cot- +cosC
E 2 2
5 Q. Co R . 1 . Since cot - = r—p; , we shall have
2 sm C
265
,,a ^'o a o ^
T, ^ 1 + cot' - cot^ - + 2 cot - cot - cos C
« £ o -E 2 2 2 2
cosec"— = 1 + cot' — = r,
2 2 sin- C
out cot" - cot -
(i-Wg(i-,i,r-J)
2 2 . „a . 2^
sill - sin -
1 — sin - — sm - + sm - sm -
2 2 2 2
sm - sm -
2 2
, ^ cos c — cos a cos Z>
also cos C = : : —
sin a sm b
l-2sin^^ - A-2sin-j'\ i\ -2sii/-^
. a . h a h
4 sm - sm -- cos - cos --
2 2 2 2
/. 2 cot - cot - cos C =
l-2sin'- - /^l-2sm-^'\ /^l-2sin^-^
2 sin - sm"-
2 2
. „ a , . r. ^ . oC ^. . oa . ^h
sin^ - + sin sm 2 sin - sin" -
2 2 2 2 2
sin - sm -
2 2
.'. by substitution we get cosec ■—
. .a . ,b . pfl . o^ . . £« • 2^ 1 • e« . -2^ -2^ ^ . 2« • «»^
sm^-siir- + 1-8111"- - sur- +sin -sm -4-sm - + sm - - sm - -2sin -sm"-
22 22222 22 22
sm -- sin - sm C
2 2
Ll
^66
. qC ^ 2« 2 ^ £ ^
1 — sm - 4 COS - COS - COS -
2 2 2 2
• 2 « . 2 ^ . 2 ^ s"i '^ s^" (*^ "■ ^) si« iS — 6) Sin (S — c)
sin - sin - sin C
2 2
, • :^ _ x/si" *^ sin {S — a) sin (8 — b) sin (S' — c)
2 a h c
2 cos -" cos — cos -^
2 2 2
The discovery of this formula is due to Cagnoli, and it has
the advantage of being easily adapted to logarithms.
60. Cor. 2. Since cos — = cot — sin — , we have from the
2 2 2
last two articles
1? , . . 7 COS^- + COS -'+COS--— 1
-c/ 1 +cos« + cos t> -r cosc 2 2 2
cos— = = ?
2 a b c a b c
4 cos-- cos- cos- 2 cos -- cos-- cos -
2 2 2 2 2 2
which is rational_, but not adapted to logarithmic computation.
61. Cor. 3. Because
E E
1 — cos — 1 — COS —
2 2
. E
sm —
2
\/(l + cos|) (l-cos|)
E
= tan — , we shall have by substitution^
2« ^b ^c a b c
1 — cos - — cos -—COS - H- 2 cos- cos- cos -
E 2 2 2 2 2 2
4 V sin S sin (5 — a) sin (5 — b) sin (*S — c)
2« ob oC a 6 c
but 1 — cos - — cos cos - + 2 cos - cos - cos -
2 2 2 2 2 2
/ 9«\ / 2 ^\ / ^' ^^ ^\'
= I 1 ~ cos - I I 1 — cos - I — I COS - COS — cos - I
\ 2/ V 2/ V 2 2 2/
267
b\' / a h
( . a . b\ /ah c\
= I sin - sin - I — I cos - cos cos -- I
\ 2 2/ V 2 2 2/
{. a . b a b c )
sin - sm - + cos -' cos cos -- > x
2 2 2 2 2 j
{. a . b a b c\
sm - sin - -- cos - cos - + cos -- \
2 2 2 2 2j
. S . /S-a\ . /-S-z^x . /S-c\
= 4 sm — sin 1 1 sin { j sm | I ,
2 V2/ V2/ \ 2 J '
if we adopt the notation of (23);
S . /S-a\ . /S-b^
A. mn
jE
tan
. S . f^-a\ . (S-b\ . (S~c\
4 sin — sm 1 j sin { J sm I 1
2 \ % ) \2/ \2/
^ x/sin *S sin (S' — a) sin (8 — Z>) sin (^ — c)
= \/tan I ,an (^") tan (^) .a» (^^) ,
which being transformed to the radius r will be adapted to
logarithmic computation.
This singular formula for the spherical excess was discovered
by Simon Lhuillier of Geneva.
62. A solid angle being the angular space included between
the several planes by which it is formed, will manifestly have
the same relation to the corresponding spherical surface whose
centre is the angular point, as plane angles have to their cor-
responding circular arcs_, and therefore the magnitudes of solid
angles may be compared by determining the ratios between
the spherical surfaces by which they are, as it were, respect-
ively subtended. Now we have seen (52) that the area of
a spherical triangle is measured by the excess of the sum of
its angles above 180°, and {56) the area of a spherical polygon
of n sides by the excess of the sum of all its angles above (n — 2)
180°; hence it follows that these quantities may be assumed as
S68
the measures of the solid angles formed by the planes whose
inclinations to one another are the same as the angles of the
triangle or polygon. The maximum limit of solid angular
space will manifestly be a hemisphere and its measure the
surface of the hemisphere in the same manner as the max-
imum limit of plane angular space is a semi-circle and its
measure the arc of the semi-circle. Representing therefore the
maximum solid angular space by the content of the hemisphere
27r
whose radius is 1, or by -;— , {Diff. Cal.) we shall have its
o
measure equal to the corresponding hemispherical surface Stt
or 360^
Ex. 1. In a cube each of the solid angles is formed by
three plane right angles, and thence we have
the solid angle of a cube _ QO^ + 90^ -f 90^ - 180° _ 90^ _ 1
the maximum solid angle ~" 36o" "" 36o" ~" 4 '
which we also know to be true from the circumstance that if four
cubes be placed together upon a plane, they exactly fill up the
angular space about a common point.
Ex. 2. In a regular right prism with a triangular base two
of the plane angles which form each solid angle are manifestly 90^^,
and the remaining one 60° : therefore
the solid angle of this prism _ 90° + 90° + 60° — 180^ __ 60° 1
the maximum solid angle 360° 360° ~ 6 *
Hence by means of the last Example, we have
the solid angle of a cube 11 6 3
the solid angle of this prism 4 6 4 ~" 2 *
Ex. 3. Let there be two regular light prisms of m and n
sides respectively, then each of the angles of their bases will be
— j 180° and I j 180" respectively : hence
269
the solid angle of the first prism \ m ) '^^ m
I ) 180^
\ m J
the maximum solid angle 360° 2m
("-^\ 180°
the solid angle of the second prism _ \ w / _ ?i — 2
the maximum solid angle 360° Q.ti
the solid angle of the first prism (m — 2) ?i
and
the solid angle of the second prism (ii — 2) m
and by means of this example, the solid angles of all regular right
prisms whatsoever may be compared.
Ex. 4. If there be any two prisms whatever, whose numbers
of sides are m and w, we shall manifestly have
the sum of all the solid angles of the first prism _ (m— 2) 180** m^2
the maximum solid angle 360° 2
the sum of all the solid angles of the second prism (n — 2) 180° « ~ 2
the maximum solid angle 360° 2
the sum of all the solid angles of the first prism m— Q
the sum of all the solid angles of the second prism ?i — 2*
63. The vertical angles of pyramids whether regular or
irregular may be ascertained and compared by the same methods :
thus if there be two regular pyramids of m and n sides
having the inclinations of two contiguous sides to each other
represented by a and /3 respectively, then according to the
principles above explained, we shall have
the vertical angle of the first pyramid _ ma — (m — 2) 180°
the vertical angle of the second pyramid 11 (^ — {ii — 2) 180°*
o4. The same principles enable us to compare the vertical
angles of cones, by comparing the areas of corresponding
spherical segments of spheres of equal radii, whose centres are
the angular points.
270
Hence since {Diff. Cal.) the surface of a spherical seg-
ment varies as its altitude, if H and h be the altitudes of two
cones corresponding to L the common length of their sides,
we shall have
the vertical angle of the first cone the height of the first segment
the vertical angle of the second cone the height of the second segment
\ L- H 1 — cos 0 vers ^
L — h 1 — cos 0 vers (p'
if 9 and (j) be the vertical angles of their generating triangles.
Ex. For the equilateral and right-angled cones we have
0 = 30°, and (p = 45^, respectively ;
1/3
the vertical angle of the equilateral cone vers 30° __ 2 _2-- V^3
the vertical angle of the right-angled cone vers 45° 1 2— 1/2
Whenever the spherical surface by which any solid angle is
measured can be divided into 7i parts either equal to one aijother,
or having any assigned ratios^ the solid angle itself can be divided
into n parts having to each other the same ratios.
This method of measuring and comparing solid angles by
means of the positions, and not the magnitudes^ of its plane
angles, was first suggested by Albert Girard in his Invention
nouvelle en Algebre published about the year 1629, and has been
■ extended and exemplified by several modern writers.
271
CHAP. V.
On the regular Polyhedrons, and on the Parallelopiped and
triangular Pyramid.
65. In every Polyhedron, the numbers of Solid Angles and
Plane Faces together exceed the ?iumber of Edges by 2.
Let :c be the number of solid angles^ y the number of plane
faces and z the number of edges ; then since every edge is
common to two plane faces, Qz will be the number of sides
of all the faces :
Within the polyhedron suppose a point to be assumed
from which to all the angular points straight lines may be
drawn, and with this point as a centre let a spherical surface
be described cutting these lines, and let the points of intersection
be joined by arcs of great circles of the sphere so as to form
as many spherical polygons as the solid has faces : let A BCD
&c. be one of such polygons, the number of whose sides is 72 ;
then by (56\ its area = A + B+ C + D-\-^c. - (n - Q) 1 80^•
and the same being found for all the polygons, we shall have the
whole surface of the sphere or 720° equal to the sum of all the
angles of all the polygons —(2 z ~2y) 180^ or the sum of all
272
the angles of all the polygons = 720^ + (2 ~j/)360^ but the
sum of all the angles about any point as A being = 360^, we
have the sum of all the angles of all the polygons = cr 360^:
whence x 360^ = 720^ + (z - y) SQO\ or or = 2 + 2 -y,
and .*. X +3/ = z -^-^l.
QQ. Cor. 1. Hence the sum of all the plane angles forming
all the solid angles of any polyhedron = {x — 2) 360^.
For if any face have n sides_, the sum of all its angles
= (2?z - 4) 90^ = (/2 - 2) 180^ ; hence the sum of all the angles
of all the faces = (2 ? - 2j/) 180^ = (z -y) 360^ = (^-2) 360°.
67. Cor. 2. In a regular polyhedron, if ?i be the number of
sides of each plane face, m the number of plane angles constituting
-\ 180° be the magnitude of
TTl ( tl — 2 )
each plane angle^ and — 180° the sum of all the plane
angles forming each solid angle ; hence retaining the notation
above used, we shall have the sum of all the plane angles
tl 1 1-1 1 m(n-'2)x 0
formmg all the solid angles = — 180 :
therefore from the last article, we get
(x ~ 2) 360° = — ^ — 180°,
n
and .•. 2wr-— 4w = m»jr — 2mj,
4n
whence the number of solid angles x = -
2 {??i -f n) — mn
, ny 4n
also smce i/ = 2+jz— a: = 2+ -^^
we have the number of plane faces y =
2 2{m -{- ii) — m n
4m
2 (m + 7i) -^ mn
273
68. Cor. 3. From what has just been proved, we imme-
diateJy obtain the number or edges 2 = -^ = • ; :
and therefore the number of solid angles, the number of plane
faces, and the number of edges of any regular polyhedron, are
to one another respectively as 4w, 4m and ^miiy or as 2?/,
2m and mn,
^^, There can he only Jive regular polyhedrons.
For^ the notation above adopted being retained, if x, y^ z be
finite positive quantities, the denominator 9.{m^rn)—mn must be
positive, and .'. 9>m-\-2n must be greater than mn, from which
m , ^
we have greater than - .
772- 2 ^ 2
First, let m = 3, or each solid angle be formed by three plane
angles^ and n = 3, or the faces be triangular, then being
3 . . • . 12
greater than - , a solid will be formed having — — or 4 solid
^ 2 ''2.6-3.3
angles; — or 4 plane triangular races; and or
^ '2.6-3.3 ^ *= ' 2.6-3.3
6 edges, and is therefore called a Tetrahedron.
3
Again, let m = 3 and « = 4, and we shall have greater
3 -2 *=
4
than - , and thus a solid will be formed having eight sohd
angles, six plane square faces, and twelve edges, and is therefore
called a Hexahedron^ which is the same as a cube.
3 5
Next, let 772 = 3 and n = 5, which gives — — greater than - ,
and therefore a solid will be formed having twenty solid angles,
twelve plane pentagonal faces, and thirty edges, and on this
account is termed a Dodecahedron.
M M
If m==3, and 7i be any number greater tb an 5, the condition
771 W
that must be orrealer than - will not be satisfied: and
thence it appears that there can be constructed only three regular
polyhedrons in which each solid angle is formed by three plane
angles.
4
Secondly, let W2 = 4 and w = 3, and we have greater
3
than - : wherefore there will be formed a solid having six solid
angles, eight plane triangular faces, and twelve edges, which is
therefore called an Octahedron.
Again, let /?2 = 4 and n = 4, or any larger number, and the
condition will no longer be fulfilled; and consequently there can
be constructed only one regular polyhedron in which each of the
solid angles is formed by four plane angles.
5 3
Thirdly, let m=5 and n = 4, then being greater than ~ ,
there will be formed a solid having twelve solid angles, twenty
plane triangular faces, and thirty edges : this solid is therefore
called an Icosahedron.
If m = 5j or any larger number, and ti be greater than 3_,
the specified condition cannot be fulfilled, and thence it appears
that there can exist five, and only five regular polyhedrons.
By supposing 2(m + ?0 — ^^w = 0, we find the values of
X, ?/, z indefinitely great : and a sphere may be considered
as a regular polyhedron of an infinite number of solid angles, &c.
70. To find the inclination of tivo contiguous faces of a
regular polyhedron to one another.
Let AB he the side common to the two contiguous faces,
C and JB being their centres, from which let CD and jEZ)
be drawn perpendicular to it, then will the angle contained
between CD and ED be the incHnation of these two faces
to each other : in the plane in which CD and TLD lie, let
275
CO and EO be drawn perpendicular to them respectively
and meeting in O, join OA, OB, OD, and from 0 as a centre
suppose a spherical surface to be described cutting the lines
OJ., OCy OD in the points p, q, r, and let these points be
joined by arcs of great circles : then it is manifest that the
angle prq is a right angle: hence retaining the notation before
used, we shall have
TT TT
Z.qpr= — 5 and /.pqr= - :
m n
«ow by Najyier's Rules, the spherical triangle pqr gives
TT
cos —
m
cos qpr-=-i\\\pqr cos qr^ and .*, cos ^r = ;
TT
sm -
n
CD F
but cos qr = cos COD = sin CDO = sin
2 '
. CDE ""'l
/. sm — - — = 5
2 . TT
sm -
;i
from which the required angle CDE may be found,
(1) In the Tetrahedron m = 3 and 7i = 3,
. CDE cos60« 1 . ^rir' 1
" sm — : — = — ^ — ^TTfT = ~r~ ^ and cos LDE = - .
sin 60^ V3
276
(2) In the Hexahedron m = 3 and n = 4,
r,'s\n^^= cos_6^^ 1 ^^^ ^^^ CD£=0, or CjDE=90^
2 sm 45^ /2 '
(3) In the Octahedron m = 4 and w = 3,
. CDE cos 45^ ./a" , ^^^ 1
.*. sui = — — ^-77 = v - 5 and cos CDi^= — 7; .
2 sm 60^ ^3 3
(4) In the Dodecahedron m = 3 and 71 = 5,
. Ci)E cosGO** 2 , ^T^T^ l-'^^
.'.sin = -77= — .. .■■ . == . and cos CiJil/=- 77.
2 sin 36' ^10-2^5 5-V5
(5) In the Icosahedron m = 5 and w = 3,
. CjDE cos 36' 14-/5 , ^^^ . ^5
.*. sm = -: — ;;-7r = 7 — , and cos CiJil/ =- .
2 sm 60' 2 V^3 3
71. To ^/zc? //te radii of the spheres inscribed in and
circumscribed about a regular polyhedron.
Retaining the construction and notation of the last article,
it is manifest that CO and £0 are respectively perpendicular to
the planes ABCf ABE and equal to each other; and the same
being true of any other two contiguous faces, it follows that C
is the centre of the inscribed and circumscribed spheres whose
radii are OC and OA respectively :
OC TT 'TT , ,^ . ,
now -— — =: COS pq = cot qpr cot pqr = cot — cot - , by Napier s
Rules: but if a side of one of the faces = a, we have
from (204)
2
CA = --^—, and /. 0A^= 0C"+ ""
9
rnr . o tt
2 sm — 4 sm'' -
n n
wherefore r and R representing these radii, we have the two
following equations ;
277
cot — cot — , and Br — r^ —
R m n . . 2 '""
4siu —
n
to determine their values.
72. Cor. If the angle CDE be found as in the last
article but one, we shall manifestly have
a TT CDE , ^, a tt CDE
r— - cot -- tan and K= - tan — tan .
Ex. 1. In the Tetrahedron W2 = 3 and 11 = 3,
r 1
.*. — =cot 60° cot 60°= " , whence R = 3r\
K 3
also R^-'r^= — . <2 ^ n = — i fiom which are obtained
4 sn/ 60° 3 '
a 3a
r = r:; und it =
2 V"6 21/6
Ex. 2. In the Hexahedron and Octahedron, since in the
former m = 3 and n = 4f, and in the latter tw = 4 and 71 = 3, we
r 1
shall have — = cot 60° cot 45° = — 7—; which shews that if
it V 3
these two solids were inscribed in one sphere, they might both be
circumscribed about another sphere, and the contrary : also, in
the former it is easily proved that r = -- , J^ = - /sj 3^ in the
latter r = — 7— and R = -7- .
Ex. 3. In the Dodecahedron and Icosahedron m = 3,?i = 5,
and m=^5, 7i = 3 respectively,
.-. ^ = cot 60° cot 36°= \/i±^^,
R 15
and the same remark may be made as in the last example :
hence also in the former r = — x/250+ 1 10 V^,
20 -
27S
R=^(\^15+ 1/3); in the latter
4
r = — ^42+18 t/5, jR= 2 ^10 + 21/5.
73. To find the content of a regular polyhedron.
From the centre 0 of the inscribed sphere, let straight lines
OAy OB, OC, &c. be drawn to all the angular points, then will
the polyhedron be divided into as many pyramids with equilateral
bases as there are plane faces^ and whose altitudes are each
equal to the radius of the sphere r : now if we retain the nota-
tion which we have before adopted^ we shall by (201) have the
area of each face = ;
TT
4 tan —
n
n c? y
and therefore the whole surface = ^— ,
TT
4 tan —
n
ni^ry
whence the content = \ whole surface x r = •
^ TT
1 2 tan -
n
Ex. In the Tetrahedron w = 3, 3/ = 4, r = — T7?> ^"^
%VQ
tan - = tan 60^ = /3 ;
n
«'
/. (1) the content of /the Tetrahedron = f= = — '^2.
2^18 12
Similarly,
(2) the content of the Hexahedron = c^ :
a^
(3) Octahedron = — 1/2 :
(4) Dodecahedron = ~ x/470 + 210.^^ •
4 ^^
c ^.3 /
(5) , Icosahedron = vl4 + 6>y/5.
279
By means of the last proposition there will be no difficulty
in expressing the content in terms of the radii of the inscribed
or circumscribed spheres.
74. Given the three edges of a parallelopiped which meet,
and the angles included between theniy to find its perpendicular
altitude.
Let OA, OB, OCj the three edges of a parallelopiped
meeting at the same angle be represented by a, b, c ; draw CD
perpendicular to the plane AOB, join OD, and let the angles
AOB, AOC, BOC be called a, /3, 7 respectively, then
CD=OC sin COD'-, now to find the sine of the angle COD,
with centre O and radius = 1, suppose the surface of a sphere
to be described cutting the edges OA and OC in p and cj^ and
OD in r\ then if 2»S = a + /3 + 'y, we shall manifestly have
sin COZ) = sin <5r7' = sinp^ sin qpr, by Napier's Rules^
= sin B— : — 7;: ^/sin -S' sin (5' — a) sin (S— 3) s\n(S — y)
sm a sm p ^ '
^y sin S sin (5 - a) sin (S — /3) sin (6' — 7X
sm a
whence CD = -, ^sin 5 sin (5 - a) sin (S — 3) sin (S - y\
sm a '
75. Cor. The whole surface of the parallelopiped will
manifestly = Q {ab sin a-\-ac sin (3 -{-be sin 7}.
76. On the same hypothesis, to find the content ofi the
parallelopiped.
280
The content = area of the base x the perpendicular altitude
==0A, OB sin AOB. CD
Qq ._ - -■-...
= ab sin a — \/ s'lnS s'm{S —a) s'm{S — Q) sin^S — y)
sni a
= 2a he ^y «in «S' sin(i5^ — a) sin (5-/3) sin(«S — 7).
77* On the same supposition^ to find the diagonal of the
parallelopiped.
Let 2 CD (which, if the figure were completed, would be
the diagonal of one of the faces) = d, then
d" = a'^-\-b'^ + 2ab cos a :
also if D be the diagonal of the parallelopiped, we shall easily
perceive that
D^=d'' + ^-h2cd cos COD=:a^ + b^ + c^-\-2abcosa + 2cdcosCOD:
now the angle COD is manifestly measured by the arc qr,
and cos ^ r = cos p q cos pr + s'm pq sin p r cos qp r
^ . n • fcos'V — cosa cos/3)
= cos p COS pr + smp sm pr < : : — y: r
i sm a sm p )
^ . . rcos 'y — cos a cos /3)
= cosp cos pr-\- sm pr{ : >
I sin a )
cos 7 sin pr cos /3 (sin a cos/? r — cos a sin pr)
— ; 1 ^
sm a sm a
__ cos 7 sin pr cos fi sin (a — pr)
sm a sm a
sin AOD , ^ sin BOD
= cos 7 : h cos p : :
sm a sin a
but in the parallelogram whose diagonal is 20Df it is evident
that
sin.iQD _ OB _b sin BOD _ OA _a
sm a SOD J' sma 2 0£) d
281
.*. cos qr or cos COD=^ - cos /3 -|- - cos 7;
a a
whence by substitution we get
D = /sj a^ -{- h^ + c^ -\- Q, a h cos a 4-2«c cos fi-\-2bc cos y.
78. Cor, By proper substitutions in the last expression,
the other diagonals are determined : and thence it may easily be
shewn that the sum of the squares of the four diagonals of
a parallelopiped is equal to the sum of the squares of the twelve
79* Given three edges of a triangular pyramid which meet,
and the angles which they make with each other, to Jind its
content.
Since by Euclid xii. 7? every prism having a triaiigular base
may be divided into three pyramids that have triangular bases
and are equal to one another, it follows that the content of the
triangular pyramid will =§• of the corresponding parallelopiped:
that is, retaining the notation of the last article, we shall have
the content of the triangular pyramid
abc
= — /y/sin S sin {S — a) sin {S — /3) sin {S — y).
80. Cor. The sum of the areas of the three sides of the
pyramid manifestly = j {ah sin a-\-ac sin /3 + ^c sin 7}.
81. Given the six edges of a triangular pyramid, to Jind
its content.
Retaining the notation of the preceding articles, and in
addition thereto representing the sides AB, AC 3 BC of the
base by h, k and / respectively, we have
a'+b'-'h^ ^ a''-{-c^-k' b'-^-c'-'l'
cos a = — ' , cosp = — , cos'y= ; ;
2ab ' ^ 2ac ' ^ Qbc '
hence the content = y^sin S sin {S ~ a) sin {S - /3) sin {S - 7)
N N
2S2
ab(
= — — >/ 1— cos^a— cos^j3— cos*7 + 2cosacos^cos7;asin(6l),
6
— ^ V^ .2/2 .22 .729"+'
HKL
the quantities i?, K, L representing the numerators of the values
of cos a, cos /3 and cos y respectively.
82. We shall conclude this chapter with a few remarks
upon certain consequences resulting from Euler's Theorem
demonstrated in (65), extracted and slightly altered from Note 8,
of the Elements de Geometrie of M. Legendre, to which work
the student is further referred for the geometrical construction of
the regular polyhedrons treated of in the preceding part of the
chapter.
(l) Let yg be the number of triangles, 3/4 the number of
quadrilaterals, 3/5 the number of pentagons, &c. which form the
surface of a polyhedron : then the whole number of faces is
3/3 + y4 + ys + 3/6 + Sec,
and the whole number of their sides is
37/3 + 4^4 -f 5^5 + 6^6 + &c.
which is also twice the number of edges : hence as in (65) if x
be the number of solid angles, y the number of plane faces,
and z the number of edges, we shall have
y =y3 +y4 +3/5 +^6 + ^^-^ and 2z = Sys -f 4^4 + 5i/^ + %6- + &c-
but since by (65), .r + y = 2; + 2, we get
2x = 4-\-2z- 2y = 4+y3 + 2y4 + 3y5 + 4y6 + &c.
from which we conclude that the number of faces of a polyhe-
dron having odd numbers of sides is always even.
283
(2) Since 2 = ^^ + 2 2/4 + ^ + 3j/,. + 8cc.
and ^ = 2 + — + 3/4 + -~^ + ^j/g + &c.
= 2 + i {^3 + 2y4 H- 3^5 + 4i/6 + &c.}
it follows that % cannot be less than | y, nor x less than 2+^y.
Also the whole number of plane angles being Sz, and the
number of solid angles ar, it is manifest that the mean number
22 42
of plane angles constituting each solid angle = — = —
^ 2(3^3 + 4y4 4- 5^5 + 6^6' + ^^') .
4 + 3/3 + 2j/4 + 3^5 + 4^6 + &c. '
also since no solid angle can be formed by fewer than three
2z
plane angles, — cannot be less than 3, or 22 cannot be less
X
than 3 a? : hence
3^3 + 4j/4 + 5y5 H- 6j/6 4-&c. cannot be less than
6 + ^ + 3^4 4- 5|^ + 6y, + &c.
nor 61/3 + 8^4 4- lOys + 12^^ + &c.
less than 12 + Sys + 63/4 + 9^5 + l^y^ + &c.,
and therefore Sys 4- 2y4 + ys cannot be less than
12 + y7 + 2y8 + 3y9 4- &c.
from which we learn that ys, y4, y^ cannot all be zero at the
same time ; or in other words, that there cannot exist a poly-
hedron all of whose faces have more than five sides.
284
* From what has been proved, it appears that y cannot be
less than
4 + ^ ,
and therefore x not less than
4t -4- ,
3 '
and consequently 3; not less than
6 + y4 + 2^5 + 3y, + &c.
also since 3y is not less than 12+3/4 + 2y5-|-3j/g+ &c. we shall
evidently have Qy not less than JC + 4, and 3y not less than
z-j"6^ for all polyhedrons whatever.
(3) If we suppose 2 2 greater than 4x, so that all the solid
angles shall be formed by four or more plane angles, we shall
manifestly have
3/3 +3/4 +3/5 +3/6 + ^c. not less than 8+3/4 + 23/5 + 3J/6 + &C.
or 2/3 not less than 8 -|- 3/5 -+■ ^y^,- + Sy^ + &c. and therefore we
infer that such a polyhedron must have at least eight triangular
faces.
This likewise gives x not less than
6 + 2/4 + 2^5 -f 3 y, + &c.
and z not less than
12 + 2?/4 + 4j/5 + 8j/tj + &c.
from which it follows that :r is not greater than y - Q, and z not
greater than 2y — 4.
(4) Let 22 be greater than 5x, so that all the solid angles
shall be formed by at least b plane angles, then we have
3/3 +3/4 +3/5 +3/6- + ^c. not less than 20 + 3j/4 + 63/5 + 93/6" + &c.
and .*. j/g not less than 20 + 23/4 4- ^3/5 + 83/5 + &c.
and thence it appears that the solid must have at least twenty
triangular faces.
285
Also on the same hypothesis x cannot be less than
nor z less than
from whence we conclude that a: cannot be greater than - (y— 2),
o
6
nor z greater than - ( j/ — 2).
(5) Smce 2^ + 2^4 + 4^5+ 6^6 + ^c. + 12
= 33/3 + 4^4 + 53/5 + 6y, + &c. + 23/4 + 4j/5 + 6j/6 + &c. + 1 2
= 12 + 3^3 + 6y4 + 9j/5 + 12^6 + &c.
==6 J2+^ +^4+ ^y5+23/,+ &c.| =Qx,
therefore it is evident that 9.Z must always be less than Qx\
or in other words, that there cannot exist a polyhedron, all
of whose solid angles are constituted by six or more plane
angles : and in fart six angles of equilateral triangles = 360°
which exceeds the sum of the plane angles forming any solid
angle whatever.
(6) If all the faces of a polyhedron be triangular, we have
^4 + 23/5 + ^ye + &c. =0, and thence we find that r = |y,
and :r = 2 + ^y.
If all the solid angles of a polyhedron be formed by five
and six plane angles, the number of the former being jCg,
and of the latter org, then j? = ^Tg + arg, and ^z = 5xr^-\-6x^f
whence 6>r — 22 = 3:5 : also z = |y, and .r = 2 + \y, from which
it follows that 2'5 = 6r — 2z= 12, and this shews us that the
number of solid angles formed by five plane angles will always
be twelve, and Xq being indeterminate proves that the number
formed by six may be any whatever.
286
CHAP. VI
On the Variations of the Sides and Angles of Spherical
Triangles.
83. In the practical applications of Spherical Trigono-
metry to philosophical subjects_, wherein certain parts of
spherical triangles are determined from instrumental observation,
and the remaining parts deduced from them by arithmetical or
logarithmic calculation^ it is manifest that the effect of any
error however small in the observed part or parts, will be
entailed upon the results as determined by computation. If
then we suppose the instrumental or original error to be of
given magnitude, we may by means of the Calculus of Finite
Differences or by Taylor's Theorem be enabled to determine
what relation the resulting error bears to it, when these errors
are of considerable magnitude, and the operation of Differentia-
tion will be sufficient for the same purpose when they are
very small. Thus, since in every spherical triangle there are
six distinct parts, any three of which would be sufficient for
the determination of all the rest^ we may suppose two of
them to remain constant, and then find the ratios of the
simultaneous changes of all the rest. We will illustrate these
principles by an example of each method, and then proceed
to the consideration of such particular cases as are most
commonly met with in practice.
. cos a — cos h cos c
84. Assume the equation cos A = — — -: — ; — : -, and
sm 0 sm c
suppose that in consequence of a becoming a + A a, il becomes
il + A il, whilst bf c remain constant, then we have
,. . ,^ cos (a + A ft) — cos ^ cos c
cos (J.-I- A A)= ^ r—. — : :
sm b sm c
.*. by subtraction, cos {A-\- A ^)-- cos A =
287
cos (a-f A a) — cos a
sin b sin c
A a
sin ( a -i ) sin —
A^\.aA V 2/ 2
or sm f A H I sin
. / Aa\ ,
sin I a H I SI
. / . AJ\ , aA V 2/
r sm I A 4 ) sin -' — = : — - — : —
V 2 / 2 sin 6 sm c
and if one of the quantities A a, A A were given, the other
might be determined by the solution of this equation ; but it
may be observed that it would be no easy matter to disentangle
either A A or A a from the other quantities with which they are
combined.
Again, since Taylor s Thebrein gives
cos {A-\- A A) — cos A = — sin il A A — cos iL -— -f &c.
1 . 2
and cos (a + A a) — cos a = — sin a A « — cos a -— + &c.,
1 . 2
{AAf ^ « Aa + cos a -j^ -&c.
sin -^AA+COS J. — &c. = -- : — ; — : ?
1 . 2 sm o sm c
in each side of which the number of terms is indefinite, and
consequently in this form the difficulty of determining the ratio
A A
' is in no degree diminished, and recourse must finally be
Aa
had to some such method as approximation.
If however any dependance can be placed upon the accuracy
of instrumental observations, it will follow that the errors above
alluded to are very small quantities, so small indeed, that they
may be neglected when connected with finite quantities by the
operations of addition or subtraction, or at most, that one
or two terms of such expressions as those above given will
ensure a sufficient degree of correctness.
^88
Thus, on the first hypothesis, we obtain
. Aa . aA
^ sni a sin sin
. ..AA 2 J 2 sma
sin A sin = — : — ; — : , ana .'.
2 sm 6 sin c . A a sm b sin c sm A
sm
2
A ^ sin «
or
A a sin 6 sin c sin J. ^
since the arc and sine are ultimately equal ; and this is the same
result as would be obtained by retaining on each side only the
first terms of the expansions given by Tai/Ior's Theorem.
A greater degree of accuracy will however be ensured on
the second hypothesis by retaining two terms of the expansions,
and rejecting all the powers of the increments above the second,
so that
sin a Aa 4- cos a
' A A . A (^^) 1 • ^
sm A A A-i- cos A ~
1.2 sin 6 sin c
A A
from which the value of may manifestly be obtained by the
solution of a quadratic; and if three or more terms were retained
aA
on each side_, the ratio — — might be still more accurately found
A a
by the solution of an equation of three or more dimensions.
. A A sin a
Reverting to the equation = -: — ; — : : — -, we
° A a sin 6 sm c sm A
observe that it is immediately derived from the proposed one by
the operation of differentiation ; that is, replacing A by d we have
dA sin a
da sin h sin c sin A
and for the reasons above assigned, the ratio of the differentials
may in all practical cases be substituted for the ratio of the
errors introduced.
85. Let ABC be a right-angled triangle whose sides and
angles are a, by c, A, B, C respectively ; then if any one of
289
these quantities remain constant, we may find the ratios of the
small contemporaneous increments of the rest.
We shall divide this into the three following cases :
(1) When a is constant:
(2) When c is constant :
(3) When A is constant.
86. Let one of the legs a remain constant*
Since cos A = cos a sin B, we get
— sin Ad A = cos a cos BdB;
dA cos a cos B cos A cos B cot B
dB sin A sin B sin ^ tan A
Since sin 6 = tan a cot ^j we shall have
cos hdh — — tan a cosec^ A JA ;
JjI cos b
db tan a cosec^ A
cos 6 sin 2 A
...(1).
sin 6 tan A cosec ^ 2 tan b
Since sin « = sin c sin A, .*. sin c = sin a cosec A^
whence cos cdc— — sin a cosec A cot Ad A;
dA cos c _ cos c tan A
dc sin « cosec A cot tI sin c cot -4. tan c
Since sin a = tan b cot Bj .'. tan /; = sin a tan B,
whence sec^ bdb = s'm a sec^ BdB;
dB sec" b sec^ b sin2jB
db sin a sec" B tan 6 cot B sec^ J5 sin 26
Since cos c = cos a cos b, .'. we have
sin cdc = cos a sin bdb ;
6?c cos a sin 6 cos c sin b tan Z>
(2).
..(3)
.(4).
db sine cos b sin r tan c
Go
.(5).
290
87- Let the h^/pothenuse c be considered cojistaut.
Since cos c = cot A cot B, .*. cot A = cos c tan B,
whence — cosec^ Ad A = cose sec^ BdB;
dA_ coscsec^B_ cot A cot _B sec^ -B _ sin 2 A
dB cosec^ J. cosec^ ^ sin 2 B
Since sin a = sin c sin A, we obtain
cos «c?a = sine cos Ad A;
dA cos a cos a sin A tan A
da sin c cos A sin a cos A tan a
Since cos il = tan 6 cot c, we have
— sin A dA = cot c sec" 6 J/^ ;
dA sec^hcotc sec^ b cos A _ Qcot^l
dh sin A sin J. tan 6 sin 26
Since cos c = cos a cos b, we have cos a = cos c sec b,
whence — sin ada=: cos c sec b tan bdb,
da cos c sec b tan 6 cos a sin 6 tan 6
.(1).
.(2).
.(3).
db sin a sin a cos b tana
88. Let one of the angles A remain constant.
Since cos A = cos a sin B, we have sin B = cos A sec a,
whence cos BdB = cos y^ sec a tan aJfl ;
^B cos A sec a tan « cos a sin B sec a tan a tan B
da cos B cos B cot a
.(4).
(1).
Since cos B = cos b sin il^ we get sin BdB = sin A sin 6(^6 ;
dB _^ sin 6 sin A _ sin b cos B _ tan Z»
c?6 sin B sin B cos 6 tan B "
Since cos c = cot A cot B, we have sin cf/c = cot A cosec^ BdB ',
dB ^ sine _ sin e cot B _sin2B
dc cot A cosec" B cos c cosec^ B 2 cote
(4).
291
Since sin b = tan a cot A, we have
Qosbdb = cot -^ sec ada\
da cos h cos /; tan a sinSa
db cot J. sec^a sin ^ sec^ a Stan 6
Since sin rt = sin c sin tI, we have cos a Ja = sin ^ cos c Jc ;
da sin A cos c sin a cos c tan a ^ .
.-. _- = = ; — = (5).
d c cos a cos a sni c tan c
89. Thus the ratio of the evanescent increments has been
determined on each of the suppositions above made, and the
last three articles include all the ditferent cases that can occur
in right-angled spherical triangles^ the right angle not being sup-
posed to undeigo any change ; and it may be observed that the
two parts for which this ratio has been found increase or
decrease at the same time when the differential coefficient is
positive, and the contrary when it is negative.
Moreover, if the small change in any one of the parts be
considered given and constant, the greatest or least values of the
contemporaneous small changes in any of the rest may be deter-
mined by putting the corresponding differential coefficient equal
to zero, according to the principles of Maxima and Mhtima.
90. Let ABC be any spherical triangle whatever whose
sides and angles are a, &, c, A, B, C respectively : then the
consideration of the corresponding small changes in the parts
may be comprehended in the four following cases :
(1) When A and a are constant:
(2) When A and b are constant :
(3) When a and b are constant :
(4) When A and B are constant.
91 . Let the angle A and its opposite side a remain constant.
cos J. 4- cos 5 cos C
bmce cos a = : — — — : — — , w'e have
sin B sm C
cos a sin jB sin C = cos A -f cos B cos C ;
292!
dB sin B cos C cos a + cos B sin C
c/ C cos B sin C cos a + sin 5 cos C
cos A + cos B cos C
but sin B cos C cos a =
and cos B sin C cos a =
tan C
cos J. -j- COS B cos C
tan jB
therefore by substitution we obtain
dB sin B fcos J3 + cos A cos C|
{COS J3 + C0S A cos C) cos 6
cos C + cos il cos B f cos c
dC sin C
sin B sin J. , . „ sin J. . ,
Since —. — — = -; , we have sin B = -; — - sin 6 ;
sin b sin a sin a
.*. cos BdB = -^ cos hdbj whence
sin a
dB sin A cos 6 sin B cos 6 tan B
(1).
c?6 sin rt cos B cos B sin b tan 6
cos a — cos h cos c
(2).
:os A =
sin 6 sin c
we have sin h sin c cos tI = cos a — cos b cos c ; whence
db sm 6 cos c cos A — cos 6 sin c
dc cos 6 sin c cos A — sin b cos c ^
from which by substituting for cos A, its value, and reducing, we
get
db cos B ^ ^
dc cos C
dB tan i^ J6 cos B
Since -77- = , and — - = — , we have
db tan 0 dc cos C
<?B tan jB cos B sin B
92. Let the angle A and its adjacent side b remain constant*
cos B -\- cos A cos C
Since cos b =
sin A sin C
293
we have sin A cos 6 sin C = cos B + cos A cos C ;
dB _ sin j1 cos 6 cos C + cos A sin C
d C sin B
(cos B + cos ^ COS C) cos C + cos A sin^ C
sin B sin C
by substitution, whence
dB cos A + cos B cos C
d C sin B sin C
— cosa (1),
sm jB sin^l . ■ ^ • a ■ .
Since — : — ~ = —. , we have sma sui ^ = sin A sni b ;
sin 6 sm a
dB sin B cos a tan B
whence -r— = — — : — = (2).
da cos B sm a tan a
^. . cosa — cos 6 cos c
bmce cos A = : — - — : ,
sin o sm c
.'. cos A sin 6 sin c = cos a — cos b cos c ;
6^a cos A sin 6 cos c — cos 6 sin c
whence -r- = : —
dc sin a
(cos a — cos b cos c) cos c — cos 6 sin^ c
sin a sin c
by substitution,
da cos 6 — cos a cos c
f?c sin a sin c
cos B (3).
dC I dB tanB da
Since -r-r: = , ~r~ — > ^^^" ~T~ ~ c<^s i), we have
dB cos a oa tan a dc
dC tan B cos ii sin i^
-T- = = (4).
dc tan a cos a sin a
93. Xe^ the two sides a a?ii/ b remain constant,
sin A sin a , . ^ sin a . ^
Since -^ — =: = -. — r, we have sin A = -: — - sm B:
sm B sin 6 sm b
294
dA sin a cos B sin A cos JB tan A
dB sin 6 cos A sin 5 cos A tan jB
(1).
^. „ cos c — cos a cos 6 . • ^i 7/-1 sincere
bnice cos C = : -. — ; , we have sm LdL = — : — r ;
sin a sni 0 sin a sm 6
. dC sin c 1 1 . .
whence =: — ^ "• ; = -7- ;;— — = — ^ ^ . •(2),
t?c sin « sin b sin C sin b sin A sin a sin B
_. . cos « — cos 6 cos c
Since cos A = : — - — : ,
sm b sm c
.'. cos A sin b sin c = cos a — cos b cos c ;
c?^! cos J. sin 6 cos c — cos Z> sin c
whence -—r- = : — ; — : : — 7—
dc sm b sin c sin J.
cos a cos c — (cos b cos^ c 4- cos 6 sin c)
sin 6 sin^ c sin A
6? J. cos B sin fl cos B cot B
dc sin 6 sin c sin J. sin ^ sin c sin c
or
(3).
^. c?J. cot 5 ^c . ,
Since — — - = : , and —7; = sm a sin B,
dc sin c aC
, J^ cot 5 sin a sin B cos B sin a
we have — — = : = : .... (4).
dL sin c sm c
94. Let the two angles A and B remain constant.
cos A -f cos B cos C
Since cos a = : — r——. — ~ _,
sin B sin C
we shall have as before
dC sin a sin B sin C sin a sin B sin^ C
rfa cos a sin ^ cos C + cos B sin C cos J5 + cos A cos C '
c?C sin a sin 5 sin C sin 6 sin C sin C
or-— = : — = — — = (1).
da sin A cos 0 cos 0 cot b
295
^. cos C+ cos A cos i? . _ sin CdC
bince cos c= -. — - — : — — - — — j .'. sin cac=^
sin A sin 5 ' ' ' sin A sin 5 '
dC sin A sin JB sin c . . . .
whence — r~ = : — — =sina sin jB=sin b sm A,,, (2),
dc sm C
^. sin a sin A , . sin A . ^
bnice -; — r = . , we have sin a = -: — - sm 6 ;
sin o sm JD sin B
da sin ^ cos b sin « cos b tan a
whence -— = -r—- = -7—^ = r (3).
do sin Jd cos a sin o cos a tan 6
6?a cot b dC
oC sin C dc
Since 3-^ = -: — ~ , and -7— = sin b sin J., we shall have
da __ cot b sin J. sin b cos 6 sin A
dc sin C sin C
95. Cor. The ratios which have been deduced in the
same manner as the last in the preceding article^ might like the
rest have been found by an independent process,, and all the
ratios determined in the last three articles may be expressed
in different terms according to the nature of the case in which
they are employed : thus in (93), we have seen that
dA
cot B
dc
sin c
35) cot B =
cot b sin c
cose
sin A
tan A
dA cot
dc tan
c cot b
A sin A *
This instance has been selected because it includes the
solution of an important astronomical problem^ but it is clear
that in every one of the other cases similar substitutions might
have been made.
296
The principles explained in this chapter may with great
facility be applied to compare the corresponding small changes
in the parts of plane triangles ; and indeed the observations
made towards the ends of articles (239), (240), (241) and (244)
in the Plane Trigonometry, are merely examples of the same
principles without introducing the notation of the Differential
Calculus.
This subject was first treated of by Roger Cotes, in his
tract entitled Estimatio Errorum in mixta Mathesi, &c. which
is the first of his Opera Miscellanea^ and may be found at the
end of the Harmonia Mensurarum,
297
CHAP. VII.
Containing some miscellaneous Propositions.
96. To expi^ess the sum of the angles of a spherical triangle
in terms of the sides.
_. , / ^ E 1 f cos a-fcos 64-cos c
Since by (58), cot —
2 o ^sin »S sin (S—a) sin (S—h) sin {S—c) '
/A + J5+C\ /E o\ ^
we have tan ( \ ~ tan ( ~ + 9^ ) = ~ cot —
l 4- cos « + cos ^ -{- cos c
2 yy/ sin S sin {S — a) sin (S — b) sin (6' — c)
97- To express the perimeter of a spherical triangle in terms
of the angles.
Resumin"^ the notation of (15) we shall have
sni
/a + b-^c\ _ . ^TT-A' + ^~ B'-\-7r-C)
= «'" 1"^ o / = - '^' { o )
_ ^ - cos S' cos (S' - A) cos (S'-B) cos jS' - C)
. A . B . C
2 sni — SHI — sm —
2 2 2
by substitution and reduction,, as in (59).
Pp
298
98. To express the excess of the sum of tivo angles of
a spherical triangle above the third in terms of the sides.
Retaining the notation hitherto used, we have
A + B\ C a b
,.„ (i±^)
/Ai-^\ c a 0 . ^
tan( ) —tan ~ tan - tan - +cos C
\ 2 / 2 _ 2 2
A^B\ C ~ sin C
l+tanl — Jtan —
V 2 / 2
1 4- cos c — cos a — cos b
Ay/ sin »S sin (5 — «) sin {S — b) sin («S — c) '
by substitution and reduction as in (58).
Similarly, tan ( ) may be expressed in terms of the
angles.
99* ^0 express the perpendicular from an angle of a
spherical triangle upon its opposite side, in terms of the sides
and angles respectively^
By Napier's Rules, sin CD = sin BC sin B
2 /—. : : :
= sin a ~ : — /s/ sin S sin (S — a) sin {S — b) sin (S — c)
sm a sin c
^ 2 ^smS sin (S-a) sm(S—b) sm (S - c)
sin c
Alsoj sin CD — sin B sin BC
2
=sin B V -cos .S' cos {S'-A) cos {S'- B) cos (S'- C)
SHI 15 sin C \ / \ / \ /
__^ 2 x/ - cos y cos (Y - .A) cos (^^ - B) cos (8^ -- C)
sin C
299
100. To Jijid the position of the pole of the small circle
of the sphere, Ziihich may he inscribed in a given spherical
triangle.
Referring to the figure of (185) PL Trig, and supposing
all the lines employed to be arcs of great circles instead of
straight lines, we have by means of the same construction,
Ah = Ac, Ba = Bc, and Ca = Cbj
whence Ab + Ba-\-Ca = Ac+ Bc-{- Cb, oiAb-ha = c + Cb,
.'. Q.Ab-{-a = c + Ab+Cb = b + c, and A5= ^^~^ ^S-a:
2
similarly_, Bc = S — b, and Ca = S - c.
Now by Napier s Rules, we have sin Ab = tan ob cot oAb,
from which if r be the circular radius required, we get
tan r = sin A b tan o ^ 6 = sin {S — a) tan —
2
. _, , . /sin(^- 6)sin(^^-c)
= sin (S-a) y — . c • /c — ^; —
sm o sm (o — a)
sin {S — a) sin (S — b) sin {S — c)
sin <S
_ />/ sin iS sin {S — a) sin {S — b) sin (S—c)
sin S
w
Hence also the segments of the sides are found, and thus the
position of the pole is determined.
101. Tofnd the position of the pole of the small circle of
the sphere, which may be circumscribed about a given spherical
triangle.
Referring back to (188), PI. Trig., and making the same
supposition as in the last article, we have
/.BAo=zABo, aBCo=lCBo, zACo=/.CAo,
.'.by addition, Z BA o-\- C = B-\- Z CA o, whence
2zJBlo + C = J5+ zClo4- zBAo = A + B,
300
and .*. zBAo= =6— C:
2
similarly, zCBo= S'-A, and ^ACo=:S'-B.
Now by Napier's Rules, we obtain
c
cos BAo = cot Ao tan - , whence if Ao = R,
2
we have tan jR = sec {S* — C) tan - :
2
but sec' (S'- C) = 1 + tan^ (.S' - C)
!1 + cos c — COS a — cos h \^
^ l~cos a — COS o~cos c -|- 2 cos a cos 6 cos c)
( 1 + cos (0 ( 1 — cos «) ( 1 — cos h)
J by reduction.
2 sin »S sin (8 — «) sin («S'— ^) sin (-8 — c)
4 cos - sm - sin -
2 2 2
^ sin S sin (^ - «) sin {S - b) sin {S - c)'
c . a . b
2 cos -- sin - sin -
,o' r\ 2 2 2
. . sec (o — C)
V sin 5 sin (6'— a) sin (^ — b) sin (-S— c) ^
. a . b , c
2 sm - sin -- sin -<
2 2 2
and tan it = — ^ . ^ . . — •
^sin aS sin (»S -- c) sin (S—b) sin (<S— c)
The segments of the angles have been found above,
and thus the pole is determined.
The articles of the Plane Trigonometry just alluded to,
readily show how great is the similarity of the formulas found
in this and the preceding articles to those investigated there :
and it is manifest that the methods here pursued would lead
immediately to the results before obtained.
301
102. Given tivo sides and the included angle of a spherical
triangle, to find the angle contained between the chords of these
sides.
Let the two given sides and included angle be a, h^ C, and
let a, /3, y be the chords of the sides a, b, c respectively, then
from (21) we have
cos c = sin a sin b cos C + cos a cos b, that is,
. o c .a a . b b
1 — 2 snr - = 4 su) -' cos - sni - cos - cos C
2 2 2 2 2
+ (l^2sin^g(l-2sin=^),
or 1 = ai3 cos - cos - cos C + 1 ^~- -\ ~ ,
2 '^ 2 2 2 2 4
. a'+t^'-^Y n a b a^0'
2 2 2 4
but if C be the angle contained between the chords a, (3, we have
, «°-+/3°-7^ a h ad
COS C = — r = COS - COS - COS L -j
2a/3 2 2 4
a h , a . b
= COS - COS -« COS C + sni - sm ~ .
2 2 2 2
From this it is not difficult to shew that C is greater than C
when it is an acute angle_, but less when it is either an obtuse or
a right angle.
103. Cor. If D be the pole of the circle circumscribed
about the triangle ABC, then the angle ADB will be measured
by the arc of the circle included between A and B: also the
302
angle between the chords of JC, BC stands upon the same
circumference, and therefore by Euclid in. QO, C'= i / ADB:
and hence it also follows that the angle ADB at the centre is
greater or less than twice the angle at the circumference
according as that angle is acute or obtuse.
104. Given the chords of two sides of a spherical triangle
and the included angle, to find the angle contained between
the sides.
r.. -n/ a b . a . b
Smce cos C = cos - cos - cos C -\- sni - sin - , we have
2 2 2 2'
^/ . a . b
cos u — sm - sui -
r 2 2
cos C =
a b
cos - cos "*
2 2
^^^ ^ ~ "I" A r> a
4 _ 4 cos L —ap
in which the radius of the sphere is supposed to be ] .
Ex. If a = ^5 and .*. a = /3_, we shall have
, 2« /^, -2^ J /^ 4cosC' — a*
cos L = cos - cos C + sm -^ , and co-s C = ;; .
2 2 4 — a"
105. Given the oblique angle contained between two given
objects above the horizon, to find the horizontal angle.
Let a and n be the objects whose angular distance an
is observed from the point O in the horizon : then if straight
lines were drawn from a and n to O, the angle aOn would
be the oblique angle^ and AON is the corresponding horizontal
303
angle, which is the same as the spherical angle APN \
let aw = c, Aa =^ H, Nn = h, then by (2o) we have
APN
sin
cos H cos h
from which the horizontal angle may be found.
106. Given two sides of a spherical triangle very nearly
equal to quadrants, to find the difference between the remaining
side and the measure of the included angle.
Retaining the notation of the last article,
let c + ^ = AN= Z A ON to the radius ] ,
TT TT
then a = - — H. and 6 = - — h:
2 2 '
whence, H and h being very small, we have
cos c — sin H sin h cos c — Hh
cos (c 4- 0) =
H
cos c
-^Hh
' (-4)(-f)
(cose— iy/0{ 1 + 1- (^- + /i')} very nearly:
l-^{H' + h') ' ^' ^
that is, cos c — 0 sin c = cos c — Hh -J- i- (i/ '+ AO cos c
since 0 is very small,
Hh~-i(H' + lr)cosc
.-. 0 =
sm c
304
__ <2.Hh - (H' + U') cose -
2 sin c
QHh(cos'^ + sin' I) - (H' + h') (cos^l - shrj)
. c c
4 sm — cos —
2 2
{H^ + 2 JT//, + h') sin^ - - {H' -2Hh + h') cos' -
2 2
. c c
4sin — cos —
2 2
/J/4-A\' c /H-h\' c
= ( I tan ( ) cot—.
V 2 / 2 V 2 / 2
This expression is called the approximate reduction to the
horizon, and was first given by M. Legendrc in 1787.
107. The sides of a spherical triangle being small with
respect to the radius of the sphere, it is required to find the angles
of a plane triangle whose sides are of the same magnitudes.
If a, b, c be the sides of the proposed triangle to the radius r,
the sides of a similar triangle to the radius i , will manifestly be
r r r
now cos A
a be
cos -- ~ cos - cos -
r r r
. b . e
sHi - sm -
r r
a a' a^
but cos- = 1 ~ -\ X nearly,
b e
and similarly of cos - and cos - ;
r r
, . b b // . . . c
also sm- = ^nearly, and similarly of sin -
r r l,2.Sr' "^ ^ r
805
therefore by substitution and reduction we get
, // + c' - a' a'-\-b'-\- c" - 2 (a' b^ + «^c'^ + b'' c")
cos il = j ;; -„
%bc Q4bcr'
= COS A — ~5 sni J ,
if A' be the corresponding angle of the plane triangle whose
sides are a, h, c :
but if il = ^1 -f- 0, we have cos A = cos A' -—O sin //, nearly ;
_ be . , be sin A' 1
•'• t7 = —-s sin A = —-J— = — - area,
6r 2.3;- 3r^
whence A' = A'-'0 = A — — ^ area :
similarly B'= B — 0 = B :; area, and 0=C — 9=C — — j area :
^ Sr- 3r'
also since 7r = A' -^ B' -{■ C' = A •\- B-jr C ^ -^ area, we have
r
— area = ^+2^fC — 7r=£, the spherical excess,
r-
.\A'=A-~, £r = B --, and C' = C - -:
3 3 3
that is, a spherical triangle under the above-mentioned circum-
stances may be treated as a plane triangle having the same sides,
and each of its angles less by one-third of the spherical excess,
than the corresponding angle of the proposed triangle.
The discovery of this beautiful Theorem is also due to
M, Legendre, and is alike remarkable for the simplicity and
conciseness of its application, and the accuracy of its results.
108. Cor. If we suppose the radius r of the sphere to be
indefinitely increased so that any finite portion of its surface
may be considered as a plane, we shall have for a plane triangle
Qq
306
cos A = , as in (l65), PL Trio;.
Also since the arc^ sine and tangent are all ultimately equal,
it may be observed generally that all formulae for spherical tri-
angles involving the sines and tangents of the sides, will be true
for plane triangles, when for the sines and tangents are substi-
tuted the sides themselves.
Since the sides and angles of spherical triangles are measured
by arcs of great circles on the surface of the sphere, if the ra-
dius be indefinitely increased, the measures of both sides and
angles will become right lines, whereas we have always supposed
angular magnitude to be measured by a circular arc. On this
account when we suppose the radius of the sphere infinite, the
angles must still be measured by arcs of a circle of finite radius
which may remain the same whatever be the magnitude of the
sphere on which the sides are described. This circumstance
will therefore render the magnitudes of the sides indeterminate
when only the angles are given : and, in fact, whenever the magni-
tude of the side of a spherical triangle has been expressed in
terms of its angles, the radius of the sphere has been supposed
finite and determinate. This additional element constitutes the
whole difference between plane and spherical triangles; for if the
radius of the inscribed or circumscribed circle of a plane tri-
angle, or any other line given in species, be known, the sides of
a plane triangle may be determined by means of the angles.
APPENDIX I.
CONTAINING MISCELLANEOUS THEOREMS AND PROBLEMS
IN PLANE TRIGONOMETRY.
On chap. I.
I. Theorems.
1. Ir-the radii of two circles be R and r, and the arcs A
and «3 the corresponding angles will be as Ar : aRo
2. The sum of the angles of any polygon of « sides
= (;i- 2) 180° English, or (?2 — 2) 200° foreign.
3. Each of the angles of a regular polygon of 2w sides
= C!—\ 180° English, or (^^) 200° foreign.
4. The greater the number of sides of a regular polygon,
the greater is the magnitude of each of its angles.
5. The ratios of the lengths of a foreign and English
degree, minute and second are expressed by the fractions
3.33.3^ 3.3^
, r and 5 respectively.
6. The difference of an arc and its complement is equal
to the complement of twice the arc.
m — 71 71
7. The complement of — — 90° English = 180°.
m + /^ 771 -{■ 71
8. The supplement of ^^-^ 200° foreign = —7— 400°.
9. Sin ( b Aj = cos^, and cos ( - -f Aj= —sin A.
10. Sin(7r + ii)= — siutI, and cos (tt 4-^)= —cos ^.
11. Sin ( — ± ^) = — cos^, and cos (~r~± -^ ) = ± sin^.
308
12. Sin^(4Ai + ]) ^ ± J\= cos A,
and cos M 4« H- 1 ) - ±Aj— + sin ^.
13. Tan({4n+S) | ± ^^ = + cot .1,
and cot f (4/?H-3) - ± ^ j= =Ftan A.
J-
14. S'm (^(3n+ I) '^±a\ = ± cos il,
and cos K 3 /< 4- 1 ) - ±^)= T sin il.
15. Tan(^(3^^-l)^±^^= Tcotil,
and cot nSw— 1) ^+ ^ j= + tan il,
^n o- A COS A tsin A cos ^ sec J. tan il cot ^
16. hmA= -= - = — = -— ,
cot A sec A cosec A cosec A
sin A cot A sin jl cosec A tan A cot J.
17. Cos^ =
18. Vers A
tan A cosec ^ sec A sec j4
tan^l— sin^ cosec Jl — cot j1 sec A — 1
tan A cosec A sec A
_ sec ^ sin A cosec A cos A sec A
19. Tan A = r = = z
cosec A cot A cot A
cos
A
20. CotA =
sin A cot~ ^
cosec A sin A cosec A cos il sec A
sec A tan A tan A
sin y4
cos A tan' A
21. SecA =
309
tan A cosec A sin A cosec A
sin i4 cot J. cos A
tan A cot A
'22. Cosec yI =
cos A
cot A sec A cos A sec Yi/ tan yl cot A
cos A tan A sin A sin J.
23. Vers {(-^^^tt- a\ +vers (f— ^V + ^ [ =vers7r.
(\77i + /// ^ {\m-\-n/ )
24. Chd A chd (7r-A) =
{c.,d| + chd(|-A)|{cM|-chdg-.)}.
25. ChdM = 1 +sin'^lcotM - 2sinyl cotl +cos^il tan^il.
26. 1-2 sin^ 30° = 2 cos"" 30° - 1 = sin 60°,
and (sin 30° + cos 30°) (sin 60° — cos 60°) = sin 30°.
27. 2 sin 30° cos 30° = sin 60°, tan 60° = 2 sin 60° = 3 tan 30°
= chd 120°, and cosec 60^ = 2 tan 30^.
28. Sin 45° + cos 45° = chd 90°,
and 2 sin 45° cos 45° = sin 90°.
29.
sin 45° - sin 30° _ sec 45°- tan 45°
sin 45° + sin 30° "" sec 45° + tan 45° '
sin 60° - sin 30° tan 60° - tan 45°
and
sin 60° 4- sin 30° tan 60° -f- tan 45°
30. If «S and s be the sines of two arcs, C and c their
cosines : then
310
31. If *S and s be the secants of two arcs, T and t their
tangents, then
32. If S and 5 be the sines or secants of two arcs,
T S
T and t their tangents, then is — greater or less than - ,
t s
according as T is greater or less than f.
33. If C and c be the cosines or cosecants of two arcs,
T
nd t their cotangents, then — will be
according as T is greater or less than t.
T C
T and t their cotangents, then — will be less or s^reater than - ,
t c
II. Problems.
1. Compare the magnitudes of two angles, when the
arcs which subtend them are inversely as the radii.
2. If to the radius r an angle be measured by an arc whose
length is a, required the length of the arc which will measure an
r
angle m times as great to the radius -- .
^ n
3. One regular tigure has twice as many sides as another,
and each of its angles greater than each of the angles of the
other in the ratio of 4 : 3 : find the number of sides of each.
4. The number of sides of one regular polygon exceeds
the number of those of another by 1, and an angle of one
exceeds an angle of the other by 4^ : find the number of sides
of each.
5. The interior angles of a rectilinear figure are in arith-
metical progression, the least angle being 120 , and the common
difference 5^ : find the number of sides.
6. Represent in the foreign scale, the English arcs :
if 15', 22' so', 30', 45", 60', 75^ and 78' 45^
311
7. Express in the English scale, the foreign arcs: 10°,
20', 33|-^ 40^ 50^ 66f, 70^ and 95'.
8. Find the complements of the following arcs in the
English and foreign scales : 15^ 18^ 2£° 30', 36°, 54^ 72^ 75^
95^ and 120^
9. What are the supplements of the following arcs in the
English and foreign divisions of the circle : 33^ 45^, 78^ 30',
150^ 180^ 2]0^ and 270^?
10. Point out where the sine increases or decreases, and
shew that it changes its algebraical sign only when it passes
through 0.
11. Trace the increase or decrease of the tangent in each
of the four quadrants, and prove that this line changes its sign
either by passing through 0 or 00 .
12. Trace the changes of algebraical sign in the secant
of an arc, and find whether sec A and sec (tt + A) have the
same or different signs.
13. Given the algebraical signs of the sine and cosine of
A in each of the four quadrants, to determine the sign of the
tangent, co-tangent, secant, and co-secant.
14. Transform from the radius 1 to the radius /-, the
formulae,
A ^ A ^ A ^
tan A = 7 , sec ii = , cosec A =
cot A ' cos A ' sin A '
chd A=- ^2—2 cos A^ sin^ A = vers A vers (-/r— /I),
chd- 0 -I- ^^chd'^'^ ~ a\ =4 vers (^ 4- a\ vers (^ - a\
15. Deduce the sine, cosine, &c. of loO^, 22o°, 270°
and 315^
16. Given any one of the trigonometrical lines defined
in this chapter, to deduce all the rest, and adapt them to
the radius r.
312
17. Given 2 siii A = tan ^, to find the sine, cosine,
&c. of A,
18. Given vers (- — A\ = v, to find all the rest, and
adapt the results to the radius r,
19. Given chd (7r—A) = k, to find all the rest in forms
adapted to the radius r.
20. If sin A cos u4 = m, find the values of sin A
and cos A.
21. Given m sin A = n cos'"^ A, to find tan A and cosec A.
22. Given m sin A + ?i cos A = p, to find tan A
and sec -4.
23. If m chd il + w cos A = p, what is the value of
chd A ?
24. If sin A (sin J. — cos A) = m, find the value of
sin A.
25. If m vers A±_n vers (tt — ^) =/>_, what is the value
of vers A ?
26. Given tan ^-|-cot ^ = 4, to find the value of tan A.
27. Given m tan A + ^i cot A =p sec ^1, to find cosec A.
28. If sin A + sin JB— m, and sin ^ sin .B = jt, required
the values of sin A and sin jB.
29. Given sin A sec B^m, and cos A cosec B = )/, to
find the value of sec A and sec B.
30. Given sin A + cosi> = y/z and sin L^-f ccs it = ??, to
find sin A and cos B.
31. Given tan A + tan B = ??2, and sec tI — sec B = //,
to find tan A and sec B.
32. If in a right-angled triangle the sine of one of the
acute angles be given, it is required to find the versed sine
and chord of the other.
313
33. Give two general formula?, one including all the arcs
whose sines are positive, and the other all the arcs whose
cosines are negative.
34. Express the radius of a circle in which the length
of 45 = L, in terms of the radius of a circle in which the
length of 60^ = /.
35. Given chdA = m and vers A = n, find vers B when
chd B=2j, the radius being unknown.
On chap. II. AND III,
I. Theorems
Involving the Trigono met viced Functions of one Arc and of some
of its Multiples and Submultip/es.
/
. Sin A =
2 sin —
2
2cos ~ 2
2
sin"
2
2cos^ —
2
A
A
cosec —
2
tan
2
A
cot-
^ A
2 tan -
2
A
2 tan ~
2
A
2cot —
2
A
2 cot —
2
.A ^A .wi .A
sec— l + tair— cosec— 1 + cof
o o. o o
1 1 1
A A A ^ . A cosec A
tan l-col— tan r^ot /i cot — - cot yl
2 2 2 2
,1 tI • 4 -'^ sin 2 ^ cosec ^
V 2. Cos yl = cos sin —
2 2 sin J 2 cosec 2 J.
314
A J A . A A A
2 cos sec ~ cosec 2 sin — cot — — tan —•
2 2 2 2 2 2
A
sec —
2
.A
1 - tan^ —
2
2
=
cosec
.A
cor - -^
A
2
• 1
=
~ A A
cot- +tan^
sin A
I
cor- +
1
1
3.
1 +
2 sin
Tan A =
1-
A
2tan —
2
tan
A
2
2s
c
il tan —
2
J
cos —
2
in^ —
2
2cot —
2
sec
.1*
2 sin —
2
sec
' 2
. it A
— 2 sin — tan —
2 2
2
2^
2 — sec —
o
.^1
osec- — -
o
■ 2
A A
cot - — tan —
2 ^
2 + 2 tan A tan — .
2 .V ^ 1
= ( 1 + sec vl) tan — =
tan hcot —
Q Q
4. 2 tan A =(sin A 4- tan A) sec^ — = sec^ A sin 2^.
^2
6. Tan 21 -tan A
2 cot il
A
2
sin J. 2 sin A
^ 6. Tanil=cot^ — 2cot2l:=
2cos''7l — cos A cos 4-+ cos 3 A
sin 2 A 1 —cos 2 A
1 + cos 2 A sin 2 A
_ . /] —cos 2 A
I -f cos 2 A
315
,„ A sin Q.A
laii — =
2 ' ' —'
cos A
I -^ cos 2 A 1 4- cos A
V 8.
^ 9.
10.
v^ 11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
2 sill
o A
^1 ^ • 2 ^
— silvers A = sin ^.
2
(2 sin ^ + sin ^J) tan" — = 2 sin A — sin 2yi.
■i2
/ A\ . ^ , A
I 1 — tan — J tan A = 1 — sec A + tan -
V 2/ 2
(1 -i-secyl).
Sin 3 A
sin tI = sin^ Q.A — sin" yl.
Sin 5 A sin J = sin* 3A - sii/ 2 A.
did — did ~ = chd"- A - clid^- - .
2 2 2
_ J. 3 A cos^ — cos2il
1 an — tan — - = — — ~ ■ — 7 .
2 2 cos tI +COS 2A
_ , , tan^ 2^ - tan^yi
Tan 3 A tan A = 77—. 9—
1 - tan" A tan 2
tan" A tan^2A
„ , , COS A + sin A
Tan 2A +sec2A= -- — : — 7.
COS A ~ sm A
1 — 2 sin^ A
— 2 sin^ A _ 1 - tan A _ 1
1 + sin 2 A "" 1 + tan A tan2J + sec2A*
Cos~ 2A — sin^A =cos A cos 3 A.
r^ o . o ^ sin 3 A sin A
Tan' 2 A ~ tan- A = ^-- ^ .
cos 2 A COS" A
sin' 2^ — 4 sin' A cos^2A--4 cos" A+3
rp 4 J 3^ .
"~ sin* 2 A + 4 sin" A - 4 cos" 2 yl + 4 cos' A— 1
Cos^ (1— tan 2 A tan ^) = cos 3 A (1 + tan 2 A tan A).
S16
22. Sin A = cos^— {^l + tan - + sec - j
/ .4 yJ\
I 1 + tan — — sec — ) .
23
. 2 sin
^=4/ -
-sin SA + >/sin^
'31-
-1
- sin3i4 -
- Js\v^
3A-
- L
. 2 cos
:os3l +
^y cos^
31-
1
24,
4- V cos 31 — ^ycos"37l — 1.
11. Theorems
Juvolvhig the Trigoiwjnetrical Functions of two Arcs and of
some of their Multiples and Submultiples,
1 . Sin A = cos (30' - 1) - cos (30^ + A )
= J^ ^sin(30^+l)-sin(S0'-l)^
= -^ ^sin(45Vl)-sin(45'-l)^ =2sin-(^45'+ -A- 1
l^tan-(45--^)^l-cot-(45"+^)
l + tan^(45'-^) 1 + cot'^ (43« + ^)
tan (45 V I) -tan (45' - I)
tan (45'+ ~\ + tan Ud^ - -\
= sin (60" 4- 1 ) - sin ((JO" - 1) =
cosec
317
2. Cos A = sin (30^ + ^) + sin (30^ - A)
= —^ {cos(45^ + ^) + cos(45°-J){
= Csin(45M-|)sin(45«-^) =
tan /^45*^ + -') -f cot ^45*^+—^
3. Tanl
'l/S sec A
1 sin (30^ + A)- sin (30^ - A)
sin (45" + ^) -cos (45
l/3 sin (30^-1- .4) + sin (30" -A)
sin (45^ + A) + COS (45^.1)
,,„(45« + |) + cot(45»-4)
A , A
tan h cot —
2 2
sin (60^ + A) - sin (60^ - A) 1
cos (60^ + A) + cos (60^ — 1) cot it
4. Sin(30^± A) = cos(60^ + il) =
cos tI 4- V^3 sin yl
5. Sin {45' ±J) = cos (45^ T A) = \/i±^!ili .
6. Tan (30"± A) = 1 (cot ^30" + ^) - tan (3(f + ^\\
7. Tan (30" + - ) tan (30" - -) = r .
318
,,. , n .V COS ^ . /l 4- sin 2^
Tail (45^ + A) = — _ . , = V - - ^ A •
^ - 1+ sin it ^ l+sm2^
1 4- sin J
r,. / n ^\ 1 + sin /J . .
Q. Tan ( 45^ + - ) = -=^ = sec A + tan A.
^ \ - 2/ COS A -
10. Sin A = 4 sin — sin ^(iO'' \ sin ^60" + -^ .
11. Cos A = 4 cos - sin (sO^ \ sin / 30^ + ~\ .
1 2. Tan A = tan ^ tan (60*' - ^^ tan (go' -\- ~\
13. 3 cot il = cot- - cot (go'' ^ + cot ^60° + -V
14. Tan ^45*^ + -) - tan (45° - ~\=z<2 tan A.
1 5. Sec (45'' + -~\ sec Ud"" - -\=2secA,
16. Sec iL = tan(45^+-)-tani = cot ^45^ - -j-tauA.
,>. ^ /TT-f A\ /ir — Ax ^ ..
17. 2 vers ( j vers ( I = vers (tt — A),
1 8. Sin B = sin (tI + B) cos 1 — cos {A + i^) sin ^1.
2cos 1 j cos(A^-5).
20. Sin(A + jB)sin3(yl-jB)=sin-(2A-L')-siir(2jB--4).
2 1 . Sin {A±B) = cos A cos J5 (tan 1 + tan B),
22. Cos {A ± L') = cos /I cos L' (I T tan A tan jB).
Sl^
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
Vers (A - B) vers {tt - (.4 + R)} =(sin A - sin B)".
Sin^ (1 4- B) - sill' (^ - i^) = sin 2 A sin 2 jB.
Cos' (i + £;) - sin' A=cosB cos (2 A + jB).
Sin-(il+ B)=sin~7l+sin^i^ + 2sin A sin Bcos{A + B).
Cot A + tan B = - — ^ .
sni A cos B
Tan ^ + tan B
Cot it + tan B
Sin2(A + E)
= tan A tan (^ + _B).
cos {A + Jj) cot B — tan A
Sin 2 J. + sin 2 jB cos (tI — U) cotii+tanA'
sin (A + B) sin (A - .B)
1 an~ A — tan- B = 3—; ;7— r .
cos A COS" B
TanU + -B) + tan(J~i^) __ tan A (l+tan-.B)
Tan (.4 -f- i*) - tan {A- B)~ tan ^' (1 + tan" A) *
1 4- COS 2 A COS 2 B = 2 (sin^ A cos^ i? + cos' A sin" jB).
If
m tan (A ~ J5) n tan B
cos^ B
cos'U-B)
, then will
34.
tan {A-2B) = tan A,
n ■\- m
If cos B {m 4- cos A)=\ + m cos ^_, then will
/A B\
H { tan tan — I .
V 2 2/
A , i>^
tan — + tan —
2 2
i^ 1-tan^A .
35. If tan — = — ; ^— - , then will
2 1 + tan^ A
2coi2A = A^tan B + sec B + ^U\\\ B — sec B.
320
III. Theorems
Involving the Trigonometrical Functions of three or more Arcs
and of some of their Multiples and Suhmultiples.
1. QhdnA chd(7r- J) = chd(;i+1) J + chdO/- 1)J.
2. Sin-4 + cos J5=2sin(^45M ~—) cos ^45^ V
3. Cos^l+sin J5=2cos(45^ + — ^^^ cos /^ 45*^ ^V
Cos 7) -sin A / 0 A-^ B\ / . A- B\
4. 7\ 7^ :— : = tan ( 45" — j tan ( 45' )
Cosi3 + sniy4 V 2 / V - 2 /
5. If ^ -f- /^ + C = TT, then
sin A + sin j5 + sin C = 4 cos — cos ~ cos — .
2 2 2
6. On the same supposition,,
sin 2^1 + sin 2 i3 + sin 2 C = 4 sin A sin B sin C.
7. The same hypothesis remaining,
_^ , . ,B , . ,C ^ . A . B , C
sm" h sm h sm f- 2 sm — sm ~ sm — = 1 .
2 2 2 2 2 2
8. If iH-B-f C = 45°, then
tan A + tan B + tan C— tan A tan 7i tan C
= 1 — tan A tan i? — tan A tan C — tan B tan C.
9. If ^ -f- B f C = (2?« + l)^, then
cos 2^ + cos 2 B + cos2C = 4cos A cos /> cos C.
321
10. If^-fi^+C=^, then
tan A + tail B -{• tail C = tan A tan B tan C -f sec A sec B sec C.
11. If the arcs Ay B, C be in arithmetical progression, tlien
sin A — sin C = 2 sin (J — B) cos 73.
cos^ — cos C sin ^ — sin C
12. Again, tan (J— 23) =
sin yi + sin C cos ^4 + cos C
tan 73 sin A + sin C
13. Also,
tan ( 73 — C) sin yi — sin C /A - O
tan '
(^)
14. And sin /i sin C
= {sin J5 + sin(yi-13)} {sin B-sin(^-B)}.
15. lfA + B + C=-, and A, B, C be in arithmetical
4
progression, then
VS — tan A=i\ + VS tan A) tan C.
16. If J. +73 + C==7r, and the sines of A, 73, C be in
arithmetical progression, then
. A . /B-C\ . C . /A-Bx
sin — sin ( I = sm — sm ( i .
2 \ 2 / 2 \ 2 /
17. On the same snpposition, if the cosines of yl, B, C
be in arithmetical progression, then
A . B C ^B
2 cos — sm — cos — = cos — .
022 2
18. Co.s-(\/?)-cos-'(J^ + ^J = 30«.
19. 2tan-^ ('-,') + tan-' G)=^'^''
322
21. If tan ^ = — — , and tan B — —y — , then
sin {A'\-B):= sin 60° cos 36^
22. Sin(^-jB)sinC-sin(A-C)sini5 + sin(JB-C)sini4=0.
23. If cos ^ = cos B cos C, then
/^+^\ /A.-B\ eC
tan { ) tan ( J = tan — .
V 2 / \ 2 / 2
24. If sin {A + i^) cos C = 2 cos (B — C) sin ^, then
cot ^ — cot B = 2 tan C.
25. Sin(^+B) sin(B-f-C + D) = sinl sin (C + D)
+ sin 5 sin ( A + JB + C + D).
26. Sin 2(^ -C) + sin 2(B - C)-- sin 2(^-5)
= 4 cos {A - C) sin (i? - C) cos (yl — B).
27. Cos(^-B-C) + cosU-.B + C)H-cos(^ + B^C)
+ cos ( J + -0 + C) = 4 cos ^ cos jB cos C.
28. If yi, B, C, D, &c. L, be any arcs, then
sin (A + B) sin U - B) + sin(B + C) sin (B-C) + &c.
+ sin {L+A) sin (L - il) = 0, and cos {A + B) sin (A - B)
+COS ( B + C) sin (B - C) + &c. + cos (L + J.) sin (L - .1) = 0.
cos A — sin B sin C cos JD — sin E sin P
29. If ^ 7^ = Ti T. ^
cos B cos C cos i^ cos ±
cos B cos C vers A — vers (B — C)
cos £ cos F vers D — vers (-E— F)
323
IV. Theorems
Involving the Numerical Values of the Trigonometrical Functions
of certain given Arcs to the Radius 1 .
1 . Sill 7^ 30
0 n,J
2 V^2
V^3+ 1/5 - Jo - Vb
2. Sin 9°
4
3. Cos 11° !■>- ^ ^^^"+^.
2
21/2— t/3~ 1
5. Vers 15^ =
6. Vers 78^ 45' =
21/2
, _ 2 — x/2 - x/ "2+1/2
7. Chd 30°
1/3—1 ^2
1/2 1 4- 1/3
8. Chd 67" 30' = J 2 - JT'~ V'2.
9. Tan 18"*
10. Tan 37° 30'
11. Cot 78° 45'
2 V" 2~+ t/3"
""' 4 H- 1/6 - t/2 *
2 + ^2+^/2
324
V3+ V5 + Vo -\/5
,3. S.c2f ^-^y^V-5-273-V-5
1 + V 5
14. Sec 5fr 15
15. Cosec 22^30' =
l6. Cosec52''S0'
0 .,/ _ 2n/2 4- J%- V'2
V^-l- V^S
^2- 1/2'
17. Sin ^'=.-^'^{V5^-\)^~j5-V5,
Vs , . 1
18. Sin 12^=-- — (1/5-1) +^^yM^^5.
19. Sin2f=-'^(i/5+l)+^x/I^V^.
20. Sin39^^=-^^(V^5 + 1)-^V^::75'.
21. Sin 4£' = - i (•5- l)+ j^ >/^'+"v^-
22. Sin57^ = - :^$^(V^5-l)+^^^7?+1^
8 K 2 o
1 V^S
23 . Sin m' = ^ (1/ 5 + 1 ) + -—r-^ Jo-\/5.
8 ^ 4\/2
1/3-1 /
-— ; (V/5- 0 +
8t/2 ^ 8
24. Sm 87"= -—7--(v/5-l) + -— ^5+t/5.
S25
V. Problems
Involving the Trigonometrical Functions of one or more Arcs.
1. Express each of the Trigonometrical Functions of A
in terms of sin 2 A and sin ~ , and adapt the results to radius r.
2
2. In terms of cos Q.A and cos — .
2
3. In terms of vers 2 A and vers — .
2
A
4. In terms of chd 2 A and chd ~ .
2
5. In terms of tan 2 A and tan — .
2
6. In terms of cot 2 J. and cot —.
2
7. In terms of sec 2 J. and sec — .
8. In terms of cosec 2 J. and cosec — .
2
9. Given tan 2 A = 3 tan J, to fmd A.
_ . 1 + tan A . , ....
10. Given = ^ sec2A, to hnd tI.
1 — tan A ^
11. From the equation_, tan — = cosec vl — sin Ay find the
value
of A
326
12. Find the value of A which satieties the equation,
1 + tan A tan — ) .
13. If sin 3^-2 sin2A + sin''il +4sin^l = 0, find the
value of sin A,
14. Given sin 4A-\- 4 sin 3 A cos A =0, to find tan A,
15. Given sin 2 A cos 2^4 + 3 sin A cos S A = 0, to find
cos A,
16. Find general formulae including all the values of A
which fulfil the conditions of the equation,
2 sin' SA + sin^ 6A = Q.
17. If sin{A-B) = cos{A+B) = ^, find the values of
A and B.
18. Given sin {A + B) + sin (A - B) = cos (A -f B)
-f- cos {A — B\ to find the value of A.
19. Given sin (A + B) - sin {A ~ B) == tan 60" sin jB, to
find the value of A.
20. If sin {9. J -r B) - sin (2.1 - B) = sin (J. + ^)
— sin (A — jB) — sin B, find the value of sin A,
21. Given tan A + tan B = sec A, find the relation
between the values of A and B.
22. Given tan C + 2 tan (1 - C) = tan (A + jB — Q, to
find tan C.
23. Given 2 tan C + tan (A - C) = tan (C - B), to find
ttin C.
24. Divide a given angle into two parts A aud B so that
sin A cosec B m
sin B cosec A ?i *
327
cos A sec B m
25. So that
cos i> sec A n
26. So that
27. So that
28. So that ™" " "■" ,
tan h cot A n
sec A m
cos
B
sec
A '
vers
A
m
vers
B
11
chd A
m
chd B
n
tan
A
cot
_B
29. So that
tan B n
30. If tan A ^ tan i^ + tan C = tan A tan B tan C, it is
required to lind the general value of (7I + i^ 4- C).
31. If sin A — 2 sin jB + sin C = 2 sin jB vers (A— B), find
the relation between A, B and C
32. Given cos A -{-COS JB = w, and cos 5 A + cos 5 B = ?/,
to find A and B.
33. If ^ -f 5 + C = 180°, and tan A tan 5 = m, tan .4
tan C = )t, find the values of tan A, tan B and tan C.
On chap. IV.
I. Theorems.
1. In a right-angled triangle, if A, B, C, r/, ^, f be the
angles and sides, and C the right angle, then
in — = V . cos — = V , tan — = V .
sin
2ab , h^-cr ^ 2nb
2. Sm 2 A = -2-7TT . cos 2^1= ■ ^, tan2^=^,
828
3. Sin (45' ± A) 4 ~ (^)' '''' (4^°± ^)
= -4-(-^^). tan (45^ ± J) =7^.
4. Sin (A -73) = r— , cos (1-jB)= -x" .
tan ( J[ - JB) = -— — .
2ab
5. The area = — = '' J c-a^ = - Jc'-lr=^ ^ sin Svl
2 2 7.2
^ . nr^ « ^ b
— — s\n2B = — tan jB = — tan ^4.
4 2 2
6. If «', 6' be the segments of the h^pothenuse made by
a line bisecting the right angle, then
' jf 2,72
ah {a + hf
7. If the lines drawn from the acute angles to bisect the
opposite sides be a, /3, the tangents of these angles are
8. The radius of the insciibed circle is
(a + b-c)
, (a -f c - 6) (Z> + c - fl) ah
^(a-\-h — c) = -k :
a -\- b + c a + b + c'
9. The radius of the cirumscribed circle is
a be a be
(a + b + c) (a + b — c) (b + c — a) (a -{- e - b) 5'
10. The sum of the diameters of the inscribed ami ciicum
scribed circles is a ^- b.
329
11. l( A, Dj C be the angles of a triangle, and 2 cos B
sin A
sin C
, the triangle is isosceles.
12. If ^^" ^ = ^"\ - , then will the triangle be either
tan B sni" x>
isosceles^ or right-angled at C.
13. In any oblique-angled triangle, if A, B, C, a^ h, c be
the angles and sides respectively, then
C ^ . ^
cot — tan — + tan —
a+h 2 c 22
, and
a-b r^-JW CL-b A H
tan I I tan tan —
(^')
Vers A _ a(S-b) vers(^+B) _ (S—a){S-b)
^^' Vers B~biS- a)' """' vers C 6'OS-c)
15. 1 an — tan — = — - — , and — = — .
2 2 S B ^ - a
tan —
o
16. Sn. (-^) = (— )-«i ' -^ (-^) = (~) ""i'
2 ;2
and sin r.4 - B) = (^-7-) «i" (^-^ "h B).
/^-f JB-C>
tan
_ ^h'-c^ __ tan B
^'^"^ A + C- B\ " ,r + c'^-/r ~ tan C
tan
(^^^)
Tt
880
18. The perpendicular from the angle C upon the opposite
side
_ a + b -\-c __ c cos (A — B) — cos {A + B)
"" I B ~i sinC *
cot h cot —
2 2
19. The distance of the perpendicular from the middle of
the base
c sm(A—B) c tan^ — tanB
~ 2 sin C "" 2 tan A + tan B *
c" sin J. sin B
20. The area of the tnanole = -
i sin C
c" cos(A- B)~cos(A+B) 2a he ABC
: ■— — = ~_ cos — cos — cos — .
4 sm{A-^B) aH-/; + c 2 2 2
21. Four times the area of the triar.gle
- (« + ^ + g)^ {cos A -(-cos ^4- cos C~l}
sin A+sin i^ + sin C
= n/
4(a^ + //4-cV(«' + ^'' -f c"T a2 + Z^- + c"
cot" A + cot- i^+cot" C cot J. 4- cot i? + C0I C
22. The sum of the perpendiculars from the angles upon
the opposite sides = 2 area ( j,
23. If R and r be the radii of the circumscribed and
inscribed circles, then
abc 1 111
QRr = , . , , or-— = _+— + _-.
a-f-o-^c 2Hr ab ac be
24c. The diameter of the inscribed circle
. / X A S C
= (« -f 6 + 6) tan — tan — tan —
2 2 2
331
__ 2 ^ ctV'c^ sin A sin B sin C _ rtftc (sin A + sin B 4- sin C)
25. The diameter of the circumscribed circle
a + 6 +(; 4 / a^c
.v/I
sin ^ -f sin B + sin C sin A sin B sin C
a 4- 6 + c
" T B C'
4 cos — cos ~ cos —
2 2 2
26. The sum of the squares of the distances of the centre
of the inscribed circle from the angular points
Qahc
= ab + ac-\-bc
a-{-b + c
27. If d be the line drawn to bisect the angle C, and
meeting the opposite side in E, then
. ,,^ a-rb C
tan ALL = tan — .
fl-6 2'
cos ACE = ; — , and ^ =- ttt {(a + o) — c }.
2a b {a-Vby
28. If P and p be the perpendiculars from the extremities
of the base of a triangle upon the line bisecting the vertical
angle at distances D and d from it, then
4 Pp = (a - /> + c) (6 + c - a\ and 4 D c? = (« + ^ + c) (« + 6 - c),
and the nre3. = Pd = pD.
29. The perpendiculars drawn from the angles of a triangle
upon the opposite sides meet in one point, and the rectangles
of the segments of the perpendiculars are equal in each.
30. If three straight lines be drawn from the angles of a
triangle to bisect the opposite sides^ they meet in the same point :
and the sum of the squares of the sides of the triangle is equal
to three times the sum of the squares of the distances of the
point of intersection from die angles.
332
31. If lines be drawn from the angles of a triangle to any
point, the products of the sines of the angles thus formed taken
alternately are equal, as are also the products of the alternate
segments of the sides.
32. If two angles of a triangle be bisected bylines meeting
in a point, the remaining angle will be bisected by the line
joining it with this point.
33. The perpendiculars to the three sides of a triangle at
their middle points, meet in one point.
34. If a J bj c be the sides of a triangle, and a, b, c
the perpendiculars drawn from a point within the triangle, to
bisect the sides, then
fa b c) abc
a b c ) abc
35. The side of an equilateral triangle inscribed in a circle :
the side of a square inscribed in the same circle '.: s/ 3 : /^ 2 ;
and the area of the triangle : the area of the square :: 3^3 :
36. The square of the side of a pentagon inscribed in a
circle is equal to the sum of the squares of the sides of a regular
hexagon and decagon inscribed in the same circle.
37. If a point be assumed in a regular polygon of n sides,
from which perpendiculars are drawn to each of the sides or
sides produced ; the sum of these perpendiculars : the radius of
the inscribed circle :: ii : 1.
38. If the external angles of a quadrilateral figure be
denoted by a, /3, y, S, and the sides by a, b, c, d, then
a sin a-\-b sin (a + /3) + c sin (a + /3 + 7) + J sin (a + i3 + 7 + S)=0,
a cos a + b cos (a + jS) 4- c cos (a -f- /3 + 7) + ^Z cos (a + /3 + 7 + ^)=0.
39. The area of a regular polygon inscribed in a circle is
a mean proportional between the areas of an inscribed, and of a
circumscribed regular polygon of half the luimber of sides.
333
40. The area of a regular polygon circumscribed about
a circle is an harmouical mean between the areas of an inscribed
regular polygon of the same number of sides, and of a circum-
scribed regular polygon of half that number.
41. If an equilateral polygon of ^ sides be inscribed in a
circle whose radius is 1, the side= V 2 — v 2-}-^2 + 8tc. the
radical sign being repeated v times.
42. If the diagonals of a quadrilateral whose opposite
angles are supplemental to each other, intersect at right angles,
their segments are proportional to the rectangles of the sides
which are terminated at their extremities.
43. Jn every polygon, any one side is equal to the sum of
the products of each of the other sides and the cosine of the
angle made by it with the aforesaid side.
44. In every polygon, the perpendicular upon a side from
any of the angular points is equal to the sum of the products of
the sides comprised between that point and side, and the sines of
their respective inclinations to that side.
45. The square of a side of any polygon is equal to the
sum of the squares of all the other sides, diminished by twice
the sum of the products of all those sides, taken two and two
together, and the cosines of the included angles.
46. Twice the area of any polygon is equal to the sum of
the products of its sides except one taken two and two together,
and the sines of the sums of the exterior angles contained by
those sides produced.
47» A circle is inscribed in an equilateral triangle, an equi-
lateral triangle in the circle, a circle in the last triangle, and so
on, in infinitum: then the radius of any one of these circles is
equal to the sum of the radii of all those within it.
48. If lines be drawn from all the angles of a polygon
to any point, the products of the sines of the angles so formed
taken alternately are equal.
334
49. In a right-angled triangle, a perpendicular is drawn
from the right angle to the opposite side : then the areas of the
circles inscribed in the triangles made by it^ are proportional to
the corresponding segments of the side.
50. In a plane triangle, the differences of the segments of
a side made by a perpendicular from the opposite angle, by the
contact of the inscribed circle, and by the line bisecting the
opposite angle, are in geometrical progression.
51. If through any point O within a triangle, straight lines
be drawn from the angles A, B, C to meet the opposite sides
in «_, b, c respectively, then
Oa m Oc _
52. In any triangle, the rectangle contained by the excess
of the semi-perimeter above each of the sides including any
angle, is equal to the rectangle of the radius of the inscribed
circle and the radius of the circle which touches the base and
the two sides produced.
53. If 7' be the radius of a circle inscribed in a triangle,
7'j, To, ?3 the radii of three other circles touching the sides and
1111
sides produced, of the same triangle, then — = — + h — ,
7' r, /'g r-g
and the area of the triangle = v ^'^'i^s^s*
54. If R be the radius of the circle circumscribed about
a triangle, r the radius of the circle inscribed in it, the distance
between the centres of these circles is ->/ J^^~ 2Rr.
55. If r^, To, ry be the radii of the circles touching one side
of a triangle and the two others produced, the distances of their
centres from that of the circumscribed circle whose radius is R are
jR:'-\-^lRr,, jR''-^%Rr., and V-^' + ^-Krg.
335
II. Problems.
1. Given the three sides of a triangle, to find the per-
pendicular upon one side from the opposite angle, and the
segments into which that side is divided.
2. Given the perimeter and area of a right-angled triangle,
to find the sides.
3. Given the perimeter of a triangle, to find the sides,
when a perpendicular from one of the angles to the opposite
side divides that side in a given ratio.
4. Given one angle of -a triangle, and the straight lines
drawn from each of the other angles to bisect the opposite sides,
to find the sides of the triangle.
5. Express the perimeter of a triangle in terms of two
of the angles, and the perpendicular from the remaining angle
upon the opposite side.
6. Given the lines bisecting the acute angles of a right-
angled triangle and terminated by the opposite sides, to find the
area of the triangle.
7. Given the perimeter of a triangle and the ratios of its
angles, to find the sides.
8. In a right-angled triangle, given one of the sides con-
taining the right angle and the radius of the inscribed circle, to
find the sides.
9. Given the hypothenuse of a right-angled triangle and
the radius of the inscribed circle, to find the sides.
10. Given the perimeter of a right-angled triangle and the
radius of the inscribed circle, to find the sides.
136
11. Given the area of a right-angled triangle and the
radius of the circle inscribed in it, to find the sides.
12. Given the three angles of a triangle and the radius of
the inscribed circle, to find the sides.
13. Express the area of a triangle in terms of the radius
of the inscribed circle and the three angles.
14. Given the three angles of a triangle and the radius of
the circumscribed circle, to find the sides.
15. Express the area of a triangle as a function of the
radius of the circumscribed circle and the three angles.
16. Investigate an expression for the area of a triangle
involving all the sides, and the tangents of all the semi-angles.
17. In a right-angled triangle, given the radii of the
inscribed and circumscribed circles, to find the sides and
area.
18. In any triangle, given the vertical angle, the radius of
the inscribed circle, and the sum of the lines drawn from its
centre to the angles at the base, to find the sides.
19* Given the perimeter, the area and one angle of a
triangle, to find the side opposite to it.
20. Given the area, the vertical angle and the sum of the
including sides, to find the sides of the triangle.
21. Given the radius of the circumscribed circle, the
vertical angle and the ratio of the sides containing it, to find
the sides of the triangle.
22. If a circle be described about a triangle, find the
distances of the bisections of the sides from the circumference.
387
23. Given the three straight lines drawn from the angles of
a triangle to bisect the opposite sides, to find the sides of the
triangle.
24. Express the area of a triangle in terms of two of its
sides and an angle opposite one of them : also, in terms of two
of its angles and a side opposite one of them.
25. Determine the triangle whose sides are three con-
secntive natural numbers, and whose greatest angle is double
of the least.
26. Find the angles of a triangle, when the base_, the
sides, and the perpendicular are in continued geometrical
progression,
27. Given three straight lines, to find the radius of the
circle so that they shall be the chords of three contiguous arcs
which together make a semi-circle.
28. In a given scalene triangle, it is required to draw from
one side to another produced, a straight line which shall be
bisected by the third side.
29. Given the perpendiculars from the angles upon the
opposite sides of a triangle, to find the angles and sides.
30. Given the area, the base and the sum of the angles at
the base of a triangle, to find the angles.
31. Given the angles of a triangle and the perpendiculars
upon the sides from a given point within it, to find the sides.
32. Given the angles of a triangle and the perpendiculars
upon the sides from a given point without it, to find the sides.
33. Compare the sides and areas of the squares and regular
octagons described in, and about, the same circle,
Uu
338
34. Given the ratio of the side of a regular polygon
inscribed in a circle to the radius, to find the number of sides
and the magnitude of each angle.
35. The alternate angles of a regular pentagon being
joined, it is required to compare the sum of the isosceles
triangles so formed with the pentagon.
36. Find the side and area of a regular decagon inscribed
in a given circle.
37. Determine the sides of a regular hexagon and dode-
cagon inscribed in the same circle, and compare their perimeters
and areas.
38. The area of a regular polygon of n sides in a circle :
the area of another regular polygon of 3 n sides in the same circle
:: p : (j : find the values of the angles subtended by a side
of each at tlie centre.
39. The area of a regular polygon inscribed in a circle ; the
area of a similar figure circumscribed about it :: 3 : 4; find the
number of sides.
40. Find the side of a regular quindecagon inscribed
in a circle of given radius.
41. The area of a regular polygon inscribed in a circle
being given, and the area of one circumscribed with the same
number of sides^ it is required to find the areas of the inscribed
and circumscribed polygons of half the number of sides.
42. Find the area included between two regular polygons
of the same number of sides,, one being inscribed in, and
the other circumscribed about, a circle of given radius ; and
determine the number of sides when this area has a given ratio
to either.
339
43. Given two sides and tlie included angle of a quadri-
lateral, to find the sides and diagonals, when two opposite
angles are right angles.
44. Express the area of any quadrilateral in terms of all
the sides, and two of the opposite angles.
45. Given one side of a polygon and the angles made by
it with the lines drawn from its extremities to all the other
angles, to find the area of the polygon.
46. A circle has an equilateral triangle inscribed in it;
a circle is inscribed in the triangle which also has an equilateral
triangle inscribed in it, and so on : find the sums of the
perimeters and the areas of all the circles and triangles.
On chap. V. VI. VII.
Theorems and Problems.
c = y {a - bf-\-4ab sin" - = V {a + bf-4ab cos~ ~
^—^^ — ^
V ('^ + ^)" si"'— + (« - ^f <^«s' —
= b (cos A + sin A cot jB).
2
a
2. Sin C
cos B + sin B cot C
c sin A
s/ b' + c'^ - 9.bc cos A
_ sin B {a cos B± sj b"^ - a" sin^ B
~ h
= ~ sin G J5 ± sin ii \/ 1 - C^\ sin' B.
^/■-(-:y
3. Cos C =
340
a — c cos -B
x/ a^ -^ c'^ — 2 rt c cos B
2 sin' B + cos 7i V 1 - (yj'si"' B.
c sin yl _ c sin A
''■ '^"" '^' == 7«'-cSinM ~ 6 - c cos I
Z> COS ^4" \/ ci^ — b'
sm
6 COS ^ — cot il ^ a - h^ sin^ A
1 + sec yl V /^^'^ - sin^ A
1 — cosec J.
v/(=y - .i„.
5. If /> be the perpendicular upon the side c frooi tlie
opposite angle C, then are the other two sides respectively
equal to
^J c^'\-pc cot C+ sj c^-^pc tan C,
and s/c^-\-pc cot C — n/c^ 4-pc tan C.
6. Given the angle B, the side c and the sum ot the
remaining sides, to solve the triangle.
7. Given the angle B, the side c and the ratio of the
remaining sides, to solve the triangle.
8. Given the angle B, the side a and the area, to solve the
triangle.
9. Given the area, the base and the sum of the angles
at the base^ to solve the triangle.
10. Giveji the logarithms of the three sides of a phuic
triangle, to determine the logarithms of the segments of one of
them made bv a line bisecting the opposite angle.
341
11. If the base of an isosceles triangle be c and the per-
pendicular from one of the equal angles upon the opposite
side py then
log area = log p-{-2 log c ~ 2 log 2 — i- log {{c -\- p) (c — p)].
sin ^ a
12. if tan A tan IS = 3, and -:^ — — = - , and we assume
sni B h
%ab
tan 20 = — —5 -jT , then will
3(a^ — o^)
tan
A /3a cot (b , ,^ A /3b tan d>
^ = V — -X. ^ and tan B= S/ ^
13. In finding the sine of half an arc, shew that when 0 is
small, a large error may be expected in applying the formula
. 0 . / 1 - cos e
sm - = V ,
2 ^ 2
and a small one in using tlie formula
n
sin - = ^ ^y \ -\- sin 0 - ^ ^ \ - sin 0.
14. The Sun's altitude being oO", lind the position of
a stick of given length that the shadow may be the longest
possible, and determine the shadow's length.
15. The aspect of a wall is due south and the Sun is in
the south east at an altitude of 30^ : lind the breadth of the
wall's shadow.
16. A person attempts to swim directly across a stream of
given breadth, where will he reach the opposite side, if he
swim n times as far as he would have done, had there been no
current, and what angle does his course make with it ?
17. Three objects A, B^ C form an isosceles triangle
whose vertex is B and whose angles are as the numbers 4, 1, 1:
a person walking from .1 towards C measures a base AD=a feet
and observes the angle BDC : he then advances to E, h feet
342
lailher, and linds the angle EEC the supplement of BDC :
lind the sides of the triangle.
18. Coasting along shore observed two headlands, the
iirst bore N. N. W., and the second N. E. by E. : then steering
12 miles E. N. E., the first bore N. W. and the second N. E. :
shew how the distance and bearing of the two headlands from
each other may be found.
19. A person on a tower can see the top of a pillar
of known altitude from which he wishes to know his distance
and the height of the tower : he can see also an object on
the horizontal plane from which he has formerly observed
the angular distance of the tops of the tower and pillar : shew
how he may find the required distances.
20. For determining the distance between two inaccessible
objects A and Bj two positions C and D are taken such that the
triangles ACT), BCD are not in the same plane: state the
requisite observations for determining their distance, and the
bearing and elevation of one as seen from the other, and
give the solutions of the triangles in logarithms.
21. A person wishing to ascertain the horizontal distance
of two inaccessible objects from each other, can find no point
from which they are visible together : he finds however two
stations the distance between which he can determine, from
which the objects may be separately seen: explain what obser-
vations and measurements it will be necessary for him to make,
and how they must be applied to effect his purpose.
22. The top of a tower is visible from three stations
A, Bj C in the same horizontal plane: at each of the stations
the angular distance of the top of the tow^er from each of the
other two stations is observed : given the distance between
A and B, and the height of the tower, to find the distance of C
from each of the other stations and from the tower.
23. A hill rises due north at an angle of 45^_, and a shaft
was discovered in it making an angle of G(f with the horizon,
and extending 100 feet in a north cast direction which led
into a cavern stretchini!; horizonlaliv to the north east. At the
343
foot of the hill, 300 feet in a south west direction from the
mouth of the shaft, another opening was found extending-
horizontally 120 feet due north: find the length and direction
of the least shaft that can be cut from tlie extremity of this
opening to reach the line of the cavern.
24. Prove that— = f 2^ - ^\ seconds = 52'' 44"' S'"' 45\
12 V 2V
25. If the sides of a triangle be a, b, c,
and X + - = 2 cos A, y -\ — =2 cos 7?, then -- -\- bx = c,
26. If Cq = cos a cos B cos C &c., c„ = the sum of the
products of all the cosines but n, multiplied by the sines of
those Hj then
cos (^ + 5+ C + &C.) = fo-C2 + C4 — &C.
sin {A + i3 + C + &C.) = Ci -C3 +(^5 - &c.
27. Prove that
^aJ - \ sin 71 A = (cos A + ^ -\ sin A)" -(cos .4- ^~sin^)^
and 2 cos ?iA=^(cosA + />/ — 1 sin A)" + (cosil'— ^/ — Isin^d)";
and adapt them to the radius r.
28. Shew that
2 sm
m\A= s/ysin^nA— 1 +sin tiA
V ^sin'- n yl — 1 + sin ?i A.
and
2 cos A~ \J V^cos' nA—\ -\- cos « A + -r
V Vcos^ nA—\ + cos n A
and adapt them to the radius r.
29. Find the sum of the 7?^'^ powers of the tangent and
cotangent of an arc.
30. Solve the equation a"" — 62— 4 = 0, by means of
a table of natural sines and cosines.
344
31. Solve .T^ — 49i "- 120=0, by means of trigonometrical
iormuKe.
32. Solve the equation x^ — ^ x^ — ^x •{- ^ = 0, by the
trisection of an arc,
33. Solve .r^ — Sax" — Sx -{- a = 0, by means of an arc
whose tangent is a.
34. Solve 2* -j~ 4.r^— 6:c'^-- 4 j:-|- 1 =0, by means of an
arc of 45^
35. Determine the roots of x"^ + 9,x^ — x^ — 2x -{- I =0,
by Trigonometry.
36. If cos 0= , then uill
{a'\-hj'^\r ^{a-^bj - D" = <^{d''\-bT cos- a
37. If the quadrant of a circle be divided into an odd
TV
number of equal parts so that 6 = , then to the
radius /%
sinO sin 30 sm 5^ &c. sm (2//— 3)^= ( - j
- )
38. If the semi-circumference of a circle be divided into
TT
an odd number of equal parts so that 0 = . then
(r \ " "^ ^
and cos 0 cos 20 cos 3^ &c. (/i ~ 1) 0 === ( - j
39. If the semi-circumference of a circle be divided into
an even number of equal parts so that 0 = — , then
* * 2ff
chdO chd30 chd50 8cc. did (£« -- \)0 =^/2/^
345
40. If the circumference of a circle be divided into any
number of equal parts as ii of which 0 is one, then
chd e chd 2 e chd 3 0 &c. chd (m - 1 ) 0 = // r" ' '.
41. Prove that sin 0 + sin (0 + ^) + sin (^ + 2^) + &c. is
a recurring series, find the scale of relation, and by means of it
the sum of 7i terms.
42. Prove that
sin (}-f sin 2^4-sin 304-&C. to 7Merms ^ .0
= tan ill + 1) - ,
cos6^ + cos£^ + cos 30-1-&C. to n terms 2
sin 6 -{- sin 3^ + sin 50-j-&c. to n terms ^
and 7— -^^—— = tan ni),
cos y + COS 36^ -J- cos ou -\- cvc. to n terms
43. Find the sum of vers 0 + vers Q.0 + vers 30 + &c.
to 11 terms.
44. If the circumference of a circle be divided into any odd
number of equal parts so that 0 = , then to the raduis r,
^ (2 « — 1 )
chd'0 + chd'20 + chd"30-[-&c.+chd'(2?/-2)0 = (4?i-2)r^
45. If the circumference of a circle be divided into an
even number of equal parts as 2w, then the sums of the squares
of the alternate chords are equal to each other and to S/^r*.
AQ. On the same supposition, the product of the squares
of the odd chords together with the product of the squares of
the even chords = 4r'''',
47. If the circumference of a circle whose radius is r,
be divided into 2// equal parts and from one of the points
straight lines be drawn to all the rest, the sum of all these lines
45^
2r cot
48. If Ay B, C, Df Sec. be the angular points of nn
equilateral polygon of ?n sides inscribed in a circle whose
radius is r, and P be any point in the circumference, then
PA'"-\-PB^" + PC-"+&c. = m times the middle term of
(1 +r)-"\, if ?/ be less than m.
X X
346
49. Find the sum of all the natural suies to every minute
in the quadrant.
50. Sum the series
cos 0 -\- 2 ^^s 2O -\- ^ cos SO + &c. to 71 terms.
51. Sum the series
cos^O + 2cob^20 + 3 coa'^ 30 + See. to 71 terms.
52. Sum the series
sin^ COS0 + sin2^ cos30 + sin 30 cos^c^ + ^c. to n terms.
53. Sum the series
tan^sec^^ + ^tan- (^sec- j + ^ tan - (^ sec --j + 8cc,
to n terms.
54. Sum the series
• n / . ^V ' ^ / ■ ^V ' ^ / - ^V
smtf I sm - I +2sm ~ I sm - ) +4sm -( sm- I +&c.
to n terms and to infinity.
55. Sum the series
tan^l tan- ) +2tan-( tan- | +4tan-( tan- ) +&c.
V 2/ 2V 4/ 4V 8/
to n terms and to infinity.
56. Sum sec^'^-f 4sec^20+l6sec'40 + 64sec^80 + &c.
to it terms.
57. Prove that a: sin 0 4- — sin 20 H sin 30 + &c. in inf.
2 3
if a sin 0 ■)
= tan-M ^\
U — TCOS0J
58. Sum the series
e^ sin 0 sin 20 4 sin 3 0 — &c. to infinity.
2 3 -^
59. Resolve {a"—9,ah cos 0 + 6^)*^"* into a series of cosines
of 0 and its multiples, by means of the equation
2COSW0 = x"" -\ ,
and the binomial theorem.
APPENDIX II
CONTAINING MISCELLANEOUS THEOREMS AND PROBLEMS
IN SPHERICAL TRIGONOMETRY.
Theorems and Problems.
1. In a right-angled spherical triangle wherein C is the
right angle,
sin {a-b) A -{^ B A — B sin (c - 6) ^A
-7-~ — ~ri = tan -— tan — - — , and -: — ; — = tan ~ .
sni (a + b) 2 2 sm (c + b) 2
B + A
tan ( ...
^ ^ ,« \ 2 ) /c + 6\ fc-b^
2, Ian - =
(£±£_45»)
— = tan ( ) tan ( — -- ).
2 /B-A , _n\ \ 2 / \ 2 J
.„(£^+«-)
3. Tan^- =-
2 c cos (A + B) tan A tan B - 1
2 ~" cos {A - B) tan A tan 5 + 1 '
and 2 cos c = cos (fl + />) + cos (a — b).
c +a'
/c + a\
4. Ta„(4a» + f)=-^,
tan
A^
tan —
2
tan (^^ ^\
cot(45« + -) =
tan —
2
, ... .. A'
and cot*
(«• * I)
tan
('-?)
348
2/0 ^\ 1 + s^" ^ sin A sin (C -f b)
%/ 1 - sin c sin A sin (i> — 6) '
6. In any spherical triangle whose angles and sides are
A, B, C, a, b, c,
-b\ /a -I- b
. /A-hB\ ^^A 2 ; C /A + jB\
sin I I = — - cos — , cos I I =
\ 2 / c 2 ' V 2 /
. ,A+B^ ^"%-V; C M+Bx '"'{-^J. C
.Sin — :
c 2
cos - cos ^
2 2
sin -- sin --
7* ^^'^ t^^^ sanie hypothesis,
/A-B\ /^+A
cos ( I , COS I I
. /^/-f-^\ V 2 / . c /a-^-bx V 2 / c
sni I I = sni - , cos ( | = ' cos -
\2/ . C 2' \2/ .C 2
sm - , sni —
2 2
. /a — b\_ \ 2 / . c /«~<^\ ^" \ 2 / c
^'"V"l~;~' c~^"^i' '"<^= c — ""^5'
COS - COS -
(a + b\ /a — b\
». :5m ^A-±W^ ^^-^ \-±JL sin C,
COS —
sm
349
sin (A ^ B) = sin C
sm —
2
£). Sin (rt + 6) = ^ sin c,
sin^ -
14- B\ . /A- B>
sin
(^") ■■" (^)
sin (a " b)= --, sin c.
2 ^
cos'' —
10. Sin'Csin(« + ^)sin(fl-6)=sin®csin(74 + B)sin(l-i?).
1 1 . Sin a sin c + cos a cos c cos B = sin A sin C — cos A
cos C cos b.
^. . cos C /</ s'm^ B—s'in^ C s'ln^ b + s'm C cos B cos b
1 2. Sin A = -^^^ r-Yi — ^T7^
1 — sin b SUV C
13. Cos C = - cos (A - 5) sin^ - - cos (1 + B) cos* - .
^ 2 2
14. Sin^— = cos^ I — I sin - + cos^ I ) cos
2 \2/2 \2/
15. Cos — = sin I I sin — hsin ( ) cos
2 \2/2 \2/
c c
16. Cos c = cos (a — 6) cos^ — -f- cos (a + ^) sin' — ".
2 2
sin —
2
17. Sm' - = sm^ (-^) cos= - + sm' (^- )
18. Cos- - = cos' [-^j cos- -+COS (^-^j
19. Sin' 6^=
350
(sin^ a siii^ b sin^^ c siii A sin B sin C)l
1 B C ^^
2 tan — tan — tan —
2 £ 2
^ g ^, (sni a SHI />» sm c sin A sm J5 sin C)»
20. Cos «S = ; .
a b c
2 cot •" cot - cot '-
£ 2 2
21. If a , b' be the segments of the base contiguous to the
angles A and B respectively^ made by a perpendicular arc from
the angle C, then
, cos a — cos 6 cos c , cos 6 — cos « cose
tan a = — : , tan b = -. ,
cos b sin c cos a sin c
(a-^-bx (a — b\
■a'-h\ '^" (-?-)'''" ("T-j sinU-iJ)
'a-i-b\ /a — b
tan I
/a—()\ V - , . ~ , _...,__ _,
tan I 1 = = - — , , . „, tan
V 2 /
c
c sin {A-\-B) 2
tan -
2
22. If A\ B' be the corresponding segments of the vertical
angle, then
cos jB -|- cos ^ cos C , cos A + cos JB cos C
cot A = — :— 7^ 5 cot ii = . ,
cos A sm C cos B sm C
{A + B\ /A-B\
. I w ia" I ) tan I I , , , ^
A''-B\ V 2 / _ V 2 / sin (« - ^>) C
c
2
tan
\ 2 /
C ~sin(« + Z>) ^""2
cot
23. In an isosceles triangle, wherein b = c, prove that
. a A
sin — cos —
2 . 2
sin 6 = 7 , and sin B = .
. A a
sm — cos -
2 2
24. If the two sides a, b of a spherical triangle be supple-
mental to each other, then sin 2^ + sin 2 J3 = 0.
351
26. In a right-angled splierical triangle whose right angle
is C,
A sin (c — b) sin(c— 6)
2 sin a cos 6 tan a cos c
26. On the same hypothesis
Sin - = sni - cos - 4- cos - sni - .
2 2 2 2 2
27. The sines of the arcs drawn from the angles of a
spherical triangle perpendicular to the opposite sides, are to
each other inversely as the sines of the sides upon which they
fall_, or of the angles from which they are drawn.
28. If d be the length of an arc bisecting the angle C
and terminated by the opposite side, then
2 sin a sin 6 C
tan a = —. — r- cos — .
sm (a^r b) 2
29. If -D be the length of the arc drawn from the angle C
to bisect the opposite side, then
^ sin {A + B) c
cos D = : — — COS - .
sm C 2
30. Draw through a given point in the side of a spherical
triangle, an arc of a great circle which shall cut off a given
portion of it.
31. Find the whole number of equal and regular figures
which may be described upon the surface of a sphere so as
exactly to cover it.
32. If the sides of a spherical triangle AB, AC be pro-
duced to by c so that Bb, Cc shall be the semi-supplements of
AB and AC respectively: prove that the arc be subtends an
angle at the centre of the sphere equal to the angle between the
chords of AjB and AC.
S52
33. If each of the sides of a spherical triangle be produced
till they meet, three triangles will be formed ; and if ri, r^, rg
be the circular radii of their inscribed circles, then
tan r tan r^ tan r2 tan rs
= sin 5 sin (S - a) sin (S — b) sin {S — c),
34. On the same hypothesis, if Ri, Bof ^8 be the circular
radii of their circumscribed circles, then
cot R cot Ri cot R2 cot i?3
= —co3S'cos(S'-A) cos{S'- B) COS {S''-C).
35. If a spherical triangle be inscribed in a circle whose
pole is in its base, the angle at the vertex of the triangle will be
equal to the sum of the angles at the base.
3o. If two arcs of great circles terminated by a circle on
the surface of the sphere cut one another, the rectangle of the
tangents of the semi-segments of one of them is equal to the
rectangle of the tangents of the semi-segments of the other.
37. The sums of the opposite angles of a spherical quadri-
lateral inscribed in a circle are equal to one another.
38. If a spherical quadrilateral be inscribed in a circle, the
rectangles of the sines of the semi-diagonals is equal to the sum
of the rectangles of the sines of half the opposite sides.
39. Ih a spherical quadrilateral inscribed in a circle, whose
sides are a, b, c, d^ if T> be the diagonal joining A and C,
snr —
(. a . d . b . c\ / . a . c , b . d\
sm -^ sin -^ + sni -' siu -- I I sin - sm -- + sm -- sin -- 1
22 2 2/\222 2/
. a . b . c . d
sin - sm - 4" sin -< sm -
2 2 2 2
r..
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