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/^Mj zosf./o
€^
A TEXT-BOOK
ON
ADVANCED ALGEBRA
AND
TRIGONOMETRY
WITH TABLES
BY
WILLLA.M CHARLES BRENKE, Ph.D.
ASSOCIATE PROFESSOR OF MATHEBIATIC8 IN
THE UNIVEBSITY OF NEBRASKA
NEW YORK
THE CENTURY CO.
1910
MalkZoS-^JO
BAR
i
'^^T^U^e^,^^ s^^c
COPTKIGHT, 1910, BY
THE CENTURY COMPANY
Fubli8?ied, August, 1910
Stanbope iptess
p. H. GILSON COMPANY
BOSTON. U.S.A.
•^
?
(V.
'I)
\a
I
TABLE OF CONTENTS
CHAPTER I.
Pags
The Operations op Algebra 3
(The numbers refer to articles.)
1. Letters as Symbols of Quantity. 2. Signs of Relation. 3.
The Four Fundamental Operations. 4. Rational Numbers. 5. Zero. 7.
Infinity. 8. Powers. 9. Some Important Relations. 10. Exercises. 11.
Factoring. Factor Theorem. 12. Exercises. 13. Highest Common
Factor. 14. Least Conmion Multiple. 15. Exercises. 16. Fractions.
19. Exercises.
CHAPTER II.
Involution; Evolution; Theory of Exponents; Surds and Im-
agxnaries 17
20. Involution. Negative Exponent. 21. Exercises. Zero Exponent.
22. Evolution. 23. Rational Exponent. 24. Irrational Numbers. 25.
Irrational Exponents. 26. Imaginary Numbers. 27. Reduction of
Surds. 33. Exercises.
CHAPTER III.
LOQARITHMS; BiNOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS. 28
34. Logarithms. 39. Laws of Operation with Logarithms. 41. Ex-
ercises. 42. Binomial Theorem for Positive Integral Exponents. 45.
Exercises. 46. Approximate Computation.
CHAPTER IV.
Linear Equations 37
48. Linear Equation. 50. Infinite Solutions. 51. Exercises. 52.
Graphic Solution. 55. Exercises. 56. Coordinates. 58. Use of the
Graph. 59. Exercises. 60. Problems. 61. Simultaneous Linear
Equations. 63. Graphic Solution. 64. Exercises. 65. Three Equa-
tions in Three Unknowns. 68. Four Equations in Four Unknowns.
60. Exercises and Problems.
iii
IV TABLE OF CONTENTS
CHAPTER V.
Quadratic Equations .- 54
72. Solution by Factoring. 73. Solution by Completing the Square.
74. Solution by Formula. 75. Exercises. 76. Nature of Roots. Dis-
criminant. 77. Exercises. 78. Relations between Coefficients and
Roots. 80. Exercises. 81. Graphic Solution. 82. Parabola. 84. Ex-
ercises. 85. Equations Reducible to Quadratics. 86. Exercises and
Problems. 87. Simultaneous Quadratics. 89. Nature of the Solutions.
91. Graphic Solution. 93. Standard Equation of the Circle. Exercises.
95. Standard Equation of the Ellipse. Exercises. 97. Stancfard Equa-
tions of the Parabola. Exercises. 99. Standard Equation of the
Hyperbola. Exercises. 100. Rectangular Hyperbola. 102. Exercises.
103. Solution of Two Simultaneous Quadratics. 11. Summary of
Methods for Solving Simultaneous Equations. 112. Exercises. 113.
Exponential Equations. 114. Exercises.
CHAPTER VI.
Ratio, Propobtion, Variation 88
115. Definitions. 116. Laws of Proportion. Exercises. 118.
Variation. 119. Direct Variation. 120. Inverse Variation. 121. Joint
Variation. 122. Exercises.
CHAPTER VII.
4
The T^iGONOifETRic FuNcmoNS 94
124: Trigonometric Functions. Exercises. 128. Functions of Com-
plementary Angles. Cofunctions. 129. Api^caticm of the Trigono-
metric Functions to the Solution of Right Triangles. 130. Exercises.
131. Angles of^any Magnitude. 132. The Trigonometric Functions of
any Angle. ISiT^ine Values. 135. Variation of the Trigonometric
Functions. Graphs of the Trigonometric Functions. Exercises. 136.
Periodicity of the Trigonometric Functions. 137. Relations between
tEe Functions, fts. Exercises. 142. Versed Sine and Coversed Sine.
Exercises. 143. Radian Measure. 144. Radians into degrees, and
conversely. 145. Exercises. 146. Angles corresponding to a Given
Function. 147. Use of Tables of Natural Functions. 148. Exercises.
149. Given one function, to find the other functions. 150. Exercises.
CHAPTER VIII.
Functions op Several Angles 121
152. Functidftfof (x J: y). Exercises. 156. Functions of 2 x. Exer-
qjses. 157. Functions of iz. Exercises. 158. Addition Theorems.
Exercises. 159. Exercises.
Pagb
^ABLE OE.' CONTENTS
CHAPTER IX.
Ratios and — -r—* Inverse Functions, Trigonobobtric
Equations. 133
160. Ratios and . Exercises, 161. Inverse Functions.
X X
Exercises. 164. Trigonometric Equations. Graphic Solution. Ex^
ercises.
CHAPTER X.
Oblique Plane Triangles 144
169. The Law of Sines. 170. The Law of Cosines. 171. The Law
of Tangents. 172. Functions of the Half Angles. 173. Solution of
Plane Oblique Triangles. Exercises. 179. Exercises and Problems.
CHAPTER XI.
The Progressions, Interest and Annuities 161
180. Arithmetic Progressions. Exercises. 184. Geometric Pro-
gressions. Exercises. 188. Infinite Geometric Progressions. Exer-
cises. 190. Harmonic Progressions. 191. Exercises. 192. Interest.
193. Annuities, 194. Exercises.
CHAPTER XII.
Infinite Series 171
195. limit of a Variable Quantity. 196. Infinite Series. 199. Alter-
nating Series. 200. Absolute Convergence. 201. The Comparison
Test. 202. The Ratio Test. 203. Exercises.
CHAPTER XIII.
Functions, Derivatives, Maclaurin's Series 179
204. Functions. 205. Variation of Functions. Exercises. 206.
Difference Quotient. 208. Limit of D. Q. » Slope of Tangent. 209.
Examples. Exercises. 210. Derivative. 211. Calculation of Deriva-
tives. Exercises. 215. The Derivative as a Rate of Change. Exercises.
217. Higher Derivatives. 218. Maclaurin's Series. 220. The Bino-
mial Theorem. 222. Exercises.
CHAPTER XIV.
Computation, Approximations, Differences and Interpolation. . . 199
223. Remarks on Computation. 224. Useful Approximations. Ex-
ercises. 225. Computation of Logarithms. Exercises. 227. Differ- ^
ences. Exercises. 230. Interpolation. Exercises.
vi TABLE OF CONTENTS
CHAPTER XV-
Pagb
UnDETERMIIVED COEFFICIENTB. PARTIAL FrACTIONB 210
234. Theorem of UndetenniDed Coefficients. ExerciaeB. 235. Par-
tial Frectioiis. 239. Exerdses.
CHAPTER XVI.
DBTEBiaNAMTS 217
240. Detenninants of the Second Order. 241. Determinants of the
Third Order. Exercises. 243. General Definition of a Determinant.
247. Properties of Determinants. 248. Solution of Systems of Linear
Equations. 249. Exercises.
CHAPTER XVn.
PoLAB Coordinates. Complex Numbers. De Moiyre's Theorem.
Exponential Values OF SIN X AND cos X. Hyberbolic Functions. 231
250. Polar Coordinates. 252. Curves in Polar Coordinates. Exer-
cises. 253. Complex Numbers. 256. De Moivre's Theorem. 259.
The nth Roots of Unity. Exercises. 260. Expansion of sin n^ and cos n^.
Exercises. 261. Exponential Values of sin a; and cos :r. Exercises.
CHAPTER XVIII.
Permutations. Combinations. Chance 242
263. Permutations. Exercises. 264. Combinations. Exercifses.
266. Exercises. 267. Probability or Chance. Exercises. 270. Exer-
cises.
CHAPTER XIX.
Theory op Equations 249
272. Factor Theorem. 273. Depressed Equation. Exercises. 274.
Number of Roots. Exercises. 275. To Form an Equation having
Given Roots. Exercises. 276. Relations between Coefficients and
Roots. 277. Fractional Roots. 278. Imaginary Roots. 279. Multi-
ple Roots. 280. Exercises. 281. Transformation of Equations. 282.
Synthetic Division. 285. Occurrence of Imaginary Roots in Pairs.
286. Exercises. 287. Approximation to the Roots of an Equation.
289. Exercises. 290. Cardan's Solution of the Cubic Equation. Nature
(4 the Roots. 291. Ferrari's Solution of the Quartic Equation. Exer-
cises.
TABLE OF CONTENTS vii
CHAPTER XX.
Pagk
Spherical Tbioonometry 269
292. Spherical Geometry. 293. Spherical Right Triangles. 294.
Napier's Rules of Circular Parts. 297. Oblique Triangles. Law of
Sines. Law of Cosines. 298. Principle of Duality. 2l99^ Formulas
for the Half Angle. 300. Formulas for the Half Sides. 301. Napier's
Analogies. 302. Area of a Spherical Triangle. .303. Solution of Spheri-
cal Oblique Triangles. 305. Exercises. 306. Applications to the Ter-
restrial Sphere. Exercises. 307. Applications to the Celestial Sphere.
Exercises.
Answers to Odd-Numbered Exercises 284
Index 297
Appendix A. List op Formulas 301
Appendix B. Tables I to VII 315
Protractor Inside of back cover
i
PREFACE
In a considerable number of our colleges and universities the
work of the first semester in mathematics is devoted to Algebra
and Trigonometry. Usually Algebra is taken up first and then ^
Trigonometry, or else the two subjects are studied on alternate
days. Neither plan is quite satisfactory. It has therefore seemed
to the writer that a single book, treating both subjects in a corre-
lated manner, might be of service both to student and teacher.
In the present text the principal departures from the subject
matter usually treated will be found in chapters 13 and 14. The
chief aim has been to follow a mode and sequence of presentation
which shall introduce the student who needs to apply his knowl-
edge of mathematics in his other work as directly as possible to
those facts and concepts which are most useful to him.
For this reason much stress is laid on graphic methods in the
chapters on linear and quadratic equations, and this is followed
up later as opportunity arises. It is thought that the extra time
so used will be more than made up when the student begins his
study of Analytical Geometry, because he will have become grad-
ually familiar with the fundamental idea of this subject and need
not readjust himself after an abrupt transition to a strange and
m5rsterious realm.
For a similar reason the basic idea of the Differential Calculus
is presented in a study of the derivative, and application is made
to some of the simple standard functions. Maclaurin's formula is
also obtained, and used to derive several standard expansions,
among them the binomial theorem for any exponent.
A considerable emphasis has been placed on numerical compu-
tation, that the student may have some training in ready calcula-
tions. This can be largely supplemented by. requiring students
to work out mentally in class many of the numerical exercises.
It has been thought advisable to include some matter which
may be omitted if only one semester is to be given to this course.
Just what is to be omitted must of course be left to the judgment
of the instructor.
W. C. B.
Lincoln, March, 1910.
/
ADVANCED ALGEBRA
AND
TRIGONOMETRY
ADVANCED
ALGEBEA AND TEIGONOMETRY
CHAPTER I
The Operations of Alobbba.
1. Letters as Symbols of Quantity. — In algebra, the letters of
the alphabet are used to designate quantity or magnitude. Thus
we speak of a line whose length is { feet, of a weight of w pounds,
or of a velocity of v feet per second. Here the letter used, Z, w, v,
is suggested by the quantity considered, length, weight, velocity.
When a number of different lines are considered, say n lines, their
several lengths may be indicated by Zi, Z^, Z3, . . . , fci, or by Z^^^,
ZC2)^ i(3)^ . . . , Z^'*^. Three or four different lengths may be indi-
cated by accents (called " primes ")> a^ ^'> ^"> ^'"7 • • • •
Fixed or known quantities are usually designated by the first
letters of the alphabet, as by a, b, c, . . . ; unknown quantities
which are to be determined from given data are represented by
the last letters of the alphabet, as by x, j/, 2, . . . . If a; denote
a quantity of a certain kind, other quantities of the same kind
are indicated by Xi, X2, X3, . . . (read, "a: sub-one, x sub-two,
X sub-three, etc."), or by x^^\ x^^^j x^^\ . . . (read "a; super-
script one, X superscript two, x superscript three, etc."), or by
x', x", x'", . . . (read " x prime, x second, x third, etc.").
2. Signs of Relation. — These are
= , read " equals," " is equal to," etc.;
7^, read " is not equal to ";
< , read " is less than ";
>, read " is greater than ";
<, read " is not less than ";
>, read " is not greater than ";
=, read " is identical with ";
= , read " approaches."
3
4 FUXDAMENTAL OPERATIOXS [3
Siffm ai A gg r^p U ioii, — When aevenl quantitifis are to be
treated as a m^ ooe, they are endosed by parentheses ( ),
brackets []fOr braces \ l,cfr a Ime is drawn over than, called a
▼incoltun, .
SwDS of Qualify. — These are
+, pontiYe; — , negative; 1 1, abscdnte Yalue.
The first two amply indicate opposite qualities; thus, if +v^ or
mmply v, denote a velocity in one direction, then —v denotes an
equal velocity in the opposite direction; if +i denote a tempera-
ture above zero, --t denotes an equal temperature below zero.
The third symbol is used to indicate that we are dealing simply
mth the numerical (absolute) magnitude of a quantity, without
regard to its sign.
3* The Four Fundamental Operations. — These are, addition,
subtraction, multiplication and division, indicated by the symbols
+, — , X, -^, respectively. It will suffice to recall the rules or
laws in accordance with which these operations are to be performed.
They are here given in the form of equations, and the student is
asked to state each in words.
Laws of Addition.
1. If a = 6 and c = d, then a + c = b + d.
2. If a = 6 and c ^ d, then a + c 7^ b + d.
3. a + 6 = 6 + a. {Commutative law.)
4. (a + b) + c = a + (p + c). {Associative law.)
Laws of Subtraction. — (Subtraction defined by (a— 6)+ 6= a.)
1. If a = 6 and c = d, then a — c = 6 — d.
2. (a - c) + 6 = (a + 6) - c.
3. a + {b - c) = {a + b) - c.
4. {a + c) — b = {a — b) + c.
5. (a — c) — 6 = (a — 6) — c.
6. a — (6 + c) = (a — 6) — c.
7. a — (6 — c) = (a — 6) + c.
Laws of Multiplication.
1. If a = 6 and c = d, then oc = bd.
2. If a = 6 and c 7^ d, then ac y4 bd. y
3. a X 6 = 6 X a. {Commutative law.) ^
4. a y, i}> y. c) = {a y,b) Y^ c. {Associative law.)
5. {a + b — c)Xd = aXd + bXd — cXd. {Distribuiive law.)
.1
4,6] RATIONALITY. ZERO (h\
Laws of Division. — (Division defined by (a -^ 6) X 6 = a.)
1. If a = 6 and c = d, then a -^ c = 6 -^ d, provided c, d f^ 0.
2. (a -f- 6) X c = (a X c) -^ 6, provided 6 7^ 0..
3. a X (6 -^ c) = (a X 6) -^ c, provided c ?^ 0.
4. (a -7- b) -7- c = (a -r- c) -r- by provided 6, c 5^ 0.
5. a -7- (6 -^ c) = (a -7- 6) X c, provided b, c ^ 0,
Some Working Rules. — The sign before a parenthesis may be
changed if the sign of each of the terms enclosed is changed also.
When several quantities are to be subtracted^ change their signs
and add them.
Division may be expressed as a multiplication of dividend by
reciprocal of divisor.
The sign of a product will be + or — , according as there are
an^ven or an odd number of negative factors.
4. Rational Numbers. — All positive integers can be formed by
adding +1 to itself a sufficient number of times. Through the
operation of subtraction, negative integers are introduced. By
performing the operations of addition, subtraction and multipli-
cation on the system of positive and negative integers, no new
numbers are formed. Division, however, does introduce a new
class of numbers, namely fractions, positive or negative, formed
of the quotient of two integers.
All numbers, positive or negative^ which are formed of the
quotient of two integers, are called rational numbers. They can be \
obtained from +1 by means of the four fundamental operations.-
Rational Expressions. — Let there be given certain quantities,
a, by ... X, y, ... . Any expression which can be built up
from these quantities by means of the four fimdamental operations
is called a rational expression (or function) in terms of the quan-;
titles involved.
6. Zero. — Zero is defined as that number which may be added
to any quantity without changing the vahie of the quantity. As an
equation, the definition is
a + = a.
Since (a — 0) + = a,
it also follows that
a — = a.
6 ZERO. INFINITY [6,7
6. The operation of division by zero is excluded, because, what-
ever be the number a, there is no number which represents a -^ 0.
The reason for this we proceed to consider. In the first place,
must be less in absolute value than any assignable number,
however small. For if this were not the case, we would have
a + 7^ a. Now consider the quotient t, and suppose a to be
fixed, and b to be taken smaller and smaller. As b tends toward
zero, the quotient t increases without limit and becomes larger
than any assignable number. But as b approaches zero, t takes
the form ^r and at the same time increases without limit so that
no value can be assigned to this form.
Example. Let x - 1.
Then x — o?
and 1 - X = 1 - «« = (1 -hx) (1 - «).
Dividing by 1 — x, we have
1 = !+«.
Therefore 1=2, since x = 1.
We are led to this fallacy by dividing by zero in the form of
\ — X. Since we assumed a; = 1, therefore 1 — a; = 0, and hence
division by 1 — a; must be excluded in this problem.
In any expression involving fractions, those cases in which the
denominator of any fraction vanishes must be treated as exceptional
and especially considered.
If, in a product, a factor approaches zero, while the other factors
have any assigned values, then the product approaches zero.
This is expressed by the equation
a X = 0.
7. Infinity. — A quantity which increases without limit is said
to become infinite. When 6 =^ ("6 approaches zero")? if ct is
any fixed number, r increases without limit. Such quantities,
which are larger than any assignable number, are all indicated by
the same symbol, qo (read "infinity"). As an example, consider
the law of gases, pressure times volume is constant, or
pv ^ c, or p = -•
V
8,9] POWERS. IMPORTANT RELATIONS 7
When V is very small (relative to the constant c), p will be very
large, and as v becomes still smaller, p must increase. We can
choose V so small that p will exceed any assignable quantity, or p
becomes oo when z^ = 0. This is often indicated by lim jp = go
(read " the limit of p is infinity, when v approaches zero ")•
We are thus led to write the equation,
— = 00, when a ^ 0,
This is not a proper equation, but simply an abbreviation for the
statement, " A fraction whose numerator is not zero, and whose
denominator approaches zero, becomes larger than any assignable
quantity."
Since a quantity which increases without Umit can be made as
large as we please after being increased or diminished, multiplied
or divided by any number, we have
oo+a=Qo, 00 — a = oo; ooXa = oo, oo-^a = oo.
8. Powers. — For brevity we put a X a = a^, a X a X a = a^,
and a X a X a ... to n factors = a**. The quantity a^ is
called the nth power of a. The number n is called the exponent
and a the base of the quantity a^,
9. Some Important Relations. — The following equations and
statements should be verified carefully and committed to memory:
1. (a +6)2==a2 + 2a6 + 62.
2. (a -6)2=a2 -2a6 + 62.
3.^,c^— ^ = (a + 6)(a-6).
. 4. a3+ 63 = (a + b) {a^ - a6 + 62).
5. a3- b^ = {a - b) (a^ + ah + b^).
6. {a +b + c)2= a2 + b2 + c2 + 2 (oft + oc + 6c).
7. The square of any poljuomial equals the sum of the squares
of the separate terms plus twice the product of each term by each
following term.
— 8. a** — 6" is divisible by (a + 6) and (a — 6) when n is even.
9. a**— 6" is divisible by (a — 6), not by (a + 6), when n is odd.
10. a^ + b^ is divisible by. (a + 6), not by (a — 6), when n is
odd.
11. 0^ + 6** is not divisible by (a + 6) or (a — 6), when n is
even.
8 EXERCISES [10
10. Exercises. — Simplify, by removal of parentheses aad ool-
lection of like terms :
1. (fo-i) + (l-ia).
8. (0.8<^ - 3.47 06 - 17.25ac) - (i<j? - 0.47 oft - 12fac).
». 44aj + {[48y - (fiz +3y~ 7x) +4z] -l4Sy-Sx+2z- (4a; + y)]} .
6. 6a- {4a -[86 -2a + 6] + (36 -4a)}.
Perform the operations indicated in the f ollowmg exercises and simplify the
results when posdble:
^. }A(i6^-4c» + }aP-3).
& Sa^(s^ -Zj^y + 3x1^ -2),
9. 0.6ac2d*(2a26-3cd» + Jac»-5).
10. 3jG^(6a*-46^ + 2a6»-3c^.
•IL (i:«-2x + l)0r«-3x + 2)..
12. (3a«6-2c^6^-|-a6^)(2c^-a6-562).
^ (a^-7ajV + 6V-l^)(x»-2irir^+/).
Jt4. (a+6)» + (a-6)».
^6. (§a + J)»-(Ja-J)«.
16. (a:2 + i_j^+2y)(a:2_,.l+j^_2y).
17. la? + (a + h)x + ab][s?-{a + h)x+ab].
18. [(X + a)2 - ax] [(X - a)2 + aa;].
19. a (a + 1) (a + 2) - (a - 1) (a - 2) (a - 3)
20. [x(y-l)-y(x-l)][(x+y)2-(x-y)2].
21. 31i m^p6 H- - lOf w?np\ t
22* cPbc^ -^ A o*&^. « ^ *
23. iJxV-^-iia:*!/*.
24. 3a2(6+x)3-^6a«(6+x)*.
26. 1.75 x* (x2 - 1)* -5- 25 x^ (1 - x*)*.
26. (80^6 - 24a46* + 16a768) -^ - 8a*6.
27. (8x»y-ixy*-iy5-|-22/^)-5- -tx2y3.
28. x5(a2 + 6?) -2x*(a* + &2)3^3jS(^2_,.6?)
29. (6a3x-17aV + 14ax^-3x*) -^(2a-3x).
80. (4y*- 182^+22^2 -7y-|-5) -5- (22/ -5).
31. [2sf^ + 7a?y-'9y^(x + y)]^(2x''3y).
32. (-ic^ + id^ - Ud^ +(]P) -^ i- IcP +2d).
33. (fta*-Ja36+l|a262 + ia63)-=-(ia+J6).
34. i^m^ + ^mhi- Ji wn^ + V n^) -=- (im - i n).
36. (x5-i§x* + fix5-Jx2-Wx + i)-(^-ia; + 5).
36. (2a' - 16a + 6) -^ (a + 3).
37. i^x^-x^j^ + Qxy^-^y^) ^(2x^-xy + 3y^),
38. (x* + 4x^2/^ - 322/^) -J- (x - 22/).
39. (a^ - 5a362 - Sa^ft' + h^) ^ (a^ - 3a6 + Ix^h
40. (x^ -82/3) -^(x -22/).
41. («^x4-92/2)^(J^ + 3^)^
Ill FACTORING 9
42. (27 (fh^ + Mix?u^)'i- (Sab + ^xy).
43. (aV + c«) 4- (o^ft? - o6c + c2)^
44. (u^-S2ffi)'hiu-2v).
46. {a-b + c-df.
46. (a;-§y-2w +«?)*.
11. Factoring. — To factor an expression is to find two or more
quantities whose product equals the given expression. When
two or more expressions contain the same factor, it is called their
common factor.
We shall illustrate the methods commonly used in factoring
given expressions by means of some typical examples.
(a) Expressions, each of whose terms contains a common factor.
Example, J x*i/^2* -|- J ajV« — iix^i? - i s^j^z (i xs? + i — ^afyz).
(6) Expressions whose terms can be grouped, so that each group
contains the same factor.
Example, a^ - 7a:^ + 14^ - 8v* = (x» - 8^*) - (7a:^ - 14^)
= (x-2y)(a? + 2xy-{-^y^) -7xy(x-2y)
^(X'-2y)(x^-5xy+^y^)
= {x—2y){x—y) {x — 4iy).
(c) Trinomials of the form ax^ -^-bx^ c.
Let A, fc be a pair of factors whose product is a, and m, n a
pair whose product is c. Arrange these four factors as in the
adjacent schemes ^X^ ^X^ and form the cross-products as indi-
cated. The sum of the cross-products must equal b. If this
is true in the first, scheme, the factors are (hx + n) {kx + m) ;
in the second, the factors are {hx + m) (kx + n).
Example, 12a? — 7a; — 10.
Here h, k may be one of the pairs of numbers 1, 12, or 2, 6, or 3, 4, both num-
bers to be taken with the same sign. The numbers m, n may be —1, 10, or
+1, —10, or —2, 6, or +2, —5. By trial we find that h, k must be 3, 4,
and w, n must be 2, —5. The factors are therefore (3x4-2) (4 a; — 5).
To find the factors of 12x^-- 7 xy — 10 y^, we would proceed
as above and obtain {3 x + 2 y) (4 x •- 5 y).
(d) Expressions which can be written as the difference of the
squares of two quantities.
The factors are the sum and the difference of the two quantities
respectively, / .
Example. a* + aV -|- 6* ^ a* +'2 a^ + 6^/ - a^9
' « ^^ +^- (dibf—
-^V + a^ + &^) i<^ -ab-\-l?).
xk6 FACTORING [12
(e) Expressions of the form P2+ 2 PQ -f- Q^, where P and Q
are monomials or polynomials.
The expression is then the product of two factors each equal to
(P + Q), and is therefore (P + 0)2.
Example, sf + ]^ — 2xy —4ax + ^ay + 4ta^
= (x — y)^ — 4 a (x — y) 4- 4 a*
(/) Factor Theorem. — If a polynomial in x reduces to zero
when X is replaced by A, the poljuomial contains the factor
(x — A).
Proof: Let the polynomial be
P = aox'» + aix^-^ + a2a:**-2 ^ . . . ^ an-ix + On.
Putting h for x, we have by hypothesis
aoA'* + aiA'»-^ + a2A'»-2+ . . . +an-iA + an=0.
Therefore by subtraction,
+ an-i(x — A).
But each term of the right member of the last equation contains
the factor x — h. (See 8 and 9 of 9.) Hence P is divisible by
{x - h).
Example. Factor x' + 3 a? — 4 x — 12.
If this is the product of three factors (x — ^) (x — k) (x — 0» then evi-
dently hkl — 12. Hence we substitute in the given polynomial the factors
of 12, and find that it vanishes when x = 2, x = — 2, and x = — 3. Hence
the factors are (x - 2) (x + 2) (x -|- 3).
12. Exercises. — Factor:
1- ^ -#^3"-^- ^^^ X2-X-110.
2 ^^2ax+a'-V 12. 7 + 10x+3x^.
o oJt A \\ ^' 13. x2-10a2x+9a4.
3. oX^— 4xH-l. 14 / 4. m4 «. 1
s ft^jTiolr 7,^ ^^- ar^2/^-3xyz-1022
5. 6x^ + 19x2/ -Tjr. ^^ oj-T^-ir:^
6. x2-2x-24.
7. 8x*-27x2/'
16. 2 + 7x-15x2.
17. x3-64x-x2-|-64.
a 27x* + 8x2/«. }?• ^e"^i
9. X*- 13x2 + 36. V^ f , lU^.
10. 4a*-5a» + l. ^0. (a+6)» + l.
13,14] H.C.F. AND L.C.M. 11
21. (x» + j/S) - (x + js/)3.
23. aV+acd + o6c + W.
24. 1 - aV - 6^2/* + 2 otxy.
26. a^y — a^j^ — s^j^ + x]/^.
26. a8-82a* + 81.
27. a;V-17x2y-110.
28. (a2-|-3)2-36a2..
29. x« + 9x2 + 16a; +
90. 4x*-S£^-3? + ^
13. Iffi^est Common Factor. — The highest common factor
(H. C. F.) of two or more polynomials is the polynomial of highest
degree that will divide them all without a remainder.
When each of the given poljuomials can be factored by inspec-
tion, the H. C, F. is easily determined from their common factors.
Example, The H. C. F. of 32 (x - l)Hx + l)«(x2 + 1) and 24 (a? - 1)»
{x + l)2(x2 + 1)2 is 8 (a; - l)Hx + l)2(a;2 -|- 1).
When the given poljmomials cannot be readily factored, we use
a method like that of arithmetic.
Let the given polynomials be Pi and P2 and let Q be the quo-
tient and B the remainder when Pi is divided by P2. Then
Pi = P2Q + R.
Hence any factor common to Pi and P2 is also a factor of R,
Hence it is a conmion factor of P2 and R. Divide Pj by R,
obtaining
P2 = RQi + R\'
Hence a common factor of P2 and R is also a factor of Ri, Divid-
ing -B by fix, we obtain
R = R1Q2 "h R2y
and the common factor must be present in Rzj and so on.
Rule. — If at any step there is no remainder, the last divisor
is the required H. C, F.
14. Least Common Multiple. — The least comm^on multiple
(L. C. M.) of two or more poljuomials is the polynomial of lowest
degree that is exactly divisible by each of them.
When the given polynomials can be easily factored by inspec-
tion, form the product of all the tjrpes of factors present in any
of them, taking each factor the greatest number of times that it
occurs in any of the given expressions; this product is their L. C. M.
12 H.C.F. AND L.C.M. [15^
When the given polynomials cannot readily be factored, their
L.C.M, is obtained by use of the following theorem:
The product of the H. C. F. and L.C.M. of two polynomials
equals the product of the polynomials.
Proof: Let F be the H. C. F., and M the L. C. M. of the two
poljmomials Pi and P2. Also let
-~ = Qi and -y = Q2;
then Pi = FQi and P2 = PQ2.
Hence P1P2 = P X PQ1Q2.
Since P contains all factors common to Pi and P2, Qi and Q2
have no conmion factor, and the product PQ1Q2 contains all the
factors of the types present in both Pi and P2.
... M = P0iQ2 = ^^; or, MP = PiP2.
Rule. — To find the L.C.M. of two polynomials, divide their
product by their H. C. F.
To find the L. C. M. of more than two polynomials, find the
L.C.M. of two of them, then the L.C.M. of this and a third one
of the polynomials, and so on.
16. Exercises. — Find the H. C. F. of
1. 6(a; + l)«and9(a?-l).
2. a*-6*anda*-6*.
3. 12 (a? + 2^)2 and 8 (a;* - 1^).
4. w^ — t;*^ and u^ —1^.
6. (c?x — aa?)^ and ax {c? — a?)*.
6. 27 (a* - 6*) and 18 (a + 6)2.
7. (24a2 + 36(rf> - 48ac) and (30a« + 45a% - eOah).
8. 125a^ - 1 and 35a? — 7a; +6aa; -a.
9. 4a? -12x2^+91/2 and 4a:2_ 92^
10. a? + 2a; -120 and a? -2a; -80.
11. 12a?-17aa; + 6a2and9a?-|-6aa;-8a2.
12. x3 + 4 x2 — 5 X and a;* — 6 x + 5.
13. x^ + aV + 7x + 21 and 2x* + 19x2+35.
14. a* + 7a3 + 70^ - 15a and a^ - 2a2 - 13a + 110.
16. 20x*+x2-land75x* + 15x*-3x-3.
16. X* - ax' - aV - a»x - 2a* and 3x^ - 7ax2 + 3a2x - 7a« .
17. X* - 2/*, x' + y', and x^ + 2/*.
18. a? — 2 a^ — ax, a? — 6 a^ + ax, and a? — 8 a^ + 2 ax.
19. a* + a^lf + h\ a* + a}?, and a% + 6*.
20. 3x»-7x2y + 5xj/2-2^, x^j/ + 3 X2/2 _ 3 ^^ _ ^^ and 3x» + 5xV +
X2/2— ^.
16,17] FRACTIONS 13
Find the L. C. M, of :
21. Sc^a^u^ and 12 aba?t/^,
22. 4ax^y2, Qa^xj^, and 18 a^a^y.
23. 02-62 and (a - 6)2.
24. a26a; — a62y and cibx + l?y,
26. a:2 _ 3a. __ 4 and a:2 _ 3. _ 12.
26. a?-l and a:2_|.4a._^3
27. 6ar^ + 5a;-6 and 6a:2_i33.^6
28. 12^+bx-Z axid^:i^ + :j^-'X,
29. 12x2 - I7ax + 6a2 and 9a:2^6aa._g^
30. a« - 9a2 + 23a - 15 and a2 - 8a + 7.
31. m^ + 2 m^n — mr? — 2 n* and m^ — 2 rr^n — mr? + 2 rf.
32. 0^2 — ^, (a; — 2/)^ ai^d a: + y.
33. 0^2 + 3a; + 2, ar^ 4-40; 4-3, and 0^2 + 5x4-6.
34. a:2 + 5a. 4. 10, a:3 - I9x - 30, and x* - 15a; - 50.
36. a;2_|-2x-3, a;^ 4- 3 a;2 _ 3. _ 3^ and a;* 4- 4 a;2 ^ a. _ g^
36. 6a;2_i3a;4.6, 6x2+5a;-6, and 9a;2_4
37. a;2 - 1^ a;2 ^ 1^ and a;* 4- 1.
38. a? 4- 1, a;* - 1, and x^ - 1.
39. a^ - 6^ a» - 69, and a^ - 6«.
40. a^ — j/2, a;^ 4- 2^, 25^ — 2/^, and x® + jf,
16. Fractions. — An algebraic fraction is the indicated quotient
N
of two algebraic expressions. It is written in the form yr, N
being called the numerator and D the denominator.^
When N and D have a conmion factor F, so that we may put
N_= NiFsLudD = DiF,
then the fraction may be simplified as follows:
N^NiF^Ni
D DiF Dx
When all factors common to N and D have been removed in
this way, the fraction is said to be reduced to its lowest terms.
When the common factors of N and D are not obvious on
inspection, find the H, C. F, of N and D, and remove it as above.
17. Sign of a Fraction. — By the rules for division we have,
D D -D -D
Hence the rules: Changing the sign of either numerator or denomi-
nator changes the sign of the fraction.
Changing the signs of both numerator and denominator does not
affect the sign of the fraction.
14 FRACTIONS [18
The sign of a fraction may be changed either by changing the
sign standing before the fraction, or by changing the sign of the
numerator or of the denominator.
18. An integral expression is one whose literal parts are free
from fractions.
A mixed expression is one formed from the sum of an integral
part and one or more fractions.
A complex fraction is one whose numerator, or denominator,
or both are fractions or mixed expressions.
Every mixed expression and every complex f ruction can be re-
duced to a simple fraction (or to an integral expression).
For, two or more simple fractions can be reduced to a common
denominator and then combined into a single fraction by writing
the sum of the numerators over .the common denominator. For
this purpose the simplest common denominator is the L. C. M. of
the separate denominators. This is called the least common de-
nominator of the fractions considered. In this manner we reduce
A mixed expression is reduced by the formula
„,N PD + N
^+5 D~'
Finally, a complex fraction is reduced by first reducing its
numerator and denominator separately to simple fractions. The
reduction is then completed by the formula,
N
D N ..D' ND'
N' D ^ iV' N'D
D'
Examples.
2,1/ X
1. Simplify
l^l ' 1»? 1_^
X y X y
First reduce each fraction to a simple fraction, thus:
2 2 _ 2xy
1 1 y — X y —X*
-— — ""^
X y xy
y _ y _ xy
J y x-y x-y'
X
19] EXERCISES 15
X _ __ X _ __ xy _ xy
1—5 y "• ^ y — a? x ^ y
y y
Reducing to the common denominator x — y^ we have
y — xx^yx — y x — y
x^ x^ x'^
2. x^ — ' = x' ^ ' « a;8 —
^+ 1 ^ + ^2^11 ^+ X2-,1
X <
X X
X — X (1 + i^) — x3 X X
19. Exercises. — Reduce to simple fractions or to integral
expressions:
la±x_a-x\ ^^ ^^ zj^+^
\a — X a +x/ x2 — y2 X + y
2. ^!jZ^^?Lz^. ,o 45(x~y) . 27(x~y)2 .
32(z+y) ' 1286(2 + ^)2*
a8 + 63 • 2ab ' ^•
g^ — &^ y x2 + xy + y»
X3-2/8 o2+62 • 13.
^ a^ — b^ x2 + xt/ + 1/*
a2 + 4 ox ' ox + 4 x2
a^ + ab . ah(a+ 6)2
i.
5
\y5 X6/ \y X/
11.
15.
02 + 62 • a* -64
x" + yi2 _^ x4+y4 *•• ^2t,2 - 1x4 * «t;2 + 1;8
'»• a;i2 - yi2 • a;8 —
^ X2 + 7X+12 . X2 + 6X+8 16. g+4P-±|^^^,
^' x2-x-12 ^ x2-2x-8 - P*-3p2+9 p«+27
x + l,y + l -- g6 -, y« , x2 + xy + y 2
^ ^ X ' y (x - y)2 X - y
11
X y
X — y x* —
x + y x8 +
9. 1+-^. ^®' x + y xMj^
a;+i x-y"*"x2-y2
X
* +-^^ 19. — K —
10. ^ + y ^-y - x+
» i^i+y
X — yx + y 3 — y
X(X-1)^1 -X2^X(X + 1)
x2-6x + 6 x2-4xH-3^x2-3x + 2
16 EXERCISES [19
22.
28.
24.
x-l 2(a; — 2) , x — 3
x*-5x + 6 ic2-4aj-f.3 ' x«-3x + 2
7 + 3x« 5-2x« 3-2a; + x«
4—^2 4+4a;+x< 4 — 4a;+a:*
l-2a; 2a;-3 , 1
3 (x2 - X + 1) 2 (a;2 + 1) ' 6 (x + 1)
\ftc oc ab aj \ a + h -{-c/
"• (l+§-5-i+i)(l-j}
\o X 6 yy \a X o y/
00 /t 7x\ [7x 49x2 343x» \
^' ^^ 11 W Vll y "^ 121 y2 "^ 1331 W
~ (■•«-*! ^)("^'»»t)-(lf+^.)gl-^J
V3c2 5a2y ^762 Qcy \5a^ 3c2/
jjg /4x8y2 3a;2y8 2xy< y6\ /2x2y 3xy2 3y»\
* V 5a8 2a26 ■*'3a52 6«y\3a2 5ab 2h^i
CHAPTER II
Involution. Evolution. Theory of Exponents. Sxtbds
AND ImAGINARIES
20. Involution is the operation of raising a quantity to an
indicated power.
The symbol a** represents a X a X a ... to w factors (8),
n being a positive integer. Hence, if m be a second positive
integer, we have by cancellation,
(1) — = a^-^ when n>m:
Qn 1
(2) — = when n <m.
Negative Exponent. — We now define the symboJ a-** to be
a" a X a X a ... to n factors
Then —^ = a-^*"-**) = a**-*".
We may now write,
(3) ^ = a^-^,
whether n is greater or less than m. Hence by the introduction
of the negative exponent, the two equations (1), (2), may be written
as a single equation, (3).
We now easily verify the following rules for operating with
integral exponents, positive or negative.
I. a-~=— . IV. (a«*)~=a«*~.
a**
m. a^^-ha-^^a**-^. VI. (^Y=^
17
18 INVOLUTION. EVOLUTION [21,22
21. Exercises.
L State the above rules in worda.
2. Verify the above rules by means of the definitions for a* and a~*.
8. Show that rule II contains rule III.
i. Show that rule V contains rule VI.
Perform the operations indicated in the following exercises, and express
the results in forms free from fractions:
7.
Zero ExpOfaexkt — If in rule III we put n == m, we get
But a^-TT a^=^ 1. Therefore we define the symbol aP by the
equation a^ ^ 1. Then III is true for all integral values of n
and m, equal or unequal. Hence we add to the above rules:
VIL ao = 1. 1
22. The nth root of a quaniUy a (symbol yja or a^) is a quan-
tity whose nth power is equal to a.
Evolution is the operation of finding the indicated root of a
quantity.
By definition, we have
or
^ X Va X Va . . . to w factors = ('7a)*= a,
111 / l\n
a^'Xd^Xa^ . . . to w factors ^^ya"") = a.
The last equation will be covered by rule IV (20) if we extend
that rule to the case where m is the reciprocal of a positive inte-
ger. We now extend rules I- VI and assume that m and n may
be not only positive or negative integers or zero, but also the
reciprocals of positive or negative integers.
If we let w = - and m == -^, r and s being integers, we have
23,24] INVOLUTION. EVOLUTION 19
-^ 1
or
1 1 11
Iir. or -T-a* =ar '.
IV'. U*r = a"
1 11
7 + 7 1 1
"'. ?'-"-;
These equations define the rules governing operations involving
roots.
Exercise, State the above rules in words. What is the meaning of a
negative root?
23. Rational Exponent. — By the preceding laws we now have
a meaning assigned to the symbol a** when n is any rational
number (4). For, if n = p -r- g, p and q being integers, we have
/ i\p 1
2
an ^a9=:\^a^j = (ap)«
2
that is, a« means the pth power of the gth root of a, or the gth
root of the pth power. In a fractional exponent, the numerator is
the index of the power, the denominator the index of the root.
By combining rules I- VI and F-VF, we see that the former
set of rules holds when m and n are any rational numbers. Hence
we adopt the rules of (20) as the rules governing quantities affected
with rational exponents.
24. Irrational Numbers. — By the operation of evolution we
are led to numbers which cannot be produced from integers by
means of the four fundamental operations. Thus if we attempt
to calculate V2 we are led to a non-terminating decimal. To
four decimals we have
1.4142 <V2< 1.4143,
or 14142 ^2 <i^l^.
10000 ^ ^ ^ ^ 10000
We have here two rational numbers between which V2 lies. By
going out to a sufficient number of decimals, we can obviously
obtain two rational numbers containing \/2 between them and
dififering from it by as little as we please. By taking successively
4, 5, 6, . . . decimals, proceeding as above and noting each
20 INVOLUTION. EVOLUTION [25
time the smaller of the two rational numbers, we obtain a series
or sequence of rational numbers which increase and approach
V2; by noting each time the larger of the two numbers, we obtain
a second sequence of rational numbers which decrease and also
approach n/2.
If on the other hand we consider the sequence of numbers
13 133 1333
' lO' lOO' lOOO'
4
these evidently approach the value -^y which is a rational number.
The idea here indicated is used to define irrational numbers.
Without going further into the subject here, we shall say that an
irrationcd number is one which can be represented to any degree of
approQcimation, hut not exactly as the quotient of two integers.
Such numbers may be produced in performing the operation of
evolution on rational numbers.
Real Numbers. — The rational numbers, including all integers
and quotients of integers, and the irrational numbers together
constitute the class of real numbers.
Irrational Expressions. — We now extend the idea of irration-
ality to algebraic quantities in general by the following definition:
An algebraic expression is said to he irrational when its parts
are affected hy other than the four fundamental operations.
Hence any expression involving indicated roots is irrational. As
examples, we have
VT+^; (x2- xy)'i + (xy - 2/2); J ^+2a + a\
▼ 1 — a
The last expression may be simplified. Thus,
v/
1 + 2 g + a2 Vl + 2a + a2 1 + a
1-a Vl-a VT^^
A surd expression is one involving an indicated root which can-
not be exactly found.
A surd number is an indicated root of a number which cannot
be exactly found.
26. Irrational Exponents. — What meaning shall we attach to
the expression 2 "^^ ? j^i ^^^ ^^^ as, . • . be a series of rational
l-ie^"^
^\:.*y-'^.
26] IMAGINARY NUMBERS 21
numbers approaching V2 in value. Then the quantity toward
which the series of numbers 2«», 2% 2«», . . . approaches is 2^.
Similarly we obtain a meaning for a^, when x is irrational.
We now define a=^ as a symbol subject to the foUowihg laws:
I. «"*=^J IV. (a'")*' = a*^ (no< a**');
IT. aray==a''^v; V. (a6)*= a*6*;
in. a«-^a^ = a--^ VI. (^J =^;
provided that the symbols a, 6, x, y, a^, h'^, a^, ft*' stand for real
numbers.
26. Imaginary Numbers. — When a:^ = 1, we have obviously
X = ± 1. What is X when x^ = — 1? The answer cannot be
a real number, since the square of every such number is posi-
tive. To obtain an answer to th e question, we introduce a new'
number whose symbol is V — 1, and which is defined as the
quantity whose square is —1.
Since V— 1 is not a real number, it is of ten called imaginary
and denoted by i. Hence the quantity i s \/— i is defined by the
equation i^ = — 1.
We now define V— a by the equation
I. V— a = i\/a.
(This is in accordance with our rules for exponents, since
V^^ = VaX-l= Va V^ = i Va.)
Then the product V— a X v^— 6 is determined by the equation,
n. \/^^ X V^^ = i Va X i Vft = i2 \/a6 = - Vaft.
The results of the operations of algebra, applied to any number,
are always expressible in the form a + 6i, where a and 6 are real.
Such a result may be considered as consisting of a real units and
h imaginary units, a X 1 + ft X i; it is called a complex number.
Two numbers of the forms a + M and a — 6i are called con--
jugate complex numbers.
When a = 0, the complex number a + 6i becomes hi called a
pure imaginary.
V
22 SURDS [27,28
The rules for operating with complex numbers, aside from II
above, are considered in chapter 17.
Principal Root. — There are in general n distinct quantities, the
nth power of each of which equals a given number a (see 269).
That is, a given number has in general n distinct nth roots.
Thus,
the square of + 2 or — 2 is 4;
the cube of - 2, (1 + iVS), or (1 - iVS) is - 8;
the fourth power of + 2, — 2, + 2 i or — 2 1 is 16.
The principal root of a number is its real positive root when
one exists; if not, its real negative root; when all roots are imagi-
nary, any one of them may be chosen as the principal root.
1
In this text the symbol for a root, y/a or a**, will mean the
principal root only.
Thus: \/4 = 2, not ± 2; if we wish to indicate both square roots,
we always write ± Va.
27. Reduction of Surds. — The expression ^/a is usually called
a radical, V being the radical sign, n the index of the radical and
a the radicand. When the radicand is not a perfect nth power,
the expression is a surd.
A surd is said to be in its simplest form when all factors of the
radicand which are perfect powers of the same index as that of
the radical have been taken out from the radical sign. Thus:
v^
•/8o*6« 2ab tr^
Two surds are similar when they can be expressed with the
same index and radicand. Otherwise they are dissimilar,
A quadratic surd is one whose index is 2.
28. The sum, difference, product and quotient of two dissimilar
quadratic surds are always surds.
Proof: Let the surds be Va and Vb. Since they are dissimilar,
neither ab nor a -r- b can be a perfect square. Hence the product
or quotient of the two surds is a surd.
Further, let c be a rational number, and assume that
Va±Vb = c.
29-32] SURDS 23
Squaring, a ± 2 Vab + & = c,
or ±2 ^/ab = c — a ^ b.
But a surd cannot equal a rational expression by definition.
Hence the assumption is false, and the sum or difference of
two surds is also a surd.
29. Given a relation of the form a + Vb = c + V^; then a == c
and b = d» _
For, on transposing, we have \/b — \/d = c — a; hence iib 9^ d,
we have a surd equal to a rational number, which is impossible.
Therefore b = d. Hence also a == c.
30. To rationalize the denominator of -7= ;=•
Va + Vb
Rvle. — Multiply both sides of the fraction by Va — Vh.
31. To obtain the square root of a + Vft.
Assume that V a + \/b = Vx + ^/y. To find x and y.
Squaring, a + Vb =^ x + y + 2 Vocy = x + y + V4 xy.
Hence a — x + y and b = 4ixy (29).
Then a^ — 6 = x^— 2xy + y^ = (x — y)^,
or ± Va^ — 6 = a; — y.
But a = x + y.
Therefore x = i (a ± Va^ - b) and y = i (a =F Va^ - b).
32. The index of a surd may be multiplied by any number if at
the same time the radicand be raised to the power indicated by this
number.
For, yfa^^a'^^ a*"** = "'^/a*".
In combining surds by multiplication or division this rule is
used to reduce them to surds with a common index. This is
accomplished by writing all the surds as fractional exponents and
then reducing the exponents to a common denominator.
24 EXERCISES [33
33. Exercises.
Write the following with positive exponents and in simplest form :
Reduce to radicals with the same index :
9. \JZ and \Ji, 19. \Ja, 's/by and ^/c.
10. -^32 and '75. 20. V^ and -s/^^.
11. y/2 and %/§. 21. -s/P, ^/os, and -v^.
12. \/5 and -725. 00 /^ "Z?^ j '/«
18. V5. <^. and </3. "^ V^' Vf:""'W_
14. V3, -VS, and </4. 23. i'/^, y/i, and ^/^.
17. -V?, V5, and 'VA- „. m/j" „ p/J
18. 70:3, 71, and 'VH. V^' ^' V ^'
Combine by performing the indicated additions and subtractions, reduc-
ing to similar surds when necessary:
26. 2^/3-5^/3+9^/3. 32. 2 Vi75 - 3 V63 + 5 V28.
27. 4-s/i-3-y4+2\/4. ^ 33. 3 VS + 6 V^ - V^.
28. 3 V2 + V32. 34. </^7^ - </Kc* + \/S4c8.
29. ^y2 + 3 V32 - J Vi28. 36. ^/cfix - ^/a^ + Vo^-
30. 5 ^^4 + 2 ^32 - -yioa? 36. \/a^ + -y^iap^s _ ab Md.
31. i-s/5+2i\/5 + iV40.
Reduce to the form ^Jx + V^ •
37. V4 + 2 V3. 42. VlO + 2 \^.
38. V3 + Vs. 43. V7 + 2 ViO.
39. V2 + V3. 44. V7 - 4 V3.
40. V8 + Vis. 46. Vl3 - 2 V30.
41. V5 - V2i. 46. Vll - 4 V7.
15. VJ, Vf , and Vf . ^ /^ /7
16. VI, va, and 7Rj. 2*- y 7' y ^» ^d
33] EXERCISES 25
Perform the following multiplications and divisions :
47. (3+2 V2) (3-2 \l2). 69. ^/28 ^ V?.
48. (5 +2 V3) (3-5 V^). 60. V48 -^ V3.
49. (2 ye - 3 ys) ( V3 +2 V2). 3,. ^ ^ ^,
60. (V7-V3)(^ + V5). ^^ ' 3;_
61. (-s/g - 2 ^) (4 V3 + \/2). ^^- ^^^ -^ ^^•
62. (V^+5+Va)(V^+6-V5). 63. \^43 -5- -^3. v^
63. V^TTVw Vm - V^~i- ^- "N/I^ -^ V6.
64. V^^^^ X \/^- ««• V^ - ^
y a + 1
66. Va -ya2 X V ^a. .V- . r—
66. Vo; Vx« X Va; -^/S. J ^
Kn ^n-o w »/::x:i 68; "</a%9 -^ V«463.
67. yx^y^ X Va^2/^-
"•V^xy/
c^5 69. -727^^9 -^ V6 25
a26* 70. -y8"x3y2 ^ 2a;2y8.
Express with fractional exponents instead of radicals :
71. (^/^)^' 76. (-V^)^
72. (V^)^ 77. (^/^^.
73. «/^)l 78. (V^!^)''.
74. «/^)^ 79. «/ix + y)^)\
76. (- V^)^ 80. {^1J^)\
Rationalize the denominator of :
81. -i^. 87. ^ + 2^. •
V2 2 VS3 - 3
a
82. -7^- 1
V6' 88.
^Jx + y - \Jx-y
83
-n/^ 89 •^+"^^
84.-^- ■ ^^-^2
«B « 90. ^}1 - -^ll -
^•\^^a V1O + V13
86. ^/|±^. ■ 91. ^ - '
's/a — ^b * a y/b + c ^d
92. Calculate to three decimal places the values of the fractions in exer-
cises 89 and 90.
26 EXERCISES [33
Perform the following operations and simplify results:
93. Va</52 X VV^ X \^a2 -s/a7.
94. (x-^+a;-42/-i+2/-i)(x-^-x-iy-*+y-0.
96. (2o-*-3a-* + a-i -2) (a"* - 2a-* + 3).
96. Write out the result of replacing a"* by 6 in exercise 95.
97. \a^-2a^ + 3a*/\2a*-a^ + 2/.
98. V2/^-a2/n+36y'»-c/Vy» + &yn-cyoA
99. (2a-^6-«-3a-«6-J)^
100. (a-* +6-*)*. 103. (m-2+w-*)*.
101. (a;*-y«)^ 104. (a"^ x'^ - ax^.
102. (l-2.n-J)*. 106. (a^+a^-^a^y.
106. (2oJ-36»-4c*)^
107. (a* - 26* + 3c* - 4(i*)l
108. (x*j/* - 2 x'y* + 3 a;*y - 2 x*y*)^.
109. Write out the result of replacing x* by u and y* by » in exercise 108*
110. (x-1) ^(\^J-l).
HI. (x + 1) -i-is/i + i).
112. <Vi - .^) -^ (Vi - \^).
113. (a^v-6*) -5-(aA-6A).
Hi. (x^ - xy^ + a;*y - y') -5- ( V« — Vy).
116. (aS - a'^ - 4a8 + 6a - 2 Vo) -^ (a« - 4 VS + 2).
116. (a* - 6* - c* + 2 \/63) (o* + 6* - c*). /
I
Express the following in the form a V~" !•
117. V^; ^r-25; V^^-
118. V-a2; V-^; V-x2n
119. V=^. 121. \/^=^256.
120. V^^^. 122. 'V^^^.
123. V^^^ - V^=^ + V^^T2i.
124. V^^ + V^=^ - V~4a4.
126. V- (w+?i)2 + V— (w - w)^ - V- w2.
33] EXERCISES 27
Multiply and reduce to the form a + h V— 1 : (i — V"-!)*
126. (a+&V^^)(o-&V^)- 129. (\(8 +t Vi2) (V2 +t V3).
130. (-1 +i V3)'.
128. (x + 2t)(y-3t). V V2 V2/
Reduce to the form a + W by rationalizing the denominator :
^^2» Q lo-- 135. ; — p«
133. 1+-!- 136. ^
1-* * 7 + 2 V^
134. ^±1?. 137. ^"^'
Clear the following equations of radicals:
(Example. To clear the equation Vi + Vy + V2 = 1 of radicals put
\lx-\-yly = 1 - V^;
squaring, a; + y + 2 V^ =1+2 — 2 V^>
or, x + y — 2 — 1 = — 2 V^ — 2 V^.
Squaring again, (x + y — z — 1)2 — 4 xy + 4z + 8 V^«>
or (a; + y - 2 - 1)2 - 4 (a^ + 2) « 8 V^^^;.
Squaring again, [(a; + y - 2 — 1)2 — 4 {xy + 2)12 = 64 xyz. q.e.f .
138. VF+4 = 4. 146. VxT2b - V^^ - 1 -3 = 0.
139. V2a; + 6 =3. 146. Vx - "4= = V5"^=^.
\Jx
140. </^+i = 2. ^^^^ V l5 + V2x + 80 =5.
141. </^T6 = c. j^ ^g-^P^ ^ Vl2 - 2 Vi.
142. Vi + V» = 1. 149, VTTVi = Vll-3%^.
143. V5TT - V* - 1 - 2.
144. V32+a; = 16 - ^^. V81 + I - Vs^
160. ^^&!l^=13.
/
CHAPTER III
Logarithms. Binomial Theorem for Positive Integral
Exponents
34. Logarithm. — The simple laws of operation for exponents
have given rise to a method of calculation involving the use of a
function called the logarithm. We shall first illustrate this method.
Suppose that we know the powers of 10 which are required to
produce a set of numbers, as in the adjacent table, where the
exponents are given to the nearest figure in table.
the third decimal. The exponent of 10 in each 5.00 == 10^-®®® •
equation is called the common logarithm (or 5.50 = 10^-740
the logarithm to the ha^e 10) of the number on ^'OO = iqo.tts
the UfU Thus, the logarithm of 5.00 is 0.699, 6.50 = IQP-^^^
of 5.50 is 0.740, and so on. As equations, ''^ ~ ^^«" ^.
we write ^ ^ 7.50 = lO^^^s
we wnie g^ ^ ^^0^03
logio 5.00 = 0.699, g go = ioo.©29
logio 5.50 = 0.740, 9.00 = 100.954
and so on. 9.50 = lO^-^^s
10.00 = 101000
36. By aid of such a table products of numbers (within certain
limits) can be obtained by adding the logarithms of the factors;
also, division is reduced to svbtradion of logarithms.
Example 1. Find the value of 6.5 X 8.5 X 9.5.
We have 6.5 X 8.5 X 9.5 = IQOsisx 100-929 x \^M
r= 100.818+0.929+0.978
= 102.720 = 102 X 100-720.
Now 0.720 lies almost exactly midway between 0.699 and 0.740; hence the
number corresponding to 100-720 ^^iu be midway between 5.00 and 5.50 and is
equal to 5.25. (This involves the assumption that a logarithm changes pro-
portionately to the change in the number, an assumption which is not exactly,
but very nearly, true except for numbers near zero, provided the changes in
the numbers are small.)
Therefore,
6.5 X 8.5 X 9.5 =- IO2 X 100720 = 100 X 5.25 = 525.
The exact value is 524.875.
28
36] LOGARITHMS 29
Definition. Interpolation is the process of calculating numbers
intermediate to those given in a table.
Example 2. — Find the value of — — , y, ' —
Let 10* = 6.25; 10& = 7.20; 10^ = 5.76.
Then ^'25X7.20 ^ 10^X10^ ^ ^^^,^,
^^^^ 5.75 10^ ^
Since 6.25 lies halfway between 6.00 and 6.50, we take for a the value
halfway between the corresponding exponents, so that a = 0.795 (more exactly
0.7955). To get 6, we note that 7.20 lies f of the way from 7.00 to 7.50; hence
we take for b the number lying in the corresponding position between the
exponents 0.845 and 0.875. Therefore
h = 0.845 +iX 0.030 = 0.857.
Similarly, c = 0.759.
Hence, 6^25X7^^ Iqo. 795 +0.857 -0.759 =: ioo.89S.
* 5.75
The corresponding number lies between 7.50 and 8.00, and nearer the latter.
Since our exponent, 0.893, lies H oi the way from 0.875 to 0.903, we find the
number lying in the corresponding position between 7.50 and 8.00, that is,
7.50 + ii X 0.50 = 7.50 + 0.32 = 7.82.
mu t 6.2 5 X 7.20 - Q^ . , ,
Therefore, ^-^ — = 7.82 approximately.
' This result is correct to two decimals.
36. By the aid of our table, powers and roots of numbers may
be found by applying the operations of multiplication and division,
respectively to their logarithms.
Example, Find the value of ^9.353.
We have -^9^ = (9.35)*.
Let 9.35 = 10«; then (9.35)* = 10***.
From the table, a = 0.954 + ^V X 0.024 = 0.971.
Therefore, -^g^SP = 100-728 = 5.00 + if X 0.50 = 5.35.
A more accurate value is 5.335, so that the second decimal of our result is
slightly in error.
Obviously the calculation of the last result by the methods of
arithmetic would be very tedious, and with a slight increase in
the complexity of the exponent these methods would become quite
useless.
[37-39
30 "'^^' ; LOGARITHMS
- h
We shall now consider the.jjg^eral theory of the method illus-
trated above.
37. Logarithi|^^a Number. — Let a be a certain fixed number,
n any other n^^Br, and let x be the exponent of a required to
produce n. Then x is the logarithm of n to the base a.
As equations, '|
if cT «= fi, th/en x = log« n.
We give below some Very simple tables of logarithms.
Number.
Logarithm
Base - 2.
n.
logio n.
n.
*
logio n.
i
-3
.001
-3
5.00
0.699
i
-2
.01
-2
5.50
0.740
I
-1
.1
-1
6.00
0.778
1
1.0
6.60
0.813
2
1
10
1
7.00
0.845
4
2
100
2
7.50
0.875
8
3
1000
3
8.00
0.903
88. Exercises.
lat is the value of loga 1?
lat are the logarithms of 8, 16, 64, 128 to the base 2?
'8. What are the logarithms of 8, 16, 64, 128 to the base i?
4. What are the logarithms of }, ^j, ^Ij, to the base 3? to the base i?
6. Vfjk&t are the logarithms of ^f- and ^^ to the base }?
6. .miat are the logarithms of 2, 4, 8 to the base 64?
7. wAis the base, if log2 » 1? if loga = 1?
8. WMM^he base, if log }» 4? if log 25 » - 2?
9. WHHhe base if log 49 » 2? if log .0081 » 4?
10. WhatVlog2(-4)?
11. Why would it be inoonvenient to use a negative number as the base of
a system of logarithms?
12. If n = (€*+!')*-», find logen.
15. Ifx» ^e ^(6P+«), find loge X.
14. n a = [(lO^'-O ^ ""^^
16. Show that olos.^ » z.
, find logio a*
39. Laws of Operation witii Logarithms. — Since a logarithm is
an exponent, the laws of operation for logarithms are the same as
those for exponents.
V-
40] LOGARITHMS 31
Let X be the logarithm of m, j/ that of n, the base being a.
Then
Hence
hogan =2/, (ay=n. ^
mn = a^^^ and — = a*"*'.
or, logamn = x + y = loga w + log«n,
and loga — = X — y = log„ m — logon.
We have therefore the rules:
I. The logarithm of a jjToduct equals the sum of the logarUhm^ of
the factors.
II. The logarithm of a fraction eqitals the logarithm of the numer-
cUor minus the logarithm of the denominator.
A
Also, if as before,
logo m = Xj so that m = a*,
then, if p and q be any real numbers,
X
fpp ss a^^ and y/m = a«.
Hence, loga w* == pa; = p logamn, ' .^H
I
X 1
and log« yni ^g"^~ log«^-
We have therefore two additional rules:
III. The logarithm of any power of a number equHJ^ the ex-
ponent of the power times the^ logaritftm of the number.
IV. The logarithm of any root of a number equals the logarithm of
the number divided hy the index of the root.
(Rule III contains rule IV, since the power in question may be
fractional.)
40. The following facts regarding logarithms should also be
carefully noted.
(a) In any system the logarithm of the base is 1.
For a^ = a. .'. loga a = 1.
(6) In any system the logarithm of 1 is 0.
For a^ = 1. .-. loga 1=0.
32 LOGARITHMS [41
(c) In any system whose base is greater than unity, the log-
arithm of is —00.
For if a^ = m and a > 1, then if a; is a large negative number,
m will be small. As z increases indefinitely, alwa3rs being nega-
tive, m approaches zero. That is,
a-* = if o > 1; /. log = - oo.
{d) A negative number has no (real) logarithm, the base being
positive.
(e) As a number varies from to +oo, its logarithm varies
from — « to +00, the base being greater than 1.
When the number is greater than 1, its logarithm is positive.
When the number is less than 1, its logarithm is negative.
41. Exercises. (See Appendix for tables and explanation of
their use.)
1. Discuss (c) of (40) when the base is less than unity.
2. Discuss (e) of (40) when the base is less than unity.
In the following exercises, the base is understood to be 10, and four-place
logarithms are to be used.
3. Find log 831, log 8.31, log .831, and log .0831.
4. Find log 78.03, log .073, log .00284.
6. Find the approximate value of 564.1 X .0065.
6. Calculate ^154.2 and (7.541)3.
7. Calculate 518 -^ 313 and 25.03 -^ 2.14.
8. Calculate .001022 -^ .0000513 X 1.415.
9. Calculate 17 V29 and 41 Va5i2.
10. Calculate 7^35^ X ^0.472.
n. Calculate ^'^^S^T^^
(.00346)4
12. Calculate V^214 - ^/2U,
Write as a single term:
13. log a — log 6 + log c — log d.
14. 3 log a; — 4 log y + 2 log z,
16. } log w + i log v — } log w,
16. log? + log^ + Iog^-Iogg.
17. 3 log a — log (a; + 3/) — J log (ax+b)+ log ^ax + ft.
42,43] BINOMIAL THEOREM 33
The Binomial Theorem for Positive Integral Exponents
42. This theorem is used to express (a + 6)** in expanded form.
We shall here obtain the formula assuming n to be a positive
integer; the proof for other values of n will be found in (221).
By actual multiplication we have
(a + &)2 = a2 + 2 06 + fe2^
(a + 6)3 = a3 + 3 a^b + 3ab^ + b^,
(a + b)^ = a* + 4 a% + 6 a%^ + 4 oft^ + b^.
Here we observe the following laws:
I. The number of terms is 1 greater than the exponent of the
binomial.
n. The exponent of a in the first term equals that of the bino-
mial and decreases by unity in each succeeding term. The ex-
ponerU of b is 1 in the second term and increases by unity in each
succeeding term.
in. The coefficient of the first term is 1, and of the second
term the exponent of the binomial. If the coefficient of any
term be multiplied by the exponent of a in that term, and the
result be divided by the exponent of 6 plus 1, we obtain the
coefficient of the next following term.
43. Now let
(1) (o+6)'»=o»+cia^-i6+C2a»-2624. . . . +c^_ia'»-('»-i)&'»-i
We have here assumed laws I and II and have written the ex-
ponents accordingly. Assuming also law III, we shall have
,^. n — 1 n — (m — 1) n — m
(2) ci = n; C2 = — 2" ci; Cm = ^ Cm-i; Cm+ 1 = ^-^:1 c,n.
We can now show that the same laws are true for the expan-
sion of (0 + 6)^+1.
Multiplying (1) by (a + 6) and collecting like terms we have
(3) (a+6)»+i=a'»+i+(l+Ci)a('*+i)-i6+(ci+C2)a('*+i)-262+. . .
+(cm-i+Oa^ + i-'"6'"+((v»+Cnt+i)a('»+i)-('» + i)6'^+i+. . •
The number of terms will be n + 2, since the exponent of a starts
with n + 1 and decreases to 0. Hence law I is still true. Also
law II is evidently true.
34 BINOMIAL THEOREM [44
According to the third law, we should have
(l + ci)=n + l; ci+C2=- ^''"'"^^T\ i + ci); . . .
/ I \ (n + 1) — m . , .
(Cm •+■ Cm+ 1) — fn-\-l — '^'"~ ^ ^*"^*
These equations all become identities on substituting from (2).
Therefore all three laws are true for the expansion of (a + fe)"+^
provided that they are true for the expansion of (a + 6)**. But they
are true for (a + 6)*, hence for (a + 6)^, hence for (a + ft)®, and so
on, for any positive integral exponent.
This method of proof is called proof by induction.
Writing out the values of several coefficients we have,
n (n — 1) n(n— 1) (n — 2)
ci = n; C2 = -172—; '^ = TTy^s "J " " '
_ n (n — 1) (n — 2) . . . (n — m + 1)
1 • 2 • 3 • . . . m
where c^ is the coefficient of the (m + l)th term.
In place of 1 • 2 • 3 • . . . m we use the symbol Im or ml (in
either case, read "factorial m"). Then equation (1) becomes
. , n(n-l)(n—2) • • • (n— w-Hl) « ^,^ .
\fn
When a = 1 and 6 = x we have,
44. The expansion of (a + 6)** may be reduced to that of
(1 + xY thus:
(a + br = a-(l+^A'' ^a4l+n^+ • • -l.
In place of Cm to denote the coefficient of the (m + l)th term
of the expansion of (a + b), the symbols nCm or (^) are often
used. These are called the binomial coefficients.
• •
• • •
451 BINOMIAL THEOREM 36
Table of Binomial Coefficients
w = 1
n = 1 11
n = 2 12 1
n = 3 1. 3 3 1
n=4 14641
n = 5 15 10 10 51
Example 1.
Expand (o* - 2 b^Y.
+ 4(a*)(-26T + (-26y
= a^ - 8 Jb^ + 24 a6* - 32 a*6' + 16 6^
Example 2.
Find the fifth term in the expansion of (a;" * — J y^)^.
This term will be
8 '7 ' 6 ' 5 f^^i\A\ 1 81 35 _ I 12
46. Exercises. Expand:
2. (2a-36)». ^ V5/
5. (a-i+5-«)4. 12. (l+ax)7.
; , .,/■ 14. (a«+»+o«-v)B.
6. (2p*-33*)». 16, {x^-yu)*.
7. (aa; + 6j/)8. 17. (^i + ^j^'.
8. (iM-«+2»*)T. 18. (^n/J_„v;)7.
9. (V2s-\/3y)'. 19. (e2x +ie-2x)*.
To expand a trinomial or other polynomial, proceed by grouping the terms
in two groups, thus: \
= a:3 + 3 2;2 (2, + 2) + 3 X (y + 2)2 + (2/ + 2)8.
The expansion may now be completed by the formula.
21. (x + y - 2)8. 24. {x-y + u- v)^,
22. (Vi-\^ + Va)'. 26. (l+2a; + 3a^ + 4aH»)8.
23. (H-2a+3a2)4.
36 BINOMIAL THEOREM [46
Calculate:
26. the 6th term of (3 + 2 x^)^.
27. the 5th term of (V2c + ^IZdY^.
28. the 8th term of (2 6* - i VS)^.
29. the 12th term of (3 y^ + J y*)".
30. the 10th term of (Vjo^ - Vio^)^.
46. Approximate Computation by Use of the Binomial Theorem.
— When a? is a small fraction, the tenns of the formula
(1 + a?)" = 1 + ^ + ^ ^^Z^ ^\ ^ + ' • •
rapidly decrease. In any numerical problem in which only approxt-
mate results are required, retain only enough terms of the expansion
to obtain the desired degree of accuracy.
It will often be fomid suflBcient to use the simple formula,
(1 + x)'* = 1 + na;, approximately.
Example 1. Calculate (0.997)^ to three decimals.
(0.997)4 = (1 - .003)4 = 1 - 4 X .003 = 0.988.
Exercise. Show that the terms neglected will not affect the third decimal
place.
Example 2. Calculate (2.05)^ to three decimals.
(2.05)3 =23(1 + .025)3 = 8(1+3X.025+3X.000625+ • • •)
= 8 X 1.0769 = 8.615.
Exercises. Calculate to three decimal places the value of :
1. (0.995)5. 2. (1.05)7. 3. (33^)4.
4. (2H)4. 6. (3.998)«. 6. (8.0125)2.
7. Calculate the value of (.99995)7 to seven decimals.
iO
CHAPTER IV
Linear Equations
47, If X = Yy and m ^ Uj
then X + m =^ Y + n, X — m ^ Y — n,
mX^nY, and -X = ir. '
w, n
That is, if both members of an equation be increased or diminished,
midtiplied or divided, by the same or equal quantities, the resuits
are equal.
Also if X=Y, then X» = P»,
n being an integer; that is, if both members of an equation be raised
to the same integral power, positive or negative, the results are equal.
If X=Y, then ^/X = ^/Y,
provided the corresponding nth roots of X and Y are selected.
If X + m=Y,
then subtracting m from both members,
X = r - m.
That is, a term may be transposed from one side of an equation to
the other provided its sign is changed at the same time.
When the members of an equation involve sums or differences
of fractions, the equation may be cleared of fractions by multiply-
ing both members by the L. C. D. of the several fractions.
48. Linear Equation. — If a? be an unknown quantity related
to the known quantities a and b through the equality 005 + & = 0,
this equation being called the standard form of the linear equation
in one unknown, we obtain the value of x as
b
X — •
a
37
38 LINEAR EQUATION [49,50
Every linear equation in one unknown may he soloed by reducing
it to standard form and applying the last formula.
The reduction of an equation to standard form will involve
some or all of the following steps:
1. Clearing of radicals. (33, after exercise 137.)
2. Clearing of fractions.
3. Expanding products or powers of polynomials.
4. Transposing and cancelling.
5. Collecting terms.
To verify the value found, substitute it in the given equation.
The result should be an identity.
49. Example 1. Solve for x: (1 + 6) x + oft = 6 (a + a;)+ o.
Expanding the products:
X -\-hx '\- ab = a6 + 6x + o.
Cancelling like tenns:
X — a
Check:
(1 + 6) a + a6 = 6 (a + a) + a.
Example 2.
1 2 x+2
2"^x + 2 2x
Multiplying by the L. C. D., 2 a; (x + 2):
a; (x + 2) + 4 X = (x + 2)2.
Expanding: x2+2x+4x =x2-f-4x+4.
Cancelling: 2 x = 4 or x = 2.
Check: i + i = J.
Example 3. Solve for x: V^T-20 - y/x - 1 -3 = 0.
Transposing: Va? + 20 = y/x — 1 + 3.
Squaring: x +20 = x - 1 +^y/x -1 + 9,
or, 2 = ^/x — 1.
Squaring: 4=x — 1 or x = 5.
Check: V25 - V4 - 3 = 0.
50. Infinite Solutions. — Consider the equation
z + l x — 1
Since x + 1 cannot equal x — 1 for any value of «, there is no
value of X which will satisfy the given equation.
But if we substitute in the given equation successively x = 10,
100, 1000, etc., the equation is more nearly satisfied, the larger
the value of x. We can take x so large as to make the differ-
61,52] LINEAR EQUATION 39
ence between the two members of the equation as small as we
please; for this difiference is
1 1 -2
X+ 1 X — 1 x^ — 1
For brevity we say that x =oo is a solution of the equation^
meaning thereby that as increasing values of x are substituted,
the equation is more and more nearly satisfied.
Substituting formally a; = oo, we obtain
^ ^ or = 0.
00 + 1 00 — 1
The equation of example 2 of (49) admits the solution a? = qo.
This will be evident on putting oo for x in
1 2 ^ x+2 ^ 1 1
2"^ a; + 2 2x 2^ x
61. Exercises. Solve for x, including infinite solutions when
present:
1. 5 (a — x) = 3 (6 — x). cx-\-d
2. p (aj — 1) -h a; = g — p. *q wi __ 2d
3. a (J)x — c) — ac — abx. ^
. m-x x-n d
m
^ ^ jj m n _ m —n
6.
7.
X X
a + hx a 12. Viin:5 + V^-13 = 14.
^ "t ? t 13. 3 V16X+9 + 9 = 12 \^.
g + oa; _ h
c-^-dx^d' 14. Vx - \/x^^b - V5 = 0.
o + 6 c —d
^' a + hx c-dx' 16. VV +a? = ! + ^/^.
X a+x *®' :: + - + -= 0.
*• 6 b+x
X
62. Graphic Solution of Linear Equations. — Suppose that a
given equation has been reduced to the standard form,
ax + b =^ 0.
40 GRAPH OF LINEAR EQUATION [53
The solution of the equation is that value of x which reduces the
binomial to 0. For brevity, let us represent the binomial by y,
so that
y = aa? + b.
Then we want that value of z for which j/ = 0. If now we form
a table which gives the values of y corresponding to a series of
assumed values of x, we may obtain from it by inspection the
exact or approximate value of x for which y is zero.
Example,
Let 2 a; — 1 = so that y = 2 a; — 1.
Corresponding values of x and y are :
X = - 2, - 1, 0, + 1, + 2, + 3, . . . .
y = - 5, - 3, - 1, + 1, + 3, + 5
By inspection we see that y ^ when x lies between and 1.
63. Graph of the Equation y — 2x— 1. — We shall now repre-
sent the corresponding values of x and y graphically.
Divide the plane of the paper into four quarters or quadrants
by drawing two mutually perpendicular lines, XX and YY,
intersecting at 0. (See figure.)
Adopting any convenient unit of
—i H:
i%
o'*i *a *9 length (say one-fourth of an inch, or
'IL one side of a square of the cross-
section paper), mark on XX a series
of points whose distances from O
shall equal the assumed values of x. When x is positive, the
distance is laid oflf to the right from 0; when x is negative, to
the left.
In this way all positive and negative integral values of x are
represented by a series of segments having a common starting
point 0, and ending in a series of equally spaced points on the
line XX, each of which represents an integral value of x. Non-
integral values of x are represented by segments whose end points
fall between two points representing integral values. Thus in
the figure are marked the points corresponding to a; == ±1, ±2,
±3, +2J and -If.
Now to represent the value of y corresponding to a given value
of X, mark the representative point of x on XX, and at this point
lay oflf a segment perpendicular to XX and having a length equal
54]
GRAPH OF LINEAR EQUATION
41
^
A
i
n
*s
to the value of y) this segment is drawn upward when y is posi-
tive, and downward when y is negative.
When we construct in this way the pairs of values of x and y
given in the example of (62), we obtain the figure below. We
thus get a series of points, Pi, P2) • • • > Pe, whose distances
from the line XX are the values of the binomial 2 x — 1 for the
assumed values of x. Inspection of the figure shows that as x
increases from —2 to +3, y (i.e. 2 a: — 1)
increases from —5 to +6; also that y =
between x = and 1.
Exercise, By similar triangles, show that any
three, and hence all the points marked in the figure,
lie on a straight line.
By drawing a smooth curve (in this case a
straight line) through a sufficient number of
points Pi,. P2, ... we obtain the graph of
the equation y = 2x — 1. The points Pi,
P2 ... are said to lie on this graph.
64. Graph of y = <wd + ft. — The graph of the equation y^ax + b
is a geometric picture which indicates the value of y correspond-
ing to any assumed value of x.
We shall now show that this graph is a straight line, by show-
ing that any three of its points are coUinear.
Let xi, X2, and X3 be any three values of x; let j/i, 2/2, and 2/3
be the corresponding values of y. Lay oflf the corresponding
values (a;i, yi), {x2, 2/2), and (xz, 2/3) so that (see figure)
xi =Oilfi, 2/1 =MiPi,
X2 = OM2, y2 = M2P2)
xz = OMzy 2/3 = MzPz.
But since yi is the value of y obtained
by putting x = Xi in y = ax + b, and
similarly for 2/2 and 2/3, we have
2/1 = flKTi + 6
ifi^ .fl^ A^f
2/2 == 0x2 + &
y3= CLX3 + b
2/2 - 2/1 = a (X2 - Xi),
ys - 2/2 = a (0:3 - X2).
Therefore,
2/2-2/1 _ 2/3 - 2/2 /_
X2 — Xi Xs — X2
(=a).
42
GRAPH OF LINEAR EQUATION
[55,56
But t/2 - yi = M2P2 - Ml Pi = M2P2 - M2H = HP2 ;
2/3 - 2/2 = M3P3 - Jlf2P2 = KPs;
0:2 - xi = OM2 - OMi = ilf iJlf 2 = Pii?;
and 0:3 - 0:2 = OM3 - OM2 = M2Af3 = P2K.
Substituting these in the two fractions above we obtain
HP 2 KPz
PjI~P^'
Therefore A PiffP2 is similar to A P2JfP3.
Hence the points Pi , P2, P3 lie on a straight line.
Theorem: The graph of the equation y ^ ax + h is a straight line.
Corollary: To construct the graph of the equation 2/ = cu; + 6,
construct two points on it and draw a straight line through them.
66. Exercises. Draw the graphs of the equations (each set
to the same reference lines) :
1. y==a; + l, 2y = 2a; + 2, 5a; = 5a;+5, iy = ia; + J.
2. y = 3x-4, 22/ = 6a;-8, A;2/ = 3A;x-4A;.
3. y=x + l, 2/=a;+2, y=a; + 3, 2/=a;-l.
4. y = 3a; — 4, 2/ = 3x— 2, y = 3a;, y=3a; + l.
6. 2/^a; + l, 2/ = 2a; + l, 2/ = 3a; + l, y = Ja; + l.
6. y = 3a; — 4, y = 6x — 4, y = 9x — 4, y = fa; — 4.
7. y — — x + 1, y= — 3x— 4.
8. y=x — \,y — Zx-\-^,
Explain the effect on the graph of y = oa? + 6, of :
9. Multipl3ring the equation through by a constant.
10. Changing the value of h.
11. Changing the value of a.
12. Changing the sign of h.
13. Changing the sign of a.
56. Coordinates. — Divide the plane into four quadrants by
the lines XX and FF as before, and let P be any point in the
II I plane (see figure), obtained by laying off a
p pair of corresponding values of x and y.
The position of P is completely determined
-^ as soon as x and y are given. Therefore
X and y are called the coordinates of P,
X being called the abscissa, and y the
p
y
y
y
p
p
III IV ordinate.
A point whose coordinates are x and y is referred to as f ^ point
57] GRAPH OF LINEAR EQUATION 43
The four quadrants of the plane are numbered consecutively
as in the figure, and are called the first quadrantf the second quad-
rant, and so on.
The line XX is called the axis of^x, and YY the axis of y.
It is evident (definitions of x and y in (63)) that the signs of
the coordinates in the four quadrants will be as in the following
table:
Quadrant Abscissa Ordinate
I + +
II - +
III
IV + -
67. Linear Equation in Two Variables. — If x and y are unre-
stricted, the point (x, y) may have any position in the plane. But
when a relation between x and y is given, as2/ = 2:cory = a?+l, or
2x — 3y + 4 = 0, the point (x, y) is thereby restricted to a defi-
nite path, which we have already called the graph of the equation.
A relation of the form Ax + By + C = is called the general
linear equation in two variables.
Theorem: The graph of the linear equation Ax + By-\-C^Qis
a straight line.
A C
Proof: It B f^ 0, we can write j/= — -j^x — ^, which has the
form 2/ = ox + 5. Therefore the graph is a straight line when
B f^O.
C
If JS = 0, the equation reduces to Ax + C — 0, or a; = — -j>
unless A = 0. But if A = and B = 0, then C = and the equation
vanishes identically. Excluding this, we may reduce Ax + C —
C
toa:=---j,orx = a constant. But this is a straight line parallel
to the 2/-axis. Therefore the given linear equation represents a
straight line. (Hence the term " linear " equation.)
Exercises.
A C
1. Show that the equations Ax -f By + C = and y == "~ d * *~ d have
the same graph.
2. Show that the equations Ax + By -\' C ^ and kAx -f- kBy -\- kC —
have the same graph, k being any constant.
3. How IS the graph of Ax + By + C = affected by a change in Cf
InBfinAf
44
USE OF THE GRAPH
[58
68. Use of the Graph. — When any two variable quantities are
connected by a linear equation, the relation between them can
always be represented graphically by a straight line. It is only
necessary to consider the two ^uantit^es as the coordinates of a
point.
Example 1. A man atarta'S miles aouth of A and walks due north at the
mte of 3 milefl an liour. How far is he from A at the end of x hours?
Soluliim. Let y be the required distance. Also
let y be nt^tive to the south of A, positive to the
north. Then the relation between y and t is
B-3I-5
The graph is shown in the figure. Here one square
OS the horizontal scale represents one hour, and one
square on the vertical scale represents one mile.
Exercise. By inspection of the graph, find the dis-
tance from A at the end of 0, 2, 3, 4) hours respeo
tively. Compare with the values obttyned from the
equation.
In this example negative values of x and the corre-
sponding values of j/ may be interpreted as follows:
Let the tjme be counted from the moment when the
man, supposed to be walking due north continuously
at the rate of 3 miles an hour, arrives at the point 5
miles south of A. Let time after this moment be
called poffltive, and before it, negative. Thus, 3 hours
before this moment would I>e represented by x = — 3. The conespondii^
value of V is — 14, that is, the man was
14 miles south of A.
Example 2. The relation between the
readings on the scales of a Centigrade
and a Fahrenheit thermometer is given by ■
the equation
C'HF- 32).
Draw the graph.
We shall retain the lettera F and C ,
instead of replacing them by x and y.
The graph is shown in the adjacent figure. From it the reading of either
scale corresponding to a given reading of the other may be at once read off,
with an accuracy of alxiut 1°.
Exerciae. Read off the values of C corresponding to F — — 40°, f* = 0°,
F - S7° respectively; also the values of F when C - - 30°, 0°, + 21°.
Example 3. A volume of gas expands when the temperature rises and con*
tracts when the temperature falls according to the law
59,60]
EXERCISES AND PROBLEMS
45
where
and
vo » volume at temperature 0^,
V » volume at temperature f*.
*»OCI*
Represent graphically the rela-
tion between volume and tem-
perature for a quantity of gas
whose volume at 0° is 100 cu. ft.
Replacing ^f^ by its approxi-
mate value .0037, and «o by 100,
the equation becomes
» = .37 < -h 100.
The graph is given in the adjacent figure.
69. Exercises.
1. From the figure, read ofif the volumes corresponding to the temperatures
250®, 75®, 0®, and — 273°; also the temperature corresponding to the volumes
150, 75, and 20 cu. ft. respectively.
2. Construct a graphic conversion table for converting yards to feet.
3. Construct a graph showing the relation between the circumference and
the diameter of a circle.
4. A falling body, starting with an initial "velocity of voft. per second,
acquires in i seconds a velocity given by » = gr< + wo, in which g — 32.2. As-
sume a value of vq and draw the graph of the equation.
6. Let A be the lateral area of a right circular cylinder of height h and radius
of base r. Draw the graph showing the relation between A and h when r is
fixed. Also draw the graph showing the relation between A and r when h
is fixed.
6. Same as 5, except that cone is substituted for cylinder, and slant height
for height.
Solve for x graphically:
7. 8-I-X-15.
a. 3a; = 27.
9. 2 (x - 1) = 6.
10. ix + ix ^=5,
11.
12.
x-2 ^ 6^
3x-5 19
x-1 8
n-ix 3
60. Problems.
1. If 12 be added to 7 times a certain number the sum is 54. What is
the number?
2. Find a number such that if 16 be subtracted from it and the result
multiplied by 5, the product equals the number.
3. Find a number such that if a be subtracted from it and the result mul-
tiplied by m, the product equals the number.
4. Find a number such that 3 times the number increased by 10 equals 5
times the number.
6. Find a number such that m times the number increased by a equals n
times the number.
46 SIMULTANEOUS LINEAR EQUATIONS [61
6. The age of a boy is three times that of his brother, and their combined
ages make 16 years. How old is each?
7. In what proportion must two liquids, of specific gravities 1.20 and 1.40
respectively, be mixed to form a liquid of specific gravity 1.25?
8. Two boys start together and walk around a circular half-mile track
at the rates of 3} and 4 miles an hour respectively. After how many laps will
they pass each other?
9. A can do a piece of work in 3 days, B in 5 days. How long will it take
them both to do it?
10. A can do a piece of work in a days, B in 6 days. How long will it take
them both to do it?
IL A can do a piece of work in a days, B in 6 days, and C in c days. In
how many days can they together do it?
12. At what time between 4 and 5 are the hands of a clock together?
13. At what time between 10 and 11 are the hands of a clock at right angles?
Opposite each other?
14. The sum of the ages of A, B, and C is 60 years. In how many years
will the sum be 5 times as great as it was 10 years ago?
16. Water flows into a cistern through two pipes A and B, and out through
a third pipe C. The cistern can be filled by A in 1 hour, by B in 45 minutes,
and emptied by C in 36 minutes. How long will it take to fill the empty cis-
tern when all three pipes are running?
61. Simultaneous Linear Equations. — Let there be given two
linear equations containing two unknown quantities x and y, as
ax + by + c = 0,
a'x + b'y + c' = 0.
It is required to obtain all pairs of values of x and y which sat-
isfy both equations simultaneously.
First Method — By Substitution. — Solve one of the equations
for either of the unknowns in terms of the other; substitute the
value so found in the second equation, thus obtaining a linear
equation in one unknown; solve for this unknown and then
obtain the other by substitution in either of the given equations.
Check. Substitute the values of x and y in the equation not
used Jh the last step of the solution.
Example, Solve for x and y :
^+x = 15 and ?^ + y=6.
Clearing and simplifying:
4 x + y = 45 and x -\- iy = 30.
62]
SIMULTANEOUS LINEAR EQUATIONS
47
Prom the first of these, y = 45 — 4 x.
Substituting in the second, a; + 4 (45 — 4 a;) =30.
Hence 15 a; = 150 or x = 10.
Then y = 45 — 4 a; = 5.
Check.
?^y + ^=io_z5 + 5^i.+ 5^e.
Second Method — By Elimination. — Multiply the first equa-
tion by a', the second by —a, and add the resulting equations
together. This eliminates a:, and yields a linear equation in y
alone,, from which y may be found. Similarly x is found by mul-
tiplying the first equation by 6', the second by —6, and adding.
The proper multipliers for the two eliminations are conveniently
indicated thus:
6'
a'
ax + hy + c = 0,
-6
— a
a'x + Vy + c' = 0.
Check. Substitute the values of x and y in either of the given
equations.
Example, Solve for x and y :
8a;-15 2/ + 30 = and 2a:+3y-15=0.
Indicating the multipliers:
3
15
2
-8
8x-15 2/ + 30 =
2x+ 32/-15 =
- 54 2/ + 180 = 0, or 2/ = V-,
54 X - 135 = 0, or X = f .
Check. 8 X } - 15 X -y^ + 30 = 20 - 60 + 30 = 0.
62. Exceptional Cases.
1. The given equations are not independent.
In this case one equation is a multiple of the other, so that
a = ka\ b = kb\ and c = kc\
k being a constant. Both equations are then equivalent to a single
one, and do not suffice to determine two unknowns.
By assuming any value for Xy substituting in one of the equations
and solving for j/, we obtain a pair of values which satisfy both
equations. (Why?) Hence there exists an infinite number of
solutions.
48
SIMULTANEOUS LINEAR EQUATIONS
[63,64
2. The given equations are inconsistent.
If a = fca'.and b = W, but c j^ kc^, then the given equations
are self-contradictory. For if we subtract k times the second
equation from the first, we obtain c = fcc', which is not true.
In this case there is no finite solution possible. For if we assume
X = xi and 2/ = j/i to be a solution of either equation, the other
equation will not be satisfied by these values because c j^ kc'.
63. Graphic Solution of Two Simultaneous Linear Equations.
Let the equations be
(1) ox + &y + c = 0,
(2) a'x + Vy + c' = 0.
The graph of each equation is a straight line.
Suppose L\ and L2 (figure) to be the graphs of
equations (1) and (2) respectively. Then the coordinates of any
point on Li, as Pi, satisfy equation (1), and of any point P2 on
L2 satisfy (2). Hence the coordinates of the intersection P of Li
and L2 satisfy both equations simultaneously and give the required
solution.
Exceptional Cases.
1. The given equations are not independent.
Then, as before, a == ka\ b = kb\ and c = kc\ The lines Li
and L2 will coincide and have an infinite number of common points.
2. The given equations are inconsistent.
Then a = ka\ b = kV, but c ^ kc\ The lines Li and L2 are
now parallel to each other, but not coincident. Hence they have
no common point (except at infinity). Including the infinite
solution is equivalent to the statement " parallel lines meet at
infinity."
64. Exercises. Solve for x and y, including graphical solu-
tions:
1. 2x4-2/ = 11.
3 x — y = 4.
2. 3x + 82/ = 19.
3x-y = 1.
3. 2x-\'y = 47.
a? -f y = 15.
4. 3a:+4y = 85.
5x+4y = 107.
6. 5a;-f 7y = 101.
7x — y = 55.]
6. 2a: -2/- 1 =0.
6X-32/-3 =0.
7. 15x-7y = 9.
9y-7x = 13.
8. 2x -7y = 8.
4y-9a: = 19.
65,66] SIMULTANEOUS LINEAR EQUATIONS 49
9. x-22/ + 2=0. 16. 5y-2a; = 21.
3a; - 6y + 2 = 0. IZx - 4y = 120.
10. 8x+32/ = 3. 17. i+| = 7.
12x + 9y = 3. 2 3
11. ix-\'iy-2-^0.
2x+Zy = 48.
a; + iy-3=0. 18. ^+^=34.
12. Sy-^x-l =0. Z^j.^ =, 12
18-3x = 42/. 8 "•"8
13. 2a; = ll+9y. 19. i + y=.^.
3a; -15 = 122/. ^
14. 2x + 7y-52. ^~^""5"*
3x-52/ = 16.
20 ^ = 10 - - 2/
16. 3x + 4y-5=0. . • 3 ^" 2 ^*
ia;+Jy-i =0. 4iy = 5a;-7.
Simultaneous Linear Equations in More than Two
Unknowns
66. Three Equations in Three Unknowns.
Let the given equations be,
(1) ax + by + cz + d = 0, .
(2) a'x + yy + c'z + d' = 0,
(3) a''x + V'y + c"z + 6!' = 0.
Eliminate one of the variables, say 2, from two pairs of the
equations, as from (1) and (2) and from (2) and (3). Solve the
resulting equations for x and y. Substitute the values of x and y
so found in one of the given equations and solve the result for z.
Check. Substitute the values of a;, y, and z so found in either of
the equations not used in the last step of the solution.
66. Exceptional Cases.
1. Thje given eqitations are not independent.
. (a) In this case one of the equations can be expressed as a linear
combination of the other two, with constant coefficients. Hence
any solution of these two equations is also a solution of the third.
But two equations in three variables admit an infinity of solutions.
For we can choose any value for z at pleasure, substitute it in the
two equations and obtain a pair of values of x and y.
50 SIMULTANEOUS LINEAR EQUATIONS [67-69
(b) It may happen that two of the equations can be expressed
as simple multiples of the third. Then any solution of the third
equation is also a solution of the other two. Hence again there
exists an infinity of sohdionSj since we may choose for two of the
variables any value at pleasure and obtain the corresponding
value for the third.
2. The equations are inconsistent.
In this case the equations in x and y obtained by eliminating
z are also inconsistent. Hence there is no solution (except the
infinite solution).
67. We shall not discuss here the graphic solution of three linear
equations in three variables. Interpreted graphically, each of the
equations (1), (2) and (3) represents a plane in space. In general,
three, planes meet in a single point, giving one and only one solu-
tion. The exceptional cases are:
1. (a) The three planes meet in a common line. Hence any
point in this line gives a solution.
(b) The three planes coincide. Hence any point in one of
the planes is a solution.
2. The three planes are parallel. No solution, except infinity.
(" Parallel planes meet at infinity.")
68. Four Equations in Four Unknowns. — Solution, Eliminate
one of the unknowns from three different pairs of the four given
equations. The three resulting equations can be solved for the
other three unknowns. The fourth unknown is then found by
substituting these three in one of the given equations.
Check. Substitute the values of the four unknowns in one of
the equations not used in the last step of the solution.
Exceptional cases arise, quite analogous to the preceding. We
shall not discuss them here.
The method of solution outlined above is evidently applicable
to any number of linear equations in the same number of variables.
A more convenient method involves the use of determinants.
(Chapter XVI.)
«
69. Exercises and Problems.
^- i + 6-^*- ^' 3 + 2-3*
I 3 8 2* 2"*"3 6*
69]
EXERCISES
61
3. ^+y _|. y-^ ^
3 ' 2
= 9.
2^ 9
= 5.
4. ^+i^ + ^:^ = 5.
8 ■ 6
x-hy X — y
= 10.
5. £^ + ^:z2^2^
8 ■ 5
2y-5
2x +
= 21.
40 -^ y-
2x — y
+ 22/ =
7. ,25 a; + 32/ = 10.
4.5 x — 4 2/ = 6.
8. 4.22/ + 4a; = 33.
0.77 2/ - 0.3 a; = 2.95.
9. 0.2525 X + 0.33 y = 280.
3.122X +0.0552/ = 3096.
10. 0.2 2/ + 0.25 a; = 2 (2/ - a;).
0.8 X - 3.7 2/ = - 15.3.
11. 0.1 2/ + 0.3 a; = 0.3.
0.052/ + 0.15 a; = 0.15.
12. ix-O.Qy =0.
5 (a; - 1) ^6
2 (2 2/ + 3) 6*
13 i + 1-11
1_1 ^ Jl_
X y 30'
14.? + ?= 3.
X y
15_4^
a; 2/
16. -+? = 10.
a; 2/
^+? = 20.
a; 2/
16.
2(5 -11a;) ,
• 11 (a; - 1) "•"
11-72/
3-2/
= 5.
7 + 2x 125-1442/
3 - X 36 (2/ + 5)
2.
17.
7-6a; 4
102/ -19 52/
-3a;
-11
6X-102/-17
4a; - 142/ - 5
3X-62/+2 "
2a; - 72/ + 12
18.
1
1 — a; + 2/ a;
1
2
+ 2/-1
3
1
3
1
1
4
l-x+2/ 1
-x-y
19.
1
1
a; + X
5
1
7
2/—
«
2/V a;/
1
•
20.
ox — 6y = m.
ca; + dy = n.
21.
x+2/ = 3a —
26.
a; — y = 2a —
35.
22.
a
X
23.
J? /
24.
m n
5 + ^ = 3.
r « ^
25.
rwa; p2/_^
n q
26.
x+y-l
a; — 2/ + 1
•
a; — 2/ + 1
52
5
X ^m p
K
XERCISES
[69
27.
80. 2VF
+ y-y + i -0.
y «
m n
Va; + 2/-2y-2 = 0.
^ y
a b
h +y a —x
c d
81. 1.
yx -
.1" 1.^—0
— 1 . ; f O — U.
y yx-^y
28.
3
2 +i-n
V^
/ ; 1 ■■• *'•
y \x + y
d—x c +y
-1 =
4.
82. -;-
ax +
3a;
by y
29.
yjx + l-\jy
L o — n
Va; + 1 + -N^
-1 =
2.
y
ax + hy
83.
X + 2/ = 37.
84.
aj +
y = a.
86. 2a;+3y « 12.
x + « = 25.
aj +
« -6.
3X + 22 = 11.
2/ + « = 22.
2/ +
2 = C.
3 2/ + 4 2 = 10.
88.
Ix-iy^O.
87.
IJa;
+ liy = 10.
88. a;+22/ + 32 = 32.
ix — iz - 1,
2 J a;
+ 2§ 2 = 26.
2x + Sy +z -^ 42,
iz-iy = 2,
3i2/
+ 3i 2 - 30.
Zx+ y-\'2z = 40.
89. -
y
+j-
»2.
40. -
xy _1^
: +y 5
1
X
H-i-
= 4.
X
a:z 1
+ 2 "6*
1
a;
*l-
= 6.
y
y2 1
+ 2'"7-
41.
a; + 2 2/ = 5.
42.
y + 2 + M = 2.
48. 3a; + 2/ + 2 = 4.
2/ + 2 2 = 8.
2 + M + X =3.
X + 42/+3W = 6.
z + 2u -=11.
tt + a; + 2/ = 4.
6x+2-|-3t* = 8.
M + 2 a; « 6.
X +y +z '= 5,
82/ + 32-|-5t* = 10.
44. Find two numbers whcNse sum is 1735 and difiference 555.
46. If at a given place the longest day exceeds the shortest night by
8 hours 10 minutes, what is the duration of each?
46. The sum of two numbers is 1000. Twice the first plus three times the
second equals 2222. Find the numbers.
47. The annual interest on a capital is $180; at a rate of interest li%
higher, the annual interest would be $240; find the capital and rate of interest.
48. A farmer sells 200 bushels of wheat and 60 bushels of com for $252;
60 bushels of wheat and 200 bushels of com would bring, at the same price
per bushel, $203; find the price per bushel of each.
49. Two points move on the perimeter of a circle 999 ft. long; the one point,
moving four times as fast as the second, overtakes it every 37 seconds; find
the speed of each.
60. A vat of capacity 450 cu. ft. can be filled by two pipes. If the first
pipe flows 3 minutes and the second 1 minute, 40 cu. ft. are discharged; if the
first pipe flows 1 minute and the second 7 minutes, 60 cu. ft. are discharged.
69] EXERCISES 63
How long will it take both pipes to fill the tank, and what is the discharge per
minute of each pipe?
61. How many pounds of copper, and how many of zinc, are contained in
124 pounds of brass (alloy 6f copper and zinc), if, when placed in water, 89
lbs. of copper lose 10 lbs. in weight, 7 lbs. zinc lose 1 lb., and 124 lbs. brass
lose 15 lbs.?
62. An alloy of gold and silver weighing 20 lbs. loses 1} lbs. when placed
in water. How much gold and how much silver does it contain, if gold, when
placed in water, loses i^g of its weight, and silver ^ of its weight?
63. Find the lengths of the sides of a triangle if the sum of the first and
second is 30, of the first and third 33 and of the second and third 37.
64. Find three niunbers which are in the ratio of 2 : 3 : 4 and whose sum
is 999.
66. The contents of three measures are as 4 : 7 : 6; 10 measures of the
first kind, 4 of the second, and 2 of the third together contain 20 gallons. How
much does each measure contain?
66. A vessel may be filled by each of three measures as follows: by 4 of the
first and 4 of the third, or by 20 of the first and 20 of the second, or by 28 of
the first and 3 of the third. Also, the three measures together contain 29
pints. Find the content of each measure.
67. A vessel can be filled by three pipes: by the first and second in 72
minutes, by the second and third in 2 hrs., and by the first and third in IJ
hrs. How long will it take each pipe alone to fill the vessel?
68. A and B can do a piece of work in 12 days, B and C in 20 days, A and
C in 15 days. How long will it take A, B, and C, working together, to do the
job?
69. Three principals are placed at interest for a year, A at 4%, B at 5%,
C at 6%; the interest on A and B is $796, on B and C $883, and on A and C
$819. Find the amount of each principal.
60. Two bodies move on the circumference of a circle; when going in the
same direction they meet every 30 seconds, and when going in opposite direc-
tions every 10 seconds; in the second case, when they are 30 ft. apart, they
will again be 30 feet apart after 3 seconds. Find the speed of each body and
the radius of the circle. .
CHAPTER V
Quadratic Equations
71. Suppose we wish to find two numbers whose sum is 5 and
whose product is 6.
Let X = one of the numbers;
then 5 — a: = the other number,
and a; (5 — x) = 6 or a;2 — 5 a; + 6 = 0.
To determine x we must solve this equation.
Definition. An equation of the form
where a; is a variable and a, 6, c are constants, is called the gen-
eral equation of the second degree in one variable, or, a quadratic
equation in x.
Methods for Solving the Equation nx^ + ^ + c = 0.
72. 1. By Factoring. When the trinomial ax^ + hx + c can
readily be factored, then each factor, equated to zero, gives a
value of X.
Example, x^ — bx -^-^ — Oy
or (a; - 2) (a; - 3) = 0.
a; - 2 = or x - 3 = 0.
Bience a; = 2 or a; = 3.
73. 2. By Completing the Square.
(a) The equation is reduced to the form
(a; + A)2 = fc,
whence x-\-h =± Vk, and x ^ — h ± \/k.
This reduction is effected as follows :
Given ax^ + 6a; + c = 0.
Transpose c : . ax^ -^hx ^^^ c,
64
73] QUADRATIC EQUATIONS 65
Divide by a: x^ -\- -x == —
a a
Add (;r— I to both members:
\2a)
x^
+ ^ + (^)'=-^ + (^)
or,
6V h^-^ac
x
I & . . /6^ — 4 ac .1 /r5 T —
2 a V 4a2 2 a
Hence, a? = -•
(b) The equation is reduced to the form (2ax + A)^ = fc,
whence 2 ax + A = ± "v/^, and a; = — -7^
2a
Given ax^ + &a; + c = 0.
Transpose c: ax^ + fex = — c.
Multiply by 4 a; 4 a^x^ + 4 abrc = — 4 ac.
Add 62; 4 a2x2 + 4 a6x + 62 = 52 _ 4^^^
or, (2 ax + &)2 = 52 __ 4 ^^
2 ax + & = ± V&2 - 4 ac.
rj --&±\/62-4ac
Hence, x = jr •
2a
Example, 2 a;* + x - 6 = 0.
(a) Transpose -6: 2 x^ + x = 6.
Divide by 2: x2 + } x = 3.
Add (i)2: x2 + i X + (i)2 = 3 +(1)2,
or (a + })« = il
X + 1 = ± I,
and x=±|--i = i or —2.
(b) Transpose -6: 2 x2 + x = 6.
Multiply by 8: 16 x^ + 8 x = 48.
Add 12 or 1: 16x2 + 8x + 1 = 49,
or, (4 X + 1)2 = 49.
4 X + 1 = ± 7.
Hence x = f or — 2, as before.
56 QUADRATIC EQUATIONS [74-76
74. 3. By Fonnula. In (72), by completing the square accord-
ing to either method, we obtained
-6± \/&* - 4 oc
05 =
2a
Any quadratic equation in z may be solved directly by means
of this formula, by merely inserting for a, &, and c their values
from the given equation. The Jormvla should be carefully com-
mitted to memory.
Example. 2 x* + a? — 6 — 0.
_ ~lrhVl-4X2X(-6) ^ -1±7 _ 3
76. Exercises. Solve for x:
1. x2+4a;-12 = 0. 11. 5x(x-2) + i = 1 -3a:.
2. x2-8a?= -7. 12. (1 - 3a;) (x - 6)- 2(x + 2).
3. x2 + 6x - 16. "• (^ + 1) (2a; + 3) = 4a;2 - 22.
4. ^2 + 12=7*. ^*- 7a:2-48 = 2x(a:+7).
16. 13ar2 - 30 = 6 (1 - a;)2 + 63.
16. ax-{-h ^ a;2.
17. te~262 4.aj2 =26a;.
18. a;2 + mn = — (m + w) x.
19. ca; -2c-a;2 =-2a?.
6. 14 =a;2 -5a;.
6. 5x2 -.3a; -2 =0.
7. 3a;2+5a;-42«:0.
8. 3x2-50 =25x.
.9.2.«-13x — 16. 3, ^_ax_^.
10. 3x«-7x-6-0. 2 2
t
76. Definition, A roo< of an equation is a value of the variable
which satisfies the equation.
By the formula of (73), the two roots of any quadratic equa-
tion can be obtained.
Nature of the roots of the equation ax^ + fta? + c = 0. — The
values of x obtained by the formula
^^ 2a
will be
1. real and unequal if &* — 4 ac > 0;
2. real and equal if &* — 4 ac = O;
3. imaginaiy if &* — 4 ac < o.
77-78] QUADRATIC EQUATIONS 57
For, in the first case the radicand in the formula for x is positive,
hence the square roots are real; in the second case, the radicand
vanishes, and the two values of z reduce to the common value
— 6 4- 2 a; in the third case the radicand is negative, hence both
square roots are imaginary.
•
The expression &' — 4 ac, on whose value depends the nature of
the roots, is called the discriminant of the equation ax^ + &x + c = 0.
Wfien the discriminant vanishes, the roots are equal; ax^ + &x + c
is then a perfect square.
TT. Exercises. — Without solving the equations, determine the
nature of the roots of:
1. Exercises 1-10 of (74). 6. i x^ - J x - i = 0.
2. 4a;2-|-4a; + l =0. 7. 0.1 x2 + 0.5 x + 0.8 = 0.
3. x^+x + 1 =0. 8. Iix2-6ix + 8l =0.
4. 6x2+2x-l =0. 9. ix2-ix-M=0.
6. 9x2 -f 12x + 4 = 0. 10. 0.06x2 ^ o.22x -f 0.08 = 0.
For what values of the literal quantity involved in the following equations
will the roots be real and unequal, equal, or imaginary respectively:
11. x2+2x + c = 0. 18. 2x2+4Ax-A2=0.
m
12. 4x2+4x + A = 0. 19. 2x2+4ax-a«0.
13. 3x2 - 2x - A; = o. 20. ax2 +2x + 1 = a.
14. ix2 - J X + 4a = 0. 21. a2x2 + ax + o = 0.
16. x2+26x + 4-0. 22. 2cx2 +3x - c2 = 0.
16. 3x2-4)kx+5=0. 23. (1 + A;)x2 +x - A; » 0.
17. 6x2 + - X - 3 = 0. 24. - +^x + -^ = 0.
m n 2 n + 1
78. Relations between the Coefficients and Roots of a Quadratic
Equation. — The roots of the equation
ax^ + bx + c =
«6+V62_4ac - & - V&2 - 4
are: Xi = jr- ; X2 =
a^
2a ' "^ 2a
Hence xi + X2 = — f and X1X2 = -
a a
58 QUADRATIC EQUATIONS [79-81
That is, if the equation be divided by the coefficient of z^, the new
coefficient of x, with its sign changed, equals the sum of the roots;
the new constant term equals the product of the roots.
79. Factors of the Trinomial aad^ + bx + c. — If xi and X2 be
the roots of the equation ax^ + 6x + c = 0, the trinomial is divis-
ible by X — xi and x — X2. But
(X — Xi) (X — X2) = X^ — (Xi + X2) X +.X1X2
= x^ + - x + - •
a a
Therefore a(x — xi) (x — X2) = ax^ + bx + c.
Hence to factor the trinomial ax^ + bx + c, place it equal to
zero and solve for x; subtract each root from x, form the prod-
uct of these differences and multiply it by a.
80. Exercises.
1. Find the sum and the product of the roots of the equations in exercises
1-10 of (76).
Form equations whose roots are:
2. 2, 3; 4, -1; -2, -1.
3. a, 2 o; p, g; m + n, m — n.
4. V^n:, - yT^; 1 + V^^, 1 - V^; a+b V^^, a^b yf^.
5-14. Factor the left members of the equations in exercises 1-10 of (74).
15-24. Same for exercises 1-10 of (76).
26. Show that the equation y ^x^-^bx -{-c cannot have a fractional root
if b and c are integers.
81. Graphic Solution of Quadratic Equations. — In order to
solve the equation
ax^ + bx + c = 0,
we must find the values of x which reduce the trinomial ax^ +
&x + c to zero. When a, 6, c are given numerical values, the
required values of x, when real, may be obtained, exactly or
approximately, by trial.
Consider, for example, the equation
2 a;2 + X - 6 = 0.
82,83] QUADRATIC EQUATIONS 69
Designate the trinomial on the left by y, so that
y = 2x^ + x-G.
Now form a table showing the values of y corresponding to a
series of assumed values of x :
± = . . . - 3, - 2, - I, 0, +1, +2, + 3, . . . ;
y = . . . + 9, 0, - 5, - 6, - 3, + 4, + 15, . . . .
We see that y = when x =— 2, which
gives one root exactly. Also, y must be zero ,
f^^in for a value of x between +1 and +2,
hence the other root lies between 1 and 2.
Now consider the pairs of corresponding
values of x and y as the coordinates of a
series of points and draw a smooth curve
throt^h them (figure). Scaling off the values
of X for which y = 0, we have
X = — 2 and i = 1.5 approximately.
82. Parabola. — The curve in the figure is j,^2x' + x-Q
called a parabola. It is an example of a class
of curves all of which have similar forms. The point where the
curve bends most sharply is its vertex, and a line through the
vertex and dividing the cUrve into two sym-
metrical portions is called the axis (figure).
The segments OA and OB, measured from the
origin to the points where the curve cuts the
T-axis, are called the avintercepts. The inter-
cepts are positive when extending to the right
from 0, negative when extending to the left.
83. It will be found that the graph of the
Par<A<^ equation
is always a parabola, with its ans parallel to the y-axis. (We
assume a ^ 0.)
The parabola will cut the x-axis in two distinct points, or be
tangent to the x-axis, or will not cut the x-&£\% at all according
as the equation oj? + 6x H- c = has real and unequal, or equal,
or imaginary roots. For in the first case y is zero for two distinct
60
QUADRATIC EQUATIONS
[84,85
values of x, in the second for two equal values of x, and in the
third for no real value of x.
These three cases are illustrated in the figures below.
1
y
—
\
1
•\
/
! \
/'
^
f "
\
1
\
I
A
r
j[
V
2
r
o
1
/
\
/
\
/
\
/
L
1
\
i
t
\
/
Vs^
^.
X
>
\
«
[^
X
|7
.1
/^
n
[^
y
4
\
7
)
/
S^
2
JL
y — a^ — 4x + 3 y — x* — 4x4-4 y — x*— 4x4-6
5'-4ac>0 6'-4ac-0 6»-4ac<0
84. Exercises.
1-10. Solve graphically the equations in exercises 1-10 of (74).
11-19. Draw the graphs representing the left members of the equations of
exercises 2-10 of (76).
20. On the same diagram construct the graphs of y =» x^ 4-2 x, y = x^ 4-2 x 4- 1,
and y = x2 4- 2 X 4- 2.
21. Sameasin20fory=-x2-2x, 2/=-x2-2x-l, and 2/= -x^ -2 x -2.
22. What is the effect on the graph of y = ox^ 4- 6x 4- c when c is increased
or diminished?
23. What is the effect on the graph of changing the signs of all terms of the
trinomial?
86. Equations Reducible to Quadratics.
Example 1. 2 x* - 7 x^ 4- 6 = 0.
Solve for x^ as the unknown quantity.
, 7 ± V49 - 48 ^ 3
x2 = 2 or 2*
X « db V2 or ±
Example 2, x~'-8x~*=9.
Solve for x~ * as the unknown.
v/l'
-4 8 ± 10 ^ -
: ' = — 2 — = 9 or — 1.
729
or — 1.
86] QUADRATIC EQUATIONS 61
Example 3. (2 a;2 + 5 a;)2 - 6 = 2 a;2 + 6 x.
Solve f or 2 a;2 + 5 x as the unknown.
(2 a;« + 5 a;)2 - (2 x2 + 5 x) - 6 = 0.
2a;2 + 5a; = ^-y-^ = 3 or - 2.
2 a;2 + 5 X = 3 or 2 a;2 + 5 X = - 2.
X = i or — 3; or, X = — i or — 2.
Exercise. Verify the answers in the above examples by substitution.
Example 4. x + V2 x^ + 1 = 1.
Transpose: V2 x^ + 1 = 1 — x.
Square and collect terms: x^ -f 2 x = 0.
Therefore x = or — 2.
Example 5. x — V2 x^ + 1 = 1.
Transpose: — V2 x^ + 1 = 1 — x.
Square, etc: x^ -|- 2 x = 0.
Therefore x = or — 2, as in example 4.
Exercise. Verify the answers in examples 4 and 5.
On substituting the values found in examples 4 and 5 in the given
equations, we find that th« first equation is satisfied by both
v alues of x , but not the second, provided we assume, as usual, that
V2 x^ + 1 stands for the positive square root.
: The equation of example 5 may be put in the form
X - 1 = V2a;2 + 1.
Evidently these two expressions are not equal to each other for
any real value of x. For, if x be less than 1, they are of unlike
signs; if x be gr eater tha n 1, V2 x^ is certainly greater than x,
and therefore V2a;2 + 1 > a? — 1. Hence the solution of ex-
ample 5 as above has led to incorrect results.
The reason for this is that on squaring in the second step of the
solution the sign of the radical disappears, and from that point
on we are really solving example 4 also.
When an equation is squared to clear of radicals , the answers should
be carefully verified and only those retained which satisfy the given
condition,
86. Exercises and Problems.
1. Va;2+3x-5 = VSx + l. 3. V2x2--5x + l « ^/x + 1.
2. V5x2-fl - V3(6x + 7). 4. 4x - 1 = y/7x^ -2x+4.
62 QUADRATIC EQUATIONS. EXERCISES [86
6. Va; + 3 + Vic + 8 = 5 Vx. ^^ \fo^'^^~x+^ 3
V2a;2+5x-3 4
V3a;2+x + 5 3
6. V2x+l+V7x-27«V3x+4.
7. Vx + 3 + V3x-3 = 10. "• V4a;2-xTT'"2'
8. VST17 + V^"=l = J V2i. 13 V9g2 + 6a;+"l ^ 3.
^ ^ " Vl8a;2-3a;-2 2
»-^^^+^ + ^^^^°"^- ^, V 9:c»+6x + l 3
10. Vl2+a; = V7x + 8 - 2. ' Vl8a?-3x-2 2
16. 7 +
12
y + y/4-y^ y - y/4-j^ 7
16. f J = 6.
17. ^A""^ -1+-(V57-1),
V52 + 1 2
18. ^^^ ^ + 3 V2» + l = 7 Vti.
V2« + l
19. , ^^ - Vsi = V3« + 1.
V3« + l
20. Vtt+i +^l±M - 7 V4r=^
V4 « - 3
2^ V3 g2 + 1 - V2 g2 4- 1 ^ 1
V3 x2 + 1 + V2 x2 + 1 7
(Or, by composition and division rationalize the denominator.)
22 V27x2H-4 + V9a;2+5 ^y
' V27 x2 + 4 - V9 x2 + 5
oo V5a; — 4 + V5 — a; VJx + 1 .
V5x — 4 — VS — X V4 X — 1
2^ ^^ ^ 3 V2 .
Vic +a;2 1 +x
aB.2(l+?)+3y/
^ + 9 = 14.
«- Vx2 - 16 , y — r-5 7
Va;2 --3 yx-Z
86] QUADRATIC EQUATIONS. EXERCISES 63
27.
X + m _ p — a;
X —m p + x
28.
n — a? X + p
n -\-x X — p
AA
a^ X X
X b X
32.
33 |^l_-;m4 __ 2^^ ^
y —m y+m
i--p=^4^- 84. Vo + x-Va-* = V5.
'6-ox 6c— X m + V2my — ^
t
81. ?i±^+?i:z£5 = 2^. 86. V5 + \^^^=-^-
x + a a; — a ^
37.
VS + V6 Va - X + V6^
yfx — ^ ^a -x — ^h — X
38. VS + Vo- \lax^^ = VS.
39. Va-a; + V- (a^ + aa;) = / .
^ \Ja — x-\-\lx — b ^ f^^E3.,
\la — a; — Va? — 6 \ a; — 6
41. 2x\jx-Zxd- =20.
42. / + / 7 = Vq-O'
ya — X yx — b
47. a;2 + VSa; + x2= 42 - 5x.
48. <Jx-b<J^-:- -18.
46. a;^ - 16a;* = 512. 60. x^-\- 24= 7a;- V^^'^^'TJ+IS.
^■sjlri^sj'^.-
61. ^/(l +a:)2- '^Vd - xY = ^1 - x«.
62. Find three consecutive integers, the sum of whose squares is 1202.
63. Find three consecutive even integers, the sum of whose squares is 776.
64. The sum of the squares of three consecutive integral multiples of 4 is
3104. Find the numbers.
66. A rectangle, twice as long as it is wide, has an area of 1800 square
feet. Find its dimensions.
66. How large a square must be cut from each corner of a rectangular
card 6 X 12 inches so that the remaining piece shall contain 27 square inches ?
67. As in 56, except that the original dimensions are a X 6 inches and the
remaining area A square inches.
68. What changes must be made in the dimensions of a rectangle 2 X 12
Inches to double the area without changing the perimeter ?
64 SIMULTANEOUS QUADRATICS [87,88
60. As in 58, when the original dimensions are a X h inches.
60. State some values of a and b for which exercise 59 is impossible.
61. Plnd the radius of a cylinder whose height ia 10 feet, if the total sur-
face in square feet must equal the volume in cubic feet.
62. As in 61, except that total surface equals twice the volume.
68. As in 61, except that total surface equals n times the volume. For
what values of n is the problem impossible ?
6i. What niunber exceeds twice its square root by 3 ?
65. The sum of the ages of a father and his son ia 80 years and the product
of their ages is 15 times the sum; find the age of each.
66. A number consisting of two equal digits is 3 less than 4 times the
square of one of its digits; find the number.
67. For what real values ofxisx* + 10a; + 9 positive ? zero ? negative ?
(Graph.)
66. Show that 6 + 2 a + a* cannot be negative if a is real. (Graph.)
69. Show that 3 a — a* — 5 cannot be positive if o is real. (Graph.)
70. The difference of the cubes of two consecutive integers is 127. What
are the integers ?
71. Two trains start from a station, one going due north 5 miles an hour
faster than the other, which goes west; at the end of four hours they are 60
miles apart. Find the speed of each.
87. Simultaneous Quadratics.
Definition. The degree of a monomial involving one or more
literal quantities is the sum of the exponents of such literal quan-
tities as may be specified.
For example a^x^y"^ is of degree w in x, n in y, m + n in a: and y,
m + n + p in a, X and y.
The degree of a polynomial is that of its term of highest degree.
A quadratic equation in several variables is one in which all the
variable terms are of the first or second degree, at least one term
of the second degree being actually present.
88. Solution of Two Simultaneous Equations in Two Variables,
one being Linear, the other Quadratic. — The most general forms
of such equations are:
(1) px + gy + r = 0,
(2) ox^ + bj/^ + cxy + dx + 62/ + / = 0.
Solution.
1. Solve (1) for one of the variables in terms of the other. Thus :
px-\-r
2. Substitute this value in (2), obtaining a quadratic equation
in X.
8»-91] SIMULTANEOUS QUADRATICS 65
3. Solve this quadratic for x, and let its roots be xi and X2.
4. The corresponding values of y are now found by substituting
these values for x in the first step. Thus:
pxi + r J px2 + r
2/1=-^^— — y and 2/2=-^^—
Example. (a) x + y =» 1,
(b) a;« + j^ = 4.
From (a), y — 1 — x.
Substituting in (b): x* + (1 - x)« « 4 or 2x«-2a;-3=0.
Hence a^i = i -f i V7; xa == i — i V7.
Then yi = i - i V7; yz = i + i V7.
Reducing to decimals, we have approximately
(xi, yi) = (+1.8, -0.8) and ix2, y^) « (-0.8, +1.8).
In this case there are two distinct real solutions.
89. Nature of the Solutions of Equations (1) and (2) of (88).—
The values Xi and X2 obtained in the third step of the solution in
(88) are either real and unequal, real and equal, or both imaginary.
Then the values of y obtained in the fourth step will be of the same
nature as the values of x.
Hence there are always two solutions, which may be real and un-
equal, real and equal, or imaginary.
90. These three cases may be illustrated by means of the
equations,
(1) x + y == k,
(2) x2 + 2/2 = 4.
Then x^ + (k - x)^ = 4, or 2x^-2kx + (k^ - 4) = 0.
Hence xi = i(*+ V8-A;2) and X2 = i{k - -^S^Hc^).
2/1 = i (fc - y/S^^^) and 2/2 = i (* + V8^=^).
These solutions will be
real and unequal if k^ < 8;
real and equal if k^= S;
imaginary if A:^ > 8.
91. Graphic Solution of the equations
(1) x + y = h
(2) ^2 + y2 = 4.
f
66
SIMULTANEOUS QUADRATICS
[92
Straight line x + y = 1
Circle x* 4- 2/* = 4
Considering x and y as the coordinates
of a variable point, all values of x and y
which satisfy the equation (1) give rise to
a series of points lying on a straight line
(figure).
Let us now mark some points whose
coordinates satisfy equation (2), which we
put into the form
Assuming a set of values for x, and calculating the corresponding
values of y, we have
X = 0, i, 1, U, 2, 2i, . . . ;
y= ± 2, ± i Vl5, ± V3, ± i V7, 0, imaginary.
For negative values of x we obtain the same values of y over again.
On plotting these values we obtain a series of points all of which
lie on a circle of radius 2, center at the origin.
The points of intersection of the line and the circle have coordi-
nates which satisfy both equations at once, and are therefore the
required solutions. Scaling them off from the figure we have
(^1, yi) = (1.8, -0.8) and (x2, 2/2) = (- 1.8, +0.8), as in (88).
92. Graphic Illustration of the Three Cases of (90). — In (I)
of (90), let us put successively A; = 1, 2 'N^, and 4, so that k^ < 8,
= 8, and > 8 respectively. We have then the equations,
(1) x + y = l; x + y^2^; x + y = ^,
(2) x2 + i/2 = 4; a;2 + 2/2 == 4. x^ + y^ = ^.
The three straight lines and the
circle are shown in the adjacent figure.
When A; = 1, the line cuts the circle in
two distinct points; when k = 2 V2,
the line is tangent to the circle; when
A; = 4, the line fails to meet the circle.
We may consider these three cases
as arising from special positions of a
variable line which moves parallel to
itself and occupies in turn the posi-
tions of the three lines in the figure. Circle x2+ y^ = 4
93,94]
SIMULTANEOUS QUADRATICS
67
93. Standard Equation of the Circle.—
The equation
is satisfied by the coordinates of every
point on a circle of radius r, center at the
origin, and by no other point. It is fisdled
the standard equation of the circle.
Exercises. Solve for x and y, and check care-
fully by graphs.
(x-y = 0.
. (a;2 + 2/2 = 4,
*• |2x+2/=2.
(^2 + 2/2.4,
|x-2y = 3.
6.
aJ + y'^H
Circle f radius r, center a^
. (a:2 + 2,2«9,
'• (3x+4y = 12.
(a:2.-fy2-9,
UaJ-5y = 20.
(4a:2+4y2«i,
''• |3a;-y = 1.
a;2 + y2 = 16,
2x -3y =4.
10. Determine k so that the line x -{■ y ^ h shall be tangent to the circle
a;2 + y2 = 4.
IL Determine m so that the line y = mx + ^ shall touch the circle
x2 + y2 =« 5.
12. As in 11, for the line y = mx + 2 and the circle x* + y2 = |,
94. Consider the equations
x-2/ = 1,
9^4 ^•
Proceeding as in (88), we obtain
Xi =
yi =
9 + 12.V3
13
- 4 + 12 V3
13
= 2.3 - ; X2 =
9-I2V3
= 1.3 — ; X2 =
13
- 4 - 12 V3
13
= -0.9-;
= - 1.9-.
Graphic Solution. — All values of x and y which satisfy the
first equation are the coordinates of points on a straight line.
We now plot a series of points whose coordinates satisfy the
second equation, which we solve for y in terms of x and write in
the form
1/ = ± J V36 - 4 a;2.
When X = - 3, - 2, - 1, 0, +1, + 2, + 3,
2/ = 0, ± § V5, ± I V2, ± 2, ± J V2, ± § V5, 0.
68
SIMULTANEOUS QUADRATICS
[95,96
■71
On plotting these points and draw-
ing a smooth curve through them we
obtain the curve in the adjacent figure,
called an ellipse. The line A' A is
called the major axis of the ellipse, B'B
the minor axis, and is the center.
In this case, A' A = 6 and B'B = 4;
OA = 3 and OB = 2.
Scaling off the coordinates of the
points of intersection of the two
Straight line x — y =» 1
graphs, we have as our graphic solution
(xu yi) = (2.3, 1.3); (X2, 2/2) = (-0.9, -1.9).
96. Standard Equation of the
Ellipse. — Every equation of the
form
represents an elUpse, whose major
axis is 2 a, minor axis 2 6, center at
the origin. It is called the standard
equation of the ellipse.
Exercises. Solve and check by graphs :
1.
'x2 y2 _
9 "^ 4 "■ ^'
a; + 2/ = 0.
9 ■•" 4 ~ ^'
2
8.*
9 "^ 4 " ^'
5.
6.
4x2 + 9^2 = 36,
x+2y =2.
a;2 -1-42^2 «4^
a; — y = 3.
Ellipse f semi-axea a
and b respectively
- (9x2H-162/« = 25,
'• (2x-3y = 6.
(2x2 + 3^2 « 12,
•(3xH-2/ = 2.
9.
9a;2+4y2 = i,
>-2/ = 1.
10. Determine k so that the line x — y ^ k shall be tangent to the ellipse
a;2 + 4 2/2 = 4.
11. Determine m so that the line y = mx +Z shall touch the ellipse
4x2 + 9^2 = 36.
96. Consider the equations
X - 2/ = 2,
97]
SIMULTANEOUS QUADRATlfcS
69
Yf
.'/"^
^=;2
^^
z
/
7
- 7- -7
(" -Z
. t,Z
: x^ A
z s^^
z ^^
^^
"^"--^
i
Solving as in (88), we find
xi = 4 + 2 V3, xz = 4 - 2 V3,
yi = 2 + 2 V3, 1/2 = 2 - 2 ^/3.
The graphs are shown in the figure,
that of the equation j/^ = 4 a; being a
parabola, whose vertex is at the origin
and whose axis is the x-axis.
Exercise 1. Compare the graphic solution
with that obt^ned by formula.
Exercise 2. For what value of k will the Parabola y^ — 4tX
line X — y — k be tangent to the parabola
y2 = 4 aj? Why are there not two values of A; as in the exercise of (96) ?
97. Standard Equations of the Parabola. — The equation
always represents a parabola, whose vertex is at the origin and
whose axis is the x-axis. The curve extends to the right from
when o is positive, to the left when a is negative.
The equation
always represents a parabola, whose vertex is at the origin and
whose axis is the i/-axis. The curve extends upward when a is
positive, downward when a is negative.
Parabolas
a^ =, — 4 ay
2^ = — ^ax y^ = ^ax
Exercises. Solve an4 check by graphs:
y2 =a;, ^ (2/2 = 4x,
J/ =x. ix + y ==1,
1.
3.
Ay^=x,
2a; -y = 4.
70
SIMULTANEOUS QUADRATICS
[ 98, 99
y — X
«2.
(y
3^^y,
2x.
6.
3a; + y = 3.
7.
x2 =4y,
a; + 22/ = 2.
8.
9.
2x + 6y-10.
;x2- -4y,
l2/-2a; = l.
Hyperbola^ 3» "" 2* "" ^
Straight Line, x—2y^Z.
10. Determine k so that the line 3 x + y « A; shall touch the parabola
y2-|-4a; =0.
11. Determine m so that the
line y = mx + 2 shall touch the
parabola ^ = 8 x.
98. Consider the equations
9 4^*
The graphs are shown in
the figure.
The graph of the second
equation is an hyperbola, a
curve consisting of two open branches which continually, ap-
proach the diagonals, produced, of the dotted rectangle, but never
cross them. These lines are called the asymptotes of the hyper-
bola. is the center and A' A the axis of the curve.
Exercise. Compare the solution of given equation as obtained by formula >
with that from the graph.
99. Standard Equation of the Hyperbola. — The equation
always represents an hyperbola whose axis coincides with the
X-axis, and whose center is at the origin. The curve lies between
its asymptotes, which are the diagonals, produced, of a rectangle
whose sides are 2 a and 2 6, parallel to the coordinate axes, with
its center at the origin.
The equation
represents an hyperbola whose axis coincides with the j^-axis.
/
100, 101 ]
SIMULTANEOUS QUADRATICS
71
Hyperbola, ^ - p = 1
Exercises. Solve and check by graphs:
x2
HyperboUif
r x2 - 2/^ =
\x-3y =
(x2 -2/2 =
3 p«-2^ =
9 -4 ^^^'
5 a; + 2^ = 5.
^" (4a; + 2/ = 2.
'■I
36, 3,
6.
9.
2x*-Sy' = 6,
3 a; + 2/ =6.
a;2 - y2 = - 1^
2/ - 3a; = 1.
4x2 -92/2 =-36,
(4x2 -92/2 »
j 2 2^ - X = 0.
x2 — 4 2/2 = 4,
2/ «2x -6.
10. Determine k so that the line x — 2y = k shall be tangent to the
hyperbola 4x2 - 9y2 = 3^.
11. Determine m so that the line y '^ mz — 2 shall touch the hyperbola
x2 - 2/® = 1.
100. Rectangular Hyperbola. — The equation
always represents an hyperbola whose asymptotes are the coordi-
nate axes; for the upper sign, its branches lie in the first and third
quadrants, and for the lower sign in the second and fourth quad-
rants. M P^^^^y^ilar
Red, Hyp. xy == k* Red. Hyp. xy = — ^
101. The general equation of the second degree,
ofic* + by^ + cxy + dx + ey +f = 0,
includes all the types of equation considered in the preceding
sections and always represents one of the curves there shown,
72 SIMULTANEOUS QUADRATICS [102
except in isolated cases when it can be factored into linear fac-
tors, in which case it represents a pair of straight lines, or when
it is satisfied by the coordinates of a single point only, as
3.2 _|_ y2. = ,o. The graph may also be imaginary, that is, the
equation cannot be satisfied by any real values of x and y,
as x^ + 2/^ = — 1.
The curves represented by the general equation of the second
degree are not restricted in position with respect to the coordi-
nate axes as are those shown in the preceding figures. The
center, vertices, axes and asymptotes may have any position
whatever, depending on the numerical values of the coeflScients
a, b, c, d, e.
All curves represented by equations of the second degree in
X and y may be obtained as plane sections of a circular cone. They
are therefore called conic sections.
102. Exercises. Give what facts you can about the curves
represented by the following equations, without drawing the
graphs:
1. a;2-f^ = 9. 11. x^ «4y.
2. 4x2 -}.4y2 = 16. 12. 4x2 = y.
8. 3x2 + 3y2 = 15. 13. y2 =. _ 4aj.
4. 4x2 -I- y2 = 4. 14. - 42/2 « X,
6. x2 +42/2 =4. 16. x2=-4y.
6. 16x2 + 25^2-400. 16. 4x2 « - 2/.
7. 25x2 + 16J/2 =400. 17. 16x2-25 2/2=400.
8. 2x2 + 42/2 = 9. 18. i6a.2 - 252/2 = - 400.
9. 2^ = 4x. 19. 25x2 - 162/2 = 400.
10. 42/2 = X. 20. 26x2 - I62/2 = - 400.
Construct the graphs of the preceding equations on cross-section paper.
Construct the graphs of the equations:
21. x2+2r^-6x-82/ = 0. 26. x2 -f 2 X2/ + 2/^ = 0.
22. (x-2/)2=l. 27. 5x2 +2x2/ + 52/2 =0.
23. 3x2 + 2x2/+32/2_i62/ + 23=0. 28. 4x2/ + 6x - 82/ + 1 = 0.
24. x2 - 5x2/ + 6 2/2 « 0. 29. 2/^ - X2/ - 5x + 63^ « 0.
25. 3x2 + 22/2 - 2x + 2/ - 1 « 0. 30. X2/ - 2/2 =, 1.
Solve graphically and by formula several of the preceding equations with
the equation
(a) X- 2^ = 1. (b) 2x + 32/=6.
(c) x+2/ = 0. (d) 2x-2^=2.
103-106] SIMULTANEOUS QUADRATICS 73
103. Solution of Two Simultaneous Quadratics. — When both
quadratics are of the general form, as
ax^ + by^ + czy + dx + ey +f = 0,
o'x2 + b'2/2 4. c'xy + d'x + e'y +/' = 0,
they cannot usually be solved by elementary methods. For, if
we solve one equation fpr y in terms of x say, and substitute in
the other, we obtain, after rationalizing, an equation of the fourth
degree in x. Such an equation requires rather complicated pro-
cesses for its solution. We shall therefore leave aside the general
case and discuss some special cases, such as usually arise in the
practical application of algebra. We begin with some graphic
illustrations.
104. Graphic Solution. — Since each of the above equations
represents graphically a conic section, two such curves intersect
in general in four points. All real solutions are shown by the
intersections of the graphs, and may be read off, approximately
at least, from the diagram.
Whjen the graphs intersect in less than four points (tangency is
courUed as two coincident points of intersection), some solutions are
imaginary or infinite.
The various cases which may arise are illustrated in the figures
on page 74.
We proceed to consider some special cases of simultaneous
quadratic equations.
106. Case 1. Two quadratics, one of which is factorable.
Rule: Factor the equation, put each factor equal to zero, and
solve each of the resulting linear equations with the other quad-
ratic.
{tule for factoring a quadratic. Solve for y in terms of x (or x
in terms of y); if the quantity under the radical is a perfect
square the twQ values of y are of the form y ^ ax + b and
y = a'x + b. The required factors are then
(y " ax — b) (y — a'x — V),
Graphicaliy, the factorable quadratic represents a pair of
straight lines, the other quadratic some conic. Each straight
line may cut this conic in two real distinct points, in two real
coincident points, or in two imaginary points (i.e. does not cut at
74
SIMULTANEOUS QUADRATICS
[105
9 ^4
^ + ^ = 1
Four real solutions, Four real solutions,
all distinct. two being equal.
t + t^l
9^4
{X - iy+ y' = V
Two real distinct
solutions, two
imaginary.
Two real and equal solutions,
two imaginary.
X —y =0
Two solutions, both infinite.
f + J = i;'(x-6)« + »' = f
All four solutions imaginary.
oo
xy ^1
xy = -1
Four solutions, all infinite.
The student is urged to draw, or to picture to himself mentally as
far as possible, grapJis corresponding to all equations considered.
He should he able to recognize at a glance the standard forms of
equation of the conic sections.
105] SIMULTANEOUS QUADRATICS 75
all). Heace the four solutions may be all real and distinct, or
equal in pairs, or imaginary in
pairs.
ixf -2xy -Sy' ^0,
ji*-4j/2 -4 = 0. .
The factors of the first equation are, by-
inspection.
(1 + !/) (J - 3 tf) - 0.
x+u = or x-3
Hyverbola, s=-4y' -4 ~0
SlToigH lines, ^ - 2xy -3!/*-0
Hence we have to Bolve or x-VyQandx-ZyO
{x + y = Q, [x-Zyd,
jji _4y! _4 = 0, ^*^ Jx' -4y* -4 =0.
Solving the first pair, we have
<■'•»■' -(v=i'-v^> ""■'"'-(-v^3'v=i)
These are imagiiiary. The line x + y — Q does not cut the hyperbola (figure).
Solving the second pair,
"■■■"' ■(^■;|)'"""' -(-75' -^)-
These solutions are real, and the approxiinate values may be scaled off from
the figure.
Note. An equation of the form Ax^ + Bxy + Cy* = can
always be factored. Divide by the square of one of the variables,
and solve for the ratio - or - ■
X y
The factors will be imaginary if B^ — 4 AC < 0, and in this
case the graph of the equation is imaginary. In all other cases
the graph is a pair of real straight lines, distinct if B^ — 4 AC > 0,
and coincident it B^ — i AC = 0, -
Example 2. F8Ctor2ii - 2ij/ + j/i -0.
Dividebyx*: (l)*~^(i)"*"^ " **■
.-. ^ - 1 + v^ or 1 - v:ri.
Hence the factors are
I,-(i + V^).)[»-(i-V^)xl-o.
76
SIMULTANEOUS QUADRATICS
[106
Example 3.
2x2 - y2 _ xy + 3 y - 2 = 0.
a;2 - 4y =0.
Solving the first equation for x in terms of y,
we have
Parabola, ' ic* — 4 y «
Straight lines,
2'x^ - y^ - xy + Zy - 2
or
and
1.
S.
x- y + 1 =
2x + y -2 =-
x^ + y^ = 1,
x^ + yx -22/« =0.
a;2 + y2 ==4^
x2 -2/2 ==0.
4a;2 + 9j/2 = 36,
2x2 + 5a;j/ + 3y2 = 6x + 6y.
y± V9y2-24y-hl6 _ y J: (3y ~4)
4 4 *
Hence,
x-y + l=0 or 2x+y -2 = 0,
Solving the first of these with the second
equation above, we have
(xi,^i) = (2 + 2V2,3+2V2);
(X2,y2) = (2-2^,3-2^^).
From the second equation we obtain
(xz,y3) «(-4+2V6, 10-4V6);
(0:2,^2) = (-4-2V6, 10 + 4V6).
Exercises. Solve for x and y, and check
graphically:
x2 ~ y2 = 1,
xy — 2y +x ^2.
y2-4x =0,
6x^+xy - 12 2/2 = 0.
x2 — 4 2/2 « 4^
xy — 2 y = 0.
6.
6.
106. Case 2. Homogeneous equations.
Definition, An equation is called homogeneous when all of its
variable terms are of the same degree. A constant term may be
present. (In the further developments of mathematics, the last,
sentence is omitted from the definition.)
Two homogeneous quadratics have the forms
(1) Ax^ + Biy + Cy^ = D^
(2) A'x^ + B'xy + Cy = /)'.
Solution. Multiply the first equation by D', the second by D
and subtract. The result is a new equation of the form
(3) il"x2 + B'^xy + Cy = 0,
which may be solved with either of the given equations by factoring,
as in Case 1.
1061 SIMULTANEOUS QUADRATICS 77
Graphically, equations (1) and (2) represent two conies, and
equation (3) a third conic which consists of a pair of straight lines
in case the factors are real. Conic (3) goes through the inter-
sections of (1) and (2), since the coordinates of any point which
satisfy (1) and (2) will also satisfy (3). Hence, when the factory
of (3) are real, we obtain the intersections of (1) and (2) by finding
the intersections of either of them with a pair of real straight lines.
When these factors are distinct, there are two distinct lines, either
of which may cut the conic in two real and distinct points, two
coihcident points, or two imaginary points. When the factors are
imaginary the lines are imaginary, and all four solutions are
imaginary.
Another method of solving two homogeneous equations in the
forms (1) and (2) is to put in both of them y = vx. Then divide
one equation by the other, and clear of fractions, after removing
the common factor x^. The result is a quadratic in v, whose roote
we may represent by vi and V2. Then
y = vix and y = v^x.
Substituting these values in turn
in either of the given equations,
we have two quadratic equations
in X alone.
(2x» -3«)/ + 4 -0,
{ixy - 5ff* -3 -0.
Transposii^ the conetant terms we have
2xf -3xy -^ -4.
ixy — 5 y^ = 3.
Multiplying the first equation by 3, the Hyperbolas, I ~ J^i on
second by 4, and addinB, "(3 _4 In
fliS + Ta^-Mj/a = Straight Hn«s, [oi + slJIo
or (3j-4:/){2i: + 5v) =0.
Equating each of these factors to sera, and solving with one of the ^ven
equations, we have, from the first factor,
(xi,3/i) = {4,3); fe,w)-(-4, -3);
from the second factor,
fe,y.) - (i V^ - i V^); (.X4,yt) - (- i V^, i V^).
Hence two solutions are real and two imaginary. The figure shows the graphs
of the ^ven equations and of the factors of the auxiliary equation.
78
SIMULTANEOUS QUADRATICS
[107
To solve by the second method, transpose the constant tenn as before,
then put y — vx.
We obtain
Dividing,
2a;2-3w;2 = - 4;
2a;2 -3ta2
or
4 w;2 - 5 v^x^ = 3.
2-3t; ^ 4
4»-5y2 3*
Clearing, etc., 20 1;2 - 7 1; - 6 = 0.
Hence, i; = f or — f .
Therefore y ^ \x or y= — Jx.
(These are the linear factors of tthe
auxiliary equation found above.)
Substituting these values of y in
either of the given equations, we find
X as before.
Example 2.
9 x2 + xy -h 2 y2 =60,
8 a;2 - 3 xy - y2 = 40.
The auxiliary equation is
6a:»-llxy-72/*=0,
or {2x'\-y){Zx -7y)=.0.
Solving each factor with one of the
given equations we obtain
(xi, yi) = (2, - 4) ; (X2, wl) = (- 2, 4) ;
3
EJlvpsey 9x» + a:2/ + 22/» = 60
Hyperbola^ 8 x" — 3 xy — j^' == 40
Straight lines, (2 x + y) (3 x - 7 y) =
The graphs are given in the figure.
Exercises. Solve for x and y:
{xz, yz) =
(aJ4, ^4)
\2 -N^ 2 V2/'
V 2 V2' 2 V2/
(x2 + 2^2 =9,
|x2-xy = 10.
2, ^2- 2/2 = 1,
J X2 — Xy + 2/2 = 1.
3.
4.
4x2-92/2
2/2 + X2/ =
x2 4-2x2/ =
2X2/-2/* =
= 36,
-4.
= 2,
«6.
(x2 + 0:2/ + 2/2 = 3,
(2x2 -32/=^ «6.
(2x2 + x2/-32^2 = 2,
(x2-x2/+22/2 = 1.
107. Case 3. The given equations are of the forms
ax^ + hy^ = c,
a'x^ + 6y = c'
Rule. Consider x^ and t/^ as the unknowns, and solve by the
method of linear equations.
108] SIMULTANEOUS QUADRATICS 79
Graphically, we have two conies in standard form. The four
solutions may all be real, or equal or imaginary in pairs.
Exampk. x^ ~ ijfi = i,
9i! + 16 J/* • 144.
Bj elimination we obtain,
Hence i = ± 4 VH;
V=±3 VA-
Taking dther value of x with either
value of y, we obtain the four aolutions. Uj^bola, i* - 4 y' = 4
The approximate values may be scaled Ellipae, 9a^ + I6y' =■ 144
oS from the Figure.
ExerciBes. Solve for x and y, and check graphically:
(j:! + ^ = 4, (2i2+5ys = 10, (4«*+6i/» = 20,
ia*-ys -2. )4i=+y=-4. U'-J/*=-9-
|i!+4!/! -4. U^'+S!/' = 36. (i' + i/* = 4.
108. Case 4. Symmetric and Skew-Symmetric Equations. —
A symmetric equation is one which remains unchanged when the
variables are interchanged.
A skew-symmetric equation is one whose variable terms all change
sign when the variables are interchanged. Thus
^ + y^ + x + y=-0, x^~y^ + 2x-2y='l
are symmetric and skew-symmetric respectively.
Rule. Given two such equations, put
X = u + v and y = u — v;
solve the resulting equations for u and v; then
a; = H" + ") and y = i {u — v).
Note. Equations of higher degree than the second may often
be solved by this method.
Example. x* +y* — ^^ = 9,
i' + !/= - JJ/ = 3.
Let I = « + » and y = v, — V.
80 SIMULTANEOUS QUADRATICS 1109
Subetituting and reducing:
M* + 14 uVi + »4 = 9,
t«2 + 3 1;2 = 3.
Let v^ ^ 8 and t>2 => <.
Then s^ + Ust + i^ ^9,
« + 3 « = 3.
Solving: (», = (3, 0) or (i, 1).
(If 8 and t be considered as the coordinates of a point, the equations in
8 and t represent an ellipse and a straight line respectively.)
Since tr « i V» M^d » = ± VZ,
we have (u,v)^(±y/SyO) or (±^» ^ 2 )*
where the signs are to be taken in all possible ways.
Then
aj = M+» = V3, -V3, V3, -V3, 0, 0;
y = tt - 1; = V3, - VS, 0, 0, V3, - %/3.
Here corresponding values of x and y appear in the same vertical line.
109. Case 6. Symmetric Solution. — This method of solution
is applicable to certain forms of symmetric equations, and may be
illustrated by some simple examples.
Example 1. x + y = 5,
xj/ = 4.
Squaring the first equation: x'^ -\- 2 xy -{- y^ =25.
Subtracting four times the second: x* — 2 xy + j/2 = 9,
Hence x — y = ± 3.
But X + y = 5.
X = 4 or 1; j/ = 1 or 4.
Example 2. (1) x« + xy + y* = 6.
(2) x2 - xy + 2/2 =a 10.
Subtract (2) from (1) : 2 xy = — 4, or xy = — 2.
Add xy=-2to(l): x* + 2xj/ + y2 = 4, or x+y=±2.
Subtract 3xy = - 6 from (1): x2 - 2xy + j/2 = 12, or x - y =» db 2 V3.
Hence , x=±l±V3 and y-±lTV3.
Simultaneous values of x and y are then obtidned by taking the same com-
bination of signs in these two results.
110] SIMULTANEOUS QUADRATICS 81
110. Miscellaneous methods for solving two simultaneous
equations.
These methods depend on reducing the given equations, which
may be of higher degree than the second, to one of the cases
already discussed.
1. By Substitution. — This method has already been illustrated
in several cases; in (106) we made the substitution y = vx, in (107)
we put X ^ u + V and y = u — v, and in example 2 of (107) we
put w^ = s and v^ = t. We shall give two more simple illustra-
tions.
Example 1.
'. 16.
xy
If we let - = «
X
and
1 _
y
if and we obtain,
« + < - 2,
«/ = - 16.
These may be solved by the method of (109).
Example 2.
aJ« + 1/2 + x2y2 + 2 xy = 4,
a; V - 2 a^ - 0.
Let » 4- y = « and xy ^ L Then N
«2 + ^ « 4,
^ - 2 « « 0.
The last two equations are readily solved, and give
« = + 2; - 2; 0.
f = 0; 0; 2.
The values of x and y may now be found by solving the pairs of equations,
a;+y = 2, (x + y=-2, U+y=0,
xy=0. (xy=0. (xj/ = 2.
2. By modifying or combining the given equations so as to
obtain simpler forms. In particular, a common factor may some-
times be removed by division.
Example 1.
(1) x2 - a:y « 18 y,
(2) xy -y^ ^2x.
Dividing (1) by (2), we have
?-9H or 1 =
y X
f - j === 9 or X ^ ±Zy.
82 SIMULTANEOUS QUADRATICS [111
Substituting each of these values of x in dther of the given equations, we
can solve for y and so complete the solution.
Example 2.
(1) ix^y-x=^l, (x(xy-l)^l,
(2) (x4j/2-a;2 = 3; U2(a;2y2-1) =3.
Divide (2) by (1) : x (xy + 1) = 3.
Divide this equation by (1) : ^ " = 3.
xy — I
Hence xy — 2,
Then from (2), x^ (4 - 1) = 3, or x^ = 1, or x = ± 1.
But from (1), x (2 - 1) = 1, or x = 1.
In this case the value x » — 1 must be discarded.
Hence the only solution is x = 1, ^ == 2.
Example 3.
(1) x4 + 2/* = 1,
(2) X - y = 1.
Raise (2) to the fourth power and subtract from (1):
(3) 4 x'y - 6 x^y^ + 4 x2/8 = 0.
Square (2) and multiply the result by 4 x^:
(4) 4 x'y — 8 x2y2 -|- 4 xy^ = 4 xy.
Subtract (4) from (3):
2 x2y2 = _ 4 a^^ or x2y2 + 2 xy = 0.
Hence xy =* 0, or xy = — 2.
Solving each of the last two equations with (2) we have
<...)-(.,o),(o,-«,(l±^, =l±^,(l^. =1^).
All four solutions also satisfy equation (1).
111. Summaiy of Methods for Solving Simultaneous Equa-
tions. — [Let the given equations be numbered (1) and (2).]
(a) Equation (1) linear, (2) quadratic.
Rule: Substitute from (1) in (2). Graph, straight line and conic.
(b) Equations (1) and (2) both quadratic.
Case 1. Equaiion (1) is factorable.
Rule: Put each factor separately equal to zero and solve with
(2) as in (a). Graph, two straight Hues and a conic.
Rule for factoring: Solve for y in terms of x (or x in terms of y) ;
the quantity imder the radical must be a perfect square.
Ill] SIMULTANEOUS QUADRATICS 83
Case 2. (1) Ax^ + Bxy + Cy^ = D; (2) A'x^ + B'xy + CY = D'.
Form the auxiliary equation, (1) X D' — (2) X D = 0. Factor
this and solve as in Case 1.
Second Method: Put y = vx in (1) and (2) and divide results.
Graph, two conies, centers at origin (except in case of parabola.)
Case 3. (1) Ax^ + By^ = C; (2) A'x^ + B'y^ = C\
Solve as linear equations for x^ and y^.
Graph, two conies in standard position.
Case 4. Symmetric Equations.
Put X ^ u + V and y ^ u — v.
Applicable to equations of higher degree.
Case 5. Symmetric Solution of certain symmetric equations.
(c) Miscellaneous Methods,
Exercises.
1.
a;2 + y2 = 661.
X* - y2 = 589.
8.
^ + ^=12.
x-y
16.
6x + 22/=29.
6x2/ = - 105.
2.
j/2 - a;2 = - 80.
x« - 2/* = 48.
16.
xy = 80.
a:* + 2/« = 82.
9.
5x2 + 22/2 = 373.
X = 5y.
S.
3x2-y2 = 59.
2x+5 2/ = 54.
17.
4x2-32/2= -83.
2x2 + 3y2 = 98.
10.
x2 + y2 = 10.
3x + 22/ = 26.
4.
x + 2/ = 12.
xy = 35.
11.
X - 2/ = 2.
x2 - J/2 = 120.
X + 2/ = 20.
18.
3x2-2/2 = 83.
X + 2/ = 15.
5.
x + y = 1.
xy = - 1.
12.
X2 - J/2 = - J.
x + 2/ = §.
19.
X2/ + X = 20.
X2/ — y = 12.
6.
x«+2/*=74.
13.
x2 + X2/ = 260'.
20.
2x + 32/ = 20.
x + y = 12.
xy + 2/2 = 140.
3x2/ -2/2 =38.
7.
x + y = J.
352/ = i.
14.
x2 + 2/2 = 218.
xy-y^^42.
21.
5x»-42/« = 109.
7x -52/ = 25.
22.
x+xy + y = ^7.
27.
x2 + X2/ + 2/2 =: 4.
x-\-y^l2.
X2
— xy
+ 2/2=2.
23.
0:2 + xy + 2/2 =. 217.
28.
X2
+ XI/
+ 2/2=7.
x + y = 17.
x + y + xy -5,
24.
x^y 2
29.
X2+2/2
xy = 3.
= 5(x+y).
11 5
a;2"'"2/2 36
80.
X3+2/8
xy = 4.
= 9.
26.
2x2 __3a;y4-y2 =
3.
31.
X2
-4j
f2=4.
x2 +2x2/ -3^2 =
= 5.
X2
— 2x2/ + 2x = 4y.
26.
x2 -X2/+2/2 =s37
•
32.
2x2-2
12/2+3x2/= -x-2i/.
x2 - y2 = 40.
X2
-42/2-x+2j/=0.
84
SIMULTANEOUS QUADRATICS
88. t«2 -h »* + wv « 67.
u + 1> = 9.
84. p2 + p^ + g» -= 79.
P^ - P9 + 9* " 37.
85. r^+s^+ra =» 25.
r + « = 5.
86. fJ + fi2 ^ r« - 84.
r - « « 2.
87. M4-V + w2+t;a = 162.
u-v+u^-ti^'' - 102.
88. p + g + p2+g* = lf.
89. x2+2,2_f.x+y =18.
2xy = 12.
40. ^2 _|. A:2 - fc + ^ - 32.
2Wk = 30.
41. a;2 J- 2/2 + a; + 2^ = 168.
^Jxy « 6.
42. m2 + n2 - m + n = 2400.
^/mn = 30.
48. 9M2 4-,;2+3u + t> = 3042.
^/TOvv -'48.
44. fi-^^ 1304.
r - 8 = 8.
46. p* + «* = 337.
P+9 = 7.
46. x4 - 2/* « 609.
X — y = 3.
i^l. u*-{-iA =2657.
u 4-f — 11-
48. w« + n3 = 152.
w2 -mn+n2 = 19.
49. p + g + Vp+9 = 20.
p3 4- §3 = 1072.
60. x8 + 2^ = 280.
x2 -xy + 2/2«28.
61. m2+3i;2 = 7.
7w2 — 5wv = 18.
62. p» + g» = 162.
p2g 4- pg2 « 120.
58. x» - j/8 == 335.
x^ -x^y ^ - 70.
64. «« + ^ = 855.
«< (« + = 840.
. w» - n» = 602.
mn (n — m) = — 198.
. u2»*+»2 = 17.
M»2 -j. t; =B 5.
67. xl +yi -35.
xi +yi ^ 5.
68. x22/2-18aJ3^ + 72-0.
6x2-17aJ2^ + 122^«=0.
69. a:<+x22/2+I/*=91.
a;2 _- a^ 4- 2^ = 7.
60. x» - 2/8 « 7 (x2 - y^),
x2+2/2«io(x+y).
61. «« + «*= 65.
s* + ^ = 17.
62. x2 4- y2 = o.
x2 — 2/* = 2>.
68. a; — y =« w.
xy — n2.
64. p2 4- g2 = a«.
p 4- g = 6.
66. Vi* + Viy = «•
M 4- » = ^«
66. x2 4- 2/* = o (^ - y)-
x2 4- 2/* = & (a? + 2/)- •
67. ax — by ^m.
68. 6 (x 4- y) = o (x - 2/).
x2 4- y2 = m2.
69. x< 4- 2/* » - 8.
a; - 2/ " 2.
70. p*4-«*- -9.
p - g = 3.
71. M* 4- V* = 175.
M — «; = 5.
72. r2 4- rs 4- «* = «•
r»5 4- rs3 = b.
SIMULTANEOUS QUADRATICS 85
78.
1 _1 ^1^
76.
ic* + ici/* ■■ P»
m V^V
X y a
x2^y2 6»
76.
2/8 + a;2y =- q,
nfi — n^ ^ a (m -- n).
m3+n3 =6(w+n).
74.
t«2 + to; =s m.
i;2 -|- to; = n.
77.
r5 + s5 =, 3368.
rH7S =8.
78.
a;(x + y — 2) = 1. 80.
xy
»
8«.
^82. X (x + y — «) ■
y(x + y-2) = 2.
xz
s
18 y
•
y(x + y -z) ^
z(x + y — z)^S,
yz
=
§ x.
z(x-{-y - z) =
79.
x-{-y + z ^2, 81.
xy +
X =
1.
83. X +y +z ^p.
xy = 3.
y^
+
y =
-1.
xy -^q.
xyz «= 6.
xz
+
« =
3.
xyz — r.
84.
a? (a? + y + «) * o2.
y(x + y+z) «62.
z(x + y+z) =c2.
87.
xy -{-X ^ a,
yz +y ^b.
xz +z = c.
85.
(a; +2) (y +2) =16.
88.
X2 4-2/2 = 17z.
3(x+2/) =52.
X — y = 2.
86.
(x + 2/) (a; +2) =a.
(x+2) (2/ +2) -c.
89.
•
z^+x^\l
a.
c.
Problems.
1. The hypothenuse of a right triangle is 100 ft. long. Find the other
sides, if their ratio is 3 : 4.
2. The product of two numbers is 735, and their quotient f. Find the
numbers.
3. Find two factors of 1728 whose sum is 84.
4. The sum of two numbers is 34. Three times their product exceeds the
sum of their squares by 284. What are the numbers?
6. The product of two numbers increased by the first is 180, increased by
the second is 176. What are the numbers?
6. The product of two numbers times their sum is 1820, times their differ-
ence 54Q. What are the numbers?
7. The sum of the squares of two numbers plus the sum of the numbers
is 686. The difference of the squares plus the difference of the numbers is 74.
What are the numbers?
8. The diagonal of a rectangle is 89 ft. long. If each side were 3 ft.
shorter, the diagonal would be 4 ft. shorter. Find the sides.
. 9. The diagonal of a rectangle is 65 ft. long. If the shorter side were
decreased by 17 ft. and the longer increased by 7 ft., the diagonal would be
unchanged. Find the sides.
10. The diagonal of a rectangle is 85 ft. long. If each side be increased
2 ft. in length, the area is increased by 230 sq. ft. Find the sides.
86 EXPONENTIAL EQUATIONS [113
11. The floor area of two square rooms is 890 sq. ft., and one room is 4 ft.
larger each way than the other. Find the dimensions of each room.
12. For 60 yards of cloth B pays two dollars more than A pays for 45 yards.
B receives one yard more for two dollars than does A. How much does each
pay per yard?
13. Two bodies moving around the circumference of a circle of length
1260 ft. pass each other every 157.5 seconds. The first body makes the
circuit in 10 seconds less than the second. Find the speed of each body.
14. The amount of a capital plus interest for one year is $22,781. If the
capital were $200 larger and the rate of interest i% larger, the amount in
one year would be $23,045. Find the capital and rate of interest.
15. A and B agree to do a piece of work in 6 days for $45. To finish
on time, they hire C during the last two days, and consequently B gets $2
less pay. If A could have done the work alone in 12 days, how long would
it take B and C, each working alone, to do it?
16. The quotient of a number of two digits divided by the product of the
digits is 3. When the digits are interchanged, the new number is | of the
original. What is the number?
17. If the digits of a two-figure number be interchanged, the number is
diminished by 18. The product of the originaT and the new number is 1008.
What is the original number?
18. What number of two digits is 5 greater than twice the product of its
digits and 4 less than the sum of their squares?
19. A fraction is doubled by adding 6 to its numerator and taking 2 from
its denominator. If the numerator be increased and the denominator de-
creased by 3, the fraction is changed to its reciprocal. What is the fraction?
20. A and B start at the same time from two points 221 miles apart and
travel towards each other. A goes 10 miles a day. B goes as many miles a
day as the number of days until they meet diminished by 6. How far did
each one travel?
21. The fore wheel of a wagon makes 1000 revolutions more than the
hind wheel in going a distance of 7500 yards. Had the circumference of
each wheel been one yard more, the difference between the number of revo-
lutions would have been 625. Find the circumference of each wheel.
22. Find two numbers such that their sum shall be equal to 28, .and the
sum of their cubes divided by the sum of their squares equal to 1456.
23. Two points, A on the x-axis 270 ft. from the origin and B on the
y-axis 189 ft. from the origin, move toward the origin. After 10 seconds
the distance between them is 169 ft., and after 14 seconds, 109 ft. Find the
speed of each point.
113. Exponential Equations. — An exponential equation is one
in which the unknown appears in the exponent. Thus:
114] EXPONENTIAL EQUATIONS 87
Exponential equations of the above forms may be solved by
reduction to ordinary equations by use of the principle that
if a** = a*, then u = v,
or more generally,
if a** = b^f then u log a ^ v log 6.
Example 1. Va^ « o^^-i.
X
This may be written o^ = o^x-i.
2 -2x — 1 or X = ^«
Example 2. (w* + i)* = w-2a;-2.
x2 + a;=-2a;-2 or x^ -{-3x + 2 -^ 0.
Hence x = — 2 or — 1.
Example 3. 2*+i = 32a; -i.
Taking logarithms: (x + 1) log2 == (2x - 1) log 3.
xaog:2 -21og3) =-log2 -log3.
log2+log3 ^ 1^
21og3-log2 log!*
Using common logarithms to four places,
0-7781 + ^
0.6532 + ^'^^^^ ^•
X =
114. Exercises. Solve:
1. 2^ = 8. 4. (i)« = 2i2. 7, io-x = iooo.
2. 2» = J. 6. (i)* = l. a 1000* = 100.
8. 4* = g^i. 6. (Tb)* = 25«. 9. 3«+2 = 3«.
10. -s/o^^ = a*-2. 18. 42«-8 « yz-i.
IL '>yp2x+8 = pO. * 19. a3*+2 = 62a;-l.
12. 4?*-i=28*+8. 20. 3^-*-« = 1.
18. -s/m^ =Vw3*+2. 21. 8**+2«=:512.
^** ^ " Va6-i3«. 22. 5^-2 :^25»* + i>.
16. ^''^^fai ^^"""^/d^. ^- (a*-2)8*+* = a*^»* + i>.
16. 3* = \/5. ^' (&^+^)'*-^ = &sx(x + i).
17. 3*+! = 62«. 26. </^^ y/e^^ y/^^^i - 1.
CHAPTER VI
Ratio, Proportion, Variation
116. Definitions. The ratio of two quantities is their indicated
quotient.
Thus the ratio of a to 6 is r, or as it is usually written, a : b.
The numerator of the' fraction, or the first term of the ratio, is
called the antecedent, the other term the consequent.
The ratio 6 : a is called the inverse of the ratio a : &.
Two ratios are equal when the fractions representing them are
equal.
Since r = -^f .'. a :b == ma : nib.
rnb
Hence, both terms of a ratio may be multiplied by the same (or equal)
quantities without altering the value of the ratio.
Similarly, if w ?^ n, then a :b f^ mxi : nb.
Hence, if the terms of a ratio be multiplied by unequal quar^
iities, the value of the ratio is changed.
The compound ratio of a :b and c : dis ac :bd, that is, the ratio
of the product of the antecedents to the product of the conse-
quents.
In particular the compound ratio of a : 6 and a : 6, or a^ : b^, is
called the duplicate ratio oi a to b^ a^ : b^ is called the triplicate
ratio of a to 6, and so on.
A proportion is an equality of two ratios. Four numbers are
in proportion when the ratio of two of them equals the ratio of
the other two.
Four numbers a, 6, c, d are in proportion if a :b = c : d (often
written a :b :: c : d). Here a and d are called the extremes and
6 and c the means. Also, d is called a fourth proportional to a, 6, c.
The numbers a, 6, c, d, e, . . . are in continued proportion if
a :b = b : c = c : d = d :e
88
• • • f
116] RATIO, PROPORTION, VARIATION 89
When three numbers a, b, c are in continued proportion, so
that a : 6 = 6 : c, then b is called a mean proportional between
a and c.
Since r = - or oc = 6^ we have 6 = ± Vac. Also, c is called the
6 c
third proportional to a and 6.
116. Laws of Proportion.
1. In a proportion, the product of the means equals the prod-
uct of the extremes.
2. If two products, each containing two factors, are equal,
either pair of factors may be taken as the means, the other as
the extremes of a proportion.
When four numbers are in proportion so that a :b = c : d,
then they are in proportion
3. by inversion, or b : a = d : c;
4. by alternation, or a : c = 6 : d;
5. by composition, or a + b :b = c + d : d
(,.a c ,, a,- c,. a + b c + d\.
6. by division, or a — 6 .: 6 = c — d : d;
7. by composition and division, ora + b :a — b ^ c + d:c — d,
8. Like powers (or roots) of the terms of a proportion are in
proportion, i.e.,
if a : 6 = c : d, then a** : ft** = c** : d^.
(For if ^ = |, then ^ = ^.)
9. The products of the corresponding terms of any number of
proportions are in proportion, i.e., if
a:b = c:d, a' :b' = c' :d', a" : b" = c'^ : d", etc.,
then aa'a" • • • : bb'b" • • • = cc'c" • • • : dd'd"
...
/„ ..a c a' c' a" c" ., aa'a" . . . cc'c" . . . \
(^Forif^=^.^=^,p=^,- • •.then^,j„ , =^,d" . ..7
10. In a series of equal ratios, the sum of the antecedents is
to the sum of the consequents as any antecedent is to its con-
sequent, i.e.,
ai : a2 = 61 .* 62 = Ci ^ C2 . . .
= ai + 61 + ci + • • • : a2 + 62 + ca + * • • .
90 RATIO, PROPORTION, VARIATION [117,118
For if— =»i-=«— =***=r, then ai = oar, 61 «■ 6sr, ci « car, ... .
Hence (ai + 61 + ci + • • •)=*•(««+&«+ 02 + • • • ), or
ai+hi +ci +
• • >
= r.
02 + &2 +C2 + •
11. More generally, if the ratios ai : 02, 61 : 62, ci : C2 • • • are
all equal to each other, then
> V a\ fei ci pai + gbi + rci + • • •
(a; — = -- = —=...= . — - — . . ,
02 62 C2 pa2 + qo2 + rc2+ ' • '
where p, q, r, . . . are any multipliers;
^ 02 62 C2 ▼^ 02** + 62" + C2" + • • •
Exercise. Prove 11 (a) and 11 (b). For what values of p, 9, r, . . . n
will these reduce to 10 ?
117. Example, Solve for x: =3*
X — a a
By composition and division: jr— = , «
a; = o ' *
c — tf
Exercises. Solve for x, using the laws of proportion:
^^. + l_3
2.
8.
X 2
X 5
a;-2 6
2x-3 7
6.
X H-a 6
X c
7.
a — X : X ^ p : q.
8.
X + m :a = X — m :b.
9.
a — X : X — b — a:b.
A
X -i-p a -|-x
2x + 3 6
4. 3a; -2 : 3x + 2 = 3 :4.
5, ^ "^ ^ -, ^ "i-3 • 3; _ p ■" 5 -^_ a;
* a; — 1 a; — 4'
118. Variation. — A variable quantity is one which may be
considered to assume a number of values.
A function of a variable is a quantity whose value depends on
that of the variable.
If 2/ be a function of a variable x [indicated by writing y = / (x)],
then in general, as x varies y varies with it.
Thus, the circumference of a circle depends on the radius and
varies with the radius. Hence the circumference is a function
119,120] RATIO, PROPORTION, VARIATION 91
of the radius [c = / (r)]. The functional relation is expressed by
c = 2irr.
Similarly, the area of a circle depends on the radius [A = f (r)].
The functional relation in this case is A = nr^.
Also, the cost of a piece of cloth depends on, or is a function
of, the price per yard; the running time of a train between two
stations is a function of the speed; the range of a gun is a func-
tion of the muzzle velocity.
119. Direct Variation. — A quantity y rmries directly with an-
other quantity x when their ratio remains constant.
This is indicated by writing y cc x (read " y varies directly
as x")-
If k denote the constant value of the ratio of y to x, then
y cc X is exactly equivalent to y = hoc.
The constant k will be determined as soon as the value of y
corresponding to a single value of x (other than a: = 0) is
known.
Graphically, the relation between y and x is represented by a
straight line through the origin, the inclination of the line to. the
a:-axis increasing with the absolute value of k. The line is com-
pletely determined by the origin (a: = 0, ^
2/ = 0) and one other point.
If c be the circumference and r the
radius of a circle, then «« r, for c = 2 ttt.
If we take tt = V, then c = V when r
= 1. The figure gives the graph.
Exercise, From the figure read off to Honzontal scale = 10
.1 - 'x xi_ 1 xT_ ^ • times vertical scale
the nearest unit the lengths of circum-
ference of circles whose radii are .15 in., .33 ft., 1.27 mm., .87 cm.
respectively.
1 .
120. Inverse Variation. — When y varies directly as - > that is,
X
1 k
y cc - or y = -, then y is said to vary inversely as x.
When y varies inversely as x, this may be expressed by writ-
ing xy = k.
Graphically, the relation between x and y is then represented
by a rectangular hyperbola, whose asymptotes are the coordi-
nate axes.
92
RATIO, PROPORTION, VARIATION
[ 121, 122
If t be the time, in hours, required by a train to run 10 miles,
and 8 the speed in miles per hour, then
JO..
< = — or t oz -
8 8
-*-^-»>'
JO
«<= 10
The figure gives the graph, only posi-
tive values being considered.
Exercise 1. From the figure read off to
tenths of a unit the times required to run
10 miles when s = 4.5, 7.8, and 15.6 miles
per hour respectively.
Exercise 2. Construct a curve showing the possible dimen-
sions of a rectangle whose area must be 16 sq. ft. Show that
either dimension varies inversely as the other.
121. Joint Variation. — When a quantity varies directly as the
product of two others, it is said to vary with them jointly.
Thus, if 2 oc xy, or 2 = kxyy then z varies jointly as x and y.
122. Exercises.
1. Show that the area of a rectangle varies jointly as its dimensions.
2. Show that the volume of a right cylinder varies jointly as its base and
altitude.
3. Same as in 2 for a right circular cone.
4. Show that the volume of a sphere varies jointly as the radius and the
area of a great circle.
6.- If y oc X and a; « z, show that y«^z.
6. If X oc 2 itnd y<x^Zj show that ax '\-hy<x.z,
7. If a;2 oc y and z^ oc y, show that xz oc y.
8. If X oc - and w oc - show that x oc 2.
y '^ z*
9. If X varies jointly as.p and g, and y varies directly as - , show that p*
varies jointly as x and y.
10. According to Boyle's law of gases, pressure times volume is constant.
Show th§it the pressure (p) varies inversely as the volume (v). Show graphi-
cally the relation between p and viiv^ 1 cu. ft. when p = 25 lbs. per sq. in.
11. li w ^ uVf show that wccu when v is constant, and that wacv when
u is constant.
If a : 6 = c : d, show that
12. 4a + 56:3a+26 = 4c-f-5d:3c+2(i.
13. a-26:-2a + 6=c-2d:-2c+(i.
14. ma -\- nb : pa -\- qh ^ mc -\- nd : pc -\- qd,
16. 3a + 2c:a — c^3b+2d:b-d.
16. }a-4c:J6-4(i = 2a + §c:26 + J(i«
17. a:a + c=a+b:a + b + C'{'d,
122] RATIO, PROPORTION, VARIATION
18. a2 + fl* + &2 : a2 - ah,+ b^ -^ c^ + cd + cP : c^ - cd -{- d^.
19. a + h:c +d:: %la^ +62 : VcM^.
80. Vo2 + 62 : Vc2 + rf2 = ^flS +63 : \/c8 + d«.
21. Va2 + 62 : VcM^ = \^a8 - 6^ : ^c8 - rfs.
If a : 6 » c : d and p : g » r : s, show that
22. p'^a'* : r'''6'* =» g''»c** : «*'*d'*.
23. (a+6)(p-r):(a-«(p+r)=(c+(i)(g ~«):(c - d^) (g + «).
Solve for «:
A.i o 1. L 1 1 «- a + 6 a2 — 62 (a — 6)*
24. 8a6 :x » 6c : Ifoc. 25. — ^^ : — r — = x : ^^ — r- ^•
o — ao ao
«».(^ + -*)=(^-^)-(-+^)^-.
27. The intensity of light varies inversely as the square of the distance
from the source. If the sun is equivalent to 600,000 full moons in brightness,
at how many times its present distance would it be of the same brightness as
the full moon?
28. The squares of the periods of revolution of the planets about the sun
vary as' the cubes of their mean distances. The earth makes a revolution in
one year at a mean distance of 93,000,000 miles. Venus makes a revolution
in 225 days, Jupiter in 12 years. Find their mean distances from the sun.
29. In beams of the same width and thickness the deflection due to a cen-
tral load varies jointly as the load and the cube of the length. If a beam
10 ft. long is bent \ inch by a load of 1000 lbs., how much will a load of
5000 lbs. bend a 30-ft. beam?
80. Two lights, one of which is twice as strong as the other, are 10 ft.
apart. Where on the line joining them do they produce equal illuminatiqn?
CHAPTER VII
The Trigonometric PuNmoNs "
123. Consider any number of right triangles having a common
acute angle A, as AB\Cu AB2C2, and ABzCs, in the figure.
(jg^ Since these triangles are simi-
lar, homologous sides are propor-
tional, and therefore
B\Ci B2C2 B3CS
ACi AC'
AC,
= X,
X (lambda) denoting the conmion
Bj B^ Bs ^^j^g Qf ^YiQ ratio of the side
opposite Z A to the hypotenuse in the several triangles.
Evidently, in every right triangle having an acute angle equal to
A the ratio of the side opposite /. A U> the hypotenuse has the
same value X; X depends only on Z A, and not at all on the par-
ticular triangle in which this angle may be found. For example, if
A =30°,X = i;
A = 45°, X = 4-; ^ = 60^ X = i V§.
a^
a^
Hence we see that X is a function of -4, and that to every value
of A corresponds a definite value of X.
This function is called the sine of angle A^ or
X = sine of angle ^ = sin ^«
94
I
124,125]
TRIGONOMETRIC FUNCTIONS
95
124. The ratio of the side opposite the angle to the hypotenuse
is merely one of six possible ratios which may be formed from the
three sides of any right triangle. Hence associated with every
acute angle there are six ratios, or six abstract numbers, whose
values depend merely on the magnitude of the angle. They are
called the six trigonometric ratios, or trigonometric functions of
the angle, and are named as follows:
r y ^ • ^ Opposite side
sme of Z ^ = sm ^ = -"^ — r •
hjTpotenuse
- , , , adjacent side
cosme of Z ^ = cos^ = -^r — .
hypotenuse
X i! . ^ X ^ opposite side -
tangent of Z ^ = tan^ = -" . .. •
^ adjacent side
X * , ^ . hypotenuse
cosecant of Z ^ = ess ^ = — ^ ' ^ .. .. •
^ opposite side
secant of Z ^
_ hypotenuse
"" adjacent side
X ^ X ^ J. A adjacent side
cotangent of Z ^ = cot ^ = — "^ — it — tt-*
^ opposite side
If the sides of the triangle are a, b, c, as in the figure, then
sin^
a
c
CSC^
c
a
cos A
c
sec A
_ c
tan A
a
"ft'
cot^
_b
a
126. Exercises. With the aid of a pro-
tractor (see inside of back cover), construct
triangles containing the following angles and,
by measuring the sides and dividing, calculate to two decimals
the six functions of these angles.
1. 30°.
4. 75^
7. 85^
10. 5^
2. 45^
6. 15^
8. 80**.
11. 57°.
3. 60^
6. 18**.
9. lo^
12. 38°.
Check the results of the preceding exercises by means of the
following table.
1
96
TRIGONOMETRIC FUNCTIONS
[126
Angle.
Sin.
<k».
Tan.
Cot.
Sec.
Csc.
0'
5
10
0.087
0.174
0.996
0.985
0.087
0.176
11.430
5.671
1.004
1.015
11.474
5.759
15
20
25
0.259
0.342
0.423
0.966
0.940
0.907
0.268
0.364
0.466
3.732
2. 748
2.144
1.035
1.064
1.103 .
3.864
2.924
2.366
30
35
40
0.500
0.574
0.643
0.866
0.819
0.766
0.577
0.700
0.839
1.732
1.428
1.192
1.155
1.221
1.305
2.000
1.743
1.556
45
50
55
0.707'
0.766
0.819
0.707
0.643
0.574
1.000
1.192 .
1.428
1.000
0.839
0.700
1.414
1.556
1.743
1.414
1.305
1.221
60
65
70
0.866
0.906
0.940
0.500
0.423
0.342
1.732
2.145
2.748
0.577
0.466
0.364
2.000
2.366
2.924
1.155
1.103
1.064
75
80
85
0.966
0.985
0.996
0.259
0.174
0.087
3.732
5.671
11.430
0.268
0.17*
0.087
3.864
5.759
11.474
1.035
1.015
1.004
90
/
126. Given one function, to determine the other functions. —
When a function of an acute angle is given, the angle may be
constructed by writing the given function as a fraction, and con-
structing a right triangle, two of whose sides are the numerator
and denominator of this fraction respectively, or equal multiples
of these quantities. Also, since the third side- of the triangle can
be calculated from the other two, all the other functions of the
angle may be found when one function is given.
Examples.
1.
Lay off AC
Then
Hence
4 \ adj. siqe/
4andCB = 3, ±AC.
AB = V42 + 32 - 5.
• .1 3. .4
sin A —-Z, ; cos A = vj
o o
csc A = = ; sec A = 7 ; cot A
6 4
4
3
Scaling off the angle with a protractor, we have A
from the table the angle whose tangent is .75 we have A
= 37°. By taking
« 37** as before.
SecA = 3f = f=-i?yHL^).
\ 1 adj. side/
■)l
127, 128]
TRIGONOMETRIC FUNCTIONS
97
Lay off AC — 1. With A as center and radius = 3, strike
an arc to cut the ± drawn to AC at C. This determines
the point B,
The solution may now be completed as in example 1.
Another method of constructing the triangle in this
example is to calculate CB first, and then to proceed as
in example 1.
127. Exercises. Determine the angle (approxi-
mately) and the remaining functions, when
1. sm A — T^
2. sin A
8. sin A
4. cosA
6. COSA»;r
5
13*
2
— •
3
0.6.
2
3
1
^ •
2
6. tan A
7. tan A
8. tanA
9. cot A
10. cot A
11. sec A
3
2
»4.
- V3 .
= 1.
•' 1.5.
«2.
12. CSC A
13. cos A
14. CSC A
16. tanA
3
— •
2
0.2.
2.4.
10.
16. Show that the equation sin A —2 is impossible.
17. Show that the equation cos A — 1,1 is impossible.
18. Show that the equation sec A = } is impossible.
19. Show that the equation esc A = .9 is impossible.
When A is an acute angle show that,
20. sin A lies between and 1.
21. cos A lies between and 1.
22. sec A and esc A are always greater than 1.
28. tan A and cot A may have any value from to oo.
128. Functions of Complementaiy
Angles. — Since the sum of the two
acute angles of a right triangle is 90^,
they are complementary.
By definition, we have
• r» opp. side h .
sm B = -^ = - = cos il.
hyp. c
Also, cos B = sin A; tan B = cot il; esc B=sec A; sec B = csc A;
and cot B = tanA.
Complementary Ftmctions, or Co-functions. — The co-sine is
called the complementary function to the sine and conversely..
Similarly tangent and co-tangent are mutually complementary,
and secant and co-secant.
98
TRIGONOMETRIC FUNCTIONS
[ 129, 130
The preceding equations are now all contained in the following
Rule: Any function of an acute angle is equal to the co-function
of the complementary angle.
Exercise. Verify this rule when A » 30^ 45^ and 60"^.
129. Application of the Trigonometric Functions to the Solution
of Right Triangles. — When two parts of a right triangle are
known, exclusive of the right angle, the triangle may be constructed
and the remaining parts determined graphically. By the aid of
tables of the trigonometric functions, the unknown parts may also
be calculated.
Rule : When two parts of a right triangle are given (the rt. z
excepted) and a third part is required, write down that equation
of (124) which involves the two given parts and the required part.
Substitute in it the values of the given parts, and solve for the
required part.
An exceptional case arises when two sides are given and the third
side is required. In this case we may use the formula 0^ + 6^=0^.
It will usually be better however, unless the given sides are repre-
sented by small numbers, to solve for one of the angles first, and
then to obtain the third side from this angle and one of the given
sides.
Example 1. In A ABCy given A = 40^ C = 90*, and 6 = 60\ Find the
other parts of the triangle.
To get B, we have B = 90** - A = 50".
To get a, take ^ = tan A or = 6 tan A,
Finally, c is determined from - = cos A
^ J. A "^
or c = 7 = osec A.
cos A
r^
From the table of (126), tan 40*" » 0.839 and
C sec 40** = 1.305.
Hence a = 60 X 0.839 = 50.340 and c = 60 X 1.305 = 78.300.
As a check, we should have
50.340
- = cos B or
c
0.643.
78.300
130. Exercises. ,
Determine the unknown parts of right triangle ABC^ C being 90**, from
the parts given below. Check results by graphic solution and by a check
formula containing the unknown parts. (Use the table of (126).)
131]
TRIGONOMETRIC FUNCTIONS
99
1. A = 25^ a = 100.
2. A == 70°, 6 = 150.
3. A = 51^ c = 75.
4. B = 38°, c = 50.
5. 5 = 65°, 6 = 750.
6. 5 = 10°, a = 0.15.
7. A = 48°, c = 0.045.
8. B =85°, c = 1.25.
9. B = 5r, a = 16}.
10. A = 20°, 6 = A.
11. Find the length of chord subtended by a central angle of 100° in a
circle of radius 50 ft. (First find the half-chord.)
12. Find the central angle subtended by a chord of 80 ft. in a circle of
radius 200 ft.
13. Find the radius of the circle in which a chord of 100 ft. subtends an
angle of 70°.
14. Find the length of side of a regular pentagon inscribed in a circle of
radius 500 ft.
16. Find thd length of side of a regular decagon circumscribed about a
circle of radius 100 ft.
16. From a point in the same horizontal plane as the foot of a flag-pole,
and 300 ft. from it, the angle of elevation of the top is 22°. How high is the
pole?
17. A vertical pole 75 ft. high casts a shadow 60 ft. long on level ground.
Find the altitude of the sun.
131. Angles of any Magnitude, Positive or Negative. — Con-
sider Z XPP (figure) as generated by a moving line which rotates
about from the position OX to the posi-
tion OP.
Divide the plane into four quadrants (I,
II, III, and IV in the figure below) by
means of two rectangular axes OX and OY.
In the figure, the arrows on the axes
indicate the positive directions, and
quadrant I is that included between
the positive parts of the axes. Let
a moving line start from the posi-
tion OX and rotate into the positions
OPiy OP 2, OPzi and OP4 successive-
ly, thus generating the angles XOP\,
XOP2, XOPs, and XOP4 respec-
tively.
OX is called the initial line, and
0P\ the terminal line of the angle XOPi, and similarly for any
other angle.
p, 1
Y
*V^an^
iT \
-<
r'^
.'U0*\/
(
m /
i
\'
Pa
V.
100
TRIGONOMETRIC FUNCTIONS
[132
An angle is positive when the generating line rotates counter-
clockwise (in the direction of the curved arrow in the figure)
negative when the generating line moves clockwise.
The quadrant of an angle is that quadrant in which its terminal
line lies. The angle is said to lie in this quadrant.
The initial line OX, and any terminal line, as OP2, may always
be considered to form two angles numerically < 360®, as +120®
and —240° in the figure.
When the moving line rotates from OX through more than one
complete revolution, an angle greater than 360® is generated.
Thus a rotation in the positive direction (positive rotation) through
1^ revolutions generates an angle of. 480®, lying in the second
quadrant; a negative rotation through 2| revolutions generates
an angle of —780®, lying in the fourth quadrant.
132. The Trigonometric Func-
tions of any Angle. — Let XOP be
any angle, and P any point in its
terminal line. (The four possible
cases are here shown in the figure,
according to the quadrant of the
angle.) Let OM be the abscissa
of P, MP (not PM) the ordinate
of P, and OP the distance of P.
The signs of these quantities are
taken according to the usual
convention and are shown in
the figure. We now define the
functions of angle XOP, in whatever quadrant it may be, as
follows:
ordinate (of JP) ^^ ^ distance
sin xor = ^t~^ , t -, ; CSC XOP = ri^iip ;
distance (of P) ordinate '
cos xor =
abscissa
distance '
sec XOP =
distance
abscissa '
tan xor =
ordinate
abscissa '
cot xor =
abs cissa
ordinate
When Z XOP < 90®, these definitions agree with those
of (124).
133, 134 ]
TRIGONOMETRIC FUNCTIONS
101
According to the above definitions we have the following
Table of Signs of thb Tbigonometric Functions
Quadr.
sin.
COS.
tan.
cot.
sec.
CSC.
r
I
+
+
+
+
+
t<-
II
+
.—
f-
—
— ■
+ -
III
—
+
+
•^
lY
—
+
+
—
y
Let the student verify carefully the signs in this table. He
should be prepared to state instantly the sign of any function in
any quadrant.
Observe that in the first quadrant all the functions are positive;
in the other quadrants a function and its reciprocal are positive,
the remaining four negative.
133. Approximate Values of the Functions of any Angle. —
If in the last figure the distances OP had been taken all of the same
length, all the points P would lie
on the circumference of a circle
with center at 0.
Let us draw a circle with as
center and unit radius (figure;
1 = 10 small divisions). Then for
any angle XOP we have
sin :X:OP = MP
cos XOP = OM
{-m-
(.0^.
Hence approximate values of the sines and cosines of all angles
may be read ofiF directly from the figure. The other functions
MP
may be obtained by division, since tan XOP =■ Tyrz, etc. They
may also be constructed graphically by a method explained in the
next article.
The lines OM and MP, whose lengths repre3ent the sine and
cosine of Z XOPy are commonly referred to as the line values of
these functions.
134. Line Values of the other Trigonometric Functions. —
Construct a circle as in the figure below and draw the tangents
102
TRIGONOMETRIC FUNCTIONS
[134
at S and S\ Let XOP be an angle in the first quadrant. Pro-
duce OP to meet the tangent at S in T, Then by similar triangles,
n
1
\^ —
r
A
Tm
T,
T
/\
/ /V
V\
1
3!'
\
J^stne
S
^
I /
""^^^-^
^^
^
5
/^X
T,
tan XOP =
JIfP ST
'^ OS = ^^•
In the same way,
tanZOPi = Sri;
tan ZOP2 = /SiTz.
Hence when an angle is in the
first quadrant, its tangent is
measured by the segment of the
tangent line from & to the ter-
minal line produced; the radius
of the circle is the unit of length.
By taking into account the
algebraic sign of the tangent, we find that
tan XOPz = - /ST3; tan XOP4 = - /ST4; tan XOP^ = &Tf,.
Here &*T^ and STf, are themsislves negative lines.
Hence the numerical value of the tangent of any angle equals
the segment of the vertical tangent cut off by the terminal line
produced, this segment being measured in terms of the radius as
unity. This value should be given the proper sign according to
the quadrant of the angle.
Vf^e have further,
OP OT
sec XOP =
OM OS
^OT.
By examining the other angles shown in the figure we see that
the numerical value of the secant of any angle equals the segment
of the terminal line produced from the origin to the vertical tan-
gent. The proper sign may be determined according to the
quadrant of the angle.
To obtain graphic constructions of the cotangent and cosecant,
we draw the tangents at R and 72' (figure below). Then
cot XOP =
CSC XOP =
OM RT
MP
OP
MP
OR
OT^
OR
RT;
= or.
135]
TRIGONOMETRIC FUNCTIONS
103
By examining the other angles in the figure we see that, (a) the
cotangent of any angle is numerically equal to the length of the
segment of the horizontal tan-
gent cut oflf by the terminal
line of the angle produced;
(6) the cosecant is numerically
equal to the segment of the
terminal line produced from
the origin to the horizontal
tangent.
In either case the sign is to
be determined according to
the quadrant of the angle.
136. Variation of the Trigo-
nometric Ftmctions. — In the figure of (133) suppose the point
P to describe the circumference of the circle in such a way that
the angle XOP shall vary continuously from 0** to 360°. Let us
trace the changes in the value of sin XOP = MP, In the first
quadrant AfP, and hence sin XOP, varies from to +1, in the
second from +1 to 0, in the third from to — 1 and in the
fourth frpm — 1 to 0.
Similarly cos XOP varies in the four quadrants successively
from +1 to 0, to -1, -1 to 0, and to +1.
MP
Consider next tan XOP = ^- When XOP = 0^ JIf P = and
OM
CM = 1; hence tan 0° = 0.
Now as XOP increases from 0° toward 90**, MP steadily increases
toward 1, while OM steadily diminishes toward 0. Hence tan
XOP increases from without limit, so that we write tan 90°= <»,
and say that the tangent varies from to oo as XOP varies from
0° to 90°.
Since the three remaining functions are reciprocals of the three
already considered, their variations are easily traced. Thus,
1
CSC XOP =
Hence esc XOP varies from oo to 1 in the
sin XOP
first quadrant, and from 1 to oo in the second. Now as XOP
passes through 180°, esc XOP changes suddenly from a large posi-
tive value when the angle is a little less than 180° to a large
negative value when the angle is a little more than 180°.
L
104
TRIGONOMETRIC FUNCTIONS
[135
This abrupt change in the Cosecant when the angle. passes
through 180° is expressed by saying that the cosecant has a dis-
continuity at 180°; sec 180° may be either +oo or — oo, according
to the side from which XOP approaches 180°.
In the third quadrant esc XOP is negative and varies from
—GO to —1; in the fourth quadrant from —1 to — qo. There is
another discontinuity at 360° or 0°.
The variations of the six fimctions are shown in the following
table.
Quftdr.
sin.
CSC.
COS.
tan.
cot.
I
II
III
IV
oto+1
+lto0
oto -1
-ItoO
+00 to+1
+1 to +00
— 00 to —1
— 1 to —00
+lto0
oto -1
-ItoO
oto+1
+1 to +00
-00 to -1
-1 to -00
+00 tol
Oto +00
-ootoO
O-tcr^oo
-ootoO
+00 toO
Oto-oo
00 toO
Oto-oo
The range of values covered by each of the six fimctions is indi-
cated in the diagram below.
9ec X and esc x
8in X and cos x
sec X and esc x
— oo
tan X and cat x
Exercises.
1. Trace carefully the variations given in. the above table.
2. Show that the following functions have discontinuities at the values
stated: the tangent, at 90^ and 270°; the cotangent, at 0° and 180*^; the secant,
at 90** and 270**; the cosecant at 0** and 180°.
3. Discuss the "equations," tan 90° = +oo; tan 90° = — oo. Same for
csc0° = +00 ; csc0° = — oo.
4. Draw a circle as in the figure of (133), adding also the vertical and hori-
zontal tangents. Divide the circumference into 36 equal parts, and obtain
from the diagram a two-place table of the six functions for every tenth degree
from (f to 360°.
6. By use of the results of exercise 4, trace the graph of the equation y « sin x,
[Take angle x on a horizontal scale, making one small square = 10°. On
the vertical scale choose any convenient length as 1 (» sin 90°), say 10 small
square^ At successive points x on the horizontal axis erect perpendiculars
equal to sin a;, upward or downward according to the sign. See note at end
of (143)].
136]
TRIGONOMETRIC FUNCTIONS
105
Graphs of the Trigonometric Functions
Sine Curve
(full line)
Oosine Curve
Cbroken line)
Tangent Ciurve -t-i
(full Une)
Cotangent
Curve
(broken line)
Secant Curve n
(f uU line)
Cosecant Chirve o
(broken line)
-/
r
•
J
N
y
J
7\
190*
£>•
f
90*
ZfO^
390*
f
4801*
5i0*
\
\
106 TRIGONOMETRIC FUNCTIONS [136,137
6. Trace the graph o(y ^ cos x. (On same diagram as y « sin x.)
7. Trace the graphs of 2^ » tan x and y » cot x,
8. Trace the graphs of y » sec x and y ^ cacx,
136. Periodicity of the Trigonometric Functions. — Since the
position of the terminal line of an angle x is unchanged when the
angle is increased or diminished by integral multiples of 360°, any
function of x equals the same function of x ± n.360®, n being an
integer. That is,
/(x) =/(x±n.360°),
where / stands for any one of the trigonometric functions.
Hence the trigonometric functions are periodic^ with a period
of SeO"". (See graphs on p. 105.)
137. Relations between the Functions of an Angle. — From the
general definitions of the functions given in (133) we have, putting
Z XOP = X,
sm oc; = ; cos a? = ,* tan a?
esc 35 SeCOC; . ..cot 05
ordinate
ordinate distance sin a5 . ^ cos »
tan 05= — r — ; = —r — ' = » COt 05 = -: .
abscissa abscissa cos 05 sm 05
distance
Whatever be the quadrant of angle XOP = x [figure of (132)],
we have ^, ^^. Q
(ordinate)^ + (abscissa)^ = (distance)^.
Dividing this equation through in turn by (distance) 2, (abscissa)^,
and (ordinate) 2, and expressing the resulting ratios as functions
we have
sin' oc + cos* 05 = 1,
1 + tan* 05 == sec* 05,
1 + cot* QC = CSC* 00,
All the above relations between the functions of an angle x are
true for all values of x. They form a first set of working formulas,
and should be thoroughly committed to memory. They are
collected below, as
«
C^UCic'
138] TRIGONOMETRIC FUNCTIONS 107
Formulas, Group A.
.,v . 1 ,,. . sin a? (6) sin'a? + cos' a; = 1.
(I)sma5== (4) tan 05 = ^' *
CSC a? cosa5 /^x ^ , . o o
(7) 1 + tan' a? = sec' oc,
^^^ ^^^ = iib' ^^^ ^""^ ^ '^ ^* • (8) 1 + cot'x = CSC' a..
(3) tana? = ^
cot a;
We shall apply these formulas in two examples.
Example 1. Prove that tan x + cot x » sec x esc x»
. , . sin a; , cos x sin^ x + cos^ x
tan x + cot X = h -= — = — ^
cos X ,Binx sm :i; cos x
1
sm X cos X
Example 2. Prove that
cscx
cosx.
» CSC x sec x.
tan X + cot X
cscam^ __ CSC a; esc a;
tan X + cot X sing , cos a; sin^ x + cos^ x
cos X sin a; sin x cos a;
CSC a?
1
CSC X sm X COS x « cos x.
sm X cos X
In both examples all the steps taken are true for all values of x, dnce
this is true of all the formulas of group A. Hence the given equations are
true for all values of x, and they are therefore called trigonometric identities.
The equation sin^ x — cos^ x » 1 is not true for all' values of x, but holds
only for certain special values; it is not an identity.
138. Exercises. Prove the following identities:
1. tan X cos x »= sin x. . . esc x
4. ootx —
■m sec X
O ^^__^___^ -s finn X
' cot X sec X * 5. (sin2 x + cos^ x)* « 1.
3. tanx = 6. . -. — ^ = cot^tf.
CSC X sm tan $
7. (esc $ — cot tf) (csc tf -f cot tf) — 1.
8. (sec X — tan x) (sec x + tan x) = 1.
9. (sin ^ -f cos 9)2 =* 1 + 2 sin 9 cos tf.
10. sin2 a -f cos2 a = sec* a — tan* a.
11. (sin a — cosa)2 = l— 2sinacosa. '
h
108 TRIGONOMETRIC FUNCTIONS [139
12. sin* X — cos* x — sin* x — cod* x.
13. (1 — 8in2 x) C8c2 X = cot* X.
14. OOt*0 — OOGi20 sCOt*0CO(^9.
16. tan ^+ cot » sec esc 0.
16. tan ^ sin ^ + cos (j> » sec ^.
17. sirf^sec^V = sec?# - 1. «>. (1 - sin^/S) (1 + tan*/?) = 1.
ift 8in<^ 1+cos^ 2L tan* x - sec* x « 1 - 2 sec* x.
1 -cos^ sin<^ .
. . -4-^ cosx-|-8inx 1-ftanx
^^ 14-tan*/9 Bin*/8 22. '—, — ^ ^r-^
19, ^ -r ««>" H _^ ou* p ^ cos X— sin X 1— tanx
l+oot;8/8 cofi?/8
28. (tanx — 1) (cotx — 1) =» 2 — secxcscx.
24. secd + tand = ^ r— r-
1 —smB
26. sec^sinf = (1 -|- cosd) (tand — sind).
26. tan* a -|- cot* a -|- 2 =^ sec* a esc* a.
27. sin' d + cos' B = (sin tf 4- cos B) (1 — sin d cos B).
28. (sin*^ -cos*d)* « 1 -4cofi?d+4co8*^.
29. an* B -|- cos« ^ = sin* B -f cos* ^ — sin* B co# d.
80. (sin X — cos x) (sec x — esc x) = sec x esc x — 2.
jl^ tan X — cot X __ 2 _ -
tan X + cot X ~ esc* x
82. (a cos X — 6 sin x)* -f (a sin x -|- 6 cos ^)* = a? -|- 6*.
88. cofli* ^ 4- (sin <^ cos ^)* -|- (sin (j> sin B)^ = 1,
84. tana + tan /9 — tan a tan /9 (cot a + cot /9).
139. Functions of any Angle in Terms of Functions of an Acute
Angle. — It is possible to express in a simple manner any function
of any angle in terms of a function of an acute angle. Therefore "
a table of values of the functions of angles from 0® to 90° will serve
for all angles. In fact, in view of (128), a table of functions from
0® to 45® would be sufficient, though not convenient.
1. Any angle, positive or negative, can be brought into the first
quadrant by adding to it, or subtracting from it, an integral mul-
tiple of 90®.
Thus: 760® - 8 X 90® = 40®; - 470® + 6 X 90® = 70®.
2. When an angle is changed by an integral multiple of 90®,
say n X 90®, the new terminal line lies in the same line as the origi-
nal terminal line when n is even; at right angles to it when n is odd.
3. Two angles which differ by an even mvUiple of 90® will be called
symmetrical with respect to the initial line, or simply symmetrical;
two angles which differ by an odd multiple of 90®, skevysymmetrical.
139]
TRIGONOMETRIC FUNCTIONS
109
4. When two angles are symmetrical, any function of the one is
numerically equal to the same function of the other.
From figure (a), sinx = — sinx' = sinx", etc., for the other
functions.
Angles «' and a?" are symmefrical wi^
respect to angle x
Angles x' and x" are skew-symmetrical
with respect to x
When two angles are skew-symmetrical, any function of the one
is numerically equal to the co-function of the other.
From figure (b), sinx = — cosx' = cosx", etc., for the other
fimctions.
Exercise 1. From figures (a) and (b), write down all the functions of x
in terms of functions of x' and of x".
Exercise 2. Draw figures corresponding to figures (a) and (6), when x lies
in each of the other quadrants. Then proceed as in exercise 1.
5. Rule*: Any function of any angle x is numerically equal to
,, ( same function - . , j • • r j t ^ ^^^W' 7
thel ^ .. of X increased or diminished by any < , , mul-
( co-function ( odd
tiple of 90°.
As an equation,
, \ _ ^ ± /(a? ± n • 90°), n even;
/ W - j _^ co-f(x'± n . 90°), n odd.
The sign of the resuib must he determined by noting the quadrants
of X and x ± n • 90°.
/
\
no
TRIGONOMETRIC FUNCTIONS
[ 140, 141
When the new angUy x ±n* 90°, lies in the first quadrant, give to
the resuU the sign of the given function of x, f (x).
Examples.
1. sin 680** = sin (60** +7 X 90**) = - cosSO**.
Here we diminish the given angle by an odd multiple of 90**, hence change
to the co-function. Also sin 680** is negative, hence we use the minus sign.
2. tan (- 870**) = tan (30** - 10 X 90°) = + tan 30°.
3. sec 420° = sec (60** + 4 X 90?) «+ sec 60**.
140. Relations between the Functions of +x and —a?. — The
figure shows two cases^ x in the first quadrant and x in the second
quadrant. In either ease,
sinx = — sin (— x);
CSC X = — CSC (— x);
cosx = COS (— x);
secx = sec (— x);
tan X — — tan (— x) ;
cotx = — cot (— x).
Exercise. Show that these equations
are true when x lies in the third quadrant
or fourth quadrant.
Rule: The cosine or secant of any angle is equal to the cosine
or secant respectively of the negative angle; the remaining four func-
tions qfthe angle are equal to the negative of the corresponding functions
of the negative angle. Or,
f(x) =f(—gc) when f stands for cos. or sec.
f(x) = — /(—«) when f stands for sin., esc, tan., or cot.
141. Exercises. Express all the functions of the following
angles in terms of functions of acute angles:
1. 130**.
6.
359**.
9.
- 321**.
13.
- 1060®.
2* 165^.
6.
- 25**.
10.
742®.
14.
- 401®.
4. 340®.
7.
- 125®.
11.
-665®.
16.
525®.
8.
- 250®.
12.
1100®.
16.
- 101®.
Express all the functions of the following angles in terms of functions of
angles between 0® and 45®.
17. 75®. 19. 110®. 21. -335®. 28. 790®.
18. -80®. 20. 255®. 22. 600*^. 24. -510®.
142, 143]
TRIGONOMETRIC FUNCTIONS
111
Find the values of the functions of:
26. 120^.
29. - 30**.
88.
-240°.
26. 135**.
80. - 45**.
84.
315°.
27. 150^.
81. - 60**.
86.
600°.
28. 300°.
82. ' - 120**.
86.
-510°.
142. Versed Sine and Coversed Sine. — The expressions
1 — cos X and 1 — sin x occur often enough in the applications
of trigonometry to warrant the use of special symbols fof them.
These are
1 — cos 35 = versed sine of 05 = vers x;
1 — sin 05 = coversed sine of 05 = covers-or.
Their line values are (figure), versx = MN, covers a; = HK,
X being in the first quadrant.
Exercises. Find the values of the
versed sine and coversed sine of:
1.
3o^
7.
150°.
2.
45°.
8.
- 30°.
8.
60^
0.
-120°.
4.
9o^
10.
-225°.
6.
120**.
11.
-300°.
6.
l35^
12.
- 315°.
143. Radian Measure. — The
degree is an artificial unit for
the measurement of angles. In
France, where at the time of the
Revolution an attempt was made to put all measurements on the
basis of the decimal scale, the quadrant of the circle was divided
into 100 equal parts and the angle subtended at the center by one
part called a grade. Each grade was then subdivided into 100 equal
parts called minutes, and each minute into 100 seconds. The
degree and the grade are thus two arbitrary units for the measure-
ment of angles, and any number of such units might be chosen.
There is one unit which is naturally related to the circle, and
which is as commonly used in theory as the degree in practice.
It is the central angle subtended by an arc equal in length to the
radium of the circle, and is called a radian (figure, p. 112).
Since the circumference contains the radius 2gr times, the
entire central angle of 360^ contains 2t radians, i.e.,
^ IF radians = 360"".
\
112
TRIGONOMETRIC FUNCTIONS
[144
Hence,
gr radians
^ radians
180^;
90^•
the radian is implied.
7 radians = 45®: and so on.
In dealing with angles measured
in radians it is customary to omit
specifying the unit used; it is under-
stood that when no unit is indicated
Thus, 2gr = 360^ T = 180^
- = 60®, 2i = 2J radians, and so on.
o
Note. To get the true form of the graphs of the equations y ^mnx,
y = cos X, etc., take x in radians on thea^axis, thus: a; — 0.1, 0.2, 0.3, . . . , 1,
. . . and find the corresponding values of y\ use the same unit of length for
both X and y. See graphs on p. 105.
144. Radians into degrees, and conversely.
Since 2gr (radians) = 360°,
360"^ 180° 180°
therefore, 1 radian =
also.
1 degree =
2gr
2x
3.1416-
TT
= 57°.39+ ;
gg^ (radians) = — (radians)
(radians) = .017+ (radians).
57.29+
Rule: To convert radians into degrees, mvUiply the number of
180
radians by or 57.29+.
hf
To convert degrees into radians, multiply the number of degrees
T 1
or
or .017+.
180 57.29+
By taking a sufficiently accurate value of x, we find,
1 radian = 57°.2957795 = 3437'.74677 = 206264".8.
1° = .0174533 radians.
1' = .0002909 radians (point, 3 ciphers, 3, approx.).
1"= .0000048 radians (point, 5 ciphers, 5, approx.).
. \
145,146] TRIGONOMETRIC FUNCTIONS . 113
«
The measure of an angle in radians is often called the circular
measure of the angle.
145. Exercises. Reduce to degrees, minutes and seconds the
angles whose circular measures are:
^ T 3t 5t 5t 7r
^ s' T' T' T' T'
2 12^^7
2. 1, 2, 2» 3» 4
, . T , 1 2T-f3 V'- H
S- ^ix, -li. . , ., 2 . 3 Q
^ 1 . T 1 1 2 . A^' p fy.
6.
2 T + l
t2 + 1 1-TT-l
Reduce the following angles to circular meafiure:
6. 30*», 120**, 150°, 225**, -60°.
7. 375°, - 22i°, 187°.5, 106°, 93° 45'.
8. 85°, 191° 15', 5° 37' 30", 90° 37' 30".
9. 10', 10", 0".l, 12° 5' 4", 21° 36' 8".l.
10. If the radius of the earth be taken as 3960 miles, find the number of
feet in an arc of 1" of the meridian.
11. How many radians in a central angle subtended by an arc 75 ft.
long, the radius of the circle being 50 ft.?
12. How many radians in the central angle subtended by the side of a
regular inscribed decagon?
18. A wheel makes 1000 revolutions a minute. Find its angular velocity
in radians per second.
14. If the angular velocity of a wheel is 10 t radians per second, how many
revolutions per minute does it make?
146. Angles Corresponding to a Given Function. — Let n denote
an integer, positive or negative, or zero; then 2 n is always even,
and 2 n + 1 odd; hence the angle
2 nw has the terminal line OX
(figure) coincident with the initial
line, and angle (2 n + 1) t has the x' (fn-hi)w
terminal line OX'.
Suppose now we wish to write
down all angles x such that
siiTx = i. Corresponding to a given function, there are always
(except when the angle is a multiple of 90'') two angles less than
360®; in this case they are
30** and ir - 30^
znw
114
TRIGONOMETRIC FUNCTIONS
[146
All angles with the same terminal line as either one of these will
have the same functions; all such angles are
,2ngr + 30*' and 2ngr+(ir - 30°)
••'' =(2n + l)gr-30^ '
i
[SO*
j[* (tnH}w
znw
^ Hence all solutions of the equation
sin a; = i are given by
X = 2 nir + 30° or (2 n + 1) t - 30°.
In general, if denote the smallest positive angle whose sine
is a, then all solutions of the equation
(1) sin X = a are x ^2nv + and (2 n + 1) ir — ff.
Hence also, if denote the smallest positive angle whose
cosecant is a, the solutions of the equation
(2) CSC X = a are x = 2 nx + and (2 n + 1) x — tf .
Consider next the equation
cos X = J.
The two simplest solutions are
a: =4-60° and rr=-60°.
All possible solutions are given by
x = 2n^60° and g = 2 nir ~ 60°,
or ''^ ' ^i = 2nir±60°.
In general, if be the smallest positive
angle whose cosine is a, all solutions of the
equation
(3) cos rr = a are x = 2 nv ± d.
Hence also, if be the smallest angle whose secant is a, all
solutions of the equation
(4) sec X = a are x = 2njr ±0.
Finally consider the equation
tan X = 1.
The two simplest solutions are
X = 45° and x = t + 45°,
/'
\
147]
TRIGONOMETRIC FUNCTIONS
115
and all possible solutions are
x = 2ngr + 45° and x = 27iir + (gr + 45'^),
the second set being the same asx = (2n + l)ir + 45®.
Both sets are contained in the single
equation
X — nir + 45\
the first set being obtained when n is
even, the second set when n is odd.
In general, if d be the smallest posi-
tive angle whose tangent is a, all
solutions of the equation
(6) tan rr = a are x = nv + 0,^
Hence also, if be the smallest positive angle whose cotangent
is a, all solutions of the equation
(6) cot X = a are rr = rnr + ^.
Summary of equations (1) to (6).
Let d denote the smallest positive angle having a given function
equal to a given number a. Then all solutions of the equation
I.
IL
III.
sin a; = a
CSC a; = a
are a5 = 2mr + 6 and (3n + l)'n' — 6;
cos a?
sec a;
tanx
cot a;
= a
= a
= a
= a
are a? = 2nir ± 6;
are « = mr + 6.
The angle is usually called the principal value of x.
The solutions of these equations may also be written by the
following simple rule.
Rule: Corresponding to a given value of a function, 'there are
in general two and only two positive angles less than 360°. If
these be denoted by Xi and X2, then all possible angles are given
by xi ± 2 nw and X2 ± 2 nx.
In exceptional cases there may be only one angle < 360®, as
when sin X = 1 or cos x = — 1.
147. Use of Tables of Natural Functions. — Usually the angles
corresponding to a given value of a function are not known
exactly. The angles may then be found approximately by the
J
116
TRIGONOMETRIC FUNCTIONS
[148
aid of tables of the natural functions, such as are given in (126)
and in Appendix, Table III.
These tables give the functions of angles from 0® to 90°. But
they will serve for all four quadrants, since any function of any
angle is reducible to a function of an acute angle.
When the given value of the function is not found exactly in the
table, the corresponding angle must be obtained by interpolation.
Example 1. Given sin x » —« • To find x.
The two values, xi and 23, < 360^, are shown in the figure. They are
easily found when X9, the angle whose sine is + ^ ' is known. For
xi = T -|- X3 and X2 = 2 x — a?8.
1
Since sin xs =» 5 =» .333, we find by interpolation from Table III, xz «
19** 28'. Hence, xi = 199^ 28', xz = 340^ 32'.
All possible values of x are then given by
199^ 28' ± 2nT, 340^ 32' ± 2nT.
2
Example 2. Given cot 5 x = 3.362. To find x.
From Table III, ^ x = 16* 34' or 196* 34' ( = 180* -f 16* 34').
2
Hence all possible values of x x are given by
|x = 16* 34' ± 2nT or 196* 34' ± 2 nx.
Therefore, x = 24* 51' ± 3 nx or 294* 51' ± 3 nx.
We might also write, from III of (146),
|x = 16* 34' 4- nx; hence x = 24* 51' -h^nr,
148. Exercises. Find all values of the angles which satisfy
the following equations: ^^■ "^ "^
1. cotx = 1; sinx = — J; secx = 2; cosx = 1.
2. CSC X » —^'t tanx = VS; cosx = .5; cotx = —VS.
3. sin X = — §; secx = —3; tanx = 2; cscx = 5.
4. cosx = —.257; cotx = —.998; sinx = .020.
6. tan^ - 2.500; csc^ = -3:505; sec ^ = -10.
6. vers^ = 1.450; vers^ = .605; covers^ = .750.
149]
TRIGONOMETRIC FUNCTIONS
117
149. Given one function of an angle, to find the otlier functions.
Example 1. sinx — - • Find the
other functions. ^
Take ordinate = 1 and distance — 2;
then abscissa « ± V^ (figure).
Then
cosx = db y^t tan x =» ± — jn
2 V3
cotx = ±V3.
sec X — ± — 7^ , CSC X » 2.
V3
We have found two values for each function except esc x, which is the
reciprocal of the given function. Similar results will be found in general.
Example 2.
tan X
-!(-
-3
+4
or
s)
The two possible positions of the ter-
X minal line are shown in the figure.
3 4-
Hence, sinx = ± -» cosx - ± k»
o o
* 4 ,5 .5
cot a; = — o' CSC x = ± 51 sec x « ± ^*
Example 3.
2 / +2 -2\
Then (figure),
3 2
sin 05 = ± — F^> cos x = ± — 7=>
Vl3 V3
3
tanx = -,
2
cscx = ±^, secx.±:^.
Example 4. sin x = - •
Ordinate = ^; distance = k;
hence abscissa = ± VP"—^.
118 TRIGONOMETRIC FUNCTIONS [150,151
Then cos a; = ± -'^^ 1 tan x — ± —. — > etc.
A; VA;2 - A2
Exercise 1. Construct figures for the cases when i is (a) plus; (b) minus.
Exercise 2. Is the problem possible for all values of h and A;?
a — 6 / — (a — 6)\
Example 5. tan x
/ -(a>-6) \
2VS
Here ordinate = a — 6, abscissa « 2 Va6;
or, ordinate = — (a — 6), abscissa = — 2 V^S.
In either case, distance = -|- ^J{a — 6)2-f-4a6 =|a-f6|.
TT . , a - 6 ,2 VoS^ X
Hence, smx = ± , — r^-,' cos x = ± , — p-ri' ®^'
Exercise 1. Calculate the values of the six functions when a « 2, & = 3;
when a= — 2, 6= — 3; when a = 1, 6 = 4; a = — 1, 6 = — 4.
Exercise 2. Is the problem possible for all values of a and 6?
160. Exercises. Find the other functions, given that
11. CSC ^ = •
n
12. tan e ^ a,
13. sin <f> ^ h.
14. cot^ = Vc.
16. 8ec<^=^^.
16. State for what values of the literal quantities in exercises 10-15, the
given equations are impossible.
151. To express all the functions in terms of one of them.
1. Express all the functions in terms of the cosine.
We have
cos X abscissa
cos X = — :; — = tt-t '
1 distance
Hence let abscissa = cos x and distance = 1.
Then ordinate = ± Vdist.^ — absc.^ = ± Vl — cos^ x.
1.
sin X = — J.
6.
cscx = — is.
2.
cosx = i.
7.
secx — — IJ.
3.
tanx = i.
8.
cot X =s — .75
4.
sec a: = 4.
9.
sinx » .6.
6.
cota; = Vs.
10.
, 6
cos tf — - •
c
151]
TRIGONOMETRIC FUNCTIONS
119
The figure shows this graphically when cos x is positive.
Taking into account both values of the ordinate, we have
sin X = ± Vl — cos^ x;
, Vl — cos^x
tanx= ± — — — ;
cotx=±
esc X = ±
CQSX
cosx
Vl — cos^x'
1
Vl — cos^x'
secx=
cosx
Exercise L Obtain these equations for the case when cos x is negative.
Exercise 2. Obtain the same equations directly from the formulas of
Group A.
2. Express all the functions in terms of the cotangent.
cotx =
cotx — cotx abcsissa
-1
ordinate
Hence let abscissa « cot x and ordinate = 1 ;
or let
abscissa = — cot x and ordinate = — 1.
In either case, distance = + Vl + cot^ x. (See figure, where we
assume cot x > 0.)
Hence sin x =» ±
1
Vl+C0t2x
COS X == ±
cotx
vr+coFx
etc.
120
TRIGONOMETRIC FUNCTIONS
[151
By taking each of the functions in turn, and proceeding as
above, we obtain the results shown in the following table. The
given function and its reciprocal are uniquely determined; the
other four functions are ambiguous in sign.
BIDZ
cos X
tanx
cotx
eecx
C8CX
81DX.
± Vl— sin'ic
sin re
± Vl — sin2 X
±^l—sirfix
sinx
1
sin re
C08X.
±'sJl—co^x
± Vl— cos^a;
oosx
cosx
± Vl— COfl^ x
006X
±Vl— co^x
tanx.
tanx
±VlH-tan2x
1
iVi+tSSi
tanx X
±Vl+tan2:f
lb VlH- tannic
tanx
cotx.
±Vl+cot^x
cotx
±vr+coPx
1
cotx
± VI +C0t2 X
cotx
±Vl+COt2x
iVsec'x-
-1
secx
1
secx
iVsec^x-
1
-1
±Vsec*«-
-1
secx
±Vsec?x— 1
cscx.
cscx
y/cac^x —
1
cacx
1
iVcsc^x-
-1
iVcsc^x-
cscx
-1
iVcsc^a?"
-1
• •
Exercises.
1. Express sin x cos^ x 4* sin' ^ in terms of tan x.
2. Express tan x sec x 4* sec^ x in terms of sin x.
8. Express coet^ x tan x + sin^ x cot x in terms of esc x.
1 1
4. Express
1 ^-sind'^l -sintf
in terms of sec $,
- -, costf , smtf . , .
6. Express z 1 — r + t rs ^^ terms of cos 6,
^ 1 — tan $ 1 — cot d
1
CHAPTER VIII
Functions of Several Angles
162. Fonnulas for sin (a? + y) and cos (a? + y). — Let x and y
be two angles, each of which we first assume to be less than 90*^.
Their sum will then fall in the first or the second quadrant. The
two cases are illustrated in the figures, and the demonstration
which follows applies to either figure.
Construct Z XOP = x and Z POQ = y, the terminal side of
X being taken as the initial side of y.
Mk
A^
From Q, any point on the terminal side of y, draw perpendicu-
lars NQ and PQ to the sides of angle x, produced if necessary.
Draw MP±OX and KP±NQ.
Then Z KQP = a^ and in either figure,
L/ i< -"y
sin (.x + y) =
NQ
OQ
MP + KQ MP KQ
OQ
OQ ' OQ
Hence
MP OP KQ PQ.
OP OQ ^ PQ OQ'
•
sdn (a? + y) '— sin a? cos y + cos a? sin y.
121
\
\
c^
122 FUNCTIONS OF SEVERAL ANGLES -, 1163
Also, noting that Oli in the second figure is a negative line,
( J. \-9K- OM-NM _ OM KP
cos{x + y)-^- ^ -~OQ~OQ
^OM gP_KP PQ
OP'OQ PQ'OQ'
Hence
(b) cos (« + y) = cos X cos y — sin ^ sin y.
163. In the above proofs we have assumed x and y less than
90*^. Similar proofs may be given for any other values of x and y.
We shall however use formulas (a) and (b) to verify the truth
of the formulas
(aO sin (A + B) = sin A cos B + cos A sin JB,
(b') cos (A + B) =^ cos A cos JB — sin A sin B,
for all values of A and B.
A and B will differ from acute angles by certain integral multi-
ples of 90°, say,
A = X + n • QO*"; B = y + w • 90°.
All possible quadrants for A and B (except the first, for which the
formulas have been derived) will be included by considering only
the values 1, 2, 3 forn and m.
In particular, let n = 1 and w = 2. Then
A=x + 90°; JB = y + 180°; A+B^x + y + 270°.
Hence, if formulas (a') and (b') are true, '
sin (x + y + 270°) = sin (x + 90°) cos (y + 180°)
+ cos (x + 90°) sin (y + 180°),
cos (x + y + 270°) = cos (x + 90°) cos (y + 180°)
- sin (x + 90°) sin (y + 180°).
Removing the multiples of 90° by the rule of (139) and changing
signs, these equations rec^uce to
cos (x + y) = cos xcosy — ^nx sin y,
sin (x + y) = sin xcosy + cos x sin y.
o 1l
154,155] FUNCTIONS OF SEVERAL ANGLES 123
But these are true since x and y are acute angles; hence also (aO
and (b') are true. In exactly the same way the truth of these
equations may be shown for any integral values of n and m,
positive or negative.
Using the letters x and y in place of A and B^ formulas (a) and
(b) are true for all values of x and y,
154. Replacing y by — j/ in (a) and (b), and noting that
sin (— j^) = — sin 2/ and cos (—2/)= cosy, we have
(c) sin (a; — y)= sinoseosy — cos^sinf^;
(d) cos (05 — y)= cos 05 cos f^ + sin a; sin If .
Equations (a), (b), (c), (d) are usually called the addition and
subtraction formulas of trigonometry. All the other working
formulas are deduced from them.
166. Dividing (a) by (b), we have
, , , V sin (x + 2/) sin x cos y + cos x sin y
tan \X '\- y) = 7 ; r = ; ; •
cos(x + 2/) cosxcosy — smxsmy
sin X cos y . cos x sin y
_ cos X cos y cos x cos y
^ __ sm X sm y
cos X COS y
Hence,
/ \ 4^ r \ \ tanaj + tany
(e) tan (a? + y) = r r— ^•
^ ^ ^ ^ 1 — tan a? tan y
Similarly,
/r\ i. / I \ cot a; cot y — 1
(f) cot (a? + y) = — : — r^-: — •
^ ^ ^ ^ cot a? + cot y
Also, from (c) and (d), by divisi9n,
/ \ 4, / \ tana;--tany
(g) *«°("-y)= i + tana.tany -
(h) cot(a.-y)=52*f£2tl^.
^ ^ ^ ^ cot y — cot a?
Exercises.
1. If sin X s i and sin y » }, calculate sin (x + 2/).
(Four answer: \[±^Jb± 4^^].)
H
124 FUNCTIONS OF SEVERAL ANGLES [156
2. If co8« =» i and cos j/ = J}, calculate cos (a; + y).
^ 8. If sin a » I and sin/3 » {, calculate cos (a — fi).
Show that,
4. cos (60** + x) + cos (60° - x) = cosa;.
5. sin (45** -{- 0) - sin (45** - tf) = V2 sin ^.
« X..IX J cos(tf— 0)
6. cot tf + tan^ = -r-^r ^•
sin cos 9
7. cos (A + 45**) + sin (A - 45**) = 0.
^ 8. sin ntf cos ^ + cos nd sin 6 = sin^(n + 1) 0.
/9. tan/^- j] + cot[d+^] =0.
10; From the functions of 30** and 45** p^lculate the functions of 75°.
For convenience we collect formulas (a), (b) . . . , (h) and form
Group B, numbering them consecutively with the formulas of
Group A. ,
Fonnulas, Group B.
(9)', sin(a5 + y)= sin a? cos f^ + cos a? son y.
(10) cos (05 + y)= cosajcosy — sina;sinf^«
(11) sin(a5 — y)= sinajcosy — cosajsiny.
(12) cos (05 — y)= cos 05 cos |^ + sin 05 sin If.
/io\ A / I \ tan 05 + tan y
(13) tan (o5 + y) = :; — -— ^•
. ' V ^'z 1 — tano5tany
cot 05 cot y — 1
(14) cot (05 + y) =
cot 05 + cot y
/^e\ 4. / \ tan 05 — tany
(15) tan (o5 - y) = 7-7-7 7— ^•
^ ^ V ^'z i4-tano5tany
/ic\ i./ \ cot 05 coty + l
(16) C0t(a>-y)=. ^ty_^t^ '
166. Functions of 3 05. — Putting y = a; in (9), (10), and (13) of
Group B, we have
(14) sin 2 05 = 2 sin 05 cos iv,
(15) cos 2 05 = cos^ a^ — sin* x,
= 1 — 2 sin' 05,
— 2 cos' ao— Ir
/^/?\ X « 2 tan 05
(16) tan2o5= - — - — =- •
^ ' 1 — tan' 05
For cot 2 a: use t — ?r- *
tan 2 X
157]
FUNCTIONS OF SEVERAL ANGLES
125
Exercises.
1. Verify these formulas when x is 30®; 45°; 150**; —60®.
Show that,
2^ 2 CSC 2 a; = sec x esc x.
S. cos 2 X —
sin 2x
1 +COS 2a?
1 + tan2 X
tan X.
^ 6. tan x + cotx »2csc2x.
6. Calculate the functions of 2 x when sin x —\\.
Ana, sin 2x = ±\\%\ co8 2x = \\% ; etc.
7. Calculate the functions of 2 x when tan x » f .
157. Functions of | a?. — The second and third values of cos 2 x
in (15) are
cos 2a; = 1— 2sin2a:,
cos2x = 2cos2x — 1.
Solving these for sin x and cos x respectively, we have
L 4 /l + cos 2 X
- ±y 2
sin a;
-±/
— cos 2 X
cos a:
Replacing x by i x, these become
(17)
sin|x
— cos a;
(18)
I
Dividing (17) by (18
cos I a;
1+cosa?
/
1 — COS 05 1 — COS 05
(19) ■ tanja; = ±.^ . .
» ^v 1 + COS a?
Formulas, Group C.
(14) sin 2a; = 2 sin a; COS oc. (17) sin |a;
(15) COS 2 05 = cos* a? — sin* 05
= 1 — 2 sin* Qc
= 2 cos* 05 — 1,
=±v/
(19) tan|a5 = ±y/i
(18) cos|a?
(16) tan2a; =
2 tan 05.
1 — tan* 05
— cos o&
1 + cos QO
— cos 05
+ COS 05
1 — cos a^
sin 05
sin 05
1 + COS 05
126 FUNCTIONS OF SEVERAL ANGLES [158
Exercises.
1. Calculate the functions of 15° from those of 30°.
t 2. Calculate the functions of 22}° from those of 45°.
^ 8. Calculate the functions of 7}°.
4. Calculate the values of tan (2 x — y), when sin x = J and cos y ^ }}•
Show that,
6. 8in4x » 28in2xcos22;. ^^ 14-sec0 „ « i /.
12. — ■ — T— = 2 cos2 J 0.
« • sec (?
^ o - 2 — sec^a;
0. cosji- ^^2^ 13^ sin/JcotJ/3 = 1 +C08/J.
^ cos2x ^ l-tana? 14. 1 + tan /? tan J /9 - sec /9.
l+sin2x- l+tanx' 8
cot2^ — 1
8. cos* — sin* B cos 2 0. 15. cot /9 =
2 cots
9 co6» g - 8in3 ^ 2+8in2g 2
cos tf — sin ^ 2 * ^
. l+tan^
10. coti + cscx = cot Ja;. tfl. _JE2!£__- =s ^
l-sm/J ^
11. (8in}^ + co8}^)2= 1 +sin^. ^■"^*'^2
^ ■
168. Formulas for sin «« ± sin v and for cos u ± cos v. — For-
mulas (9) and (11) of Group B are
sin (x + y) = sin z cos y + cos x sin y,
sin (x — 2^) = sin x cos 2/ — cos x sin j/.
Adding: sin {x + y) + sin (a: -^ 2/) = 2 sin x cos j^.
Subtracting: sin (x + y) — sin (x — 2/) = 2 cos x sin y.
,^ Let X + 2/ = w, and x — y =^ v;
^, lA + V J U — V
then X = — rt — and 2/ = — 9 — '
Substituting in the two preceding equations, we have
(20) sm w + sm V = 2 sm — - — cos — r —
(21) sm w — sm t? = 2 cos — - — sm -'— —
\
158] FUNCTIONS OF SEVERAL ANGLES 127
Proceeding similarly with formulas (10) and (12) of Group B,
we obtain,
(22) cos u + cos V = 3 cos — — cos — - — >
(23) cos u — cos V — — 3 sm — - — sm — - —
The last four equations, called the addition theorems of trigo-
nometry, we collect as the
Formulas, Group D.
(20) smti + sm V = 2 sm — r — cos _ ' ■
(21) sm u — sm t^ = 3 cos —-5 — sm — ~ — •
(22) cosfe + cost^= 2 cos — T — cos — :7—
(23) cos u — cos t> = — 2 sm — - — sm — - — •
X + y
Example 1. Show that "l^^-^^l^^ - ^ ■
ainx — amy . X — y
" tan — =-^
, . 2 8in— ^cos— ^
sm X + sin y 2 2
sin X — sin t/ <, x+i/.x — 1/
'^ 2 cos ^^^^-5^ sin ^^^r-^
tan54i?
tan?4i^cot^^ 2_.
Ef ? o au , , , cos 75 + COS 15® /t:
ExampU 2. Show that ^yg.I^ig, - - V3.
cos 75° + C08 15° 2 cos 45° cos 30° ,,„ ,„„. »;
cos76° -cosl6° = - 2 sin 45° sin 30° --~**^ oot30°--^^.
128 FUNCTIONS OF SEVERAL ANGLES [169
Exercises. Show that:
1. sm3a; + 8in5x=>28in4x cos x.
2. 8inlO0 + sin60 «2sin80co829.
5. cos 2 X + cos 4 a; » 2 cos 3 x cos x.
4. sin7a — 8m5a»2 cos 6 a sin a.
6. cos4(? — cos60 = 2sin50sin0.
3 X x
8. cos a; + cos 2 a? = 2 cos -^ cos 2 •
7. sin 30° 4- sin 60** = ^/2cosl5^
8. sin 70** - sin 10** = cos 40^
9. sin5a;cos3x — Hsin 8x +8in2a;).
10. 2 cos 10** sin 50** = sin 60** 4- sin 40^
4- sinA4-»nB . A •}- B
1^ T"! s = tan — J5
cos A + cos5 2
^« sin + sin 3 . ^^
12. — T-i 5-s = tan 2 0.
cos + cos 3
18. 2 cos OS cos ^ = cos (a — /5) + cos (a + ^).
14. sin4(?sin0 = } (cos3 — cos5 0).
16. cos Sx — cos 4 X B ~ 4 sin 2 x sin 3 a; cos 3 x,
16. an(2a; +3j/) + sin(2x — 3y)= 2sin2a;cos3y.
169. Exercises involving the use of formulas (1) to (23).
1. If sin X » I and sin y = J, find the value of sin (x -f y) and cos (x + j/)
when X and y are both in the first quadrant.
2. As in exercise 1, when x and y are both in the second quadrant.
8. If cos x = i and cos y = If, calculate sin (x + y) and cos (a; + y) when
X and y are both in the first quadrant.
4. As in exercise 3, when x and y are both in the fourth quadrant.
6. If sin X » i and sin y » }, calculate all values of sin (x± y),
6. If sin a = } and sin ^ » j, calculate all values of cos (a i: $).
7. If cos a » } and cos /3 » {, calculate all values of tan (a ± fi).
8. Calculate sin (x + y + 2) when anx = fV, sin y = ^, sin 2 = ^, and
X, y, 2 all lie in the first quadrant.
9. As in exercise 8, when x, y, z all lie in the second quadrant.
10. Calculate cos (x-^-y -{- z) when cos a; = |, cos y = Hi cos « « Jf , and
X, y, z all lie in the first quadrant.
11. As in exercise 10, when x, y, z all lie in the fourth quadrant.
12. Calculate tan (x -f- y) when tan x = 1 and cot y = VS.
13. Calculate all values of sin 2 (x — y) and of tan (2 x — y) when tan x » }
and tany = t^.
14. Calculate all values of cos (a + fi) when tan a ^ m and tan ^ » n.
15. Calculate cot (a — fi) when tan a = a + 1 and tan /5 =« a — 1.
X 1
16. Calculate tan (a + fi) when tan a = — r— r and tan fi =
x + l '^ 2x + l
17. If tan a » f and tan fi ^ ^^ calculate tan (2 a + ^).
159] FUNCTIONS OF SEVERAL ANGLES 129
18. Calculate sin 75% cos 75^, and tan 75^, by use of the relation (a) 75®
= ^; (b) 75** = 135° - 60°.
19. Calculate the functions of 202J°; of 7i°.
Prove the following identities:
20. an X sin (y — 2) 4- sin y sin (2 — a;) -h sin z sin (x — y) = 0.
21. cos a? sin (j/ — 2) 4- cos y sin (2 — x)-\- co82 sin (x — y) == 0.
22. cos (x+ y) cos (x — y) -f sin (y+z) sin (2/ — 2) — cos (x+z) cos (x — 2) = 0.
23. cos (x — y + 2) = cos a; cos y cos 2 + cos x sin j/ sin 2
— sinx cosy sin2 +sinx siny co62.
24. sin3x = 3sinx — 4 sin' x. «- cos (a + /8) . x z»
AC o .. « o 30. — r— ^ ^ = cot a — tan fi.
25. cos 3 X = 4 cos' x — 3 cos x. sm a cos /?
«fl X o 3tanx — tan'x -,^ cos (a — /5) x ^, ■ *
26. tan 3 x = — :; ttl — ^ S*» ^ — ^-^ = cot ^ -h tan a.
1—3 tan^ X cos a sin /8
27. cot3x = ^» ^ ~ — ^^. 32. ^^'^ ^: =« tan a — tan /?.
3 cot2 X — 1 cos a cos /3
28. ^;jm4u_^ 4tang(l -tan^g) ^^ sin (x -h y) ^ tan x + tan y
1 — 6 tan2 ^ + tan^0 * sin (x — y) tan x — tan y
29. 52ifL±^ = tan« + tani8. 34. cos (a; + y) cot x - tan y ,
COS a COS /9 cos (x — y) cot X -h tan y
86. sin (e -h ^) sin (^ - ^) = co^ ^ - cos2 0.
36. cos (w -h v) cos (w — «) = cos^ w — sin* v.
37. sin (A - 45°) = -^ (sin A - cos A).
V2
40. tan
tang — 1
tan^-h 1
tand
41. tan (a 4-5) + tanf a — ^j =
tang
8 cot a
cot2 a - 3
5t 5t 49. V2sin(g4-45°) = sing + cosg.
sin^ cos^ /
42. ^ i^ = 2V3. 60. 8in2x= ^^^^ »
sini cos^ H-tan2x
12 12
h..)-
-^ ^ CSC* X
61. sec2x —
48. tan T+g = 7^ r- csc2x-2
tanfj-gj 62. cotg - c6t2g = csc2g.
44. cos(g+|]+sin(g-|) = 0.
46. cotfg+|Wtan^g-~j = 0.
46. cot[g-|) + tan[g + |j=0.
47. cot ^ — tan f = 2. /
48. 2cos| = V2-h V2.
63. sec2gco82g = 1 -tan2g.
64. 1 4- tan e tan 20 = sec 2 g.
66. 1 — cos 2 X "= tan x sin 2 x.
66.
sec2g = — 75-T — =-
cot2g — 1
67.
sin2g ^ ^
- , « _ = tan g.
1 4- cos 2 g
68.
sin 2 g
; J7-: = cot 0,
1 — COS 2 g
^
130 FUNCTIONS OF SEVERAL ANGLES [159
69. co<^^- 1 »2cotdcot2^. g, j 1 ~co82a;
ea 2-sec«^=-.Bec*dcos2^. ^"l + co82x*
^^ cos 2$ 1 — tan $ «. oos30,8in30 oxo/»
61. =-, . c-^-T-n — 2' W. . ^ H T- = 2cot2d.
1 + sin 2 1 + tan 9 sin cos 9
g^ cos3g o,w^o* 1 M tang + cotg \^^a
08. = 2 cos 2 X — 1. 00. — r-r r r » SeC J ^.
COS X cot 9 — tan
ee. tan(46''+^)-tan(45*-^)-2tan2^.
g_ cos* 4> 4- sin< ^ 2 — sin 2 ^
cos^ + sin^ 2
^-j cos* ^ — mn* 6 < , 1 . ft 1 • • o
68. 5 J— r * 1 + J8m2x — isin22a;.
cos ~ sin ^
^^ sin X + cos X ^ ^ ^
69. ' — : — »= tan 2 x — sec 2 x.
cos x — sin X
mix -ox o 4tan«x
70. sin 2 X tan 2 X
1 — tan* X
71. coej2 e + 8in2 cos 2 ^ » cos*^ + sin^ ^ cos 2 9.
72. 1 +cos2(d-.^)cos2t^ -cos2d + cos2(d -20).
• o A fve J. sin X + sin 2 X
B sin29. 75. tanx »
1 4-cosx + cos2x
o * o VA X sin 2 X — sin X
sec2x— tan2x. 76. tanx
1 — cos X + cos 2 X
•HP o« 1 X o« • o.> cot2(? 4- tan«^
77. sec2g-itan2gsin2g- ^^^^^^,^ »
sing + cosg ^ / l+8in2g . 34. l+sec0 ^^^^^
sin^-cos^ Vl-sin2^ sec 2
79. fsin| + cos|Y = l+sin^. ^5. sec2 1 - 2 tan | esc x.
(a fl\2 OA 1 + cos 3 .30
8in|-coe|)-l-8in«. ^- ein3<l> -<»t^-
81.
, l + tan| 87. ^ If ° f -ton 671'
cobO 2 cos 45
1 — sin^
l-tan| 88. — ^4-7— s = cot (^ + |Y
2 sectf + tan^ \4 2/
1 fa^ * -j^ 1 + sm X -h cos X , X
1 — tan = 89. ttH — : ' = cot ^
f _ „^^ ^ x„^ ^ 1 + sin X — cos X 2
. — ^— ^^— =s sec X — tan x.
-- sin 2 X
sin2x
l + t"*! 90. ten5 = i/pi^^
^ 2 V 2 Sin X + SI
XX
tan X — tan = » tan p: sec X. ^^ /= . „_„ ^-^ ys
2 2 91. v3 sin 75° - cos 75"* « V2.
92. sin -^ cos 5 — sin -rt" cos -^ + cos 4d sin 2^ = 0.
93. sin 4 X + sin 2 X » 2 sin 3 x cos x.
94. sin 3 X + sin 5 X == 8 sin X cos^ x cos 2 x.
V A
169] FUNCTIONS OF SEVERAL ANGLES 131
95. — 7-T^s 7 — TVS = — :=• 97. sin 100 — an 40 = an 20 .
cos a.
cot 15** - tan 15** ^J^
9B. ^-^»^'^^^^ - cot 60-. 98. cosf^+aUcosf^-a)
1-V2C08 75** \3 / V3 /
99. cosf| + aWcos[^-a)= V2
100. cos(^ + ^)- sin (e - <f>) = 28inf^ - ^Icos/^ -^j
lOL 28in fa + ^)sin(a--^J = sin^a — cos^a.
102. sin f ^ + a j — sin (j — a j = V2 sin a.
cos a.
108. sin 40* - sin 10° = ^^^^ cos 25*
sin 75- H- sin 15'
^
104. sm32+sinx = 4sina; cos^x. 105. . »-o — . i go = v3.
sm 75 — sin 15
106. «^^ + «^y --cot^-±i^eotg^y-
COS a; — COS y 2 2
107. ^^1;:+"°^: = l. m ^^^±^° WS tan 70».
COS 70 + COS 20 sin 100 — sm40
100 (sina + sin/g)(co8a 4-oos/g) _^^ a - fi
IW. -7-. ; ^TT rrr as — COt^ n •
(sma — 8m/3)(cosa — COS/8) 2
11A (sing + sin ff) (cos g -cos/g) ♦„„2«_±_^
IIU. 7-; ; -TT ; -r- = — tan* jr •
(sm a — sin fi)(coB g + cos /5) 2
--- (sin 75- + sin 15°) (cos 75° 4- cos 15°) _„
• (sin 75- -sin 15°) (cos 75- -cos 15°)"
1 12 COS 2x •{- COS 12 a; , cos 7x — cos 3 x , 2 sin 4 a; ^
cos 6x4- cos 8 X cos x — cos 3 a; sin 2 x
118. sinx + sin2a; + sin3a; = 4 cos } x cos a; sin J x,
{Hint. Replace sinx + 8in 3a; by 2 sin 2x cos x and sin 2x by 2 sinx cos x;
from these results factor out 2 cos x and combine the remainders by the for-
mula for sin u + sin v.)
114. cos x4-cos2x4-co8 3x =4cosJxcosxcos|x — 1.
115. sin 2 X + sin 4 X + sin 6 X » 4 cos x cos 2 x sin 3 x.
116. "°^ + "°o!1"°q! - t»n 29.
cosd + cos2d + cos3^
117. cos 20- + cos 100° + cos 140° = 0.
118. cos 6 + cos 3 ^ + cos 5 tf + cos 7 ^ = 4 cos cos 2 9 cos 4 0.
119. sin^4-8in3^ + sin5^ + sin7« = 16 sin ^ cos2 ^ cos2 2 d.
120. 4 sin2 <^ cos2 ^ + (co82 .^ - sin2 ^)2 = 1.
121. (cos X cos y + sin X sin y)2 -h (sin x cos j/ — cos x sin y)2 = 1.
^^^ tan 3 X — tan X «
122. . . ^ o X ==tan2x.
1 + tan 3 X tan x
^2j tan (n H- 1) ^ - tanng ^ ^^^
1 + tan (n + 1) tf tan wd
/
132 FUNCTIONS OF SEVERAL ANGLES [159
124. tan(g + <^)-tan0 _
125. taD(g-<^)+tan0 ^^^^
' 1 — tan (^ — ^) tan (f>
126. sin 71^ cos $ + cos n^ sin =^ sin (n + 1) 9.
127. 2 CSC 4 a; — 2 cot 4 x = cot x — tan x.
1 — COS X sin ^ cos tf
& V .1 . . /a + b . , /o — b _ 2 cos a;
a' va-ft- V a + 6""-y/cos2x
180. Iftana; = ^,showthati/^^+iy^
181. 4cos*xsin3x +4sin3xcos3x — 38in4a;.
182. sinS X + sinS (120*' + x) + sin^ (240** + x) = - } sin 3 «.
138. cos 6 X = 16 (cos« x — sin« x) — 15 cos 2 x.
184. 1 + tanfi x = sec* x (sec^ a; — 3 sin^ x),
^._ 3 sin x — sin 3a; ^ ,
186. s ; jr— = tan» x.
3 cos a; + cos 3 a?
186. sin 2 a; sin 2 y = sin^ (x -h y) — sin^ (x — y).
137. sin 5 a sin a = sin^ 3 a — sin^ 2 a.
188. cos^a = J (3 -|-4cos2a -|-cos4a).
189. cos2x + cos2y + cos22+cos2(x + y + 2)«4co8(x + y)co8(y + 2)
cos (2 4- a;).
140. sin^x -h sin* y -h sin* 2 + sin* (x + y + 2) = 2 — 2 cos (x+ y)coa(y + z)
cos (2 + x).
141. cos* X -h cos* y -f cos^2 + cos* (x + y — 2) = 2 -h 2 cos (x -f y) cos(x — z)
cos (y - 2).
142. sin (x — y — 2) — sin X — sm J/ — sm 2 = 4 sin — s-^sm ^ ^°^ — '
148. 8in2a + 8in2/5 4-sin27 = sin2(a +/J + 7)+4sin(a + /3)sinG8 + 'y)
sin (a + 7).
144. sin(a+/9-7)-|-sin(a-/5 + 7) + sin(/8-|-7-a)-sin (a + /9 +7)
= 4 sin a sin fi sin 7.
145. cos(a -h/5 — 7)-hcos08 + 7 — a) -h cos (os -h 7 — /3)— cos(a+/J + 7)
= 4 cos a cos /5 cos 7.
. 146. Show that the equation sin x « a 4* - is« impossible.
147. For what values of a will the equation 2 cos x » a + - give possible
values for x ?
148. Show that 2 sin 5 = — VT+sinx — VT^^inx, provided that x lies
in the second or third quadrant.
149. Show that 2 cos 5 = — Vl + sin x + Vl — sin x, provided that x lies
in the second or third quadrant.
160. When x lies in the fourth quadrant, show that
2 sin 2 = Vl "~ sin X — Vl + sin x.
^1
\ A-^^^_
T>!.
CHAPTER IX
sift X tCLTt OC
Batios AND • Inverse Functions. Trigonometric
X
X
Equations
160, The limits of the ratios and
Let a; = Z NOP
(figure) lie between 0° and 90°; let NP be a circular arc with center
at 0, and MP and NT _L ON. Then
MP<NP< NT;
MP NP NT
nence Qp < Qp < Qp »
or sin a: < a; (radians) < tan x.
That is, the radian measure of any ^ \
acute angle lies between the sine and the tangent of the angle.
From the last inequality we have, on dividing by sin a:, \
X
1 <
sma;
< sec X,
Suppose X to decrease and approach 0. Then sec a; = 1, and con-
X
sequently also -: = 1
sm a;
Hence
and — 1.
X
„ smag ^
lim = 1.
05=0 ^
Dividing the third of the above inequalities by tan x, we have
X
cos X < ^
tana;
letting X approach zero we have
<i;
-. tanx
lim = 1
09
X
Hence, the raiio of either the sine or the tangent to the angle {in
/adians) approaches 1 as its limit when the angle approaches zero.
133
134
INVERSE FUNCTIONS
[161
When angle x is small, these ratios will be nearly equal, to 1 ;
that isy
sin a; - , J tanx - ,
= 1 + e and = 1+^1,
X X '
where e and e\ are small quantities. Hence
sin X = X + ex and tan x = x + e\X,
Neglecting the small terms ex and eix, we have
sin X = tan x = x approximately, when x is small.
Hence when x is small, sin x and tan x are nearly equal to x (tn
radians).
The degree of this approximation is indicated by the following
values:
Angle X.
Degrees radians
1^ .0174532925+
1' .0002908882+
V .0000048481 +
sinx
0174524064+
0002908882+
tan X
.0174550649+
.0002908882+
.0000048481 +
Exercises.
1. How large may z be if the approximations
sin a; == a; and tan x ^ x
are to be correct to four places inclusive? (Table.)
2. In what decimal place is the error of the apjn'oximaiwM
sin 30° = 30 sin 1** and tan 30° = 30 tan 1°?
8. How large may n be if the approximations
sin n^ ^ n sin 1° and tan n'* ^ r^ tan 1*
are to be correct to three decimals inclusive ?
4. As in exercise 3, for the approximations
sin 7i' = n sin 1' and tan n' ^ n tan 1'.
161. Inverse Trigonometric Functions. — It is often convenient
to specify an angle, not by its degree or radian measure, but by
the value of one of its functions. Thus we may speak of 30*^ as
"an angle whose sine is i-*' There is of course an ambiguity
here, since 30° is only one of the angles whose sine is \,
^
■l
r.
161] INVERSE FUNCTIONS 135
If X is an angle whose sine is a, we write c . r C .
X = sin-1 a, ^*^c - ^ r^ ";<v^ j
which may be read "a; equals an angle whose sine is a," or *^x
equals the inverse sine of a," or "x equals anti-sine a."
Similarly the equation ^^ ^^ ^CJl_ ^ 7*^'^x
is read ^'x equals an angle whose tangent is a/' or "x equals the
inverse tangent of o," or "a; equals anti-tangent a," and so on for
the other fimctions.
Obviously the equations , I <( \.. ; <
X = sin""^o and sinx = a ^ ,* .;
« *
are equivalent. Similarly for
a; = tan-^ a and tan x = a,
X = sec~*a and sec a; ~ a,
and so on.
It should be noted that ""^" in sin-* a is not an exponent; it
might equally well have been written as a subscript, sin-ix, or
in any other convenient way. The reason for writing it as above
will appear by noting that, according to the laws of exponents,
the algebraic equations
X = b-^a and bx = a
are equivalent.
When it is necessary to write sin x with an exponent — 1, it
should be written (sin x) ~*, noi sin-* x.
The smallest positive angle whose sine is a is often called the
principal value of the symbol sin-* a. Similarly for the other
functions.
If B denote the principal value of any inverse function, we have
from (146), equations I, II, III,
sin- * a = 2 nx + ^, or esc- * a = 2 nx + ^, or
(2n + l)7r- 6] (2n + l)ir-^;
cos-*a = 2nT ± ^; sec-*a = 2nir ± ^;
tan-*o = nx + ^; cot-*a = n7r+ ^.
*.
136 INVERSE FUNCTIONS [162
162. Equations Involving Inverse Functions. — In this article
we shall restrict the sjmabol for the inverse functions to mean
only the principal value of the function. Thus, sin-^ J shall mean
the angle 30° only, tan - ^ 1 = 45°, and so on.
Example 1. Show that sin-i^ — ^^Qg-i
3 4
Let X = sin-i^ and y = cos-i^J
to prove that x = y,
or that sin x = siny.
(We use the sine for convenience; any other function might be used.)
Now sin a? = - since x = sin- 1 ■= •
5 o
4 3
Also cosy = ■=; hence sin y = yl — cosy2 = - . q.e.d.
.V o o
4
Bxampie 2. Show that 2 tan-i 2 = sin- 1 -z •
o
4
Let X = tan- 1 2 and y = sin- ^-z]
o
to prove that 2 a? = y,
or that sin 2 X = sin y.
Now sin2a;=28inxcosx.
2 1
But tanx = 2; hence sina;=— p and cosx = — -- (149.)
Therefore sin 2 x — ^ = sin y, q. e. d.
Observe that if x were not restricted to be the principal value of tan-i2,
we might have sin x = —'
2
Example 3. Show that tan-i - + tan-i 2 + tan-i 8 = «■.
o
2
Let X = tan-i^; y = tan-i2; z = tan-i8;
2
then tan x = » ; tan y = 2; tan 2 = 8.
o
To prove that a? -f y -j- « == ir,
or that a; -{- y = IT — 2,
or that tan (x + 2/) = tan (ir — 2) = — tan 2.
XT J. / I \ tan X + tan y | + 2 « , .
Now tan(x + »)=j— ^^^^^ = L^=-8--tane. q.e.d.
Example 4. Show that tan-i a = sin-i , when a > 0.
Vl +a2
Let X = tan- 1 a and y — sin- 1 :
then tan x = a and sin w = , •
163, 164] TRIGONOMETRIC EQUATIONS 137
To prove that ^ — Vt
or that sin a; = sin y.
Now since z and y stand for principal values, and a is positive, both angles
are^n the first quadrant.
Then from tan x =» a we find (149)
a
sinx = ,
which is sin y, q. e. d.
Discuss the above example when the symbol for the inverse
functions is assumed to stand for all angles having the function
in question, instead of the principal value only.
163. Exercises. «
1. Show that the equation in example 4 is not true for principal values
when a is negative. (Try a —— 1.)
Prove the following:
2. tan'ii + tan-ii=T. «• cos-i J + 2sin-i J = 120».
7^ 6 4 7. 2tan-i3 = 8in-if.
3. 2tan-iJ =tan-ij. o o- ,1 • , 23
4. tan-i3+^ = tan-i(-2). « o f lo if
4 9. 2cot-i2 = csc-i|.
6. tan-i^-fcsc-iVlO = |. 10. 4tan-ii = tan-i^k + Z*
IL tan-ij + tan-i|H-tan-if^) = ir.
4n i4,. ,8,. , 13 X
12. 8.n-i^ + 8.n-ij^ + s.n-igg = 2-
13. cos- 1 ^ -{- 2 tan- ^ ^ = sin- 1 ^•
14. 2 tan-i^ — C8c-i;r = 8in-lr^•
o o DO
16. sin-i a = cos-i Vl — «^» if o > 0.
16. 2tan-im = tan-i:; -*
1 — mr
17. 2tan-i(cos2^)=tan-i(^5^^-^*^Bi?j.
164. Trigonometric Equations. — A trigonometric equation is
an equation which involves one or more trigonometric functions of
one or more angles. Thus:
sin^ X + cos X = 1; tan ^ + sec ^ = 3; cot a esc a = 2.
To find the values of the angle which satisfy such an equation,
it is usually best to use a method adapted to the case in hand.
We give here one general rule, which covers a considerable variety
of cases.
138 TRIGONOMETRIC EQUATIONS [165
Rule: To solve a trigonometric equation, express all its terms
by means of a single function; solve as an algebraic equation, con-
sidering this function as unknown; find the angles corresponding
to the values of the function so obtained. Check all answers by
. substitviion.
Examples,
1. sin2 X -{- cos a? = 1.
Expressing all terms by means of cos x, we have
1 — cos2 X + cos a? = 1, or cos* x — cos x = 0.
cos x = 0, or cos x = 1.
Hence x may be any odd multiple of ? or any multiple of 2 «-; i. e., if n be
any integer or zero,
X = ± (2n + 1) 2 ^^ X = ± 2jjjr.
Exercise. Check these answers by substitution.
2. tan^+sec^ = 3.
Expressing all terms by means of tan 0, we have
tanS ± Vl+tan2d = 3, or ± Vl + tan^ = 3 - tan^.
Squaring and reducing,
4
tan ^ = |; hence 9 = 53** 8' ± rnr.
When n is odd, these values of do not satisfy the given equation. Hence the
solutions are
^ = 53** 8' ± 2 nx.
3. cot a CSC a —2.
Then ± cot a Vl + cot* a = 2, or cot* a + cot? a = 4.
Hence cot a = ± V- J ± J Vl7.
Using the upper sign under the radical (the lower sign makes a imaginary),
we have
cota = ± 1.2496+ ; hence a = ± 38*'40' ± nir.
When n is odd, the values of a must be discarded. Hence
a = ± 38** 40' ±2 TIT.
The reason for the additional values in the last two examples is that in
example 2 we really solved both the equations tan ± sec 9 = 3, and in exam-
ple 3, both the equations cot a esc a =± 2.
166. Examples Illustrating Special Methods. — These depend
chiefly on transforming the given equation by means of some
of the standard formulas.
165] TRIGONOMETRIC EQUATIONS 139
4. 2 sin^x — 3 sin x cos a? « 1.
Since 2 sin^x = 1 — cos 2 aj and 2 sin a; cos x = sin 2 x, we have
3 i 2
1 — co8 2x — s8in2x = 1, or tan'2x=— 5.
Hence 2 x = tan-if - 1^ = -33*' 41' ± nr.
X = - 16°50'.5 ±n^-
Exercise. Check these answers. Solve the given equation by expressing
cos X in terms of sin x.
6. sin 3 y — sin 2 y — 0.
By formula (21) of (168) this becomes
2 cos 2 y sm s^ =0.
5 1
Hence cos ^ y = or sin ^ y == 0.
5 r 1
^y '^ ± (?n -}- 1)^, or 2^ ^ ± nr.
!/ = ± (2n -f 1)|, or y=±2mr.
6. cos X H- cos 3 X + cos 5 x = 0.
Since cos x + cos 5 x = 2 cos 3 x cos 2 x, we have
2 cos 3 X cos 2 X + cos 3 x « 0, or cos 3 x (2 cos 2 x + 1) =0.
Hence cos 3 x = 0, or cos 2 x « — - •
3x = ±(2n + l)L or 2x-±y^±2nT
x = ± (2 n + 1) ^ or ± I ± nir.
7. tan 4 a tan 5 a = 1.
This may be written tan 4 a » cot 5 a. But when the tangent of an angle
A equab the cotangent of an angle BfA-i-B must be an odd multiple of ^
Hence 4a + 5a = ± (2n + D^
«= ±(2nH-l)j^.
Here a is any odd multiple of 10^.
dtherwise thus: tan 4 a — cot 5 a = 0; hence 3 : — = — = 0;
cos 4 a sm 5 a
sin 4 a sin 5 or — cos 4 a cos 5 a cos 9 a ^
or 1 1 — m == "• 3 : — m — ^ "•
cos 4 a sm 5 a cos 4 a sm 5 a
cos9a=0, or 9a=±(2n + l)|-
• ^ -
140 TRIGONOMETRIC EQUATIONS [165
Exercise 1. Check these answers. Draw figures for several values of a as
10", 30^ 50^ 70**. Discuss the case a = 90**.
Exercise 2. In example 7, in passing from the first equation to the second
we divide by tan 5 a, which is permissible only if tan 5 a ^ 0. Justify the
division.
Exercise 3. Justify the division by cos z in example 4.
8. a sin ^ + & cos ^ = c.
We might reduce to sin 9 or cos 9 and proceed according to the rule of (164).
A method much preferred in practice is as follows.
In place of a and h introduce two new constants m and M such that
(a=mcosM,^ , fm^Va^ + i
ir • %/ whence < ,, , ,b
(o—msuiM; /M = tan-i-»
^ a
The given equation then becomes
c *
m (sin d cos 3f + cos d sin M) = C or sin (d + ilf ) =» — •
Hence if we let sin-i a; represent all angles whose sine is x,
c c
d + M = sin- 1 — , or d « sin- 1 M,
m m
- c . 1 &
9 = sin-i —====: — tan- 1 - .
yja^ 4. 52 a
Graphic Solution. As an example, we take the equation
sin2^ + sin^ + ^ = 0.
We want the values of which reduce the expression sin 2 ^ +
sin^ + ^to zero. ^
Let y = sin 2 ^ + sin ^ + jr .
Calculate y for a series of values of ^, as ^ = 0°, 10°, 20°, . . . ,
and plot the points {By y) in rectangular coordinates. The restilt-
ing curve will show the approximate values of d for which y is zero.
Any convenient scales may be used on the axes of d and y.
Let the student read o£f the required solutions from the graph
below:^
166]
TRIGONOMETRIC EQUATIONS
141
Exercise. By means of this graph solve the equations
(a)
(b)
(c)
sin 20 + sin 9-0;
sin2d -f 8in9= 1;
sin29 + 8ind= J.
166. Exercises.
^ 2 sin2 X — Z cos x = 0.
2. 4 sin^ a + 1 — 8 cos a.
^. sin a + cos a — '^.
4. tan 9 + cot 9 — 2.
^6. tan/5 + 3cot/5 = 4.
6. 2sin2x + 3 =5sinx.
7. 2(1 — sin^) = cos9.
8. 5sin9 + 10cos9 = 11.
9. cos 2 a; — co# x.
10. 2cos2x = 1 H-2sina;.
11. 4cot29 = cot29-tan29.
12. cos B = sin 2 9.
13. tan 2 X » 3 tan x.
14. sin 2 y » cos 3 y.
15. tana = cot 3 a.
16. cot 8 — tan 0.
Solve the following equations:
17. sec 'px — CSC qx,
18. tan y — cot 6 y,
19. sin r$ — ein8$.
20. cot (30** - x) = tan (30* + 3 x).
21. 1 + tan/3
= tan(|-ffl).
22. sin 4 a = cos 5 a. _
23. sin(60*'-x)-sin(60*H-x)=iV3.
24. sin29 + sin49 =: >^cos9.
26. sin<30**+9)-cos(60**H-d)=-i\/3.
26. sin 4 a — cos 3 a + sin 2 a.
27. sin 3 /5 H- sin /5 = cos /8 — cos 3 /5.
28. sin X + sin 2 X + sin 3 X — 0.
29 sin X H- sin 3 X H- sin 5 X = 0.
80. cos X + cos 2 X = cos i x.
Solve some of the above equations graphically, in particular 1, 2, 4, 5, 7, S,
12, 13, 14, 16, 26, 28, 29.
L
142 TRIGONOMETRIC EQUATIONS [167
167. Simultaneous Trigonometric Equations. — We shall now
give some examples to illustrate methods for solving a system
of simultaneous trigonometric equations for several unknown
quantities. To express answers concisely, we shall now use the
symbols for the inverse functions to mean aU the angles deter-
mined by the given function.
Examples.
1. Solve for r and 0: r coe $ == x,
r sin d = y.
Squaring and adding, r^ — x^ + jfl;
hence r — ± \J3^ + j/2.
Divide the second equation by the first,
tan B — -: hence 6 = tan-i -.
x' X
SL Solve for a and /9:
a sin a + & sin /3 B c,
d sin a + c sin /8 = /.
Solve for sin a and sin /9 as unknowns; hence get a and /8.
Exercise. Carry out the solution of example 2. Is the solution possible
for all values of a, 6, . . . ,/? (62.)
8. Solve for r and B:
or sin 5 + frr cos ^ = c,
a'r sin ^ + h'r cos 6 = c'.
Solve for y sin 6 and y cos 6 as unknown; then proceed as in example 1.
Exercise. Carry out the solution in example 3.
4. Solve for X and ^:
• y — mix,
^ — sin 2 X.
Subtracting, sin 2 x — sin x » or 2 sin x cos x — sin x = 0.
Hence sin x =* or cos x = J.
X = ± mr or di 60** ± 2 mr.
y = or ± J V3.
Exercise. Solve example 4 graphically.
6. Solve for y and i: y = a sin (nf + &),
y ^ a! sin (n^ -j- 6').
Equating the values of y, and expanding,
a (sin rd cos 6 + cos rd sin 6) = a! (sin ni cos 6' -j- cos rd sin 6')-
Dividing by cos vd and solving for tan nt,
a' sin 6'— a sin 6,
tan rd = r ; n •
a cos — a cos o
This determines a set of values of rd. Then y is obtained by substituting
in either of the given equations.
168] TRIGONOMETRIC EQUATIONS 143
6. Solve for r, 5, and 0: x — r coe $ coa <ft,
y =' r cos sin (f>,
z — rsin 0,
Dividing the second equation by the first, we have
- = tan 0; hence <f> = tan-i -•
Squaring the first two equations and adding,
a;2 4- y2 « r2 coB^O; hence r cos d = ± V^J* + ^.
Combining this result with the third equation, as in example 1, we have
tan d = . ; hence — tan- 1 -
r2 = x2 + j/2 + 2^.
168. Exercises.
Solve for r and 0:
1. r cos ^ = 3,
r sintf = 4.
6. rsin^^ + |)=2,
8. rcos5 = 12, rcosfd-^j=l.
r sin 9 = — 5.
• ii o 7. r «sin(d + T)
8. rcos5 « — 9, \ 4/
2r = sin(5-^)
r sin ^ « — 40.
4.rcos^H-27sin5 = 3, . .
rsind^l. 8. r = 2sinf 25 -|)i
r (sm5 + 4cosd)= 1. \ o/
Solve for r, 5, and ^:
9. r cos5 cos ^ = 3, 10. r cob5 cos ^ = — 1,
r cos 5 sin = 4, r cos 5 sin ^ = 1,
rsin5 = 5. rBin5= — 2.
Eliminate from the following equations:
11. X = r cos (?; y = r sin 5.
12. a? = o cos 5; y = 6 sin 0.
18. X '^ a^ cos^ 0; y — h^ainP 0.
14. ?co8 5 + rsin5 = 1: - sin 5 — ^cos5 = — 1.
o o
16. Eliminate 5 and ^ from the equations
x = r cos cos ^; y = r cos 5 sin ^; 2 =* r sin 5.
16. The same for the equations
X « a cos cos ^; ^ = & cos 5 sin ^; z — c sin 9.
CHAPTER X
Oblique Plane Triangles
169. Between the six parts of a plane triangle there exist,
aside from the angle-sum equal to 180°, two other fundamental
relations which we proceed to obtain. Additional relations will
then be derived from these.
The Law of Sines. — In any plane triangle, the sides are pro-
portional to the sines of the opposite angles.
Let ABC be the triangle, CD one. of its altitudes. Two cases
arise, according as D falls within or without the base (figures).
Then in the first figure,
from A ACD, A = 6 sin 4 ;
from A BCD, A = a sin B;
equating the values of h,
6 sin -4 = o sin B, or a : 6 = sin A : sin B,
In the second figure,
from A 4 CD,
from A BCD,
A = 6 sin (it — -4) = 6 sin A;
A = a sin B;
equating the values of A, we find the same result as before.
' 144
170, 171] OBLIQUE PLANE TRIANGLES 145
By drawing perpendiculars from the other vertices and com-
bining results we have the Law of Sines,
(1) a : & : c = sin ^ : sin JB : sdn C.
170. The Law of Cosines. — In any plane triangle, the square
of any side eqiwls the sum of the squares of the other two sides, minus
tvnce their product by the cosine of their included angle,
' In the above figures let AD = m. Then
First figure. Second figure,
in AACD, m=b cos -4; m=6 cos(7r— ^) = —6 cos -4;
inABCD, a^^h^+ic-m)^ a^=h^+(c+m)^
=A2+c2-2cm+m2 =h^+c^+2cm+m^
=62+c2-2cm. =62+c2+2cw.
Replacing m by its value above, we have in either case,
(2) a^=^b^ + c^-2bc cos A.
(20 SimUarly, b^ = a^+c^-2ac cos B,
(2'0 c« = a« + &« - 2 od cos C.
171. The llkw of Tangents. — In any plane triangle, the dif-
ference of two sides is to their sum cw the tangent of half the difference
of the opposite angles is to the tangent of half their sum.
From the law of sines we have,
a __ sin A
b sinfi'
By composition and division, and subsequent reduction we have,
a — b _ sin A — sin B
a + 6 "" sin A + sin fi
^ 2 cos H-^ + ^) sin H^ -B)
2 sin H^ + B) cos H^ -B)
= cot i{A+B)t8JiHA- B).
That is,
(3)
g-ft tanlJA-B)
a + b ^t&nUA + B)
^
146 OBLIQUE PLANE TRIANGLES (172
Similarly,
(30
a-e taii|(-4-C)
(3'0
a + e tanlU + C)
6-c _ tan|(JB-C)
The symmetry of these formulas makes them easy to remember.
In actual practice, they are used in slightly modified form. Thus
the first of them is written,
tim^(A-B)=^tim^(4+B).
Similarly for the other two.
172. Functions of the Half-Ancles. — When the three sides of
a triangle are known, its angles are best calculated by the formulas
now to be derived.
From the law of cosines we have,
62 + c2 - a2
cos-4 =
2hc
In practice this formula is not convenient unless a, b, and c
happen to be small numbers. Now
. 1 . . /l — cos-4 /---, , , , /l — cos4 „\
smgii = y 2 y^y ^^^ ±V — 2 /
But l-cosii = l-^^-t^^-=^
2 6c
^ 26c~62-c2 + a2
26c
^ gg - (6 ~ c)2
26c
_ (o + 6 — c) (g — 6 + c)
2 6c
Let 2 « = g + 6 + c, or s = J (^ + ^ + c).
Then g + 6 — c = 2(a — c), and g — 6 + c = 2(« — 6).
Tk^^ 1 A 4"(s — 6) (s — c)
Then 1 — cos A = —^^ -^-^ -i
2 he
and
(4) sin|-4 = Y/
{S -b)(8- C)
2 y be
i
i
172] OBLIQUE PLANE TRIANGLES 147
Simllnrly,
Observe that the sides appearing explicitly under the radical
include the angle to be calculated.
To obtain cos i -4, we have
u-^i
+ cosA
1)2 4. /^ _ ^2
But 1 + C08A = 1+ VV,
^ (6 + c + g) (6 + c -- g)
2 6c
4 a (g -- g)
^ 26c •
Hence
(5) cosj^ = V-^^^
Similarly,
(50 COS - JB = V -^ ^•
Dividing sine by cosine we have
*
(6)
(6')
(6")
2 ▼ «(« — &)
/8 " » (» — c)
If r = ■ /(« - g) (a - 6) (a - c) ^
148 OBLIQUE PLANE TRIANGLES [173,174
then
(7) taiii^ = — ^»
(70 taiiiB = — ^f
(7") taiiic7= ^
2 8 — c
All these formulas for the half-angle should be memorized,
preferably in verbal form, so that a single statement contains all
three formulas of any one set.
173. Solution of Plane Oblique Triangles. — A triangle is deter-
mined, except in such cases as will be specially mentioned, when
three parts are given, of which one at least must be a side. The
calculation of the other parts is called "solving the triangle."
Four cases arise, according to the nature of the given parts.
I. Given two angles and one side. f^
n. Given two sides and their included angle, ^^ -f.
in. Given two sides and an opposite angle. > ^^ - ^
IV. Given three sides.
The method for treating each case will now be considered.
174. Case I. Given two angles and one side, as Aj Bj a.
Formulas for finding the other parts, C, 6, c.
C = 180° - (A -h B).
From the law of sines,
, sin B sin C
o = a— — -r*, c = a~ — T.
sm A smA
Check. It is important to have a check on the accuracy of the
calculated parts. For this purpose use any formula involving as
many as possible of these parts.
In this case we use
6 sin B r • x^ • r>
- = -; — 7c , or sm C = c sm B.
c sm C
Example. Given A = 50°, B = 60°, a = 150.
To find C, 6, and c.
174] OBLIQUE PLANE TRIANGLES 149
Solution by Natural Functions.
C = 180° - (50° + 60°) = 70°.
, sinB 150 X >8660 .„_^
^°^^E:A° .7660 -^^^^^^
sinC 150 X .9397 ^^^^.
C = a-, J = r=^^^r = 184.01.
sm A .7660
Check. 6 sin C = c sin B,
or 169.58 X .9397 = 184.01 X .8660,
or 159.35 = 159.35.
Logarithmic Solution.
C = 180-(A+B).
6 = a - — 7 : log 6 = log a + log sin B + colog sin A.
smA
• j^
c = a - — 7 ; log c = log a + log sin C + colog sin A.
sm -A
Check. 6 sin C = c sin B; log 6 + log sin C = log c + log sin B.
We now write out the following scheme:
A+B =
C =
180°
-(4 + B) =
loga =
loga =
log sin B =
log sin C =
colog sin A =
colog sin il =
log 6 =
(
logc =
6 =
c =
Check. log b =
logc =
log sin C ==
log sin B =
Now turn to the tables and take out all the logarithms required,
inserting them in their proper places. Add to obtain log 6 and
log c. Insert these in the check and add. If the sums in the
check agree, or differ by only a unit in the last figure, the numerical
work is correct. Then look up b and c.
150
OBLIQUE PLANE TRIANGLES
[174
On making these calculations with the data in our example the
scheme appears as below.
C = 180° - 110** = 70°
log a = 2.1761
A+B = 110°.
log a = 2.1761
log sin B = 9.9375
colog sin A = 0.1157
log 6 = 2.2293
b = 169.6
Check. log 6 = 2.2293
log sin C = 9.9730
2.2023
log sin C = 9.9730
colog sin A = 0.1157
log c = 2.2648
c = 184X)
log c = 2.2648
log sin B = 9.9375
2.2023
Remark. In calculating with four-place logarithms, three sig-
nificant figures of the resulting niunbers are usually correct. The
fourth figure should be retained, but may be one or more units in
error. It is rarely worth while to retain more than four significant
figures.
A similar remark applies to 5-, 6-, and 7-place. tables. See
chapter on numerical computation.
Graphic Solution of Case I ; given
A, B, and a.
Calculate C = 180° - (il + B).
Lay off a line segment equal to a
and at its extremities construct an-
gles B and C, prolonging their free
sides until they meet at A (figure).
Scale off the lengths of 6 and c. The
figure shows the triangle already
^ solved above. From it we have
ao
ScoIb
loo
■
/59
6 = 167, c = 181.
No solution is possible when A •\- B > 180°.
Exercises. Solve the following triangles, including graphic
solutions.
1. A = 55*
B = 72°
a = 1000.
2. A = 65° 25'
B = 78° 23'
a = 4. 245.
3. C = 34M8'
A = 100° 17'
6 = 0.5575.
4. B = 115M0'.5
C = 40°22'.3
c = 0.00276.
6. B = 88° 20'
C = 105° 30'
a = 10.
li.
.75]
OBLIQUE PLANE TRIANGLES
151
176. Case n. Given two sides and the included angle, as
a, b, C
To solve the triangle we calculate i {A + B) as the comple-
ment of i C; then i {A — B) is calculated by formula (3). Angles
A and B are then determined and hence all the angles are known.
We can then compute c in two ways by means of the law of sines.
The agreement of the two values of c furnishes a check on the
computations.
Fonnulas.
a — b
tajii{A-B) =
c = a
a + b
sin C
tan H^ + B),
sinC
sini4
= 6
sinB'
Scheme for Logarithmic Solution.
a = log (a — 6) =
6 = colog (a + 6) =
log tan i{A+B) =
a + b--
a-b--
HA-B) =
A =
log 6 =
log sin C =
colog sin B =_
logc =
c =
log tan H-^ - B) =,
log a =
log sin C =
colog sin il =
• log c =
c =
Graphic Solution. Construct
angle C and on its sides lay
off lengths a and b, starting
from the vertex. Complete
the triangle, and measure c, A,
and B (figure, constructed for
example below).
A solution is possible pro-
vided C < 180^
Example. Given b = 12.55, a = 20.63, C = 27° 24'. Solve the
triangle.
SciUe
152
OBLIQUE PLANE TRIANGLES
[176
Logarithmic Solution.
H^ +B)= 90° - iC = 90^-13° 42' = 76° 18'.
a = 20.63 log (a - 6) = 0.9074 H-4 + B)= 76° 18'
6 = 12.55 colog(a + 6)= 8.4792 i jA-B)^ 44° 58'.4
a + 6 = 33.18 log tan H-^ + g)= 0.6130 il =121° 16U
a - 6 = 8.08 log tan H^ - 5) = 9.9996 B = 31° 19'.6
loga= 1.3145 log 6= 1.0986
log sin C= 9.6630
cologsinil = 0.0682
logc= 1.0457
c = 11.11
log sin C= 9.6630
cologsinB= 0.2841
logc= 1.0457
c = 11.11
Graphic Solution. This is shown in the figure above. Let the
student scale off the known parts.
Exercises. Solve the following triangles:
La- 1500,
2. b ^ 15.25,
3. a - 1.002,
4. & » 6238,
6. a » 16. 21,
b -750,
c = 12.65,
b » 0.8656,
c = 4812,
c =22.48,
C = 58^
A = 98** 40'.
C = 130^ 48'.
A = 75'* 22'.
B = 36^ 54'.
176. Case m. Given two sides and an opposite anj^ei as
ttf &| ^.
This is known as the ambiguoua case. We begin by stud3dng the
graphic solution.
Lay off angle A and on one of its sides take AC=h, With C as
center and radius equal to a, strike an arc of a circle. The figi^es
show the various possibilities arising in the construction, the first
three for A < 90°, the last three f or A > 90°.
B A
'.-/
B^-^A
a/
V
\
176] OBLIQUE PLANE TRIANGLES 163
In each case the perpendicular from C on the other side of angle
A is equal to & sin il. Inspection of the figures then shows that
when A < 90° and a < b sin A, no triangle is possible;
when A < 90° and a = & sin A, a right triangle results;
when A < 90° and 6 > a > 6 sin -4, two oblique triangles result;
when A < 90° and a > 6, one oblique triangle results;
when A > 90° and a = 6, no solution is possible;
when A > 90° and a > 6, one oblique triangle results.
It is always possible therefore to state in advance what the
nature of the solution in a given case will be.
Formulas. Given a,bj A.
.J, 6 . - fC = 180°-(il+B).
a lC'=180°-(il + B').
B' = 180°-B.
smC ,smC
C = a— -r = 0— — zz*
&mA 8inB
, sin C , sin C
sm A sin B'
Check. The agreement of the values of c and d as calculated
from the two expressions for each of them furnishes a partial
check on the calculations. It does not guard against an error in
log sin C, which may be checked independently. A complete
check is furnished by (6) of (172).
In carrying out the calculations according to the formulas above,
the various cases shown in the figures are indicated as follows:
(a) log sin £ =0; no solution, or right triangle.
(b) retain both B and B'; two solutions.
(c) A + B' > 180°, hence reject B'] one solution.
(d) log sin B = 0; no solution.
(e) A+B> 180° and A + B' > 180°; no solution.
(f) As in (c); one solution.
In a given numerical example the nature of the solution always
becomes apparent during the progress of the computations.
154
OBLIQUE PLANE TRIANGLES
[177
Example. Given a = 602.3, 6 = 764.1, A = SS"* 17'.3.
Logaritfamic Solution.*
log6 = 2.88316 log a
colog a = 7.22019 log sin C
log sin A = 9.79217 colog sin A
IogsinB = 9.89552 logc
B = 51° SO'.O c
B' = 128° lO'.O log a
A + B^ 90° 7'.3 log sin C
il +B' = 166°27'.3 colog sin il
C = 89° 52'.7 log </
C = 13° 32'.7 d
2.77981
0.00000
0.20783
2.98764
971.9
2.77981
9.36960
log 6
log sin C
colog sin B
logc = 2.98764
2.88316
0.00000
0.10448
log b = 2.88316
logsinC' = 9.36960
0.20783 cologsinB' = 0.10448
2.35724 log c' = 2.35724
Graphic Solution. This is shown in the figure, from which the
unknown parts may be scaled oflf.
Exercises. Solve the triangles whose
given parts are:
1. a = 29.95, h = 37.17,
A = 42** 24'.
2. a = 1756, h = 746,
A = 67*^ 30^.
8. 6 = .728, c = .542,
B = 105° 44'.
4. 6 = 6.174, a = 2.614,
B = 32° 22'.
•
Seals
I I ^^m^-m^m^
177. Case IV. Given the three sides, a, &, c.
The angles may be calculated from either the sine, cosine, or
tangent of the half-angles. When all three angles are wanted,
it is best to use the tangent. There is no solution when one side
equals or exceeds the sum of the other two.
Formulas.
8
1 / I 1. I \ . /(s — a) (
-a)(g-b)(g~c).
1 r
tan 2-4 =
1 r
; tanp^B = ^;
—a 2 8—0
1 r
tan^C =
2 8 — c
Check. i(A + B + C)= 90^
* The JBfth figure is carried to avoid accumulation of error. This is advis-
able if all possible accuracy is desired.
178]
OBLIQUE PLANE TRIANGLES
155
Example. Given a = 428.6, 6 = 806.2, c = 542.4.
Logarithmic Solution.
a
b
c
428.6
806.2
542.4
2^ 1777.2
8
s — a
8-b
8 — C
888.6
460.0
82.4
346.2
cologs 7.0513
log(s-a) 2.6628
log(s-6) 1.9159
log (s - c) 2.5393
2 4.1693
log r 2.0846
log tan M 9.4218
log tan iB 0.1687
log tan iC 9.5453
iA 14M7'.7
iB 55^51'.5
i C 19° 20'.5
Check 89°59'.7
A 29'' 35'.4
B 111° 43'.0
C 38° 41'.a
Check 1777.2
Graphic Solution. This is shown in the jBgure.
we find A = 29°, B = 112°, C = 38°.
179° 59'.4
By measuring
Sie.4
MOO
1 I 1 I I ii I
SoaU
Exercises. Solve the triangles whose given parts are:
1. a = 6192, h = 4223, c = 7415.
2. a = 156.21, 6 = 300.15, c = 410.32.
3. o = 0.00245, h = 0.00405, c = 0.00536.
4. a = 52.76, 6 = 22.84, c = 28.41.
178. Areas of Oblique Plane Triangles. — Referring to the fig-
ures of (169), we see that h is the altitude drawn on side c as base.
Hence if K denote the area of the triangle, we have
(8) X = i Ac = I ac sin ^. (A = a sin 6.)
Hence, the area of a plane triangle equals half the product of
two sides by the sine of their included angle.
The area is also expressible in simple form in terms of the sides.
In the formula above replace sin B by 2 sin i B cos \ B. Then
X = ac sin i jB cos J B
= ac\J
is — a)(s — c) 4 A (s — 6)
ac
v/'
ac
156
OBLIQUE PLANE TRIANGLES
[179
by (4') and (50 of (172). Hence,
(9) K = V« («-«)(«-&)(«- c).
When the given parts of the triangle are such that neither of the
above formulas applies directly, it is usually best to calculate
additional parts so that one of these formulas may be used.
179. Exercises and Problems.
1.
2.
3.
4.
a » 183. 9,
a
= 1.925,
a = 42.31,
a = . 41409,
6 » 584.9,
b
= 2.243,
b = 71.70,
6 = .49935,
c = 166. 6.
c
= 7.25.
c = 71.35.
c = . 18182.
6.
6.
7.
8.
a = 183.7,
a
= 283.6,
a = 783,
c = 22,504,
A =36* 55'. 9,
A
= 11* 15',
B = 42* 27',
B = 55* 11',
C =70* 58'. 2.
B
= 47* 12'.
C = 55* 41'.
C = 45* 34'.
9.
10.
11.
12.
6 = 3069,
b
= 100.2,
a = 3186,
a = . 8712,
B = 15* 51',
B
= 48* 59',
b = 17156,
6 = . 4812,
A = 58* 10'.
C
= 76* 3'.
A = 147* 12'.
A = 24* 31'.
13.
14.
16.
16.
a = 1523,
A
= 61* 16',
a = .39363,
6 = 147.26,
b = 1891,
a
= 95.12,
c = .23655,
'c = 109.71,
A = 21* 21'.
b
= 127. 52.
C= 22*32'.
A = 41* 15'.
17.
18.
19.
20.
h = .5863,
a
= 10.374,
6 =6.4082,
6 = .8869,
a = .8073,
c
= 9.998,
c = 18.406,
a = 3.0285,
C = 58* 47'.
B
= 49? 50'.
A = 33* 31'.
C = 128* 7'.
21.
22.
23.
24.
a = .8706,
a
= 20.71,
A = 41* 13',
a = 4663,
b = .0916,
b
= 18.87,
a = 77.04,
6 = 4075,
c = .7902.
C
= 55* 12' 3".
b =91.06.
C = 58*.
25.
26.
27.
28.
a = 43031,
a
= 16082,
a = .00502,
b = 2584,
c = 31788,
c
= 13542,
b = . 00558,
c = 5726,
A = 19* 12'. 7.
C
= 52* 24'. 3.
c = .00466.
A = 27* 13'.
29.
30.
31.
32.
b = 37403,
a
= 6148,
a = .01520,
b = 8204,
a = 49369,
c
= 7512,
6 = . 03366,
c = 9098,
A = 81* 47'.
A
= 133* 30'.
c = .02114.
A =62* 9'. 6
179]
OBLIQUE PLANE TRIANGLES
157
a
b
C
h
a
A
33.
532,
704,
73^
37.
1482,
1284,
27* 18'.
34.
36.
36.
a = 290,
a » . 000299,
c = 7025,
c « 356,
c = .000180,
& = 8530,
C = 41* 10'.
A =63*50'.
C = 40*.
38.
39.
40.
a = .2785,
B = 50* 20' 54",
C= 49* 47' 26"
b = .2275,
a = 235.64,
c =725.52,
B = 65* 40'.
b = 284.3L
b = 950.04.
In any triangle ABC, whose sides, opposite angles A, B, C, respectively,
are a, &, c, show that:
B A
41. 6 (s — 6) cos2 — ^ a(8 — a) cos^ ^ •
42. a = 6 cos C + cos B.
43. (a - 6) (1 + cos C) = c (cos B - cos A).
.. cos A , cosB , cosC a2-f-62_|-c2
44. 1 £ 1 — = -
a
2abc
A B
46. (b -j-c ^ a) tan ^ = {c -\-a — b) tan -^ •
46. (6 + c) (1 — cos A) = a (cos B + cos C).
47. (a2 - 62 + c») tan 5 = (a* + 6« - c«) tan C.
48. cot -^ + cot ^ -h cot rt^ = cot ^ cot ^ cot ^ .
49. The radius of the inscribed circle is i /
-b) (8 - c)
60. The diameter of the circumscribed circle is a esc A.
Calculate x in terms of the other quantities in each figure below, where a
right angle is indicated by a double arc; in each case find the value of x for an
assimied set of values of the literal quantities:
71 T
158
OBLIQUE PLANE TRIANGLES
[179
63. Find the lengths of diagonals and the area of a parallelogram two of
whose sides are 5 ft. and 8 ft., their included angle being 60^.
64. Two sides of a parallelogram are a and &, their included angle C; show
that the area is ab sin C.
66. The sides of a triangle are 4527, 7861, 6448; find the length of the
median drawn to the shortest side.
66. The sides of a triangle are in the ratio of 2 : 3 : 4; find the cosine of
the smallest angle.
67. The angles of a triangle are as 3 : 4 : 5; the shortest sidens 500 ft.;
solve the triangle.
68. The angles of a triangle are as 1:2:3; the longest side is 100 ft.;
solve the triangle.
69. From a station on level groimd due south of a hill, the angle of eleva-
tion of the top is 15°; from a point 2000 ft. east of this station the angle of
elevation is 12®; how high is the hill ?
70. The angle of elevation of the top of a building 100 ft. high is 60°; what
will be the angle at double the distance ?
71. A flag-pole on a building subtends an angle of 7° 40' at a point on the
ground 500 ft. from the building; on approaching 100 ft., the pole subtends
an angle of 7° 50'; find the height of the pole and the building.
72. On level ground, 250 ft. from the foot of a building, the angles of ele-
vation of the top and bottom of a flag-pole surmounting the building are
38° 43' and 31° 2' respectively; find the height of the building and the pole.
73. From level ground the angle of elevation of the top of a hill is 11° 30';
after approaching 3000 ft. up an incline of 3° 27', the angle of elevation of the
top is 21° 32'; how high is the hill ?
74. From a level plain, the angle of elevation of a distant mountain top
is 5° 50'; after approaching 4 miles, the angle is 8° 40'; how high is the moun-
tain ?
179] OBLIQItE PLANE TRIANGLES 159
76. From a point 60 ft. above sea level the angle between a distant ship
and the sea horizon (the offing) is 20'; how far away is the ship ? [ConEoder
the surface of the sea as a plane, and the distance to the horizon 10 miles.
See (226) ex. (4).]
76. From a point on level ground the angle of elevation of the top of a hill
!b W 12'; on approaching 1000 ft., the angle is 17"" 50'; how high is the hill ?
77. A building surmounted by a flag-pole 20 ft. high stands on level ground.
From a point on the groimd the angles of elevation of the top and the bottom
of the pole are 53® 5' and 45® 11' respectively. How high is the building ?
78. On approaching 1 mile toward a hill, the angle of elevation of its top
is doubled; on approaching another mile, the angle is again doubled; how high
is the hill ?
79. A and B are two points neither of which is visible from the other. To
determine the distance AB, two stations C and D are chosen and the following
measurements made: CD = 500ft.; ^ACD-:30®25'15";Z ACB= 85*40'20";
Z BDC = 35* 14' 50"; Z EDA = 80® 20' 25"; find AB.
80. In a chain of three non-overlapping triangles, the following data are
known:
AB = 1000 ft.
A ABC, AACD, ZCDE,
ZA^W 36', Z A = 56® 32', Z C = 55® 30',
Z C = 40® 0'; ZC ^ 50®20'; ZE ^ 77®02';
Calculate DE. (Express DE in terms of AB and the necessary angles by
the law of sines.)
81. In a chain of four non-overlapping triangles, the following data are
known:
AB ^ 11289 meters.
A ABC, ACBD, ADBE, ADEF,
Z A = 58® 10' 35", Z B = 86® 50' 0", Z D = 79® 12' 8", Z D = 50® 41' 5",
Z B = 69®55'0"; ZC ^ 46®48'0"; Z B = 73®29' 10"; Z ^ = 45® 20' 40";
calculate EF.
82. In a chain of five consecutive triangles, each having a side in cbnunon
with the preceding, as ABC, CBD, BDE, DEF, EFG, express FG in terms
of AB and the necessary angles.
83. A tower 50 ft. high stands on the edge of a cliff 150 ft. high. At what
distance from the foot of the cliff will the tower subtend an angle of 5® ?
84. The sides of a triangle are 100, 150, 200 ft. At the vertex of the
smallest angle a line 100 ft. long is drawn perpendicular to the plane of
the triangle. Find the angles subtended at the fsirther end of this line by
the sides of the triangle.
86. A right triangle whose perimeter is 100 ft. rests with its hypotenuse
on a plane, the vertex of the right angle being 10 ft. from the plane. The
angle between the plane of the triangle and the supporting plane is 30®. Find
the sides of the triangle.
160 OBLIQUE PLANE TRIANGLES [179
86. An equilateral triangle 50 ft. on a side rests with one side on a plane
with which its plane makes an angle of 60^. How far is the third vertex
from the plane ?
87. As in exercise 86, if the triangle, instead of beingequilateral, has sides
45 V5
40, 20, 30 ft. and rests on the shortest side. Ans. . •
88. The rades of a triangle are as 1 : 2 : 3, and the longest median is 10 ft.
Find the sides and angles.
89. The following measurements of a field A BCD are made: A to B, due
north, 10 chains; B to C, N 30^ E, 6 chains; C to D, due cast, 8 chains; cal-
culate ADf and the area of the field in acres. (1 chain = 4 rods.)
90. The following measurements of a field ABODE are made: ^ to ^, due
east; 25.52 chains; B to C, E 40** 26' N, 22.25 chains; C to D, N 48^ 26' W,
33.75 chains; D to J^, W 31"* 15' S, 18.32 chains; calculate EA and the area of
the field in acres.
91. In the field of exercise 89 how much area is cut off by a line due east
through B ?
92. In the field of exercise 90 where should an east and west line be drawn
so as to bisect the area ?
93. In the field of exercise 90 where should a north and south line be
drawn to cut off 30 acres from the western part of the area ?
94. If P be the pull required to move a weight TF up a plane inclined to
the horizontal at an angle i, and m the coefficient of friction, then
p ^ ^ sin t+/* cost
cos i — M sin i
Calculate P when W = 1000 lbs., i = 30^ m = 0.1.
95. In exercise 94, what is i if P =* i TF and m = 0.1 ?
96. If I be the length of a plane inclined to the horizontal at an angle i,
M the coefficient of friction and g the acceleration due to gravity (32.+ ft.
per sec. per sec.) the time in seconds required by a body to slide down the
plane is / —
T'^ a/ ^i
y g (sin i — /* cos i)
What is T when I « 25 ft., i = 20^ m - 0.1 ?
97. In exercise 96, find i when I = 100 ft., fi «0.1, T « 5 sec.
\Sy 98. When light passes from a rarer to a denser medium, the
j/ index of refraction m is determined by the equation
L - y. _ smt
yZy^tyyZ, "" sin r
""-1-ij-?^— "^ When M = 1.2, what must be i (angle of incidence) to give a
'*' deflection of 10°?-
99. Find the total deflection of a ray which passes through a wedge whose
angle is 30® and index of refraction 1.4, if the ray enters the wedge so that the
angle of incidence is 25®, and moves in a plane ± to the edge of the wedge.
100. Solve exercise 99 when the angle of the wedge is a, the angle of inci-
dence i, and the index of refraction m.
CHAPTER XI
Thb Progressions. Interest and Annxhties
180. Arithmetic Progressions. — Let a,b,Cy , . . , ik, { be quan-
tities such that the difference between any one of them and the
preceding one is constant. Then the quantities are said to form
an arithmetic progression. (We shall abbreviate this into A. P.)
The quantities a, 6, c . . . , fc, i are called the terms of the pro-
gression, a and I the extremes^ and b, c, . , . ,k the means. The
constant difference between consecutive terms is called the
common difference.
Let a denote the first term,
I denote the last term,
d denote the common difference,
n denote the number of terms,
S denote the sum of the terms of any A. P. Then
the second term is a + d,
the third term is a + 2 d,
• • • • •
the last or nth term is a +(n — 1) d; that is,
(1) ? = a + (n - 1) d.
Also
S = a+(a + d) + (a + 2d)+ • • • +(a + 7r^d);
S = Z +(Z -d)+(l -2d)+ ' ' ' +{l -n^^d).
Adding,
2S=(a + Z)-h(a + 0+ . . . +(a + l)^ n(a + 1).
Hence
(2) 8 = ^(a + l).
Putting for I its value from (1),
(20 s = n(a + ^!^dy
161
162 THE PROGRESSIONS [181, 182
We shall refer to the five quantities a, I, d, n, S, as the elements
of the A. P. When any three elements are given, the other two
may be found by use of the preceding formulas.
181. Problem. To insert m arithmetic means between two given
qwmtUies, a and I.
Since there are 2 extremes and m means, the total number of
terms is m + 2. Hence if d be the common difference,
i = a + (m + 2 - 1) d;
hence
J I — a
a = 7*
m + 1
Then the required means are
a + d, a + 2d, . . . , a-^-md,
^ , When m ~ 1 we have only a single mean, called the arithmetic
mean. It equals \{a + l).
182. Example8.
1. Find the sum of all the integers from 1 to 100 inclusive.
Here 5 « 1+2 +3 + • • • +100.
Then a = 1, Z = 100, n = 100,
and 5 =|(a + = i X 100 (1 + 100) = 5050.
2. How many terms of the progression 3, 0, ~3, . . . are required to
make the sum equal — 27.
Here a = 3, d = - 3, 5 = - 27; to find n.
From (2'), - 27 =n ^3 - ^^^ XsY or n2~3n-18=0.
Hence n = 6 or — 3.
Since n must be positive we discard the second value.
I 3. Find four numbers in A.P., such that the siun of the first and last shall
be 12 and the product of the middle two 32.
Let the numbers be a — 3 rf, a — rf, a + rf, and a + 3 rf, with a common
difference 2 d.
Then a-3d + a + 3rf = 12
and (o - rf) (a + d) = 32.
Hence o = 6 and rf = ± 2.
Therefore the numbers are
0, 4, 8, 12, or 12, 8, 4, 0.
183,184] THE PROGRESSIONS 163
183. Exercises. Find the last term and the sum of each of the
following arithmetic progressions:
1. 7, 11, 15, . . . , to 13 terms; 6. 63, 58, 53, . . . , to 8 terms;
2. 5,8, 11, . . . , to 12 terms; ^ aj,x+2j^,a;+4 j^,. . . , tolOterms;
3. 2, 2i, 3, . . . , to 25 tenxis;
4. 1,1.1,1.2, . . . ,to200terms; ^- P^P "i^^P-Q* - - .,to20terms.
Find the other elements of the A.P., given that:
5. o = 10, n = 14, 5 = 1050; 16. . n = 35, fif = 2485, d = 3;
9. o = 3, n = 50, /S = 3825; 17. n = 50, 5 = 425, d - §.
10. a = - 45, n = 31, 'S = 0; 18. w = 33, fif = - 33, d = - f;
11. Z = 21, n = 7, /S = 105; 19. S - 624, a = 9, d = 4;
12. Z = 49, n = 19, 5 = 503J; 20. S = 2877, a = 7, d = 3;
13. Z = 148, n = 27, S = 2241; 21. S = 623, d » 5, Z = 77;
14. Z = - 143, n=33, fif^ -2079; 22. ,S = 682.5, d = 1.5, Z = 45;
16. n= 21, /S = 1197, d « 4; 23. fif = 95172, d = - 7, Z - 567.
* 24. Find the sum of the first 100 odd numbers.
26. Find the sum of the first 50 multiples of 7.
26. A body starting from rest falls 16 ft. during the first second, and in
every other second 32 ft. more than during the preceding. How far does the
body fall in 12 seconds;.how far during the 12th second ?
27. According to the rate of fall in exercise 26, how loi% will the body take
to fall 1600 ft ?
28. A body which is projected vertically upward loses 32 ft. of its initial
velocity each second. If the velocity of projection is 320 ft. per second, how
high will the body rise ?
29. If 100 apples are laid in a straight line, 3 feet apart, how far must a
person walk to carry t^em one at a time to a basket standing beside the first
apple ?
184. Geometric Progressions. — If the numbers a, 6, c, . . . ,
kj I are such that the ratio of any number to the preceding number
is constant, the numbers form a geometric progression. (We
abbreviate by writing G. P.)
The expressions '^ terms f" "means," "extremesj" are used here as
in the case of A. P. The constant ratio of any term to the preced-
ing is called the ratio of the geometric progression.
If a, Z, n, and S have the same meaning as in the case of the
A. P., and if r denote the ratio of the G. P., the first n terms are,
a, ar, ar^, ar^, . . . > ar^-^.
164 THE PROGRESSIONS [185, 186
Hence
(1) I = ar*-^
Also 8 ^ a + ar -{-ar^ -{- • • • + ar^"^
and rS^ar + ar^-^- • • • '\'ai^'^ '\'ai^.
Therefore rS — S = ar^ -- a,
or (r - 1) S = (r~ - 1) a.
Hence.
(2) 8 = a 7 = a
Substituting from (1) in (2) we have
(20 ^-T^t-
r — 1
When any three of the five elements are given, the other two
may be obtained by use of two of the preceding formulas. In
some cases this involves the solution of an equation of nth degree
or of an exponential equation.
186. Problem. To insert m geometric means between two given
numbers a and I.
The total number of terms being w + 2, we have, if r denote the
ratio, •
m+1/7
Z = ar'"+2-i or r=y-.
The required geometric means are then
ar, ar^, . . • , a^"*.
When w = 1, the resulting single mean between a and I is
VaZ. The square root of the product of two quantities is called
their geometric mean.
186. Examplea,
1. Find the sum of the first 10 terms of the G. P. 2, 2^, 2», . . . .
910 — 1
Here a = 2, r = 2, n = 10; hence5 = 2 = 2046.
2. How many terms of the G. P. 1, 2, 4, . . . are required to make the
sum 63 ?
Here a = 1, r « 2, <S = 63; to find n.
fn — 1 2** — 1
From <S = a r- we have 63 = -75 r- ; or, 64 =» 2**.
Hence n = 6.
187, 188] THE PROGRESSIONS 165
3. Four numbers are in geometric progression. The sum of the first and
last is 18, the product of the second and third 32. Find the numbers.
Let the numbers be a, ar, ar^ and ar^.
Then
(1) a +ar3 = 18; (2) a2r3 = 32.
Multiply (1) by o and in the result replace a^r^ by 32.
Then o* + 32 « 18 a; hence a == 16 or 2.
Substituting the values of a in (2) we find r » ^ or 2. Hence the numbers are
16, 8, 4, 2; or 2, 4, 8, 16.
(We disregard the imaginary values of r.)
187. Exercises. Find the last term and the sum of the terms
of the following geometric progressions:
1. 4, 8, 16, . . . , to 7 terms. 4. 9, 3, 1, . . . , to 11 terms.
2. 2,6,18, . . . , to 9 terms. 6. 1, }, ^s, . . . , to 10 termiB.
3. 1, 4, 16, . . . , to 7 terms. 6. 8, 2, i, to 20 terms.
7. a, a(l +x),a (1 +x)*, . . . to 8 terms.
8. w^, mn, m-^n*, . . . , to 9 terms.
9. Insert 3 geometric means between 8 and 10368.
10. Insert 5 geometric means between 2 and 31250.
11. Insert 5 geometric means between 36 and s\.
12. Insert 6 geometric means between 3 and 49152.
13. Insert 4 geometric means between 48 and ^f.
14. Insert 5 geometric means between 81 and V]^.
Calculate the unknown elements, giyen:
15. r= 128, r=2, n=7. 22. a==l, Z =2401, 5=2801.
16. Z =78125, r=5, n=8. 23. a = 10, Z =ft, 5 = 19^.
17. i=A, . r = i n = 5. 24. a=3125, Z =5, 5=3905.
18. a=9, ?=2304, r=2. 25. a=3, r =3, 5=29523.
19. o=2, Z=64, r=2. 26. a=8, r =2, 5=4088.
20. a=3, i = 192V2, r= V2. 27. r=2, n=7, 5=635.
21. a =2, Z = 1458, 5=2186. 28. Z = 1296, r =6, 5 = 1555.
188. Infinite Geometric Progressions. — Consider a line segment
AB oi unit length, and bisect it atAi, then bisect A iB at A2, AzB
at A^ and so on (figure).
The points of bisection Ai, A2, A^, . . . "l ± A' f ^^.f
continually approach B and the sum of
the segments AAi + -41-42 + A2AZ + • • • approaches AB or 1.
But the sum of these segments is represented numerically by the
series
' 1.1,1, 1 1 1 1 1 .
2 + 4 + 8+'*'' "^^ 2 + 2^ + 23+*'''
166 THE PROGRESSIONS [189
and hence by taking n large enough we can make the sum
iS« = -4- — 4- — 4- • • • 4- —
«n 2 ^ 22 ^ 23 ^ ^2'*
differ from 1 by as little as we please. Hence we take
111
2 + 4 + 8 + • • • to mfinity = 1.
The sum Sn above is a geometric progression with r = § and
a = J. Its sum to n terms is therefore
^'•"2i^nf
As n increases, (J)** approaches 0, and Sn approaches the value
^ I 7 = 1, as found above.
A geometric progression in which the number of terms increases
without limit is called an infinite geometric progression.
For the sum of n terms of any G. P. we have
y.n _ 1 1 - r"
/q^ =a r- = a-z •
^ r — 1 1 — r
If now r < 1, then r** approaches when n approaches oo, and the
formula for the sum of an infinite G. P. is
8 = _ J provided \r\ < 1.
(When r = 1, or when r > 1, S is infinite.)
Example. A ball is thrown vertically upward to a height of 60 ft. On
striking the ground it always rebounds to one-third the height from which
it fell. How far will it travel ?
The distance covered during the first rise and fall is 120 ft., during the sec-
ond rise and fall, J X 120 ft., during the third, ^ X 120 ft., and so on indefi-
nitely. We have an infinite G. P., with a — 120 and r » J. Hence the total
distance will be
5 = i^=: 180ft.
189. Exercises. Sum the following infinite geometric progres-
sions:
1. 8, 2, }, . . . . 3. 5, 3, 1 6. 1, -}, +1, -I, ....
*• 1» *i le> .... 4. 2, y, fgf ... . o. o, 1, 7» »* ... •
190, 191 ] THE PROGRESSIONS 167
7. If in the example worked above the ball requires 4 seconds for the £irst
rise and fall, and half as much time for any subsequent rise and fall as for the
preceding, how long before the ball will come to rest ?
8. How far has the ball in the above example traveled at the 10th rebound ?
190. Harmonic Progressions. — If the numbers a, 6, c, . . . , fc,
I axe such that their reciprocals form an arithmetic progression,
they are said to be in harmonic progression (abbreviated to H. P.).
Problems relating to harmonic progressions are solved by reduc-
tion to A. P.
If a, by c form a H. P., then b is called the harmonic mean between
a and c. Let the student show that we then have
, 2ac
= — ; —
a + c
191. Exercises.
1. In an A. P. the sum of the 9th and 12th terms is 40; the difference
between the squares of the 15th and 11th terms is 400. Find a and d.
2. In an A. P. of 10 terms, the sum of the terms is 65 and the sum of their
squares 1165. Find a and d,
3. In an A. P. of 20 terms, the sum of the 3rd and 12th terms is 30, the
product of the two middle terms is 725. Find a and d,
4. In an A. P. of 14 terms, the product of the first and the last is 276 and
the product of the middle two is 1326. Find a and d.
6. Find four numbers in A. P. such that their product is 840 and their
sum 11.
6. Find four numbers in A. P. such that their product is h and the sum of
their squares is k.
7. Find five numbers in A. P. such that their product is a, their sum 5 6.
8. The sides of a triangle form an A. P. with a common difference 2. Find
the cosine of the largest angle, if the longest side is twice the shortest.
9. Find the angles of a triangle if they form an A. P. with d = 5°.
10. Between every pair of consecutive terms of the G. P. 1, 2, 4, 8, . . .
insert a new term so that the result is again a G. P.
11. As in exercise 10 for the G. P. a, ar, ar^, ....
12. In a G. P. of 10 terms, the sum of the even terms is 30 and of the odd
terms 60. Find a and r.
13. Find four numbers in G. P. such that the product of the first and last
is 400 and the quotient of the middle two is 14.
14. Find three numbers in G. P. such that their sum is A, the sum of their
squares k,
16. If a tree, now 4 inches in diameter, increases its diameter 5% each
year, how thick will it be in 20 years ?
16. A seed yields a plant from which 4 new seeds are obtained. How many
seeds are available from the 10th generation of plants ?
168 INTEREST AND ANNUITIES 1192
17. An Indian potentate offered to reward the inventor of the game of chess
as follows: one grain of wheat for the first square on the chessboard, 2 for the
second, 4 for the third, and so on, doubling each time for the 64 squares. What
would be the cash value of this reward, with wheat at $1.00 a bushel, allow-
ing a million grains to the bushel ?
18. A right triangle has a hypotenuse 2 ft., angle 30°. From the vertex
of the right angle a X is dropped on the hypotenuse, foiming a new right
triangle which is treated similarly, and so on indefinitely. Find the simi of
all the Js so obtainable.
19. The altitude of an equilateral triangle is a. A circle is inscribed in it»
and in this circle a new equilateral triangle. The operation is repeated on the
new triangle, and so on indefinitely. Find the sum of the altitudes and of
the perimeters of all triangles so obtainable.
20. Find the sxun of the perimeters and of the areas of all the circles in
exercise 19.
Interest and Annuities. — This subject affords a simple and use-
ful application of the theory of progressions.
192. Interest. — Let P denote a sum of money loaned, or
principal^ and r the yearly rate of interest expressed in fractions
of a dollar. Then the amount of P dollars in one year is
Ai=P(l + i-).
If principal plus interest for one year is allowed to run a second
year, the amount at the end of the second year is
A2 = ili(l + r)=P(l+r)2,
and so on.
Hence if An be the amount of P dollars in n years, interest at
rate r compounded annually, we have
(1) A„ = JP (1 + vY.
If interest is compounded every t years instead of annually, then,
after n compoundings, the amount is
(10 A,, = J> (1 + rt)^.
Thus if we want the amount of $100 at the end of 2 years, inter-
est 4 per cent compounded quarterly, we have, '
P = $100; r = ^U) t = i; n = 8.
Then An = 100 (1 + .04 X i)* = $100 (1.01)8 = $108.25.
193] INTEREST AND ANNUITIES 169
193. Annuities. — An annuity is a sum of money payable yearly,
or at other stated periods.
Let A be the amount of each payment, r the yearly rate of
interest, n the number of payments to be made.
Assuming the first payment now due, and that each payment is
put at interest, compounded annually, what is the total amount
accrued when the last payment has been made?
The first payment is at interest n — 1 years, its amount
A (1 + r)**-^; the second n — 2 years, its amount A (1 + r)'*"^;
and so on, to the payment next before the last, which is at inter-
est one year, its amount A(l + r); the last pajrment amounts
to A. The total amount S is therefore
S = A+A{l+r)+A(l+r)^+' - • + A (1 +i-)'»-i, or
(2) ^^^ (i+rr-i ^^ (i+.r-i .
^' 1 + r — 1 r
Present Worth. — How much cash in hand, placed at interest
compounded annually, will amount to the sum S just obtained
when the last payment is made, that is, in n — 1 years?
Let Q be the amount required, called the present worth of the
annuity.
Let Q\ be the sum which with interest will yield in n — 1 years
the amount of the first pajrment, or il (1 + r)**-^. Then
Qi(l + r)'»-i=il(l + r)'»-i or Qi = 4.
Let Q2 be the sum which with interest for n — 1 years will yield
. the amount of the second payment, or A (1 + r)'*-^. Then
Q2(l + r)'*-i =A(l + r)'*-2 or Q2= ^
1 +r
Similarly if Q3, O4, . . . Qn be the present worths of the 3rd,
4th, . . . last payments of the annuity we have,
Qs = /I , xo , Qa = ,. . x3 > • • • ' On =
Hence
170 INTEREST AND ANNUITIES [194
The sum in the parentheses is a G. P. with ratio t— — Apply-
ing the formula and reducing,
(1 + !•)* — 1
(3) Q=^ /7V\n~i '
V / ^ !• (1 + r)* *
194. Exercises.
1. Find the amount of $1412 in 19 years at 4%, interest compounded
annually.
2. Find the present worth of an annuity of $100, there being 20 annual
payments of which the first is now due.
3. Find the amount of $1000 in 10 years at 4%, interest compounded
quarterly.
4. Find the amount of $1000 in 20 years at 4%, interest compounded
semi-annually.
6. In how many years will a sum of money double itself at 5% simple
interest ?
6. In how many years will a sum of money double itself at 5%, interest
compounded annually ?
7. An annuity of $100 is to begin in 10 years from date and to run 10
years. Find its present worth if money brings 5% compound interest.
8. Find the present worth of a perpetual annuity of A dollars, compound
interest r%, the first payment now due. (Q = Qi + Q2 + Qs + • • 'ad inf.).
9. As in exercise 8, except that the first payment falls due in m years.
CHAPTER XII
Infinite Series
196, Limit of a Variable Quantity. — When a variable quantity
changes in such a way thai it approaches a fixed numerical value, so
thai the difference between the variable and the fitted quantity becomes
and remains less than any assignable magnitude, however small, then
the fi:ced quantity is called the limit of the variable.
For example, as x varies the variable quantity 1 + x can be
made to dififer from 1 by less than any small quantity e, by simply
taking | a: | < e, and the nearer x is to 0, the nearer will 1 + a: be
to 1. Hence, as x approaches 0, the limit of 1 + x is 1. As an
equation this is expressed by
lim (1 + x) = 1. ( = is read "approaches. ")
x-O
Exercise. Show that :
(a) lim^^ = 1; (b) lim(l + J) = 2; (c) limlog(l + x)= 0;
1
(d) lim (l - ^)= 0; (6) lime' =1; (/) lim (l + JV = 1.
»-10\ W/ x-O n-oo\ n/
196. Infinite Series. — A sequence or succession of terms, ui, U2,
Uz, . . . , Un, » ' ' 9 unlimited in number, is called an infinite series.
The sum of the first n terms of a sequence we denote by Sn-
Then
Sn-Ui+U2 + U3+ • • • +Un.
As n increases and we form the sum of more and more terms of
the sequence, one of three alternatives is open to Sn, namely:
(a) Sn approaches a fixed limit S, which is then called the sum
of the infinite series, and the series is said to converge.
(b) Sn increases without limit; the infinite series then has no
sum and is said to diverge.
(c) Sn oscillates; the infinite series has no sum but oscillates,
and is again said to diverge.
171
172 INFINITE SERIES [197
Examples.
Ill 1
(a) 2 + 2^ + 23+ • • • +2^+ ' • ' •
Sn = 2 + 2^+ • • • +2^~2 i-1 ~-^""W • ^^^-^
lim Sn = I = S, The series converges to the value 1,
n * 00
or, ^ + ^+ • • • +^+ •••=!. [(188), figure.]
(b) l + 2 + 3+---+n+---.
iS» = 1 + 2 + 3+ • • • +n; then obviously Sn increases with-
out limit as more and more terms are added. Hence the given
series has no sum, and diverges.
(c) 1-1 + 1-1+---.
Here Si = 1; iSf2 = 1 - 1 = 0; Sa = 1 - 1 + 1 = 1; S4 = 0; and
so on indefinitely. Sn oscillates from to 1 as n varies, the series
is oscillatory and has no sum. We say that it diverges.
197. To show that an infinite series converges, it must he shown
that Sny the sum of its first n termsy approaches a definite limit as n
increases indefinitely. When such limit does not exist, the series is
divergent.
The direct method of determining whether a given series con-
verges or diverges is to form the sum of its first n terms Sn, and let
n increase indefinitely. This method is applicable only in the
few cases where a formula for Sn is available. The standard case
is that of the infinite geometric progression.
. • •
a + ar + ar^ + • • • + af^'^ +
1-r"
Here Sn = a + ar + ar^ + • • • + ar^-^ = a
1-r
When r is numerically less than 1, i.e., \r\ < 1, then r" approaches
as n increases and
lim Sn = a ^ = S.
n— 00 1 r
When r = 1,
Sn = a + a+ ' • • +a ^ na.
198, 199] INFINITE SERIES 173
Hence Sn increases without limit when n increases. When
I r I > 1, r** increases indefinitely with n; hence /S„ does the same.
Therefore, the geometric series, a^ ar-\' ar^-\' • • • > converges when
I !• I < 1, and diverges when \r\ i 1.
Putting a = 1 , we see that the simple power series, 1 +x+x^+ • • • ,
converges when | a? | < 1 and diverges when | a? | ^ i .
198. We next consider indirect methods for establishing the
convergence or divergence of a given infinite series.
Theorem 1. When an infinite series converges, its nth term ap-
proaches zero as a limit when n increases.
Proof. Let the convergent series be 1/1+1^2+^3+ • • • +Un+ • • • .
Then Sn = Wi+ti2+ • • • +Wn and /Sin-l = Wi+t^2+ • • • +Wn-1.
Hence Wn = Sn — Sn-i.
By taking n large enough, both Sn and Sn-\ can be made to
differ from the sum of the series and hence from each other by
as little as we please; hence their difference, Um can be made to
differ from zero by less than any assignable small quantity.
lim Un = 0.
n — 00
This is a necessary condition for the convergence of any series.
Test for Divergence. — From Theorem 1 we infer that an infinite
series diverges whenever lim Un 9^ 0.
00
199. Alternating Series. — A series whose terms are alternately
+ and — is called an aUemaiing series.
Theorem 2.. An aUemating series converges provided thai (a)
each term is numerically less than the preceding, and (b) the limit
of the rUh term is zero as n increases indefinitely.
Proof. Let the series be
ui —u^ + uz — u^ + U5'-ue+ • • • .
Write this in the two forms,
(ui - U2) + (m3 - U4) + (us - -oe) + • • • ;
Ui — (U2 — Uz) — (U4 — Us) —
...
Each set of parentheses incloses a positive quantity according to
condition (a) of the theorem; hence assuming that ui, U2, Us, . . .
are themselves positive quantities, the first form shows that the
174 INFINITE SERIES [200,201
sum of the series is positive, i.e., > 0, and the second that the
sum is less than the first term ui. Also, since lim Un = 0, the sum
cannot oscillate. Hence the series converges to a value between
and its first term.
Example. The alternating series,
i-§ + i-i+ • • •
converges to a value between and 1.
200. Absolute Convergence. — A series is said to converge abso^
Ivtely when it remains convergent if aUits terms are taken positively.
Thus if ui, U2, W3, . . . be in part negative and in part positive,
the series
converges absolutely provided that the series
I Wl I + I W2 I + I t^3 I + • • •
converges.
Exercise. Show that the series
1 + X + x^ + • • • and a + ox'^' ax^ +
both converge absolutely when | a: | < 1.
201. The Comparison Test.
Let wi + W2 + W3 + • • •
be a series known to converge absolutely or to diverge.
Let »i + V2 + f 3 +
. • *
...
be a series to be tested for convergence or divergence. Then,
(a) If the Urseries converges absolutely and, for all values of n, Vn is
numerically less than Un, the v-series also converges absolutely;
(b) If the Vrseries diverges and Vn is numerically greater than u^
and if all the terms of the v-series have the same sign, the v-series also
diverges.
Proof. •
Let Un^\ui\ + \U2\ + \Us\+ ' ' ' +\Un\
and Vn=\vi\ + \V2\ + \V3\+ ' ' ' +\vnl
Then by condition (a), Un approaches a limit, say U,8ian ^co^
and also, Vn < Un- Hence, since Vn must increase steadily with
201] INFINITE SERIES 175
n, but is always less than Un, it must approach a limit V, less than
U. Hence the v-series converges.
Under condition (b), Un increases without limit, and also,
Vn > Un- Hence Vn also increases without limit and the t^«e^ies
diverges.
Standard Test Series. (For use in Comparison Test.)
(1) a + ax + ax^ + • • • + aa;** +•••,) Conv. when | a: | < 1;
(2) l + x + x^+ . . . +x''+ . . . , i Div. when I a; I ^ 1.
(3) 1 + 22 + 23+' ' '+2^"*"* ' ' ' Convergent.
(4) l + 2 + g+- ••+-+•••, Divergent.
(5) 1 + 1 + 1+...+1+. .., jConv.whenp>l;
W pTgp-rgp-r ^n^^ ' i Div. when p ^ 1.
The first three of these series are geometric progressions and have
already been considered.
Series (4) can be shown to diverge by grouping its terms thus:
i+i+a + i)+(i+j+ ++i)+a+iV+ • • • +^js)+ • • • .
We can form in this way an infinite number of parentheses, each of
which is > i. Hence the sum is infinite.
Series (5) is, term for term, greater than or equal to (4), when
p = 1 ; hence for these values of p the series diverges, by condition
(b) above. When p > 1, the series is shown to converge by
grouping its terms as follows:
1 + (1 + L) + (1+. . . +1.)+(1+ . . . + J-U . . . .
IP ' \2^ ~ 3py ' \AP ' "^ jpj ' \gp ' ' 15P/ ~
Considering each group of terms as a single quantity, we see that
this series is less, term for term, than the series
1+2 4 8
•^ 2^ 4^ 8^ '
-,1,1,1,
or 1 + 2^rn + 4^31 + g^ri + • • • .
But this is a G. P. with ratio o^zt> aud hence converges. There-
fore the given series converges.
176 INFINITE SERIES [202
Examples,
1. The series l+p4-oi+* * ' '*"n»"^* ' * converges; for it is less,
term for term, than (3).
2. The series 1 + -, ^ + -, 5 + • • • + i h • • • diverges: for
logio2 logio3 logion^ * '
it is greater, term for term, than (4).
202. The Ratio Test. — The aeries U1+U2+U3+ . . • +w„+ • • •
converges absoliUely if, beginning at some point in the series, the
ratio Un -r- Wn-i becomes and remains numerically less than a fixed
positive number which is itself less than 1.
Proof. Assume that
< r < 1 f or all values of n > JV,
Wn-l
N being a fixed positive integer.
Then • | Wn | < ^ I Un-i \ when n> N.
Hence putting n = iV + 1, iV + 2, . . . , we have
\uN+i\<r\uN\;
I un+2 \<r\ un+i \<r^\uN\;
un+zI <r\uN+2\ <r^\uN\;
Adding, we have
I UN+i I + I un+2 I + un+ sI + • • • < I wjv I (r + r2 + r^ + . . .)-
Writing the given series in two parts,
(U1+U2+ ' • ' +Un)+ (Un+1 + Un+2 + UN+3 +'''),
we see that the first part, formed of N terms where iV is a fixed
finite integer, must have a finite sum. The second part cannot
exceed the left member of the last inequality above, hence is less
than the right member of that inequality. But the series r'\-7^ +
r^ + • • • converges and has a finite sum, since it is a G. P. with
ratio r < 1. Hence the sum in the second pair of parentheses has
a finite limit, and the given series converges.
Similarly it can be shown that the series diverges when the test-
ratio Un -^ Un-1 becomes and remains greater than 1, or even
when it approaches 1 from the upper side.
203]
INFINITE SERIES
177
When the test-ratio Un -J- Wn-i is at first less than 1, but
approaches 1 as n increases, this method gives no information
about the series.
Examplea,
1
1. 1+ J_+-4^+
• • •
1-2 ' 1.2.3
+
1 .2. 3
• • •
n
+
Here
Un
~ - , which approaches as n *^ oo . Hence the ratio test
Un-l
is satisfied and the series converges.
2. sin X -h 2 sin* a; -f- 3 sin' x +
+ n sin'* X +
Un
Un-l
n
n sin** X
(n — 1) sin'*"!^
n
n - 1
sin a; [ .
As n * 00, — ^^ •*■!, and if we chopse x differ^it from an odd multiple of
ni so that I sin :c I < 1, we can take n so large that the test-ratio will be
less than r, where r is less than 1. We need only take x < sin-i r
n
Hence the series converges for any value of x which is not a multiple of r
»• i+i+i+
+1 +
n
Un
Un-\
fails.
n - 1
n
, which approaches 1 from the lower side. Hence the test
*• 2^3+4 +
+
Un
Un-1
n
n + 1
+
n
n - 1
n
n2
n2 - 1
Here the test-ratio is greater than 1, approaching 1 from the upper side
as n *»■ 00. Hence the series diverges. This series may also be shown to
diverge by comparison with (4) of (201).
203. Exercises. Test the following series:
1. 1 +22 +P +
2. l+J+f +
n -f- 1
a 1 + 2 a; -h 3 x2 -f. . . . -^.' (^ + 1) x» -h • . •
4. cos X + cos2 X + • • • + cos** »+•••.
6. tan X + tan2 a; + • • • -f- tan" x + • • • .
6. sin-ix + (sin-ix)2 + •. • • + (sin-ix)** +
7. logio X + (logio x)2 + • . • + (logio x)» + •
8. Ll2 ,2^ ,3^ . ...
178 INFINITE SERIES [203
fi I.I.I.
• • •
*
10.* |1 a; -h [2 a;2 + [3 a;» + • • • + [n a?« +
4A x' , a;* a:^ ,
«A 1 ic^ x^ aj*
^^ ^"|2+"[i"|6+ ••••
iA 1 1 .1-3 1'3«5 , . I.3»5»7 ^
16. X — }a^+Jx3 — Ja4-j- . . .
* [n s 1 . 2 • 3 • • • • »»•
CHAPTER XIII
Functions. Derivatives. Maclaurin's Series
204. Functions. — Let x denote a variable quantity and y a
quantity whose value depends on that of x. Then y is said to be
a junction of x. Thus
2/ = x^ + 1, 2/ = ^^ 2/ = sin {ox + 6)
are all functions of x.
As an equation, we indicate that i/ is a function of x by writing
When a body is dropped from rest, the space s (ft.) fallen
through in the time t (seconds) is s = § g^. Here s is a function
of <, or
When a train is running at 30 miles an hour, the space s (miles)
covered in the time t (hours) is s = 30 L Hence
s=f{t); f(t)=-SOL
When the relation between y and x is given by an equation of
the form y = f(x), y ia called an explicit function of x.
Suppose the relation between x and 2/ to be given in the form,
x2 + 2/2 = 1.
Here y is not given directly in terms of x, but nevertheless the
value of y depends on that of x; for when we substitute for x first
one value and then another we get in general different values of y
on solving the equation. In such case y is called an implicit
function of x.
As other examples, we have
2/2 = 4 x; sin (x + 2/) = 1 ; a' + a^ = b,
206. Variation of Functions. — Consider the relation y = z^.
When X = a, then y = a^; when x = a + h, 2/=(a + A)2.
179
180 DERIVATIVES [206
As X changes from a to a + A, i/ changes from c? to (a + A)-.
The total change in x is A, and the corresponding change in y is
(a + /i)2 - a2 or 2 aA + A2.
Let us designate a change in a: by Ax (read " increment of x,"
or " delta x ") so that in this example Ax = A; let the corre-
sponding change in 2/ be Ay, so that we have in this case
A2/ = 2aA + A2 = 2aAx + Ax^.
In general, if y = f{x), then to the values x and x + Ax of the
variable x correspond the values / (x) and / (x + Ax)* of y. Hence
the change in y, corresponding to the change Ax in x, is
^y =/(^ + ^)-/(^)-
Continuous Function. — When Aj/ = with Ax, y is called a
continuous function of x. We assume all our functions to be
. continuous unless the contrary is stated.
Exercises.
1. Given y ^ x^. Calculate Ay, when x = 2 and Ax = 0.1.
2. As in exercise 1, when y — Vx.
3. As in exercise 1, when y — x^.
4. As in exercise 1, when y = 10*.
6. Given y = sin x. Calculate Ay, when x = 45® and Ax = 5®.
6. As in 5, when x = 30° and Ax = 1°.
7. As in 5, when x = 1 and Ax = 0.01.
206. Difference Quotient. — The fraction
change in y Ly -
, y or ■T'— >
change m oc Lac
is called the difference quotient of y relative to x.
Thus, if 2/ = x^j then Ay = (x + Ax)^ — x^ = 2 x Ax + Ax .
Hence the difference quotient is
Ay 2 X Ax + Ax . -. , .
-^ = —^ = 2 X + Ax.
Ax Ax
We shall abbreviate Difference Quotient by writing D. Q.
Exercises. Calculate the D.Q. in the exercises of (206).
* / (x + Ax) stands for the result obtained by replacing x by x + Ax in /(x).
207, 208]
DERIVATIVES
181
M* X
207. The D.Q., t^, geometrically. — Let the curve in the fig-
ure represent a part of the graph of the
equation y = fix).
Let P be a point on the curve hav-
ing coordinates (x = OM, y = MP)^
and P' a second point (x + Ax = OM',
y+Ay = M'P'),
Let the secant PP' make an angle 6' with the x-axis.
Draw PQ II OX. Then from A PQP\
tan^' = ^.
Ax
Slope. — The tangent of the angle which a line makes with the
X-axis is called the shpe of the line.
Hence, the difference quotient, ^— , is the slope of the secant drawn
through the points (a?, y) and (x + Aa?, y + Ay).
208. Limit of D. Q. = Slope of Tangent. — Let the point P'
move back along the curve and approach the point P. Then Ax,
and in general also Ay, approach 0.
( Suppose now that as Ax approaches the D. Q. approaches a
definite limit, m.
Then the line through the point (x, y) having the slope m is
called the tangent to the curve y = fix), (x, y) being the point of
contact.
In the figure, as P' approaches P, the secant line PP' gradually
rotates about P and approaches a limiting position PT, which is
defined to be the tangent to the curve at P.*
If 6 be the angle which the tangent to the curve at P = (x, y)
makes with the x-axis, then
tan 6
Aa!.!.o\Aa;/
read, '^tangent of 6 equals the
Ay
limit of ^.as Loo approaches 0."
When -p approaches a definite limit a tangent is thereby deter-
mined. When such limit is indeterminate, the tangent does not
exist, or several tangents may be drawn at P. We shall consider
only cases where a single determinate tangent exists.
182
DERIVATIVES
209. Examples.
1. y = fl;2.
y + Ay = (a? + Ax)*
= x2 +2ajAaj + A?.
/. 'Ay=2xAaj + A?
and -r- =2a;+Ax.
Aj;
[209
Hence
lim ^=2a; = tan^.
y «" x*
(an d ^2x
Here the slope of the tangent at any point
equals twice the abscissa.
2. y^ifxK
y + ^y ^ir{x+ ax)»
= irVCa:^ +3fl;2Ax+3a;A? + A?).
Ay = A- (3x2 Ax + Sx Ax2 + ^8)
and || = 3rV(3x2 + 3xAx + A?).
Hence
lim T? = qX2 = tan e.
(an 9 » i X2
8. y = x2 -2x.
y + Ay =(x + Ax)2 - 2 (x + Ax)
=x2+2xAx+A?-2x-2Ax
=x2-2x+ (2x-2) Ax + Aa?.
Ay =(2x-2)Ax + A?
and
y
tanO
x2-2x
2x-2
f^ = (2x-2)+Ax.
Ay
lim — ^ = 2x — 2 — tan^.
4. y2 = X. Here y is an implicit func-
tion of X. Solving, we have
y = db Vx.
The upper sign gives that part of the curve lying above the x-axis, the
lower sign the part below the axis. We consider first the upper sign only.
209]
DERIVATIVES
183
Then
y = V5 and y + Ay «* ^x + Ax.
Ay « *^x + Ax — V^J.
Multiplying and dividing by
^x + Ax+ Ax, we get
Ay
(Va;4-Aa; — %^)(Va;+Aa;+Vg)
V35 + Ax + ^
Ax
Vx4- Ax + Vx
y* =» X. ton ^ = ±
2Vi
Hence ^ = ^ p.
Ax V 3J + Ax + V^
and lim -r^ = — — — tan ^.
ax^qAx 2yx,
For the lower part of the curve^ replace Vx by — V^.
6. x2 + y8 = 100.
Solving for y, we get
>t 4% "\
X
M
y = ± VIOO - x2.
Considering first only the upper half of
the circle (figure) we have
y = VlOO - x2 ;
y + Ay = VlOO - (x + Ax)*.
.-. Ay = VlOO - (x + Ax)2 - VlOO - x2.
Multiplying and dividing by the sum of
the two radicalis,
100
X
-2xAx — Ax?
Vl00-x»
Hence
and
^ VlOO -(x + Ax)2 + VlOO - aJ2
- Ay 2x + Ax
^~ VlOO -(x + Ax)2 + VlOO - x2
Aa:-.oAX 2 VlOO - X? VlOO - x2
At any point on the lower half of the circle, tan =-\-
SB tan^.
X
VlOO - x2
In all these examples the slope of the tangent at any ^ven point may be
obtained by substituting the abscissa of the point in the value of tan 0.
Exercises. Calculate the slopes of the tangents at any point (x, y) on the
following curves:
1. y = J x». 4. y2 = 4 X. 7. x* - y« = 1.
2. y = 2 x2 - 3 X. 6. y2 = - 9 X. 8. 9 x* + 16 y« = 144.
8. y = x» - X. 6. x2 + y2 = 1. 9. 4 x* - y2 a 4.
Calculate the slope in each of these examples when x » 1. Note the
results in exercises 6 and 7 and explain.
184 DERIVATIVES [210, 211
210. Derivative. — The expression lim f -pj occurs so frequently
in mathematics that a special name is applied to it. Starting
with y as any given function of x, say /(x), we can derive from
this a second function of x as follows. Calculate/ (x + Ax) — / {x)
or Aj/, divide by Ax, and pass to the limit by allowing Ax to approach
zero. Call the new function of x so obtained /'(x), so that
^'«=i:?.(B)
This is called ihiQ first derived function off{x) or the first derivative
of fix), and the expression
r-0 \AX/
lim
Ax
is called the first derivative of y vrith respect to x. It is usually
written in one of the forms
.!i?o(sl)=^-^'- =
dp
Hence the slope of the tangent to the curve j/ = /(x) at a point
(x, y) is
dy
tane = l>«,y = -T--
211. Calculation of Derivatives. — We have already calcu-
lated the derivative of y with respect to x in a number of cases.
We now obtain a few simple formulas for the calculation of deriva-
tives. Three steps are involved in every case: (1) the ccdcukUion
of Ay, (2) division by Ax, (3) evaluation of the limit as Ax = 0.
We shall assume that such a limit exists.
Formulas for Calculating Derivatives.
I. l>a, (c) = 0, c being a constant.
(1) For if c is a constant its change is 0, hence Ac = 0.
(2) Therefore T^ = 0-
I iX
Ac
(3) Hence lim =^ = or D, (c) = 0.
^^J^Q UX
211 ] DERIVATIVES 185
II. !>«, (cy) = c I>»y^ c being any constant.
Proof.
(1) The increment in y being Ay, the increment in cy will be c Ay,
(2) Dividing by Ax, the D. Q. of cy relative to x is c-^-
At/
(3). Let Ax = 0. Then c does not change, while -r^ becomes
Djf (y). Hence
Dx (cy) = lim c -r^ = c D:,y.
III. When y is a sum of several functions of x, as
y = u + v + U) + ' * * , where u, v, w, . . .
are functions of x, then
Da^y = n„u + n„v + n„w + • • • .
Proof, When x takes an increment Ax, let the corresponding
changes in w, v, ti?, . . . be Au, Av, Aw, . . . respectively. The
total change in y is, therefore,
(1) Ay = Au + At; + Aw +
• . '-
/•<>^ Tk»« Ay Au Av Aw
(2) Then ■^ = :^ + ^ + ^ + ' ' ' '
Ay
(3) Let Ax = 0. Then by definition (210), -p approaches Dxy,
— approaches DxU, etc. Hence
Dxy= DxU + Dj^ + Dxif? + • • • , when y = w + t; + ii?+-««.
IV. Let y be the product of two continuous functions of x,
say u and v.
y = u^v.
When X is changed to x + Ax, let u change to ti + Aw and v to
V + Av. Then
y +.Ay = (u + Au) (v + At;) = ut; + w At; + 1; Au + AuAt;.
(1) Hence Ay = u At; + 1; Au + Au At;.
/r»\ mv ^2/ At; , Au . ^ At;
(2) Then a^ = "Zi + ^A^ + ^^Ai-
186 DERIVATIVES [212
(3) Let Ax = 0. Then xf ' T"' T~ approach D^-y, D^u and D^t;
respectively. Also Aw = 0, since we assume u to be a continuous
function of x (206). Hence (2) becomes
J^rnV = w J^m'^ + V I>^u^ when y = w • r.
V. Let y = -, w and v being continuous functions of x.
Then y + Ay = — nr'
^ ^ r + Aw
,^. J . w + Aw w vAw — wAr
(1) and Ay = — r— : = —0-1 — r~'
^ '^ *^ r + Aw t; t;^ + 1; Ay
Au At;
f> u —
,^. XT Av Ax Ax
(2) Hence ir = ""^^ — ^—
^ ' Ax r-^ + rAr
(3) and l>«,y = lim -r^ = — 2 — ^ SL..
VI. Let y be a function of w, where u is a function of x. Thus
y — v? + 2w; w = 2x2 + 1.
When X changes to x + Ax, u changes to w + Aw and y to
y + Ay.
Now 4» = ^.^.
Ax Aw Ax
Hence l>»y = J>uy • ^»«*-
Collecting our formulas we have:
(A) l>«c = 0.
(B) 1>« {cy) = c l>«y.
(C) l>a, (tt + t; + f«; + • • • ) = -^»«* + -^«v + -'^o)^ + • • • .
(D) l>a, ( W V) = W l>a,V + V l>a,t*.
(E)X>,(g) = --^'----^-- .
(F) l>«,y = l>uy • l>a,tt.
212. We next derive the following standard formulas:
(G) y =»^; !>«?/ = na?^-^
(H) y = logi»; ^«y = |-
212] DERIVATIVES 187
(I) y = a*; D^y * a*log a.
( J) y = sinfic; l>«,y = cosa?.
(K) y = cosap; l>«y = — sina?.
(G) 2/ = x"; assume n to be a positive integer.
(1) Hence A2/ = na;~-i Ax + ^^^^-^^ • . . + Ax".
(2) Then ^ = nx«-i +'?^^P^^a:«-2 Ax + • • • + Ai"-\
Ax 1-2
(3) Let Ax = 0. All terms on the right of the last equation
vanish except the first, and
lim ^ = D,2/ = wx**-!.
Ax^O Ax
The proof when n is not a positive integer will be given after
formula (H) is derived.
(H) y = logx; y + Ay = log (x + Ax).
(1) A2/ = log (x + Ax) - log X = log^-^ — = logf 1 + — j-
1 JC_
<^)i=i-('+f)=-('+fr=^'-('+f:
(3) Let Ax = 0. We must evaluate
;
,. , / AxW
lim log ( 1 H I •
Ax
X
Let 2 =-r~5 then 2 = oo when Ax = 0, provided x 5*^ 0. [x =
Ax
is excluded by our standing assumption of continuity (206).] We
must now evaluate
lim (1 + iy.
Let 2 = 1, 2, 3, . . . , n. The corresponding values of f 1 + -)
are 2, 2.25, 2.37, ...,fl + -j. Asn increases, these values
188 DERIVATIVES [212
steadily increase, but always remain less than 3, no matter how
large n may be. For, by the Binomial Theorem,
V^n) *^"n^ 1-2 n2^ 1.2-3 n^^
to (n + 1) terms
_^ ,^ rV n/ \ n/\ n/ Vto(n+l)
"^■^ 1-2 "^ 1-2.3 "^''"henns.
As n increases, each term of the expansion increases as well as
the number of terms. Also all the terms are positive. Hence
their sum increases with n. Further compare the above expansion,
leaving out the first term (=1), with the geometric progression
■•^ ' 2 ' 2^ ' ' ' ~r 2» - 1 '
/ 1 _ (l)n \
whose sum is less than 2. [S = _ v *)
For all values of n, however large, our expansion is less, term for
term, than the progression. As n = oo , the sum of the progression
approaches 2, hence the expansion, excepting its first term, ap-
proaches a limit less than 2. Adding the first term, the limit is
less than 3.
This limit is an irrational number denoted by the letter e, and
has the approximate value
e = 2.7182818 + • • • .
We have now the result that
s('+.^)'=
lim
2
when 2 approaches infinity through positive integral values. The
same is true when z increases continuously, but we shall not stop
for the proof, which may be found in texts on the calculus.
Then lim log f 1 + - j = log e,
and hence Dx (log x)= - log e.
212 ] DERIVATIVES 189
Let us now take e as the base of our system of logarithms, so that
log a; shall mean logeX. Then
loge = log^e = 1.
Hence Dx (log x)= --
X
Logarithms to the base e are called natural or Naperian loga-
rithms. In the theory of mathematics natural logarithms are in
general use, common logarithms, to the base 10, being utilized only
for numerical computation.
We can now derive formula (G) without any restriction on the
value of n.
From 2/ = x**
we have log j/ = n log x. (Base e.)
Hence D^ (log y) = D^ (n log x).
Now in formula (F) replace y by log y and u by y» It becomes
2)x(log y) = DyQog y) • D,y = ^Z>,(2/), from (H).
If
Also Dx(n log x) = - > from (B) and (H).
X
y X
• •
or Dxy = — » where y = x**.
X
nx^
Hence Da^** = — = nx^"^.
X
a) y = «".
Taking logarithms, log y = x log a.
Hence Dx (log y) = Dx {x log a).
But Dx (log y) = -Dxy (see above)
and Dx {x log a) = log a.
Hence - I>x2/ = log a,
2/
or Dxy = y log a, where y = a*.
I
190 DERIVATIVES [213
i
Therefore D^* = a* log a.
(J) y = sin x; y + Ay = sin (x + Ax).
(1) Ay = sin (x + Ax) — sin x = 2 cos f x + -^ j sin -^ . (158.)
^ / , Ax\ .Ax .Ax
. 2cos(x + -TT-lsm-^r- , ^ vSm-jr-
W -7— = T = cos I X H tr- I 7
Ax Ax \ ^ 2 / Ax
(3) Let Ax = 0. Then
. Ax
sm
(Ax\ * 2 / Ax \
X + -^ ) = cos X, and —r — = 1. f IfiO. Replace a; by-^- 1
2
A^/
AX i^
lim-r^ = !>«, sina? = cosa;.
(K) y s= cos x; y + Ai/ = cos (x + Ax).
(1) Ay =cos (x +Ax) — cos x= — 2 sin (x + -rt" ) sin -h-* (168.)
. Ax
2
(3) .*. lim -r^ = !>«, cos a? = — sin a?.
Ax-O Ax
By suitable combinations of formulas (A) to (K) the derivative
of any function may be calculated.
213. Examples.
1. Calculate D, (4x3 + 3 re).
Dx(4a^ + 3x)=/>x(4a;8)+Dx(3fl;) (C)
= 4 Dxx^ + 3 Dxx (B)
-12x2+3. (G) ,
2. Calculate
^nr+fc)
I
214, 215] DERIVATIVES 191
n ( ^ \ _ (1 + log a?) Z>xe» - e»Da^ (1 + log g) .-.
^^U+logx/"" (l+logx)« ^^^
(1 + log a;) e* — e* -
- (i+iogx)« "^ a).(c).(H).
^^^ a?(l+loga;) -1
a; (1 + log x)^
3. Calculate Dx(3sin*x).
Dx(3 8in«fl;) =3/>xsin«x (B)
s 6 sin X />x sin a; (F) ; (ti » sin x) *^
» 6 sin a; cos x. . ^ '*• v ^U ^
214. Exercises. Calculate D^j^ when:
1. y = 3a^'+5a:S. 10. y = log (x + 2).
2. y = a:» + ^.
1. IL y = log(3x2-l).
4 x4.il 12, y = <r* log x;
_i _ 1 18. y » sinxlogcoBX.
4.y = x'-2x'.
J J 14. y = cBin*.
Vx -VS 15. y = tanxf = ?!I^Y
6. y = 8inx + e»*. ^ /^^^^
7. y^e^, 1^ y = cotx.
8« y^a^. 17. y a logtanx.
9. y = co8^ + ^^;^- 18. y^secx.
216. The Derivative as a Rate of Change. — The difference
quotient-^ gives the average rate of change of y relative to x when.
X changes by an amount Ax. The smaller Ax, the more nearly will
the D. Q. represent the actual (or instantaneous) rate of change
of y relative to x. Hence the limit of the D. Q. as Ax = is taken
as the actual raie of change.
Rule. To find the raie of change of one quantity relative to another,
calculate the derivative of the first quantity voith respect to the second.
Examples,
1. y = ai*. Then Dxy = 2 x. *
Hence y changes 2 x times as fast as x.
192 DERIVATIVES [216,217
2. In the case of a falling body, if s be the space and t the time and the
body starts from rest, we have
Then Dts — gt = velocity at time t
8. Find the rate of change of the volume of a sphere relative to the radius.
7 = jTr»; />rV = 4irr2.
That is, the volume of a sphere changes 4 nr^ times as fast as the radius.
216. Exercises. Calculate the rate of change of:
1. y relative to x, when y == x^ -\- z^,
2. y relative to a;, when y ^sinx.
3. y relative to x, when f = sin x cos x.
4. y relative to x, when y — sin^ x + co82 x,
6. y relative to x, when y = e*.
6. the volume of a cube relative to its edge.
7. the surface of a cube relative to its edge.
8. the surface of a sphere relative to its radius.
9. the volume of a cylinder relative to its altitude.
10. the volume of a cone relative to the radius of its base. ,
11. the area of a circle relative to its perimeter.
12. A body starts when ^ » and moves so that the space described in
time t (seconds) is s = 16 ^24- 10. Find its velocity when t - \0\ t — b) t — 0,
13. The space-time equation being a ^21? -{-Zt — h^ find the velocity at
any time t\ what is it when t — 10; t = 1; < = ?
14. As in 13, when 8 = 10 sin ( 3 < + j)«
16. Given two sides and the included angle of a triangle. Calculate the
rate of change of the third side relative to each of the ^ven sides and to the
given angle.
217. Higher Derivatiyes. — When j/ is a function of x, D^y is in
general a new function of x; the derivative of this new function
is called the second derivative of y with respect to x and is written
Z)|y. The derivative of the second derivative is called the third
derivative, written D^y, and so on.
Examples.
1. y^x?. Dxy^Sx^; D|2/'=6a:; I>x2/ = 6 ; D^y^O.
2. 1/ = sin X. Dxy = cosa;; D^y == — sin x ; Dly = — cos a; ; etc.
3. 2/ = x». 2)x^= nxn-\; Dly = w (n - l)x'»-2 ; . . . .
Dly = n (n - 1) . . . 1 = [n.
• • •
218] MACLAURIN'S SERIES 193
218. Maclaurin's Series. — Suppose that a given function of
Xf f(x)j can be represented by a converging power series in x, thus:
To find the values of the coefficients Co, ci, C2 • • • . Puta: =
in (1) and we have Co determined by
/(0)=Co.
To get ci, calculate DJ(x) ov f\x) from (1);
(2) f\x) = Ci + 2 C2X + 3 C3X2 + . . . + ncnx""-^ +
Put a; = in (2) and we have ci determined by /'(O) = cj .
From (2) calculate DJ'{x) or /"(x);
(3) /"(x)=2c2 + 2.3c3X+ • • • +n(n-l)a;'»-2 +
Put X = in (3) and we have
r(0)=2c2 or C2 = ^r(0).
Calculating Dxf'ix), or /'"(x), we have
(4) /'"(x) = 2 • 3 C3 + • • • +n(n-l)(n-2)a:^-3 +
When X = 0, /'"(O) = 2 • 3 cs ; C3 = gTs/'^W.
• • •
• • •
Similarly,
Hence
• •••••
c« = -7 — A- T /^"KO) = r /^"^(o)-
n(n — l)...l' [n
/w2 /j^S op**
/(a:) = /(0)+x/'(0)+|/"(0)+|/"'(0)+ • • • +^/<»H0)+ • • • .
Here /^"KO) is found by diflferentiating /(x) n times in succession
and putting x = in the result.
The above result is called Maclaurin's series for the function
/(x). In obtaining it we have tacitly assumed that, if /(x) be
represented by a power series, the derivative /'(x) can be calcu-
lated by differentiating the series term by term.
194 MACLAURIN'S SERIES [219
219. Examples.
1. Develop e' in a i>ower series in x.
f(x)^€^; rix)^^; /"(x)-e»; . . . ; /<n)(rc) = e».
Putting X I- 0, we have
/(0)-i; AO)-!; r(0)-i; . . . ; /^»>(0)=i.
Hence
aJ2 X* x^
This series converges for all values of x, and is used for calculating the value
of e' to any desired degree of approximation.
When X = 1,
*-i+i+ii+ji + --- +||+
from which e can be found approximately by taking a few terms of the series.
2. Develop sin x in a power series in x.
/ (x) - sin x; f(x) » cos x; /"(x) = — sin x; /"'(x) = — cos x, . . . .
When X = 0,
/(0) = 0; /'((i^«l; r(0) = 0; r'(O) = - 1, etc.
Hence
x* , x*^ x^ ,
This series converges for every value of x, and may be used for finding sin x
to any degree of approximation. Thus, put
X = 10** = ^ radians.
Then
sm
^'""-^-likhM^J-
Note. In computing xcUh an altemating aeries {signs aUematdy + and ^),
the error committed in using only a few of the first terms of the series is altoays
numericaliy less than the first term neglected.
Thus the error in sin 10° as obtained from the three terms written above is
less than
040 V18/'
^^. ,^. .or less than .00000000098.
Hence the error is less than 1 unit in the ninth decimal place.
Exercise. Show that
- X* , x* afi .
«**"i-(2+ii-[6 + "" •
Calculate cos 10° to five places.
220] BINOMIAL THEOREM 195
8. Develop log (1 + x) in powers of x,
/(O)=logl-0.
/'(0)=1.
f(.x)'
log(l+x);
rix)"
1
1+x'
fix) =
1 .
(!+»)«'
r'ix) -
2 .
(.1+x)*'
/^(x)-
-2-3
/"(O) = - 1.
/'"(O) = 2.
/»v(0) = -2.3.
^ 26^ X^
log(l + a;)= ^-2^+-3~-4+ ' ' ' •.
This series converges only when — 1 < x ^ 1, and hence can be used only
when X lies between —1 and +1 and for x = +1.
Since the base of the logarithm system in log (1 + ^) is under-
stood to be e, the last series enables us to calculate the natural
or Naperian logarithms of numbers from to 2, exclusive of 0.
For 1+x ranges from to 2 when x ranges from —1 to +1. Jn
particular, when x = 1 we have
log, 2 = l-J + i-i+---.
This is a convergent alternating series. Since in such a series
the error committed by neglecting all terms after a given one is
less than that term (199)*, 1000 terms of the series would be required
to give log 2 correct to three decimal places. The series therefore
converges too slowly for practical use. A more serviceable series
will be considered in the next chapter.
220. The Binomial Theorem. — When n is a positive integer, we
have
(l+g)~ = l + na: + ^^^"^^^ g2+ • • • +x\
We shall now derive the formula for expanding (1 + ^)^ in
powers of x for any value of n, positive or negative, integral or
non-integral.
Let f{x)=(l + x)\
* Apply (199) to the neglected part of the given series.
196 BINOMIAL THEOREM [220
Then
/'(x) = n (1 + x)»- » ; /'(O) = n.
/"(x) = n (n - 1) (1 + a;)"-2; /"(O) = n (n - 1)
/'"(z) = n (n - 1) (n - 2) (1 + z)""' ; /'"(O) = n (n - 1) (n - 2).
/<«) (x)= n (n - 1) (n - 2) . . . (n - m + 1) x**"";
/^«)(0)=n(n-l)(n-2) . . . (n-m + 1).
Hence by Maclaurin's series,
(l+a?)** = l+na?+ ^^^ ' oi? -^ 1.2.3 ^a^+- • '
, n (n — 1) • • • (n — wt + 1) ^ ,
' 1 •» • . . . m
. • .
»
provided that the series on the right, called the Binomial Series,
converges.
Convergence of the Binomial Series. — Denote the mth term
of the series by Um, the (w + l)th term by Wm+i- Then
_ n(n~l)(n-2) . . . (n-m + 2) ,.
^'^ ~ 1 . 2 . 3 • . . . (w - 1) ^ '
_ n (n - 1) (n - 2) . . . (n - m + 2) (n - m + 1)
^'"^^^ 1.2.3. . . . (m-l).m "*" •
Applying the ratio-test (202), we have
Wto+1 n — m +
Urn. m
"tn
'-'-H'-^h
The quantity in the last parenthesis is numerimUy less than 1,
when m is larger than n + 1 ; to secure this we simply start far
enough out in the series to make m > n + 1. Then the ratio
Wm+i "^ w,n will be numericdlly less than x, and hence, if x be
numerically less than 1, the series converges. When x is numeri-
cally greater than 1, the series diverges. For the ratio Wm+i "^ ^w
equals the product of two factors, [ Ij and x. As m
increases the first factor approaches — 1 as a limit. Hence if
|a:| >1, the product will also ultimately be greater than 1 nimier-
ically. Finally, when x = ± 1 our binomial reduces to 2** or
respectively and we need not consider the series at all.
2211 BINOMIAL THEOREM 197
We therefore use the binomial series for (1 + xY only when
|X|<1.
221. Binomial Series for (a + J^)"* — We have
(o + 6)» = a» (l + ^)"
\ o^ 1 . 2 o2^ ^ 1 . 2 .... m o"^ /
or,
(a + 6)" = o" + ««»-» 6 + 2i2_ila— »5« + . . .
_^ «(n-l) . . ♦ (n-w + 1) ^^_^ ^^
1 • 3 • . . . «»
• • •
h
a
The series converges when
less than a.
The wth term of the expansion is
< 1, that is, when h is numerically
'''""■ 1 . 2 .... (w - 1) ^ ^ •
Examples,
1. vr^-(i-.)»-i-^.+i^^-ii^i^i^^+
• • •
2. find an approximate value of V-98.
ViOS - VI - .02 « 1 - i (.02)- \ (.02)2 _ . . . « .990+.
The neglected part of the series is less, term for term, than the G. P.,
(.02)2 -f. (.02)3 + . . . + (.02)n + • • • ,
whose sum is
S = Y^^ " -^^^ approx.
3. Plnd the 7th term of the expansion of v (2 — 3 \lxY in powers of x,
^{2 - 3 yfiY = (2 - 3 Vi)*.
Hence a * 2, 6 = — 3 V^:, n = 1, m = 7.
■^ «. - *"'\'.r/.';;;."'" ^'-(-W'>'-'-^''-
In this case the expansion converges if
3 \^ I < 2, or I 9 a; I < 4, or I jr I < J.
For negative values of x the expansion would involve imaginary terms be-
cause of the presence of V^*
198 EXERCISES . [222
222. Exercises. — Write the first four terms of the develop-
ments in series of the following functions, and give the values of
X for which the series converge.
1. tanx. 15, _
2. secx.
8. sin**. W-
1
»• v'('-i')*
4. sinaJ2. 1 +x
6. e««. * l~x
e. e-x. ^ (l-2x)-§.
7. e*+c-*.
8. el 20. (a;2-l)-i.
21. (2-x3)^.
9. c""«. 22. (V2- V9"*.
11. sinx + cos*. «- /I x^ \l.
X __»
10. e« + e"«.
12. sin ax. \ V3 V2/
18. Vr+x. 26. (2ai + 3xi)"*.
14. Vr=^. 26. (ai+Sx^)*.
By use of the binomial theorem calculate to three decimal places inclusive
the values of:
27. VlO. 81. V0j096.
28. -n/SO. 82. ^8^.
2«- ^^ 88. -^62475.
80. -^1121.
Calculate to five decimal places inclusive the values of:
84. sin 25°. 41. c"^
36. sin 5°. 1
36. sin 1°. * ^
«7- «i°10'- 48. log 1.1.
38. cos 50°. ,, , , ^
-««« 44. log 1.2.
89. cos 100*. *
t 46. log (.75).
40. -.
e
CHAPTER XIV
Computation. Approximations. Differences and
Interpolation
223. Remarks on Computation. — (1) In a series of similar
computations, perform similar operations together. If the same
number is to be added to each of several others write it on the
edge of a slip of paper and hold it over or under each number in
turn.
(2) When a result is wanted to say three decimals, computa-
tions should be carried to four places so as to avoid accumula-
tion of errors which would vitiate the third place.
(3) As a general rule, 4-, 5-, 6-, and 7-place logarithm tables
will yield respectively not more than 4, 5, 6, or 7 significant figures
of a number.
(4) Results should be stated with an accuracy commensurate
with that of the data. Thus, if a line be measured 10 times to
0.01 ft., the mean of the 10 measures should be given to 0.001 ft.
More than three places in the mean would be a useless refine-
ment. Do not state an angle to seconds when it results from
computations which render even the minute uncertain.
224. Useful Approximations. — Let the student verify that,
when X, y, u, v are small decimals, we have approximately:
1. (l+x)il+y) = l+x + y. ^
2. (l+x)(l-t/) = l+x-j/. 6. )^zr^^l+x.
1 a 1 -faJ ^ ,
8. (1 - x) (1 - J/) = 1 - x - J/.
8. (1+ a;)'* = 1 + nx.
As special cases of (8) we have
9. ^/m^l + ix. ,, 1 - 1
. 11. . = 1 — s X.
10. vr^=^ = i -ix. VI +x 2
199
^
200
12.
APPROXIMATIONS
[224
\^r\x.
18. (lH-x)2-lH-2a:.
14. (1 - a;)2 = 1 - 2 ».
16. e* « 1 H- X.
More accurately:
21. flinx = X — 1 x*.
Examples. ^
t. .987 X .993 = (1 - .013) (1 - .007) = 1 - .013 ~ .007 = .980.
The error is .013 X .007 = .000091.
1 1
C06X
16. loge (1 + X) « X.
17. logio (1 + x) « .43 X.
18. sin X ~ X (radians).
19. tan X » X.
20. cosx s 1.
22. tan x = x + § x^.
1 - i x2.
.987 1 ~ .013
= 1 + .013 « 1.013.
8. V:987 = (1 - .013)* =1 - I (-013)
» .9935, correct to four places.
4. Find the range of vision from a point h ft.
above the surface of the earth.
Let A be the station of observation (figure),
AB^h ft., BC^DC^R=- 3960 miles.
Then
R » 3960 X 5280 ft.
x = y/iR + h)i-Ri = V2i2A+A2= %J2MJi +
2R
For moderate elevations, ^^5 is small and the second radical = 1 approxi-
mately.
Hence
X = \j2Rh approximately.
The error in this value of x is -r^ x approximately.
Exercises.
.965
1.
.982'
1
Calculate the approximate values of,
.85X1.12 . ^^5-20-
^ 1.15 X. 92' '• ^^•^"'
1
6.
6. (1.15)2.
- 1.125' - V:976'
7. Prove the last statement of example 4.
8. How far can an observer see from a mountain one mile high ?
9. What is the distance to the horizon as seen by an observer on the sea-
shore with his eye 6 ft. above the water level ? (Three-mile limit.)
10. If the range of a gun on a warship is 10 miles, how high should the
lookout be stationed to detect objects coming within range?
225] COMPUTATION OF LOGARITHMS 201
11. What is the error in each of the approximations
(i) . . . (23) whenx, y, w, t; « 0.1? When x, j/, w, »« 0.01?
12. Calculate to four decimal places sin 130^ and cos (— 100^}. (Reduce
to functions of angles < 45^.)
13. Calculate a 4-place table of natural sines, from 0^ to 45**, at intervab
of 5^
14. As in exercise 13 for a table of natural cosines.
226. Computation of Nattiral Logarithms.
We have log (1 + x) = x - -^ + g- - -|- +
Replace a; by — x:
• • •
/f2 <|*3 /j^
log (1 - X) = - X - 2- - g- - J • • . .
ic+ g-+ g-+ • • • h
provided — 1 < x < 1.
But log (1+x) - log (1 - x) = log^ "'■^
T^f ^+^ w_+l. 1
Let -z = ; or, x =
1-x
1-x n ' ' 2n+l
Then log (1 + x) — log (1 — x) = log (n + 1) — log n
and
^°g^"+^)=''^g"+427ri:i+ 3(2n+l)3 + 5(2n+l)» + - • J-
By means of this equation log (n + 1) can be calculated when
log n is known. The series on the right converges rapidly and
for all positive values of n. Putting successively n = 1, 2, 3, ... ,
we obtain in turn log 2, log 3, log 4, . . . .
We will now obtain an estimate of the maximum error made in
stopping at any term of the series.
Let fc = 2 n + 1.
Then the mth term of the series is
1
^"* (2m-l)fc2w-i'
and the remainder of the series will be
"• (2m+l)fc2'«+i"^(2m+3)A2"*+3-f (2in+5)fc2«+fi'^' ' ' *
202 COMPUTATION OF LOGARITHMS [225,226
Then B„ is certainly less, term for term, than the series
1 r, , 1 , 1 ■ i_ 1 1
(2m + l)A:2"'+iL^'^Jk2'rjfc4-l- ' ' * J" (2m + 1) F-+1 _ J_'
A;2
since the series between the brackets is an infinite G. P. with ratio
1^. Also, since
fc = 2 n + 1 and n = 1, .'. k>2 for all values of n. Hence
1
and therefore
o
Rm <
<2
(2m + 1) F'^+i (2m + 1) (2n + l)2^+i
If we now include the factor 2 which stands before the bracket
in the equation giving log (n + 1), the total error is less than
4
(2m + l)(2n + l)2'^+^
when log (n + 1) is calculated by using only the first m terms of the
series.
Thus in calculating log 5, we have n = 5 and the error in stoi>-
ping with the mth term is less than
4
(2 m + 1)112^+1*
4 . .
Hence when m = 1, the error is less than ^ ^^ ' that is, if we use
only the first term of the series, log 5 will come out correct to 3
decimal places inclusive. When m = 2, the error is less than
4
v--rz , so that the first two terms will give log ,5 correct to 5 places,
6*11
and so on.
Exercises.
1. What is the error in log 7 when only one term of the series is used? When
two terms are used ?
2. How many terms of the series are required to give log 7 correct to
10 places?
3. How many terms of the series are required to give log 17 to 20 places ?
4. Calculate a four-place table of natural logarithms of the numbers from
1 to 20 inclusive.
226. Common Logarithms. — When the natural logarithm of
a number is known, its common logarithm may be found by
227] DIFFERENCES 203
multiplying by a certain constant factor called the modulus of the
common system of logarithms. We shall show that this modulus,
or multiplier, is
M = logio e = 0.4342945 ....
Let the natural logarithm of any number be x, its common loga-
rithm y. To express y in terms of x. We have, if n be the number,
loge n = X and logio ^ = 2/;
or, n = e* and n = 10^.
Hence lO*' = e^.
To solve for y, take logarithms of both members to the base 10.
Then y = x logio e,
which proves our statement. To find the value of logio ^> we need
only calculate loge 10 and take the reciprocal of the result.
Exercises.
1. Calculate the modulus M to 5 places.
2. Calculate logio 101 to 10 places.
3. Calculate logio 11 to 10 places.
4. Calculate a four-place table of common logarithms of the numbers
from 1 to 20 inclusive.
227. Differences. — Consider a sequence of quantities wq, wi,
W2, . . . , Wn, . . . , and form the differences, Awo = wi — wo>
Aui = W2 — Wi, . . . , Ai^n-i = Wn — Wn-i, • • • , Called the first
differences. Form next the differences of these differences, called
the second differences of the original sequence, and so on. We
obtain in this way the entries in the following difference table,
where the successive difference columns are denoted by Ai, A2, A3,
. . . and the original sequence by Ao.
A2 A3 ...
U2-2UI+U0 ^3_3^2 + 3^^^^
Us -'2U2+Ui ...
Ao
Ai
Uo
Ui —Uo
Ui
U2 — U\
U2
Us — W2
Us
•
•
•
'Wn-2
Un-l — Un-2
...
204 DIFFERENCES [228
We observe that the coefficients follow tiie binomial law. Let
the student prove by induction,, that this law is followed in all
the successive difference columns.
22§. The nth tenn of the sequence, in terms of its first term
and the first terms of the first n difference columns.
Let the first term in the fcth difference column be denoted by
Aifetio. Then we have
. . • • . .
• • • • •
Solving successively for Vi)^ Ui, 112, . . . , we have
Wo = t^,
ui = Uo + AiUof ^
^2 = Wo + 2 Aiw© + A2W0,
1^3 = Wo + 3 Aiiio + 3 A2W0 + Aaiio,
. • • • .
Here the coefficients again follow the binomial law, and there is
suggested the formula
(1) Wn = Wo + nCiAitio + «C2A2Wo +' ' ' ' + AnW©.
rr Assuming the formula true for u^^ we can show that it holds
for Wn+i. For apply formula (1) to the nth term of the first
order of differences, which is i^n+i — Wn- We obtain
Wn+l - Wn = Aitio + nClA2Wo + uG^^ZUq + • • • + A^+lWo-
Adding equation (1) to this we get
Wn+l = Wo +(«Ci + 1) AiT^o +(nC2 + n^l) A2W0
+ (nCa + «C2) A3W0 + • • • +An+lWo.
But .
nCi + 1 = n+lCi, rf^2 + n^l = n+lC2, n^S + rf^2 = n+lC^3> • • • >
as is easily verified by substituting in the values of the binomial
coefficients. Hence
Wn+l=Wo + n+lCiAil^o + n + lC2A2Wo+n+lC3A3tio+ ' • * +A„ + illo.
Hence, if (1) holds for w„, it also holds when n is replaced by
n + 1, that is, for Wn+i. But we have shown that it holds for u^\
hence it holds for 1/4, hence for tis, and so on.
229] DIFFERENCES 205
229. The sum of the 'first n terms of the sequence, in terms
of its first term and the first terms of the first n — 1 difference
columns.
From the equations just preceding formula (1) we have, by
addition,
uo^uoy
uo + ui = 2uq + Aitio,
tio + wi+ti2 = 3tio + 3 Aii^o + A2iio,
iio + wi+W2 + W3 = 4wo + 6 Aitio + 4 A2W0 + AatiQ.
The coefficients on the right are respectively those of the expan-
sions of (1 + a;)^ (1 + xYy (1 + a:)3, and (1 + xY, the first term of
the expansion being omitted in each case. Let 8^ denote the smn
of the first n terms of the sequence;
Then by analogy with the preceding equations we assmne that
(2) S» = vPlVii + nC2^\UQ + nC^a A2W0 + nCA^^zVi) H h ^n-\UQ.
We show by induction that (2) holds for all values of n. Adding
(1) of (228) to (2) and noting that Sn+i = Sn + Um we have
Sn+l==(nCl + l)iA) + (nC'2+nCl)Ait/o+(nC3 + nC2)A2iA)H \- K^O
- n + lCiiA)+n + lC2AiWo+n+lC3A2t/o+ ' ' ' + An^o-
Therefore (2) is true when n is replaced by n + 1. But we veri-
fied above that (2) is true when n = 4. Hence it is true when
n = 5, hence when n = 6, and so on.
When the rth order of differences is zero, all following orders of
difference are also zero. Hence any term of the sequence and the
sum of any number of terms can be expressed in terms of the first
term of the sequence and the first terms of the first r — 1 difference
columns. For then formulas (1) and (2) both stop with the term
involving A^-iiioj and we have
(3) t^ = llo + nClAlUo + nC2^2Uo + ' ' ' + n^r-l A^-ll^.
(4) Sn ^nClUo + „C2Aitio + nCs^JiUo + ' ' • + nCrA^-iiA).
Example, Find the sum of the squares of n consecutive integers beginning
with 10.
«n = 102 + 112 + 122 + . . . + (10 -f-n - 1)2.
206
INTERPOLATION
[230
Our difference table is as follows:
Ai
21
Ao
100
121
144
169
196
23
25
27
A2
2
2
2
A3
&ence r = 3.
Sn-
Then
nCiUo + nC2Ait/o + nCzAittQ
^XlOO+^^te^X21+5^(^-^^(^
-2)
1-2
«J(2n8 + 57n2 + 541n).
1-2.3
X2
Exercises.
1. Find the sum of the squares of the integers from 1 to n inclusive.
2. Find the sum of the cubes of the integers from 1 to 20 inclusive.
3. How many balls in a square pyramid whose base has n balls on a side.
4. As in exercise 3 for a triangular pyramid.
6. Find the sum of n terms of the sequence a, a + d, a + 2 d, . . . .
6. Find the 10th term and the (n + l)th term of the sequence 50, 72, 98,
128,162, ... . Ans, 392; 2n2+20n + 50.
230. Interpolation. — Suppose the terms of the sequence i^, wi,
i/oj • • • to be the values of a function / (x) for a series of equally
spaced values of x. Thus:
U2 =f{xo + 2h),
•X
Wn =f{xo + nh).
These values are shown graphically in the figure, as ordinates of
the curve y=f{x). From the equally spaced ordinates given,
we wish to calculate intermediate ones. This is called inter-
polation.
Replacing the w's in (1) of (228) by their values above, we have
(5) / (xo + nh)^f (xo) 4- nAi/ (xo) + ^^^f;^^ A2/ (xo).
n(n- l)(n-2)
+
1-2.3
A3/(Xo) +
230] INTERPOLATION 207
This formula has been derived when n is a positive integer.
It is also true for fractional values of n, provided the series on the
right converges. We shall not stop for the proof, but merely
give some simple applications. In practical cases the successive
diflferences Ai/Cxq), A2/(xo), . . . become rapidly small, so that
first differences ar e_u suaUy sufficient , second differences are occa-
sionally needed, while third and higher differences are required
only in theory or in the calculation of extensive tables.
For fractional values of n, formula (5) gives values of the func-
tion intermediate to those in the table. Thus when n = 2|, we
get / (xq + 2^h), which is the ordinate to the curve y ^ f{x) falling
midway between the ordinates f(xo + 2 h) and / («o + 3 A).
Example 1. Given the values of log 100, log 101, . . . , log 109 to five
decimal places, to calculate log 100.7 and log 107.35.
Here /(x) « log x; xo - 100; A « 1. To calculate log 100.7 we put n = .7.
Our difference table is,
fix) Ai/(x) A2/(a;)
log 100 = 2.00000 , QQ432
101 = 2.00432 ' 428 "" -^^^^^^
102 = 2.00860 424 ^
103 = 2.01284 421 3
104 = 2.01703 4jg 5
105 = 2.02119 4J2 ^
106 = 2.02531 4Q7 5
107 = 2.02938 ^q^ 3
108 = 2.03342 ^^ 3
109 = 2.03743
Then
/(xo + nh) = 1(^ 100.7 = log 100 + .7 X .00432 - '^^j'^Jl^^ X .00004+
....
1X2
= 2 + .00302 + .00000 = 2.00302.
Here the second differences are so small that they can be neglected^ arid
0HLI£§3dUa.tliat. nhtiflJnfidi>3LQrdmaix or lineaw-^interpolation. Graphically
this amounts to replacing the curve y —fix) by its chords.
To calculate log 107.35, it is best to consider log 107 as the first term, or
/ (xo), and put n = .35. (We might take / (xo) = log 100 and put n = 7.35.;
We find
1<^ 107.35 =log 107 + .35 X .00404- '^^.^ff"^^ X .00003 + • • • =2.03079.
i X ^
Here also second differences are negligible.
All ordinary tables are constructed so that linear interpolation is sufficient.
208
INTERPOLATION
[231
Example 2. Given sin 10*, sin 15*, '. . . , sin 45*, to calculate sin 17* 20'.
The tabular numbers and their differences are given below:
fix)
Ai/(x)
Mfix)
^f{x)
mn 10* = 0. 1736
15* « .2588
+ .0852
-.0020
20*= .3420
832
26
-.0006
25* = .4226
806
32
6
30* = .5000
774
38
6
35* - .6736
736
44
6
40* = .6428
692
49
5
45* = .7071
643
22 22
Here«o = 10*; A = 5°; then 17* 20' «= xo + tf ^ and hence n = ^
10 10
Then
22
(i - ')
22/22
15
mn 17* 20' = sin 10* + ^ X..0852 - ^^^
X.0020
?2/22_ \/22_ \
15^15 J\15 I
1 X2X3
X .0006 +
«.2979.
Here the amount contributed by the second difference is .0003, so that
linear interpolation would have been inaccurate.
231. Exercises.
1. From the table of example 1 calculate log 104.6.
2. From the table of example 2 calculate sin 12* 30', sin 27* 30', and sin
36* 15'.
3.
0.6745
7*
Vn (w - 1)
10
0.0711
15
465
20
346
25
275
30
229
35
196
40
171
45
152
50
136
Calculate the tabular number
when n = 22; when n = 33.6.
4. Altitude.
Refraction.
10*
5'13".l
12*
4' 22". 5
14*
3' 45". 2
16*
3' 16". 6
18*
2'54".0
20*
2' 35". 7
22*
2' 20". 5
24*
2' 7". 6
26*
1'56".6
Calculate the refraction for alti-
tudes 14* 40' and 21* 26'.
232] INTERPOLATION 209
S» Greenwich Moon's Moon's
meantime. right ascension. declination.
h h m 8
6 14 32.14 IS** 47' 37". 7
2 6 19 49.41 18* 49' 15". 9
4 5 25 6.62 18** 50' 20". 6
6 5 30 23.69.' 18*» 50' 51". 7
8 6 35 40.59 . 18' 50' 49". 4
10 6 40 57.26 18'' 50' 13". 7
Calculate the moon's right ascension and declination at 0^ 35" 20* Green-
wich mean time.
6. From a four-place table take log 310, log 320, . . . , log 400. Hence
calculate log 317.5.
232. Differences as a Check on Computed Values. — When a
number of values of a function are calculated for equal intervals of
the argument, the differences should, ordinarily, vary in a regular
manner. An irregularity in one of the difference columns indi-
cates an error in the tabular values, and often enables the com-
puter to determme the amount of the error and so correct it.
Example,
log 70 - 1.8451
75 = 1.8751
80 » 1.9030
85 » 1.9284
90 » 1.9542
95 = 1.9777
100 = 2.0000
105 = 2.0212
The irregularity in A2 causes us to examine Ai; here the differences .0254
and .0258 are probably incorrect, which throws suspicion on the tabular number
standing between them, namely 1.9284. This number should evidently be
larger, and by trial we find that 1.9294 is probably the correct value.
Exercises. Correct the following tables:
1. tanl5'= .268 2. ^
16** « .287
17*= .306
18*= .325
19''= .344
20*= .369
21**= .384
22**= .404
23'= .425
24*= .446
Ai
A2
+ .0300
279
-.0021
254
25
258
4
235
23
223
12
212
11
n
3. Altitude.
Refraction.
2.0
.250
10°
5' 13"
2.2
.207
11*
4' 46"
2.4
.174
12*
4' 22"
2.6
.158
13*
4' 2"
2.8
.127
14*
3' 45"
3.0
.111
15*
3' 34"
3.2
.098
16*
3' 16"
3.4
.087
17*
3' 4"
3.6
.077
18*
2' 54"
19*
2' 35"
'
7
CHAPTER XV
Undetermined Coefficients. Partial Fractions
23 3. A useful method fo r ^TFffaP^ing f^rti**^" oyppcgirkng m
seriea^^ epends on the following Theorem on Power Series.
If the equation
(1) -00 + CtlX + Ct2^^ + • • • + (hi^^ + . . . =
is true for all values of x from x = U> x = xq inclusive, where
xqj^O, then all the coeflBicients are zero, that is,
Oo = 0, ai = 0, a2 = 0, . . . , ttn = 0, . . . .
Proof. Since (1) is true when x = we have, putting for x,
Oo = 0.
Then (1) reduces to
aix + 02^2 + . . . + anX^ + • • • =0,
or
(2) x{ai + a2X+ • • . +a'*a;'*-i+ . . . )=0.
This must be true for all values of x from to a?o* Choose for x a
value £ between and Xq. Then
e(ai+a2e+ • • • +a'*e'*-i+ • • • )=0.
Then, since £ p^ 0, we must have
ai + a25+ • • • J|-«n«'*~^+ • • • =0,
or,
ai = — £ (a2 + a^e + • • • + fln^** ^ + • • • )•
The series in the last parenthesis converges, and therefore has
a finite sum S. For, putting x = e in,(l), and omitting the first
two terms, we have left the convergent series
a2e^ + as£^ + . . . + ans"^ + • • • ,
and this remains convergent after division by e^. Hence
ai =— eS
where S depends on e, but is finite for all values of e between
210
/
234] UNDETERMINED COEFFICIENTS 211
and Xo- Assume now that ai is not equal to 0; say ai = h. We
can now take e so small that eS shall be numerically less than h;
hence ai cannot equal A. /. ai = 0.
Then (1) reduces to
a2X^ + asx^ + • • • + dn^^ + • • • = 0,
or, x^ (a2 + asx + . . . -fanx^~^ +•••)= 0-
fJhoose for x a value e (not necessarily the same as e above) between
and ocq. Then
Hence, since e ?^ 0, we have
O2 + flSS + • • • + ^n^**"^ + • • • = 0,
or, a2 = — € (as + . . . + a„£'*~2 +•••)= 0-
Here again the series in parentheses converges and has a finite
sum. Hence by taking e suflSiciently small we can show that 02
cannot equal any number A, however small. /. 02 = 0.
Similarly we show that each coefficient must be zero.
234. Theorem of Undetermined CoeflScients. — If two power
series in x a re equal to each otb^^for^ vaJto SjL^LJXQmx =.Q,tO
X = oco inclusive, then the coefficients of like powers of x in the
^wo series must be equal.
Hypothesis:
(1) Oo + aiX + (l2X^ + • • • + CLnX^ +
60 + J>ix + b2X^+ • • • + bnX^ + • • • when 0^ x = xq.
Conclusion:
Proof. From (1), by transposition, we have
Oo — bo + (ai—bi) x+ {a2 — b2) x^ + - • • +(an— 6n)a:**+ • • • =0.
Hence by the preceding theorem,
Oo "■ 60 == 0, ai — 61 = 0, a2 — 62 = 0, . . . , an — 6n = 0»
Hence the conclusion stated above.
CoroUary. The theorem remains true when either or both
of the infinite series reduce t o polynomials. We consider a poly-
nomial of m terms as an infinite series in which all coefficients
after the mth are zero.
• • •
i212
UNDETERMINED COEFFICIENTS
[234
1 — X2
Example 1. Develop y^ — zr~2 ^^^ * power series.
Assume
1 -a;«
1 +x - x»
oo + aix + ojx* -f ojx* +
Clearing, and writing the coefficients of like powers of x in vertical columns, we
have
1 — X* = Oo + oi
Oo
X +02
x^ + az
+ Ol
+ 02
- OO
-Ol
x^ +
Equating coefficients of like powers of x, we have
Oo — 1, or, Oo B 1,
ai + oo = 0, Ol « — 1,
02 + Ol — Oo = — 1, 02 = 1,
08 + 02 — Ol ** 0, OS = — 2.
Hence
1 -x2
1 + X - X2
= l-x+x2-2x» +
Example 2. Develop « — __ . ^ , 3 into a power series.
If we put
1 +2x , , 2 ,
6x-5x2+x3 = 00 + OlX + 02X2 + . . . ,
clear of fractions and equate coefficients, we have to begin with 1=0. This
absurdity results from the fact that we have not taken a proper form for the
development. By inspection we see that the quotient of 1 + 2 x divided by
6 X — 5 x2 + X* should start with ^— . To obtain the development we put
ox _
l+2x 1 l + 2x
X 6 — 5 X + X*
•"
6 X - 5 x2 + x3
Developing the last fraction as in example 1,
Hence
1 + 2 X ^ 1 I 17 ,79^ 24..M 84.
6-5x+x2 6 "^36^ "^216^ "^1296^ "^
1 + 2 X ^ 1 17 79 293 ^
6x-5x2+x» 6x"^36"^216^"^ 1296^ "^
• • •
Exercises.
1. In example 1, find chi, in terms of on-i and on-2*
Find the first four terms of the expansions of:
1 + X . X
2.
3.
1 +X +X2
1 -x
1 -X -X2
4.
6.
2 - X + 3 x2
2 x« + 3 X
x2 + 2X + 2
6.
•7.
1 +X2
1 +3x+x»
2 - 3 X + x2
3x + 4x2 -x*'
235] PARTIAL FRACTIONS 213
236. Partial Fractions. — It is sometimes desirable to resolve a
given rational fraction into a siun of simpler fractions, called par-
tial fractions. This can be done when the denominator of the
given fraction can be factored. Several cases arise, according to
the nature of these factors.
For reasons which will presently appear, the methods to be ex-
plained apply only to fractions in which the degree of the numerator
is less than the degree of the denominator. When this is not the case,
divide numerator by denominator until a remainder of less degree
than the denominator is obtained.
Case 1. The denominator can be factored into linear factors
of the form {az + 6), no two factors being equal.
^^XJ0 The fraction can be resolved into a sum of simple frac-
tions, of the form r-r , equal in number to the factors of the
ax + 6
given denominator. Here A is a constant.
-, , 5a; -1 5a; -1 A . B
Clearing: 5 a; - 1 « A (a; - 5) + B (a; - 1),
or, 5a; -1 =(A +B)a; -(5A +B).
Since the given fraction must be equal to its partial fractions for all values
of X except a; « 1 and x » 5, the last equation must be true for all such values
of x; hence we equate coefficients of like powers of x (233, Corollary). We
obtain
5 = AH-B; -1=-.(5A+B).
Hence A=-l; B = 6.
5a; -1 -1.6
x2-6a; + 6 a; - 1 ' a; - 5
A shorter method for finding A and B is as follows: consider again the
equation
5a; -1 =A(x-5)+B(a;-l).
Let X = 5; 24 = 4 B; B = 6.
Let x = l; 4 =-4 A; A=-l.
We can justify the use of the values x ^ 1 and x » 5, for which the given
fraction and one of the partial fractions become infinite. For the equation
5x-l A B
I*
x2-6x + 6 x-l'x-6
must hold except when x » 1 or x » 5.
Hence
6x-l =A(x-5)4-B(x-l)
214 PARTIAL FRACTIONS [236
is true for all values of x, except perhaps re « 1 and x ^ 5. It is therefore
true when x ^ 1 + 1, however small t may be ; that is,
(1) 5 (1 + f) -,1 = A (1 + f - 6)+ B (1 + f - 1).
Suppose our equation is not true when x « l; let the two members differ by
a quantity k, so that
6Xl-l-A(l-6)+B(l-l)+A,
or, 4= — 4A+^.
From (1) we have
4 + *--4A + fA + *B.
From the last two equations, by subtraction, etc.,
Since A and B are fixed numbers, k can be made as small as we wish by taking
9 small enough. Hence h cannot equal any number except 0.
236. Case 2. — The denominator contains a linear factor repeated
r times, as (ax + by.
Rule. Corresponding to the factor {ax + by, take a set of par-
tial fractions of the form
Ai . A2 . , Ar
(ax + b) ' (ax + b)^^ ^(ax + by
This is the m ost general set^of fractions hflvinff rnnfltflnt numer-
ators and common denom inator Cox A - by.
'- ui i'i 111! I r ir-| — I T 1 1 1 ■ 11 ——-———'■— —'^ ■*"'—* " '
Example,
Sx^-x + l A , B , C , D
"TZ 5 + 71 5\5 "T
(x + 2)(x - 3)» x + 2^x-S^ (x- 3)2 ^ (x - 3)3*
Clearing:
3aj2-x + l=A(a;-3)« + B(x + 2)(x-3)2 + C(x + 2)(a;-3)+D(x + 2).
Let X = 3; then 25 = 5 D; D = 5.
Let x=-2; then 15 =-125 A; A =- sS.
Since no other factors are available to furnish other values of x for substitu-
tion, we choose any convenient values, say x = and x = 1.
Put x=0; 1 =-27A + 18B-6C + 2Z).
Put x = l; 3=- 8A + 12B-6C + 3D.
Substituting the values of A and D already found, and solving for B and C,
we have
Hence
3x2 -x + 1 _3 3 12 . 5
(x + 2)(x-3)3 25(x + 2) ' 25(x-3)^6(x-3)2 ' (x-3)»
237J PARTIAL FRACTIONS 215
237. Case 3. — The denominator contains a quadratic factor,
{(ufi + bx+ c), which cannot be resolved into real linear factors.
Rvie.^ Corresponding to a quadratic factor (cu^ + bx + c)f take
a partial fraction of the form
Ax + B
ax^ + bx + c
The reason for this assumption may be illustrated by a simple
example.
Example, Resolve 7 rw-5— ttx ii^to partial fractions.
If i = V— 1> the factors of a?2 + 4 are a; -f 2 i and z — 2 1. Suppose now
we assume
2a; -1 ^ A B C
(a;-l)(x2 + 4) a;-l"^x + 2i"^a;--2i'
Combining the last two fractions into a single one, we have
B
^ C {B + C)x + 2(C'-B)i
:• "T - — TTi' = _« . A •
xH-2i ' a;-2i 0:2 + 4
If now we introduce two new constants Af , N in place of B, C, by the relations
we have
B + C = 2Af; i(C-B) = -2i?Ar=2Ar.
Hence in place of the fractions
-. -r - — 7^.9
a: + 2i x — 2i
where B and C involve %, we take the single fraction
Mx + AN
aj2-f4 '
where M and N are real. Then, using B in place of M and C in place of 4 Nf
let
2a;-l A Bx + C
1*
(x-l)(x2H-4) «-l ' a;« + 4
Clearing: 2 « - 1 = A(a;2 + 4t)+{Bx + C)(x - 1).
Put a; = l; then 1-5 A; A = f
Put a; = 0; then -1=4A- C;C = |.
Equate coefficients of x*; then = A+B; B=-A = — J.
Hence
2a; - 1 __^^_ -x-f 9
(a? - 1) (a:* + 4) " 6 (x - 1) '^'S (a^ + 4)'
216 PARTIAL FRACTIONS [238,239
238. Case 4. — The denominator contains a repeated quadratic
factor, {ax^ + 6a; + c)'".
Rxde. Corresponding to a repeated quadratic factor {az^ +
^^j take the partial fractions,
B^x + Ci B2X + C2 . , BrX + Cr
(ax^ + bx + c) (ax^ + bx + c)^ (ax^ + hx + cY
Example,
10a:» + 7a: + 4 _ A , Bx + C , Dx + E
(x - 2) (x2 + 3)« X - 2 ' a;2 + 3 ' (x^ + 3)2
Clearing:
10x3 + 7x + 4 = A (x2 + 3)2 + (Bx 4- C) (x - 2) (x* + 3)+ (Dx + E){x - 2).
Put X = 2; 98 = 49 A; A - 2.
Equate coefficients of x*, x*, x2, and x®:
= A + B,
10 - C - 2 B,
= 6AJ-3B-2C + D,
4 =9A -6C-2^.
Hence, B = - 2, C = 6, D = 6, j& = - 11.
Therefore,
10x»-f 7x + 4 _ 2 , -2xH-6 , 6x - 11
(x - 2) (x2 + 3)2 X - 2 ' x2 + 3 ' (x2 + 3)2
239. Exercises. Resolve into partial fractions:
1. ^ , . .^ r-7.' 7. -1 7- 13.
5.
6.
X — x*
X
x2-4x + l
x^
7.
x4- 1
8.
x6 + x4 - 8
x* — 4x
9.
5X + 12
x» + 4x
10.
1
(X2 - 1)2
11.
X3 - 1
x»H-3x
10
x» + l
3x2 + 10x + 3 x4 - 1 — x«-4x2 + 4x
2 3x~l( . ^ a;6 + x4 - 8 j^ 1^
*x2 + x — 6 * x* — 4x x*H-x3 + x2+x
, xM-6_x--8 ^ 5X + 12 ^^ 1
9. = ;j —
x* — 4x
^ 1 +a:2
x«+2x2-x-2 x(x-l)3
19. /^!-2^^ . 21.
x' — 3 x2 + 2 X
20 ^^_±A£±±, 22
x3 + 2x2 + X ^- a;4 + 3x2 + 2
16.
x» + l
16.
X2 - 1
x2 -4
17.
x2 -3
x» ~7x + 6
18.
x6 - 2 X + 1
X4 + 2X8+X2
x2 + 8
x + 4
X» + X2 -
• 4x — 4
x2 -'2x
- 1
CHAPTER XVI ,
Determinants
240. Determinants of the Second Order. — When two simul-
taneous linear equations
aix + biy = ci,
are solved for z and y, we find
X =
62^1 "~ &1C2
y =
Ci\C2 — ^2^1
To express these results it is convenient to use the notation
a2 ^1
= (^162 — 0261),
where the square array between vertical bars is simply another
way of writing the expression forming the right member of the
equation. It is called a determinant, and in particular, a deter-
minant of the second order, because there are two rows and two
columns. The quantities ai, 61, a2, 62, are called the elements of
the determinant.
The value of a determinant of the second order may be obtained
by forming the products of elements which constitirte the diagonals
of the array and giving these products the signs indicated in the
scheme below:
This process is called " expanding the determinant."
The above values of x and y may now be written in the forms.
X =
ci h
ai Ci
C2 62
y y = ■
a2 C2
di hi
di bi
a2 62
02 62
z/
217
218
DETERMINANTS
[241
1. State a rule for writing the above values of x and y.
Solve for x and y, by aid of determinants:
2. x-y=«l, 3. 4x-3y«6, 4. 8a: + 5y-6 = 0,
2x + y = 3. 2x + y = l. 4a;-fy + 4 = 0.
5. 2x+ yH-1 =0,
6x + 3y-f2=0.
6. 2a; + y4-l=0,
6x + 3y + 3 = 0.
241. Determinants of the Third Order. — We shall now define
a determinant of the third order in terms of dete^inants of the
second order by the following equation:
dl CL2 ^3
6i 62 &3
_ai
62 63
C2 C3
— a2
61 63
+ ^3
61 62
Cl C2
Cl C2 C3
where the determinants on the right are to be expanded and the
results multiplied by the quantity written in front of the determi-
nants respectively.
On performing these operations and collecting terms, we have
= a\h2Cz + 02^3^1 + ^36102 — 0362^1 "" 02&1C3 — ai&3C2«
This is the expanded form of a determinant of the third order,
and may be written out by forming the products of the terms
joined by arrows in the scheme below, eabh product to be given
the sign indicated.
a\
a2 az
61
62 63
Cl
C2 Cz
^V^K^X^f
We may now verify by direct calculation that the values of x,
y, z, obtained by solving the linear equations
aix + biy + ciz = di,
a2X + b2y + C2Z — (fe,
azx + bzy + czz ^ dz,
242]
DETERMINANTS
219
are,
X =
di
6i
Cl
^2 ^2
C2
ds
63
Cs
Ol
61
Cl
02
62
C2
as
fca
C3
' 2/ =
ai di Cl
02 d2 C2
(I3 da C3
, =
ai
61
di
02
62
^2
03
63
ds
Ol
"ET
Cl
02
62
C2
as
63
C3
3. a; — y -h 2
a; + 2y + 3« = 2,
3a; + 2y+ « = 3.
4.2a?4-2y- 2 = 2,
x+ y — 22 = 1,
a; — y + z ^ 4t,
X— y+ 2 = 2,
2x+ yH-32 = l,
2x-2yH-22 = 4.
2x- yH-22 = 2,
a:-2y + 42 = 3,
3a:-3y + 62 = 1.
fll 61 Cl
a2 62 C2
CI3 &3 C3
Exexdses.
1. Verify the last statement.
2. State a rule for solving three equations of the form just considered.
Solve the following systems of equations:
y+ 2 = 1, 5. 5x + 6y--32 = 4, 7.
4x-5y + 22 = 3,
2a;-3yH- 2 = 1.
6. 3a;-6y + 92 = 2, 8.
a: + y+ 2 = 1,
a;-2y-f-32«2.
9. Show that a determinant of second or third order vanishes when the
elements of a row or column are equal respectively to those of another row or
column.
10. Show that a determinant changes sign when the signs of all the elements
of any row or column are changed.
11. Show that, if the elements of any row or colunm be multiplied by a
factor kf the determinant is multiplied by k:
242. Inconsistent or Non-independent Linear Equations. —
Consider the equations
aix + biy = Cl and kaix + kbiy = C2.
^ These are inconsistent if C2 7^ kci; they are dependent if C2 = kci,
since in this case the second equation is k times the first.
In either case the determinant of the coeflBicients of x and y
is 0. On solving by the determinant method, we find
X = 00 and y = 00 , when the equations are inconsistent;
x = jr and
y =" K' when the equations are dependent.
That is, t he incon sistent equations have no (fini te) s olution, while
the solution is indeterminate in case of dependent. ©JdU^ttions.
reometncally, the equations represent two straight lines which
.^e parallel, and distinct if C2 f^ cik; they coincide if C2 = cik.
220 DETERMINANTS [243
Hence the infinite values of x and y above are equivalent to the
statement, '* Parallel lines meet at infinity." In the second case,
when the lines coincide, the coordinates of any point on either
line satisfy both equations. Hence there are an infinite nimiber of
solutions, and hence x and y appear above as indeterminate forms.
[See exercises 5 and 6 of (240).]
Exercises. 1. Consider the equatioDs
' aix + biy + ciz = di, ka\x -f A^iy 4- kci^ = ^fe, aaa; + 6sy + c*? = <is.
The first two are inconsistent if d^ i^ kdi, and dependent when di — kdi.
Show that in the first case the only possible solutions of the three equations
are infinite, and in the second case there is an infinite number of solutions.
2. Show that the equations
aix + 6iy = and aax -f 62^ =
have one solution (0, 0), or an infinite number of solutions, according as the
determinant of the coefficients is dififerent from or equal to 0. Discuss also
'^g»5Ciaetrically.
3. Show that the equations
('
oiix + 6iy 4- ci« = 0, oax 4" ftzy + C22 = 0, aax + 6«y + csz = 0,
have one solution (0, 0, 0), or an infinite number of solutions, according as the
determinant of the coefficients is different from or equal to zero.
(Hint J Eliminate 2 so as to get two equations in x and y and discuss these
1^ in exercise 2.)
4. Show that the equations .
*2X'-Sy + 5z=-0, x + y-z=^0, Zx-'7y + llz==0
are not independent. What is the relation between them?
(Hint. To find the relation between the equations, find ki and k^ such that
hi tim^ the first trinomial plus k^ times the second shall equal the third.)
243. General Definition of a Determinant. — The array of n
rows and n columns,
(ll <l2 CLZ ' • ' CLn
61 62 &3 • • • &n
Ci C2 C3 . . . Cn
. • . •
ll (2 ^3 • • • h
is called a determinant of order n. The quantities forming the
array are called the elements of the determinant.
244,245] DETERMINANTS 221
If we form all possible products of n elements, each product to
contain one and only one element from each row and colunm,
and if these products are given proper signs, as will presently be
indicated, and added algebraically, the sum so obtained is defined
to be the value of the determinant.
Each product of n elements so obtained is called a term of the
expanded form of the determinant.
The elements ai, 62, C3, . . . , Zn form the principal diagonal.
The term 016263 . . . Zn is called the principal term of the ex-
pansion.
.244. Every term of the expansion of the determinant can be
formed from the principal term by rearranging the subscripts, leav-
ing the letters in their natural order.
For every term contains all the letters and all the subscripts,
and each only once, since it is a product containing one and only
one element from each row and each column. Hence if the letters
in any term be arranged in their natural order, the subscripts will
form some arrangement o' the numbers 1, 2, 3, . . . , n.
Conversely, every rearrangement of subscripts in the principal
term, the letters being left in their natural order, yields a term of
the expansion, since it contains one element and only one from
each row and each column.
Therefore all the terms of the expansion can be obtained by
forming all possible arrangements of subscripts in the principal
term.
We shall use the symbol An to indicate our determinant of
order n. Then we can write the equation
An = 2 ±0162^3 . . . Zn> (2 = sigma)
where the symbol S (sign for a sum) means that we are to form
the algebraic sum of all terms which may be formed from the term
written by forming all possible arrangements of the subscripts;
the signs of the terms so formed remain to be determined.
245. Number of Terms in the Expansion of A„. — The num-
ber of terms in the expansion of a determinant of order n is
1 X 2 X 3 X • • • X n, or | w.
Proof, We need only show that the number of possible arrange-
ments of the subscripts 1, 2, 3, ... n, is In.
^ I
222 DETERMINANTS [246,247
Starting with the natural order, and interchanging 1 in turn
with 2, 3, . . . , n, we form the n arrangements
1 2 3 ... 71)
2 1 3 . . . 7i|
2 3 1 ... 71,
....
2 3 4 ... 1.
In any one of these, keep 1 fixed in its position, and interchange 2
with 3, 4, . . . , n. In this way we form w — 1 arrangements in
which 1 occupies a given place. Treating each of the n arrange-
ments written above similarly, we obtain altogether n (n — 1)
arrangements. Each of these gives rise to a group of n — 2 ar-
rangements, including itself, by interchanging 3 with 4, 5, . . . , n.
Hence we obtain n (n— 1) (n — 2) arrangements. Proceeding simi-
larly we find the total number of arrangements to be [n.
246. Signs of the Terms in the Expansion of A„.
Inversion. An arrangement of the numbers 1, 2, 3, . . . , n
is called an inversion. An inversion is even or odd according as
the number of times a greater number precedes a lesser number,
is even or odd.
Thus, the possible inversions of 3 numbers are
123, 213, 231, 321, 312, 132;
of these the first, .third, and fifth are even, the others odd.
Further, the inversion of the subscripts in the term a4b2Czdi is ^
even. For 4 precedes 2, 3, and 1, and 3 precedes 1, making a
greater subscript precede a lesser one 4 times.
We now define the sign of each term of the expansion of A„ by the
rule that the sign shall be plus when the inversion of the subscripts is
even, minus when the inversion is odd.
Our determinant is now completely defined.
Exercise. Write out the expansion of
a\ 02 03 <H
h\ &2 &3 b\
Cl C2 CZ Ca
di 6,2 dz di
247. Properties of Detemunants.
1. A determinant is unchanged in value when its rows and coir
umns are interchanged.
A4 =
d
y"
247]
DETERMINANTS
223
For the expansion retjoiains unaltered.
2. Interchanging two adjacent rows or columns changes the sign
of the determinant.
For each term of the expansion will change sign, since two
adjacent subscripts will be interchanged; hence even inversions
change to odd, and vice versa.
By repeated application of this rule it follows that if any two
rows or any two columns be interchanged, the sign of the determinant
changes.
8. // all the elemenis of a row or column are 0, the determinant = 0.
For each term of the expansion contains a zero factor.
4. When all the dements of a row, or column^ contain a common
factory this may be taken out avfi written as a factor of the whole
determinant.
For each term of the expansion will'contain this factor.
It follows that, to multiply a determinant by any factor, we need
only multiply the elements of any row or column by this faxAor.
5. // two rows or columns are alike, the determinant = 0.
For by interchanging them we would have An = — An; .'• An = 0.
6. // the elements of two rows or columns differ only by a common
factor, the determinant = 0.
For by taking out the common factor the two rows or columns
become equal.
7. // in the expansion of An we collect the terms which contain the
several elements of any row or column, say the jth row, we have
u-
An =
a\ 0^2 az • ' * an
bi b2 bs . , . bn
Jl J2 J3
Jn
ll I2 I3 . ' ' In
= jlJl + J2J2 +
. . •
Jn^n*
Here Ji is called the cof actor of the element ji, and similarly for
«/2, . . . , Jfi*
8. A determinant is unaltered in value when the elements of any
row are increased by a constant multiple of the corresponding elemerUs
of another row. Similarly for columns.
224
DETERMINANTS
[247
For suppose that we add to the elements of the first row k times
the elements of the second. We obtain the determinant
ai + A;6i, 03 + kb2, . . . , an-}- kbn
&i &2 • • • &n
An' = Ci C2 . . . Cn
h
h
In
Let Ai, A2, . . . , -An be the cofactors of the elements of the first
row, so that
An' = (ai + fc6i)ili+ (02 + kh2) -42 + • • • + (On + kbn) An.
= {aiAi + 02^2 + • • • + a^An) +
fc(Ml-hM2+ • • • +hryAn).
ai a2 * • ' dn
&1 &2 • • • &n
• • • •
+ k
h h . . . In
61 62 • . . 6n
61 62 • • • &n
• • • •
L\ I2 » • • (fn
The first of these determinants is An, the second equals 0.
An' = An.
It follows that we can add to the elements of any row any linear
combination of corresponding elements of other rows.
Example, Without expanding, show that
102 104 106
99 98 97
12 3
= 0.
Subtract the second row from the first. The new form is
3 6 9
99 98 97
12 3
This is zero, by 6.
9. // the cofactors of any row or column be multiplied by the ele-
ments of any other row or column, the sum of the products is zero.
247]
DETERMINANTS
225
For we have
ai Ct2 ' • • Ctn
&1 &2 • • • &n
h I2 . . . In
= aiili + 02-^2 +
• • •
+ a»|An.
Replace the a's by the elements of any other row, as the second.
The result is
bi &2 • • • &n
&1 &2 • • • ^n
h h ' • ' In
= biAi + 62-A2 +
• • •
+ 6^n = 0.
10. If we strike out from An the jth row and kth column^ the remain-
ing determinant, of order n^ 1, is designated by Aj^ky o/nd is called
the minor of (he dement standing at the intersection of the row and
the column struck out.
Thus the minors of a\, 02, and a^ in the determinant
are, respectively,
Ol
02 az
61
&2 63
Cl
C2 C3
62 &3
)
61 63
, and
61 62
eg C3
Cl C3
Cl C2
We shall now show that, except as to sign, the minor of any
element equals the cofador of that element. We shall consider a
determinant of third order, although the argument will apply to
determinants of any order. We have
= aiili + 02-4.2 + fl3^8>
ai a2 az
A3 = 61 62 63
Cl C2 C3
where A\, -4.2, -4.3 are the cof actors of ai, 02, 03, respectively.
Then Ai = Ai.i.
For, since aiili contains all the terms of A3 which involve a\, and
since the expansion of A14 contains all possible products of ele-
226 DETERMINANTS [ 248
ments, one from each column and each row except the first, there-
fore ill and Ai,i must be identical. Now interchange the first
two columns, so that A3 becomes —A3. Then
— A3 = 62 61 ^3 = ~ 02^12 "" fli-^i " flsiis.
02 Oi
^3
62 61
63
C2 Ci
cs
61 63
The expansion
The minor of 02 is unchanged, namely
Ci C3
of this multiplied by 02 gives all the terms of the expansion of
—A3 containing 02.' But these are also contained in ^0^2^
Hence Ai,2 = — -4.2, or -4.2 = -- Ai,2.
In the same way, by moving the third column into first place
by iwo successive interchanges, which does not alter the sign of
the determinant, we find Ai,3 = ^13.
Let Aj,k denote the cof actor of the element standing at the
intersection of the jth row and A;th column of A^; we can bring
this element to the intersection of the first row and column by
j — 1 + fc — 1 successive interchanges of rows and columns.
Hence A„ will become (--iy+*-2 . ^^ qj. (~iy+*An, since (— 1)"^
a 1 ; hence by reasoning as above we find
11. We can now expand An according to the elements of its first
row in the form
An=aiAi,i — a2Ai.2+08Ai,3— a4Ai,4+ . . . +(— l)'*~^Ai,n.
To expand An according to the elements of any other row, we
can move this row into first place and then apply the last formula.
By this rule we can express a determinant of order n in terms of
determinants of order n — 1. Hence by repeated application of
the rule we can write out the complete expansion.
By a similar process the determinant can be expanded ac-
cording to the elements of any column.
248. Solution of Systems of Linear Equations. — We shall illus-
trate the method of solving a system of n linear equations involving
n unknowns by considering three such equations with three un-
knowns.
248]
DETERMINANTS
227
Solve for x, y, and z the S3rstem of equations
aix + biy + ciz^^^di,
a2X + 622/ + C22 = cfc,
Ozx + bsy + Czz = ^3.
Let the determinant of the coefficients be denoted by A, so that
ai 61 Ci
A = 02 &2 C2
cts &3. C3
Let the cofactors of ai, 02, 03 be ili, -4.2, ^3 respectively.
Multiply the given equations in order by ili, A2, A^, and add the
results. We obtain
{aiAi + a2A2 + 03^3) x +(hiAi + 62^2 + bsAs) y +
{CiAi + C2A2 + C3A3) Z = diAi + ^2^2 + ^3^3.
From (7) and (9) of (247) we see that the coefficient of a? is A,
and of y and z zero. Hence we get
X =
d^Ai + d2A2 + dzAs
di
61
Cl
^2 &2
C2
da
&3
C3
ai
61
Cl
02
&2
C2
03
63
C3
Similarly by multiplying by the cofactors of 61, 62, &3 and
adding we get y, and by multiplying by the cofactors of Ci, C2, C3
and adding we get z. The results are as given in (241).
In precisely the same way we can solve n linear equations in
n unknowns.
The exceptional cases which arise when A, the determinant of
the coefficients, is zero, have been considered in (242) for the case
of two and three equations. A similar discussion applies to the
case of n equations.
When the equations are homogeneous (i.e., di = 0, ^2 = 0,
ds = . , .), and A 7^ 0, the only solution is x = 0, 2/ = 0, 2; = 0,
. • . ; when A = 0, there exists an infinite nimiber of solutions.
^
228
DETERMINANTS
[249
249. Exercises. Evaluate the following determinants:
a 13
oH-1 2 2
aH-2 3 1
4
2 10
3 2 4
6 2 3
2.
a h g
h g f
9 f c
3.
6.
1115
12
3 4-32
1-14 5
6.
3
1
2
1
5
2
4
1 2
8.
2 5
4
3
4 1
6
1 6
1 -3
7
-5 3
s
9.
o a b
—a o c
—6 — c o
6 2 8 2
2 2 8 5
16 4 2
3 2 5 3
4 4 4
-1 -9 -1 9
-17 1-1
9 16 27 23
I
10.
o
■a
h
c
a
o
'd
-e
h c
d e
o f
-f
11.
ai o o o
02 bi o o
OS &3 cs o
04 &4 CA di
12. Show, without expanding,
that
6 1-7
5 -10 5
4 3-7
= 0.
13. Show that
1 1 1
X y z
ic2 J/2 z2
= (y - x) (2 - x) (z - y).
14. Show that
18 36 58 50
26 39 80 78
17 39 55 45
9 16 27 23
4 4 4
-1 -9 -1 9
-1 7 1-1
9 16 27 33
16. Give two pairs of values of x
and y which satisfy the equa-
tion
x y 1
3 1 1 =0.
1 -2 1
16. Give the coordinates of two
points on the line
X y 1
1 1 1 =0.
2 3 1
17. Trace the graph of
■
X y I
3 1 1 =0.
1 -2 ]
18. Give the coordinates of two
points on the line
= 0.
X
y 1
ffll
5i 1
at
ht\
249]
DETERMINANTS
229
19. Give three sets of values of x, y,
z which satisfy the equation
X y z 1
3 1-21
1-221
-14 11
0.
20. As in 19, for the equation
X y z I
ai Oi 09 I
bi bi bt 1
Cl C2 Ci 1
«0.
Prove the following identities:
21. cob(x + y)
. cos 2 re
a
0-
cos a; sin rr
sin ^ cos 1/
coex sin re
•
sin
X cos a;
. sin (x — y)
Binx coax
aiay cosy
24.
26.
= 0.
26.
27.
sinx sm^ sin 2
COSX cosy C0S2
sin^x coei^o; 1
sin^y co^y 1
an^z coe^z 1
cos X sin re cos x cos rr (sin y ■}- smz)
cosy sinycosy cos y (sin re + sin 2)
cos 2 sin 2 cos 2 cos 2 (sin rr + sin y)
sinrr sin 2 re sin 3 2;
sin^a; sin2 22; sin^Sa;
8in22; sin4rc sin6x
o sin (y — 2) + & sin (2 — a?) + c sin (a: — y).
= 0.
s 2sinrcsin2a;sin3a; (sin 2 rr — 2 sin x).
a b c
a' 6' c'
=
d e f
+
d e f
g h k
g h k
U Show that
a-ha' b + b' c + c^
d e f
g k k
29. Show that the equations
— 4x + y + 2 =0 and x — 2y + 2 =
are satisfied by
X :y :z =^
1 1
2 1
!2a
ftre satisfied by
X :y :z —
-4 1
1 1
te -f my + W2 = 0,
X + m'y -f n'z = 0,
-4 1
1 -2
m n
•
I n
•
I m
m' n'
•
V n'
•
V m'
230
DETERMINANTS
[^9
81. Show that the equations
2x + 4y + 5«=0,
3a;H-6y + 62 = 0,
are satisfied byrc:y:2sl: — 3:2.
Solve the following systems of equations:
82. 2a; + 3y-42H-7
7x-4y-l «0,
9a;-424-l =0.
83. 20w + 2w-7«0,
4i; + 5«>-l =0,
4u-3u? + 2-0.
0, U. -r + 8 + t + u^i,
r-« + < + w=3,
r + «-< + t*=2,
r-f«-f< — u = l.
36. 2a;— y — 32 + 10 = 1,
a? + 2y-f-« — it;«2,
3a;-3y-z + 2«7=— 1,
— x-y + 22-3ti; = 0.
A; + Z + m-2n = l,
2ib-i + 2m-4n = 2,
- A; + 2Z + 3w -6n --2,
fc -Z-h4m -8n »- L
CHAPTER XVII
PoLAB Coordinates. Complex Numbebs. DeMoivre's
Theorem. Exponential Values of sin x and cos x.
Hyperbouc Functions
260. Polar Coordinates. — We have made repeated use of the
system of rectangular coordinates, in which the position of any
point in the plane is defined by its abscissa and ordinate. A second
system of coordinates defines the position of a point with reference
to a single fixed line, called the initial line, and a fixed point on this
line, called the origin or pole.
In the figure, let OX be the initial line aaid the pole. We shall
consider OX as the positive direction of the initial line. Let P
be a point in the plane to be
considered. The position of
P is then fixed by its distance
OP = r from 0, called the
radius vector, and by the
angle XOF = ^, called the
vectorial angle. Then r, d
are called the polar coordinates of P, and the point is indi-
cated by (r, ^). Similarly Pi is the point (ri, ^i). The coordi-
nate ^ is positive when measured counter-clockwise from OX;
r is positive when measured from along the terminal side of B\
it is negative when measured from along the terminal side of B
produced back through 0. Thus the points (5, 30°) and (—5,
210°) coincide. Similarly with (135°, -3) and (-45°, 3).
Exercise. Plot the following points:
(45», 1); (45^ -1); (60°, 3); (-60^ 3); (|. 4); (- ^, 2); (^, |);
Calculate the rectangular coordinates of each of these points, taking as
origin and OX as the o^-axis.
231
232
POLAR COORDINATES
[251,252
261. Relation between Polar and Rectangular Coordinates. —
Let be the origin and OX the initial line of a system of polar
coordinates (figure). Let OX and OY
be the axes of a rectangular system of
coordinates. Then
(x = rcos^, [r-^x^ + V^
^y = rsin^; 1^ = tan-^^-
262. Curves in Polar Coordinates.
— When r and B are unrestricted, the
point (r, B) may take any position in the plane. When r and B are
connected by an equation, the point (r, B) is in general restricted
to a curve, the equation between r and B being called the polar
equation of the curve.
Example 1. Trace the curve whose polar equation is r » sin ^.
Aflsume a series of values for $, calculate the corresponding values of r and
plot the points whose coordinates are corresponding values of r and 0.
«»o**,3o^ 6o^ 9o^ l2o^ l5o^ l8o^ 210°, 24o^ 27o^ 3oo^ 330% 360**.
r-O, .5, .87, 1.0, .87, .5, 0, -.5, -.87, -1.0, -.87, -.5, 0.
The graph is shown in the figure.
For values of ^ > 360®, and for
negative angles, no new points are
obtained. The curve is a circle,
with radius » i.
Example 2. Trace the curve
r ^2$,
Here $ is understood to be in
radians.
- ^ X X 3 X ^^
'i' 2' T' *"' * * •
^ X 3x rt
r=0,2, X, "Y* 2x, . . . 00.
For negative values of d we
get corresponding negative
values of r. The curve is
the double spiral in the fig-
ure, the branches shown by the full line and the dotted line being obtained
from the positive and the negative values of $ respectively.
253]
COMPLEX NUMBERS
233
9.
r
-e^
10.
r
- logio
11.
r
-4.
12.
X
8S — *
Exercises. Trace the following curves:
1. r « 2 sin 6, 6. r = sin- 1 0,
2. r — OOB0. 6. r = tan-i ^.
8. r » tan 0. 7. r^ = 1.
4. r » sec 0. 8. r = 2*.
253. Complex N umb ers. — Let a and b denote any two real
numbers and i = V— 1. Then the quantity a + ib is called a
complex number. It may be considered as made up of a real
units and 6 imaginary imits, a X 1 + 6 X i.
Real numbers can be represented by points on a straight line.
To represent complex numbers ^y
geometrically, we require a plane.
Let OX and OF be a system of
rectangular axes, and P a point
in their plane having coordinates
(a, 6) (figure). Then P is called
the representative point of the
complex number a + ib.
When 6 = 0, P lies on the a^axis, and the complex number
reduces to a real number. Thus all points on the a;-axis corre-
spond to real numbers, and this
line is called the axis of real
niunbers.
Let P (figure) be a point (x, y)
in the plane, and let z be the com-
plex number represented by P.
Then
« = 05 +
Now take OX as the initial line and as the pole of a system of
polar coordinates. Let the polar coordinates of P be (r, B). Then
Hence
X = r cos B] y = r sin B.
z = x + iy = r (cos B + i sin B).
Here r is_ called thQ modulus and B the angje pf the. complfix.
niunber z.
234
COMPLEX NUMBERS
[264,255
When r is fixed, and d is changed by integral multiples of 2 ir,
we obtain a set of complex numbers of the form,
2 = r [cos (^ + 2 nir) + i sin (^ + 2 rnr)];
n = 0, ±1, ±2, . . . .
All these nimibers have the same representative point.
264. Addition of Complex Numbers. — The sum of two comr
plex numbers,
z s x + iy and z' = x' + iy\
we define by the equation
z + z' ^ {x + x')+ i (y + y').
We proceed to consider this
sum geometrically. Let P, P'
(figure) be the representative
points of 2, 2' respectively. On
OP and OP' as adjacent sides con-
struct the parallelogram OPQP\
Then Q is the representative point
of z + z\ For the coordinates of
Qare (x + x', y + y').
The difference of the two complex numbers z and z' we may
define by the equation
z-z' = (x-x')+i{y- I/O-
Exercise. Give a geometric construction for the representative point of
z — z'.
266. Multiplication of Complex Numbers. — The product of
the two complex numbers,
z =zr (cos 6 + i sin B) and 2' = r' (cos 6' + i sin 6'),
we define by the equation
zz' = rr' (cos 6 + i sin 6) (cos 6' + i sin $').
256, 257] DE MOIVRE'S THEOREM 235
Miiltipl3ring out the product of the two binomials we find
zz' = rr' [cos B cos ff — sin ^ sin ^' + i (sin ^ cos B' + cos B sin %')\
= rr' [cos (^ + ^0 + i sm (^ + ^Ol-
Therefore the modulus of the product zz' equals the product of (he
moduli of z and z\ and the angle of zz' equals the sum of the. angles
of z and z'.
By repeating this process we find
zz'z'' . . . = rrY' • • • [cos (^ + ^' + ^" + ; • • )
+ isin(^ + ^' + ^"+ • • • )]
for any finite number of factors 2, p', 2", ....
When the factors are all equal this reduces to
«*• =r f* (cos n6 + i sin n6)/
n being a positive integer.
Exercise. Show that the above definition of the product zz' is the same as
where z ^x •\-iy and 2' = a;' + ij/.
266. DeMoivre's Theorem. — When r == 1, then z = cos ^ +
i sin B. Hence by the above result we have
(cos 6 + < sin 6)* = cos n6 + i sin n6.
This equation contains what is known as De Moivre's Theorem.
267. Definition of «^. — Let p be any real number, positive or
negative, rational or irrational. Then by analogy with the result
for z^ when n is a positive integer, we define z^ by the equation
s^ = r^ (cosjpO + i sinjpO),
where 2; = r (cos ^ + i sin B).
Then, if q also be real, we have
29 = r^ (cos qS + isin qB)j
and
«»«« = rP+fl[cos {p + q)B + i sin (p + q)B]^ »p+«.
236
COMPLEX NUMBERS
[258
/
Hence the rules for exponents will be the same when the base is a
complex number as when the base is real.
Examples,
1. Find the modulus and angle of 2 ~ 3
-4i.
Here
3 = r COB d; — 4 = r sin ^.
r = V32 + 42 = 5; tan $ =
-4
or.
$ = tan
-(-I)
The angle lies in the fourth quadrant. *
2. Express 2 (cos 150** - i sin 150°) in the
form X + iy.
2 (cos 150° - i sin 150°) = 2^ - i V3 - |] = - V3 - i.
3. Find the value X)f (1 + i)* (2 - 3 i).
. (l+i)« = l + 2i + i» = 2i.
(l+i)«(2-3i) = 2i(2-3i) = 4i - 6i^= 6 + 4t.
Exercises.
1. Find the modulus and angle of
l-i;_4 + 3i; -5 + lli; 2i; 2; (l+t)(l-i);
3V3+3i; (3V3-3i)^ (l +tV3)(V3 +i).
Give figure for each case.
2. Find the value of:
(l + t)3; (l-t)4; (l+t)2(i+2i)2; (3-3i)2(V3 + i)'; (l-iVs)*.
268. Theorem. If P and Q are any real quantities and if
P + iQ=0, then P-^0 and Q==0,
Proof. By hypothesis, P + iQ = or P = — iQ.
Squaring, p2 _ _ q2
Now P2 and Q^ must be positive, hence the last equation states
that a positive quantity equals a negative quantity. This is
impossible unless both quantities are zero.
P = and Q = 0.
This theorem is used to replace a given equation of the form
P + iQ =
by the equivalent equations
P = 0; Q = 0.
t/
259]
ROOTS OF UNITY
237
?
/
As a coroUary we have, if
P + iQ^P' + iQ',
then P-=P' and Q = Q'.
For the given equation is equivalent to (P — P') + i (Q — QO = 0.
269. The nth Roots of Unity. — To solve the equation
X" — 1 = 0, or x" = 1,
replace 1 by its value cos 2kw + i sin 2 fcir, k being an integer.
We obtain
x" = cos 2 fcir + i sin 2 kir,
. 1
Taking the nth roots of both members we have, by putting p = -
in (267), "*
2fcir , . . 2kT
X = cos h I sm •
n
n
Here A; may be any
integer; letting k
= 0, 1, 2, . . .
n — 1, we obtain n
distinct values of
X, that is, n dis-
tinct nth roots of
1 . For other values
of k we obtain the
same roots over
again.
Geometric Rep-
resentation of the
nth Roots of Unity.
— The nth roots of
1 are,
A; = 0; Xi = cos + i sin 0=1,
= 1: X2 = cos h I sm — >
n n
k = 2; X3 = cos h ^ sm — >
n n
fc = n — 1; Xn
2 (n - 1) TT , . . 2 (n - 1) IT
cos — ^^ — h 2 sm— ^^ ^—'
n
n
/
238
EXPANSION OF SIN nS AND COS nS
[260
The representative points of Xi, X2t 0:3, .. . a:„ are ob-
tained as n equally spaced points on a circle of radius 1, the
coordinates of the first point being (1, 0) (figure).
To obtain the nth roots of any number a, we need only multiply
one o f its arithmetic nth roottby the nth roots of unity.
Example, Find the cube roots
of unity.
These are given by
ic«co8— ^ + tsin— ^; k— 0,1^2,
k^O; XI = COB 0° + i sin 0° = 1.
* = !; X2 = cos 120° + i sin 120**
A; =2; 0:3 = cos 240° + i sin 240°
" 2 2 ^'*-
To find the cube roots of 8, we have y/S ^ 2 ^/l =^ 2; -1 + i V3; -1 - i V3.
(We here use Vs to denote any cube root of 8, not merely the principal root.)
Exercises.
1. Solve the equations x' — 1 =0 and x' — 8 = algebraically and com-
pare with above results.
Solve the following equations by the trigonometric method and give a figure
for each case:
2. x^^l; 4. a;6 = l; 6. a;« = 1;
3. a:* = 81; 6. a;6 = 32; 7. x* = 27.
260. To express sin n6 and cos n6 in terms of powers of sin
and cos 6, n being a positive integer.
We have
(cos 6 + 1 sin 0)^ = cos n^ + i sin ruB.
Expand the left member by the binomial theorem, reduce all
powers of i to ±1 or ±i, and group the real terms and those
involving i. The above equation then becomes
n (n — 1)
cos 71$ + i sin rid
= [003"^ —
+ i(n cos'*" ^ ^ sin ^ —
I?
cos'»-2^sin2^ +
)
n(n- l)(n-2)
[3
cos^'^^sin^^ +
..).
261 ] EXPONENTIAL VALUES OF SIN X AND COS X 239
This equation has the form
Hence by the corollary in (258) we have I « ^
cosn^ = cos"^ - ^ \Z cos^-2 $ sin^ ^ + . . . .
if
sin n^ = n cos'*"^ ^ sin ^ — n{n— a^^"" >^ cos^"^ 6 sin^ ^ + . . . .
Examples,
sin 4 ^ = 4 cos* ^ sin ^ — 4 cos d sin3 ^.
cos 5 ^ = cos^d — 10 cos^d sin2^ + 5 cos sin*^. ^
Exercises. Expand in powers of sin and cos 0:
1. sin 39; 3. cos 40; 6. sin 60;
2. cos30; 4. sin 50; 6. cos 70.
261. Exponential Values of sin x and cos ac. — We have the
expansions, (219),
x^ . x^
• • >
X^ . x^
...
In the first series replace x by ix and define the result to be e**;
noting that
i^ = — 1, 1*3 = — i, i* = 1,
• • .
we obtain
/ii*2 /|[*3 /i*4 /if*o
Hence
c** = cos 05 + i sin a?.
Replacing x by — x ;
^-i» _ ^Qg a? — i sin a?.
A
240 HYPERBOLIC FUNCTIONS [262
From these equations we find .
cos » = — ; sina? = -
2i
These formulas are useful in many applications of the trigono-
metric functions.
Exercises. Using the exponential values of sin x and cos x, show that:
1. sin2 X + co^ X = 1. 3. cos 2 X = co# x — sin^ x.
2. sin 2 x = 2 sin X cos x. 4. cos* x — sin* x = cos? x — sin* x.
262. The Hyperbolic Functions. — In the expansions for sin x
and cos x given at the beginning pf (261) replace xhyix and define
the results to be sin ix and cos ix respectively. We obtain
amix^i\x + T^ + j^+ . . . 1;
cosic = 1+^+14+ • • • •
These equations we consider as defining t he sine and cosing of
the imaginary quantity ix.
Multiply the first equation by i and subtract the result from the
second. We obt ain
cos ix — i sin fx = e*.
Change x to — x;
cos fx + i sin ix = e~*.
(Note that sin ix = — sin (— ix) by the definition of sin ix.)
Combining the last two equations by addition and subtraction,
we find
6* + e~*. . . .6* — e"*
cos IX = ^ ; sin tx = I — s — •
2 ' 2
\ We now define
Hyperbolic cosine of x ( = cosh x) = cos ix;
Then
Hyperbolic sine of x ( = sinh x) = - sin ix.
cosna? = r — I smhx^ —
262] HYPERBOLIC FUNCTIONS 241
These functions are related to the hyperbola somewhat ss the
circular functions to the circle.
The remaining hyperbolic functions are defined by the equa-
tions
tanh X = — a-— ; coth x = r — r— ;
cosh X tanh x
sech X = — r — I csch x =
cosh X sinh x
Exercises. Show that:
1. sinhO«0; coshO«l. 5. 008h(— i;) « ooshx.
2. 8inhTi = 0; coshTi--!. g^ cosh* x - sinh* « = 1.
8. sinh^=i; cosh^=0. 7. sech* x = 1 - tanh* a;.
4. sinh (— x) = — sinh x. 8. — csch* x = 1 — coth* x.
Draw the graphs of the equations (see tables) :
9. y = e*. IL y ^ coshos.
10. y s 6~'. 12. y » sinhx.
CHAPTER XVIII
Permutations. Combinations. Chance
263. Permutations. — A permutation is a definite order or
arrangement of a group of objects, or of part of the group.
Let there be a group of n distinct objects. The number of
possible arrangements/ taking r of these objects at a time is called
the number of permutations of n things r at a time, and is denoted
by nPr.
Theorem 1. The number of permutations of n things rata time is
nPr=^(^ — 1) ...(» — !• + 1).
Proof. Evidently nPi = w-
Now with each of the n objects we may pair any one of the remain-
ing n — 1 objects.
Hence n^2 = n{n — 1).
With each one of these n (n — 1) permutations containing 2 objects
we may associate one of the remaining n — 2 objects.
Hence nPs = n (n — 1) (n — 2).
Proceeding in this way we obtain the formula stated.
When r = n we have
nPn = |_^-
Exercises.
1. How many numbers of four figures each can be formed from the digits
1,2,3,4?
2. How many 3-figure numbers can be formed from the digits 1, 2, 3, 4, 5?
3. How many numbers greater than 1000 can be formed from the digits
1, 3, 5, 7, 9?
4. How many changes can be rung with 8 bells, 4 at a time?
264. Combinations. — A combination is a group of objects,
without reference to their arrangement.
242
265] PERMUTATIONS AND COMBINATIONS 243
The number of different groups or combinations of n objects,
each group containing r objects, is called the number of combina-
tions of n things r at a time, and is denoted by JJt-
Theorem 2. The number of combinations of n things r at a
time is
^ __n£r _ n (n — 1) » ' ' (n — y + 1)
\r \r
Proof. Suppose all the combinations of the n things r at a time
to be written down. Each group so written will yield, by per-
muting its objects in all possible ways, jr permutations. Hence
there are \r times as many permutations as combinations, or
^ ' jT^g^ ^ ^P^ = ^ (^ _ 1) ^ ^ ^ (^ _ ^ ^ 1).
Hence the theorem.
Exercises.
1. How many triangles can be formed from 6 points, no three points being
collinear?
2. How many tetrahedrons can be formed from 12 points, no four points
being coplanar?
3. How many conmxittees of 3 persons each can be formed from a club
of 10 persons?
4, Show that nCr = nCn-r'
(This is a convenient formula when r is nearly as large as n. It is then
shorter to calculate nCn-r-)
5. Show that nCo + nCi + nC2 + • • • + nCn = 2'».
(Expand (1 + z)^ and put x — 1; nCo is defined to be 1.)
6. How many committees, consisting of from 1 to 9 members, can be
formed from a club of 10 persons?
7. Find the value of 20C18.
266. Theorem 3. — The number of permvMions of n things n at
a timej when p things are alikej is
\n
\P
Proof. Let P be the nmnber of permutations sought, and sup-
pose them written down. If now the p things in question were
unlike, by permutating them among themselves each of the P
permutations would yield \p permutations; the total number of
permutations so formed would be [p P and must equal „P» or |n.
Hence the theorem.
244 CHANCE [266,267
Similarly, the number of permutations of n things n at a time,
when p things are all of one kind, and g of a second kind, will be
and so on.
266. Exercises.
1. How many permutations of seven letters each can be formed from the
letters of the word "arrange"?
2. How many permutations of 11 letters each can be formed £rom the
letters of the word " Mississippi " ?
3. How many words, each containing a vowel and two consonants, can
be formed from 4 vowels and 6 consonants?
4. How many even numbers of four figures each can be formed from the
digits 1, 2, 3, 4, 5, 6?
5. How many elevens can be chosen from 20 players if only 6 of the 20
are qualified to play behind the line?
6. As in 5, if in addition, only 2 men are qualified for center.
7. How many sums can be formed with one coin of each denomination,
from a cebt to a dollar?
C As in 7, except that there are two coins of each denomination.
9. If two coins are tossed, in how many ways may they fall?
10. As in 9, for 10 coins.
11. If two dice are thrown, in how many ways may they turn up?
12. As in 11, for 3 dice.
267. Probability or Chance. — If a bag contains 4 white and
3 black balls, and a ball is drawn at random, what is the chance
that it be white ?
In order to solve this problem we first define chance or probi^
bility.
Definition. The measure of the probability of the occurrence of
ap event is taken to he the giwtient,
number of favorable ways
total number of possiblejg
In the problem above, since there are 7 balls altogether, there are
7 possible ways of drawing one ball; of these 4 are favorable, since
there are 4 white balls. Hence the chance that a white ball be
drawn is 1
3
Similarly the chance for a black ball is =•
267] CHANCE 245
If an event can happen in a ways, and fail in b ways, then, by the
definition, the chance that it will happen is — r-r, and that it
a + o
will fail is
a + b
Since the event must either happen or fail, the probability
for which is — j-r H rr = 1, we have 1 as the mathematical
a+b a+b
symbol for certainty.
If p is the probability that an event will happen, 1 — p is the
probability that it will faiL
Example 1. From a bag containing 4 white and 3 black balls, 2 balls are
drawn at random.
(a) What is the chance that both be white?
Number of favorable ways: 4C2 = 6.
Number of possible ways: 7C2 = 21.
Hence the required chance is: P ~ oT ~ 7*
(b) What is the chance that at least one be white?
Favorable cases: both white, 4C2 =6;
one white, other black, 3 X 4 = 12.
.'. Total number of favorable cases is 18.
Number of possible cases, as before, 21.
XT 18 6
Hence P ~ 21 ~ 7 *
A shorter method is as follows: The probability that both balls be black
is ?^ = — = - . Hence the chance that at least one be white is 1 — = = = •
7C2 21 7 . « «
Example 2. From 12 tickets, numbered 1, 2, . . . 12, four are drawn at
random.
(a) What is the probability that they bear even numbers?
Since 6 tickets bear even numbers, the number of favorable cases is 6^4.
The total number of ways of drawing 4 tickets from 12 is 12C4. Hence
6^4^ 6«5'4>3 ^1
^~i2C4 12.11.10.9 33'
(6) What is the chance that two bear even, the other two odd numbers?
We can select two tickets bearing even numbers in 6C2 ways; also two bear-
ing odd numbers in 6C2 ways. Combining any one of the first with any one
of the second gives 6C2 X 6C2 favorable ways. Hence
6C2 X 6C2 ^ 5
^" 12C4 11'
246 CHANCE [268, 269
268. Exercises.
L If 5 coins are tofised, what is the chance of three heads?
2. If 5 coins are tossed, what is the chance of at least two heads?
3. If 3 balls are drawn from a bag containing 5 white and 4 black balls,
what is the chance that all three are white?
4. In exercise 3, what is the chance of drawing 2 white balls and one
black baU?
5. In exercise 3, what is the chance of drawing at least one white ball?
6. What is the chance of two sixes in a single throw of two dice?
7. What is the chance of throwing three sixes in a single throw with three
dice?
8. Three dice are thrown. What is the chance that the sum of the
numbers turned up is 11?
9. As in 8, except that the sum is to be 7.
10. Six cards are drawn from a pack of 52. What is the chance of three
aces?
11. Six cards are drawn from a pack of 52. What is the chance that all
are of the same suite?
269. Compound Probabilities.
Definition. Two events are said to be independent when the
occurrence of one does not affect that of the other.
Theorem 4. The chance that both of two in devendent events shall
haDpen is the moduct of their separate yrooab ilities .
Froof, Suppose the first event happens in a ways and fails in
6 ways, out of a a + fe possible ways, and that the second happens
in a' ways and fails in V ways, out of a total of a' + V ways.
Combining each of the a favorable ways of the first event with
each of the a' favorable ways of the second, we have aa' favorable
cases. The total number of possible cases is (a + fe) (a' + 6').
Hence
aa' a ^^ a'
(a + fe) (a' + feO a + fe " a' + fe'
which is the product of the separate probabilities of the two events.
As an immediate extension, we have the
Theorem 6. // the probabilities of several independent events be
Pif V2) - * ' Pny the probability that all vxill happen is
-P = 1>1 X 1>2 X • • • XPn*
Example, From a bag containing 4 white and 3 black balls, 2 balls are
drawn in succession. What is the chance that both are white?
269] CHANCE 247
On the first drawing the chance for a white ball is = : on the second, ^ . The
7 o
probability of both events is therefore
7^6 7'
Definition. Two events are said to be dependent when the
occurrence of one of them affects that of the other.
Theorem 6. Of n dependent events, let the chance that the first
rmU happen be pi, the ch ance that the second then follow s be P2, that
the third then follows be ps, and so on. The chance that all these
events shall happen is then
P = PiXp2Xp3 X • • ' Pn
This is an immediate consequence of the preceding theorem.
Theorem 7. If p be the chance that an event will happen in one
trial, the chance that it will happen just r times in n trials is
p = nCrP^ii - py
\—T
Proof. The chance that r trials out of n shall succeed is p*",
and that the other n — r trials shall fail is (1 — p)'*"*'. Hence the
probability of success in r particular trials and of failure in the
n — r other trials is p*" (1 — pY'^. But of the n trials, any r
may be the successful ones, which gives nCr possibilities, each
having a probability p*" (1 — p)""*". Hence the result stated.
Examples.
1. In a class of 3 students, A solves on the average 9 problems out ,of 10,
B 8 out of 10, C 7 out of 10. What is the chance that a problem, presented
to the class^ will be solved?
The problem will be solved unless all three students fail, the probability
for which is
10 ^ 10 '^ 10 500
Hence the chance that the problem will be solved is
3 497
1 -
600 500
2. Two bags each contain 5 black balls, and ^ third bag contains 5 black
and 5 white balls. What is the chance of drawing a white ball from one of
the bags selected at random?
248 CHANCE [270
The chance that the bag containing white balls be chosen is -^ . The chance
1 ^
that a white ball be now drawn from this bag is x . Hence the probability
that both events happen and that a white ball be drawn is
3^2 6'
3. A coin is tossed 10 times. What is the chance for just 3 heads?
The probability of a head in one trial is ^ . Hence
270. Exercises.
1. Three hats each contain 5 tickets, those in two of the hats being num-
bered 1,2, . . . 5, and those in the third hat being blank. What is the chance
of drawing a ticket bearing an even number from one of the hats selected at
random?
2. If in exercise 1 two tickets be drawn from a hat chosen at random,
what is the chance that both bear even numbers?
3. If each of two persons draw a ticket from one of the hats in exercise 1,
the first ticket being replaced before the second is drawn, what is the chance
that both persons draw the same number? What is the chance that both
draw blanks?
4. If a coin be tossed 10 times, what is the chance for at least 7 heads?
6. How many different sets of throws can be made with a coin, each set
consisting of 5 successive throws?
7
6. The chance that a person aged 25 years will live to be 75 is ^^ • What is
the chance that, of three couples married at the age of 25, at least one shall
live to celebrate their golden wedding?
7. A bag contains 10 white, 6 black, and 4 red balls. Find the chance that,
of three baUs drawn, there shall be one of each color.
8. A gunner hits the target on an average 7 times out of 10. What is
the chance that 5 consecutive shots shall hit the target?
9. Two dice are thrown. Find the chance that the sum of the numbers
turned up shall be even.
CHAPTER XIX
Theory op Equations
271. We shall refer to the equation
as the standard form of the equation of nth degree ; pox^ is called
the leading term and pn the constant (or absolute) term.
The coeflBcient of the leading term may be made equal to unity
by dividing the whole equation by this coeflBcient.
When all the terms written in equation (1) are present, the
equation is said to be complete; when one or more terms are
absent, the equation is said to be incomplete. An incomplete
equation may be made complete by supplying the missing terms
with zero coeflBcients.
We shall represent the polynomial forming the left member of
equation (1) by /(a?); /(a) shall denote the value of this poly-
nomial when z = a,f (b) the value when x = b, and so on.
A root of an equation is a value of x which satisfies the equa-
tion; hence a is a root of the equation / (x) = if / (a) = 0.
In the present chapter we shall consider methods of finding the
roots of the equation / (x) = 0.
272. Factor Theorem. — If a is a root of the equation f (x) = 0,
thenf(x) is divisible by (a: — a), and conversely.
Proof/ Divide /(x) by (x — a); let Q be the quotient, R the
remainder. Then
f(x) = (x-a)Q + R.
Putting X = a, we obtain 72 = 0, since f(a)= by hjrpothesis.
Hence / (x) is divisible by (x — a) without a remainder.
Conversely, assume
/(x) = (x-a)Q.
Put X == a and we have / (a) = 0; hence a is a root of / (x) = 0.
[See also (11), (f).]
249
250 THEORY OF EQUATIONS [273-275
273. Depressed Equation. — When a is a root of the equation
/ (x) = 0, we may write
/(a:) = (x-a)Q.
Any other value of x which reduces / (x) to zero must also reduce
Q to zero, and is therefore a root of the equation Q = 0.
But if / (x) is of degree n, Q will be of degree n — 1. 'Hence when
one root of an equation is known, the other roots may be found
by solving an equation of degree one less than that of the given
equation, and whose left member is found by dividing the left
member of the given equation by (x — the root). ^
The new equation is icalled the depressed equation.
Exercises. Show that each of the following equations has the root indi-
cated, and find the other roots:
1. x* - 9x2 + 26x - 24 = 0; x = 2.
. 2. X* + 3x2 ~ Sx - 24 = 0; x = - 3.
3. 3x«-14x2 + 17x-6 = 0; x=}.
274. Number of Rootg. — We assume that every equation of
the form (jQiJ27JJ^ha§ JktJssgt«02gJ88l« ^^^^ ^®» there exists at
least'one value of x, real or imaginary, which satisfies the equation.
It can then be shown that an equation of the nth degree has just
n roots.
For, let ai be a root. Form the depressed equation by removing
from /(x) the factor x — ai, and let the new equation, of degree
n — 1, be /i(x)= 0. By the above assmnption, this equation
has a root, say 02. Removing from /i(x) the factor x — 02, we
obtain a new equation, /2(x)= 0, of degree n — 2, and so on.
After n — 1 steps, by which n — 1 roots are removed, we have
an equation of the first degree which gives one more root. Hence
there are just n roots.
275. To Form an Equation Having Given Roots. — Let it be
required to form an equation whose roots are ai, a-z, aa, . . . On.
Obviously the required equation is
A (x — ai)(x — a2)(x — as) . . . (x — On) = 0,
A being an arbitrary constant.
Exercises. Form the equations whose roots are:
1. 1, 2, 3. 3. 2, 2, -2, 0. 5. ±1, i, }.
2. 1, -1, 2. 4.-1, -2, -3, -4. 6. -i, §, }.
(Write the results from exercises 5 and 6 with integral, coefficients.)
276,277] THEORY OF EQUATIONS 251
276. Relations between Coefficients and Roots. — In the case
of 2, 3, and 4 roots respectively we find on expanding,
(x — ai)(x — 02) = a;2 — (ai + a2)x + a\a2.
(x — ai)(x — a2)(x — as) = x^ — (ai + 02 + 03) x^
+ (aia2 + a2a3 + aias) x — aia2a3.
(x — ai)(x — a2)(x — a3)(x — 04) = x* — (ai + 02 + as + a4)x3 +
(...) /p2 __(...) ^ _f. aio^a^a^.
We here observe three facts, namely:
1. jTAe coefficient of the leading term is unity;
2. The coefficient of the second term is the negative sum of the roots;
3. The constant term is the product of the roots, plus when the
number of roots is even, minus when odd.
We shall show by induction that these results are true in general.
Assume them to be true for w — 1 roots; then if the equation
whose roots are ai, 02, . . . On-i, be
x^-i + gix'^-^^ . . . +g^_^ ^0,
we have by hjrpothesis,
gi = -(ai + a2+ • • • +On-i); gn^i = (-l)""^aia2 . . . an^
Multiplying the above equation by (x — a^), which introduces
the new root an, we find on collecting in powers of x:
^ +(qi - aJa:'*-^ + . . . - a^qn-i = 0,
or, x'*-(ai+a2+ • • • +an-i + aja:""^ + • • •
+ (— l)'*aia2 . . . an-ia„ = 0.
Hence if the results are true for the case of n — 1 roots, they
hold also for n roots. But they are true for 4 roots, hence also
for 6, hence for 6, and so on.
Exercise. Show by induction that the coefficient of the third highest power
of X equals the sum of the products of the roots taken two at a time.
277. Fractional Roots. — An equation with integral coefficients,
in which the coefficient of the leading term is unity, cannot have as
a root a rational fraction in its lowest terms.
252 THEORY OF EQUATIONS [278
Proof. Assume that the equation
has integral coefficients and that one of its roots is t > where a and
6 are integers, relatively prime. Putting a: = r we have,
Multiplying through by h^~^ and transposing,
a"
Here we have a fraction in its lowest terms equal to an integer,
which is impossible. Hence t cannot be a root.
Corollary . Any rational root of an equation whose coefficients are
integers and whose leading cbeffi^dent is unity must he an integer.
278. Imaginary Roots. — // the general equation of nth degree,
with real coeffi^dentSj has an imaginary root a + ib, then also the
conjugaxe imaginary , a — ib, is a root.
Proof. Assume that a + ib is a root of the equation
Then
(a + *)» + pi{a + iby-^ + P2 (a + ib)""^
+ • • • + pn_i (a + *) +Pn = 0.
Expanding the binomials, reducing all powers of t to ± 1 or ± i,
and collecting terms, we have a result of the form
Hence P = and Q = 0. (268.)
Now substitute a — ib for x and proceed as before. The result
will be
f{a-{b)=P-iQ,
since the only difference is in the sign of i. But P = and Q = 0,
hence P — iQ ^ 0, or /(a — ib) = 0. Therefore a — i6 is a root.
279-281] THEORY OF EQUATIONS 253
We may state our result as follows: Imaginary roots, if preserU
*tmHam\ iwm^ awwifc^ »J ■ w**
at all, always occur in conjugate pairs,
279. jMuitiple Roots. — When an equation has two or more roots
equal to the same value " a," then " a " is called a multiple root.
Suppose that the equation
has m roots, each equal to a. Then
f(x) = (x-a)-^Q,
where Q is a new polynomial.
Let/'(x) denote the first derivative of f(x) with respect to x;
then
fix) = (X - a)-^ + m (a: - a^'^Q.
This shows that/'(x) contains the factor (x — a)*^"^, and hence
that, if / (x) = contains a root ** a " repeated m tim^s, f(x) = will
contain this root repeated m — 1 tim£s; f (x) and f{x) will then have
the factor {x — a)^"'^ in common.
Hence we have the following rule for finding multiple roots of
the equation / {x) = 0.
Find the H. C, F, (13) off (x) andf.(x); to a factor (x - a)"^'^ of ^
the H. C, F, corresponds a factor (x — a)*" off (x).
280. Exercises. Test for multiple roots and find all the roots
of the equations: ^^^ ^^;:^5--
1. x3_3a;2+*'4 = 0. ^ ^- 5. x*-3x«-7a;2 + 15x + 18=0.
2. x3 - 3 a; - 2 = 0. ^^ 6. o^ + 4 a;« - 16 x - 16 = 0.
3. x4-2a;3-iia;2+i2a; + 36=0. 7. o^ -a;' - 3x2 + 5x - 2 = 0.
4. x4-2x8-39a;2 + 40x + 400=0. 8. 4x4 + 6x8 + 5x2 - 6a; + 4 = 0.
9. 9x4-54x3 + 80x2+6x-9=0.
10. 72x6 _ 276x^ + 278x8 + 45x2 - lOSx - 27 = 0.
•
281. Transformation of Equations. — In the following discus-
sion we assume that any missing powers of x are inserted, supplied
with zero coeflBcients, so as to make the equation formally com-
plete. We consider the equation
/ (x) s pox"" + pix^-i + p2a;"-2 + • • • + Pn-ix + p» = 0.
/
254 THEORY OF EQUATIONS [281
I. To change the signs of the roots.
Put X = — y. We obtain,
Po(-y)"+Pi(-2/)''-^+P2(-2/)"-H • • • +P»-i(-2/)+Pn = 0,
or, on multiplying through by (— l)**,
Po2/'*-/>i2/""^+P22/'*~^- • • • +(-l)'*"^Pn-iy+(-l)'*Pn=0.
Hence, to change the signs of the roots, change the signs of alternate
coefficients, beginning with the second term,
n. To multiply the roots by a constant factor, m.
Replace xby— (so that y = mx).
Then
-e)"+p.(r'+-er+---+-e)+'-=°-
Multiplying through by m**, we have,
Voy"" + mpiy'''-^ + m^pat/**"^ + • • • + m^'-^pn-iy + m^pn = 0.
Hence, to multiply the roots by a constant factor m, multiply the
y coefficierUs in order, beginnirigvrith the second J by mym^,m^, . . . m^.
When m = —1 we obtain the preceding rule for changing the
signs of the roots.
m. To increase the roots by a constant quantity, h.
Replace x by j/ — A (so that t/ = x + A). Then
Po (y - h)^ + pi (t/ - hr-'^ + p2 (y-hr-^+ . . .
+ Pn-l (y-h)+Pn = 0.
Expanding the binomials and collecting in ppwers of y, we
obtain a result of the form,
We shall now show how to obtain the coefficients Pi, P2, . . . Pn-
Replacing y in the last equation by x + A, the result must be
the original equation, / (x) = 0. Hence
f(x)=po (x + hr + Pi(x + A)~-i + P2(x + A)»-2 + . . !
+ Pn-l(x+A)+P,.
This shows that if / (x) be divided by x + h, the remainder is P^.
If the quotient be divided by x + A, the remainder is Pn^ 1 ; divid-
ing the second quotient by (x + A) , the remainder is Pn -2, and so on.
282] THEORY OF EQUATIONS 265
Hence, to increase the roots of the equation by h, divide f (x) by
x + h, then divide the quotient by x + h, divide the new quotient by
x + hy and so on. . The successive remainders are^ in order,
■* n> ■* n-l> -1 n-2; • • • ■* !•
A concise method for performing the required divisions will be
explained in the next article.
282. Synthetic Division. — When h and the coefficients po>
Pi, P2, . . . Pn are integers, the work of dividing f{x) may be
performed by the method of synthetic division. We shall illustrate
this by increasing the roots of the equation
a^-8a;-15 =
by 2.
Performing the first division at length, we have:
a:3+0x2-8a:-15
a^ + 2x^
-2x^
-8a;
-2a;2
-4x
— 4x —
15
-4a;-
8
X +2
^2 — 2 X — 4 quotient.
— 7 remainder.
We first shorten this operation by omitting to write the powers
of X, using only the detached coefficients, thus:
1 + 0-8-15
1 + 2
-2
-8
-2
-4
-4-
15
-4-
8
1+2
1-2-4
- 7
This may be written more compactly as follows:
1+0-8-15 I +2
+2-4- 8
1st quotient 1 — 2 — 4 1 — 7 remainder.
256
THEORY OF EQUATIONS
[282
Dividing the quotient by a; + 2 we have,
1-2-4 I +2
+ 2-8
2nd quotient 1 — 4 [ + 4 remainder.
Dividing the second quotient by x + 2 we have,
1-4 I +2
+ 2
3rd quotient 1 1 — 6 remainder.
The whole operation may now be written thus:
1+0-8-15 I +2
+2-4- 8
1-2-4
+ 2-8
— 7 1st remainder
1-4
+ 2
+ 4 2nd remainder
— 6 3rd remainder.
Then the transformed equation is:
a:3-6a;2 + 4x-7 = 0.
To diminish the roots of an equation by A, proceed as above
with X — h in place oi x + h. As an example, we diminish by 4
the roots of the equation
x^-5a^ + 7x^ -17X+11 =0.
1 - 5 + 7-17 + 11 I -4
- 4 + 4-12 + 20
1- 1+ 3- 5
- 4-12-60
— 9 1st remainder
1+ 3 + 15
- 4-28
+ 55 2nd remainder
1 + 7 + 43 3rd remainder
- 4
ll+ 11 4th remainder.
Hence the transformed equation is:
a:^ + 11 a:3 + 43^2 + 55 a. « 9 = 0.
283, 2g4]
THEORY OF EQUATIONS
257
In using the method of synthetic division note that the coeffi-
cient of the leading term remains unchanged.
283. The graph of the equation y =f{x)^ when
To construct the graph which shall
represent the fluctuating values of y
as X varies, we assume a series of
numerical values for z, calculate the
corresponding values of y, and plot
the points (x, y). On drawing a
smooth curve through these points,
we obtain a graph such as that in
the figure, which represents the equa-
tion
y = 7? — 2x — 1.
• • •
+ i>n-ia5+i>n.
,r
■
«
•^^^
/
^
/,
\
1
1
y « x" — 2a;— 1.
Here a set of corresponding values of x and y are:
y ^ ij ' itt, o, . . . , u, — 0, • • • •
Since the curve crosses the x-axis when y = 0, we see that the
abscissas of the points where the graph of the equation y = fix)
crosses the x-axis {called the x-intercepts of the graph) are the real
roots of the equation f(x)=0.
An inspection of the above graph shows that one root of the
equation a:3~2a; — 1 = is —1, another root lies between
— 1 and 0, and the third between +1 and +2. On removing
the factor x + 1 from this equation, the depressed equation is
x^ — X — 1 =0. Hence the exact values of the other two roots
are i (l ± V5), or approximately, +1.62 and —0.62.
284. Effect of Changing the Constant Term. — Suppose that
we add a quantity k to the constant term of f (x), bo that the
equation
y = f(x)
becomes y = f (x) + k.
Suppose the curve y =f{x) to be plotted; on adding k to each of
its ordinates, we obtain the graph of y = f{x) + k. That is, if
258
THEORY OF EQUATIONS
k be added to ihe constant term of the equation y '=f{x), the graph
ia displaced verlicaSy through the distance k, upward if k ia ptus,
downward if k is minus.
As an example, consider the equations
CD
(2)
(3)
V-ix'-T'-2x + ».
The graphs are shown in figure (a). The curves are of precisely
the same form, but (2) lies two
I units higher than (i), and (3)
I two units higher than (2).
»)
/
-^
-
\
1
\
X,
*
\
/
1
0.
\
)
X,
X,
_
Instead of displacing the curve
vertically, say upward, the same
effect is produced in the graph
by moving the x-axis an equal distance downward. Thus equa-
tions (1), (2), (3) are represented graphically by the curve in'
figure (6), y being counted from the lines 0\X\, O2X2, O3X3
respectively.
285,286] THEORY OF EQUATIONS 259
286. Occurrence of Imaginary Roots in Pairs. — We can now
consider article (277) geometrically. Thus in the first figure of
(284), graph (1) shows that the equation
^a:3-a;2-2x + 2 = b
has three real unequal roots; replacing 2 by 4, the two positive
roots become equal; that is, the equation
ia;3-x2-2x + 4 =
has three real roots, two of which are equal; finally on replacing
4 by 6, the two equal roots become imaginary; that is, the equation
^x3-x2-2a; + 6 =
has one real root and two imaginary roots.
In general, by changing the constant term in /(;r), the graph of
y ^ fix) may be raised or lowered so that one of the " elbows "
of the curve, which at first is cut by the x-axis, will become tangent
to the X-axis, and on further changing the constant the x-axis will
fail to intersect this elbow. Thus two real unequal roots first
become equal, then imaginary.
286. Exercises. Multiply the roots of the equation
1. a;8 + a;2 - a; - 1 = by 2;
2. x3 - 2 X + 1 = by -2;
3. x3- 48a; -112=0 by i;
4. a;4 + 6 a;8 + 3 a;2 - 26 a; - 24 = by -i.
Miiltipl5^ the roots of the following equations by the smallest factor which
will make all coefficients integers
6. x2 + a; +1 = 0. 8. a;3 - .1 a;2 + .01 a; = 0.
6. i x3 - a;2 + ^ = 0. 9. a;3 + i a;2 - ^V = 0.
7, x2-Jx-i=0. 10. a;4 + 1.2 a:2 - .225 a; + .015 = 0.
Increase the roots of the equation
11. a;3-3a;2 + 4 = by 2.
12. 4 a;3 - 3 x - 1 = by 3.
13. a;4 - 2 x3 - 11 a;2 + 12 a; + 36 = by -2.
' 14.^4 -2 a;3- 39x2 + 40 a; + 400 = by -4.
260
THEORY OF EQUATIONS
[287
In the following equations increase the roots by a quantity such that the
term involving the second highest power of x shall disappear.
16. x8 - 3 x2 + 2 = 0. 17. x3 - 3 x2 - 6 X + 1 = 0.
16. a;8-2ic2 + l =0. 18. x* - 4x3 - 8x + 32 = 0.
In the following equations change the constant so that two roots shall
become imaginary.
19. x3 - x2 - 2 X = 0. 21. x8 - 3 X - 2 = 0.
20. x» - 3 x2 + 3 = 0. 22. x8 - x2 - X + 1 = 0.
Solve the following equations, given one root.
23. x3-2x2+x-2 =0; x=V-l.
24. 2x4-3x3 + 5x2-6x + 2=0; x=-2 V^.
26. x6 -8x3 -8x2 + 64 = 0; x=-l-V^.
>c
X
287. Approximation to the Roots
of an Equation. — In this article
we shall illustrate a method for
obtaining, to any desired degree of
accuracy, any real root of an alge-
braic equation. As an example we
shall obtain, to four decimal places
inclusive, one of the roots of the
equation
(1) /(x)=x3-4x2 + 4 = 0.
The graph is given in the figure.
First Step, Location of Real
Roots, We first locate the real
roots by trial. As a set^ of corre-
sponding values of x and / {x) we have
X =~2, -1, 0, +1, -1-2, +3, -1-4.
/(x) = ~20, -1, -1-4, -M, -4, -5, +4.
When / {x) changes sign, the graph crosses the x-axis, and at least
one root must lie between the corresponding values of x. Hence
there is a root between —1 and 0, another between H-1 and +2,
and a third between -f-3 and H-4. But there cannot be more than
three roots, since a cubic expression cannot contain more than
three linear factors. Hence there is just one root in each of the
above intervals.
(
[V
/
\
/
\
/
\
o
\
1
\
1
\
\
\
/
\
/
\>
X
287] THEORY OF EQUATIONS 261
We shall proceed to obtain the root between 1 and 2.
Second Step, Diminish the roots of the given equation by the inter
^gral part of the root required (281).
1-4 + + 4 |-1
-1+3+3
1-3-3
-1 + 2
1-2
-1
3
+ 1
-5
1-1.
The transformed equation is
(2) 3^-x^-5x + l-=0.
Since (1) h^ a root between 1 and 2, (2) must have a root
between and 1, that is, a decimal root. To make this root an
integer, we take the
Third Step. Multiply the roots of the transformed equation by 10
(281).
The new equation is
(3) x3 - 10 x2 - 500 X + 1000 = 0.
The root of (3) between and 10 will give the first decimal of the
required root of (1). If we neglect the terms in oc^ and a^ in (3)
we get an approximate value, x = 2. Putting a; = 2 in (3), the
left member is negative; now putting x = 1, the left member is
positive. Hence the root lies between 1 and 2, and the required
root of (1) is 1.1+.
We now repeat these steps and obtain the first decimal of the
root of (3), which will be the second decimal of the root of (1), and
so on. Indicating the three steps in order by (a), (6), (c), we
obtain the successive decimals of the root as shown below, the
process of finding the first decimal being included for completeness.
(3) 3^3 - 10a;2 - 500x + 1000 = 0.
(a) Locate the root between and 10.
Neglect terms in ofi and x^; then x = 2. Try this value and the
next smaller value (or larger, if the left member of (3) does not
change sign) and the root is located between 1 and 2.
262
THEORY OF EQUATIONS
[287
(fe) Diminish roots by figure found in (a).
1 - 10 - 500 + 1000 I - 1
—
1+ 9+ 509
1-
9-509+ 491
1+ 8
1-
8
1
-617
•
U- "
Transformed equation: 7? — Ix^ — 517 a? + 491 = 0.
(c) Multiply roots by 10.
3^3 - 70 a; - 51,700aj + 491,000 « 0.
Repeat these operations on the last equation.
(a) X = 491,000 -^ 51,700 = 9+.
By trial the sign of the left member is + when x is 9 and 8, but
changes when x is 10. Hence the root is between 9 and 10.
Q>)
1 - 70 - 51,700 + 491,000
- 9+ 549 + 470,241
9
1 - 61 - 52,249
- 9+ 468
1-52
- 9
+ 90,759
52,717
11-43
x3 _ 43 a^* - 52,717 x + 20,759 = 0.
(c) a^ - 430 X - 5,271,700 x + 20,759,000 = 0.
The required root of (1) is now x — 1.19+. Another repetition
of the process gives the third decimal.
(a) X = 20,759,000 ^ 5,271,700 = 4- .
The left member has opposite signs for a; = 3 and a; = 4, hence
the root is between 3 and 4.
(6) 1 - 430 - 5,271,700 + 20,759,000
• - 3+ 1,281 + 16,818,943
-3
1 - 427 - 5,272,981
- 3 + 1,272
+ 4,940,057
- 5,274,253
1-424
- 3
lJ-421
a^ -~421 x2 - 5,274,253 X + 4,940,057 = 0.
We thus have the required root of (1) as aj = 1.193+.
287]
THEORY OF EQUATIONS
263
We may omit step (c) in our last operation and get the next
figure of the required root by neglecting 7? and x^ in the last
equation. This gives x = .9+, and our root is, finally,
a? = 1.1939 +.
A convenient arrangement of the whole operation of finding
this root is as follows:
1-4 + + 4 |-1
-1+3+3
1-3-3
+ 1
-1 + 2
1-2
-6
-1
l|-l
1 - 10 - 500 + 1000 [-1
- 1+ 9+ 509
1 - 9 - 509;+ 491
- 1+ sf
1- 8
- 1
-517
il
1- 7
1 - 70 - 51,700 + 491,000
- 9 + 549 + 470,241
-9
1 - 61 - 52,249
- 9+ 468
1-52
- 9
+ 1^,759
6?J,717
-43
1 - 430 - 6,271,700 + 20,759,000
- 3 + 1,281 + 15,818,943
-3
1 - 427 - 5,272,981
- 3+ 1,272
+ 4,940,057
1-424
- 3
- 5,374,853
1 -481
Root, 1.1939+.
264 CUBIC EQUATIONS [288^290
288. In approximating to the roots of an equation, the fol-
lowing remarks should be borne in mind. Let the student supply
proofs when needed.
(1) Every equation of odd degree has at least one real root.
(For / (x) has opposite signs when x = + qo and x = — oo.)
(2) When an even number of roots lie between x = a and x = 6,
/ (a) and / (fe) will have like signs.
(3) Whenever more than one root lies between two assumed
values of x, especial care must be used to separate them by trial.
(4) The next decimal of a root is obtained approximately by
dividing the absolute term of the last transformed equation by
the coefficient of x with its sign changed.
(5) Should this decimal be too large, the constant term of the
next transformed equation will change sign. (Observe that in
the example the constant terms of the original equation and of
all the transformed equations are of the same sign.)
(6) Should this decimal be too small, the next transformed
equation will not have a root between and 10, except when there
happen to be two or more roots of the original equation with the
same integral part.
(7) To obtain a negative root, change the signs of all the roots
and proceed as for a positive root.
289. Exercises. Calculate to four decimal places the real roots
of the equations:
1. x» - 24a; - 48 = 0. 12. 4a:« - 3a; - 1 = 0.
2. x» - 7 a;2 + 4 X + 24 = 0. 13. x^ + a;3 _ 2 a;2 - 3 x - 3 = 0.
3. x» -2x + l =0. 14. x4-2x8-8x2 + 24x-48=0.
4. x« - x2 + X - 1 = 0. 16. X* - 4 x3- 8 X + 32 = 0.
6. x» + x2 + X + 1 = 0. 16. x* + 2 x3 + X + 2 = 0.
6. x^ -6x2 + 5=0. 17. 3x4-2x«~16x2-56x+96 = 0.
7. x»-7x-5=0. 18. x» -7x -7 = 0.
8. x» ~31x - 19 =0. 19. 8x* + 16x8 + 18x2+x + 7 = 0.
9. x« - 48x - 112 = 0. 20. 7x8 + 8x2 - 14x - 16 = 0.
10. 2x«- 18x2 + 46x -30 = 0. 21. 2x4 - 5x8 - 32x + 80 = 0.
11. 7x8 - 9x + 6 = 0. 22. 2x5 - 4x8 + 3x* - 6 = 0.
290. Cardan's Solution of the Cubic Equation. — As in the case
of the quadratic equation, so the equations of third and fourth
290] CUBIC EQUATIONS 265
degree may be solved by means of radicals. This camiot be done
for equations of degree higher than the fourth. We give here a
solution of the cubic equation
(1) ooo^ + 3 aix^ + 3 a2X + as = 0.
We first obtain a new equation containing no term of second
degree. To do this, put
x = 2/ + A.
Expanding and collecting in powers of y,
OoV^ + 3 (ooh + ai) y^ + 3 (aoh^+ 2 aiA + 02) y + Ooh^
+ 3 aih^ + 3 o-iA + as = 0.
The term in y^ drops out if
aoA + ai=0, or A= — ^•
do
With this value of h the equation becomes
3 , 3 ( 0002 ~ ai^) ^, , 00^03 ~ 3 ooaiaa + 2 ai^
^y ■• n y ■• W^ ~ "•
cio oo
Putting y — —9
we have
2? + 3 (0002 — ai^) z + (oo^fls — 3 aoaia2 + 2 ai^) = 0.
Let
H = 0002 — oi^; G = 00^03 — 3 O0O1O2 + 2 Oi^.
Then the equation becomes
(2) ^ + 3Hz + G^ 0.
To solve this equation let
Then ^_ ^
or, 2^ — 3 's/rs • 2 — (r + s) = 0.
If this is to be identical with (2), we must have
%/rs = — ff , and r + s = — G;
or, rs = — H^, and r + s = — G.
266 CUBIC EQUATIONS [290
Solving for r and 8,
' 2 ' ' 2
Then
Vr
Let the three cube roots of r be cri, a2, and 03. Then the three
values of z are
H. H, H
Zi — ai 1 22 = «2 > Zs == az •
CKi a2 as
The corresponding values of x are then found from
, , ai z ai z — Gi
do Oq Oq Oq
Nature of the Roots. — The following criteria serve to deter-
mine the nature of the roots:
(a) G® + 4ff3 < 0, three real distinct roots;
(b) (?2 + 4ff3 — 0, three real roots, two being equal;
(c) CP + 4:H^ > 0, one real root, two imaginary roots.
By direct calculation, for which we shall not take space, we find
(zi - Z2) {Z2 - «3) (23 - zt) = V-27(G2 + 4ff3),
or,
(21 - «2)2 {Z2 - ZzY (zs - 2l)2 - - 27 ((?2 + 4 H3).
When the roots are all real, their differences are real, hence the
left member of the last equation is positive; therefore G^ + ^H^
must be negative. When two roots are equal, their difference is
zero; hence G® + 4 ff^ = 0. When two roots are imaginary, they
must be conjugate imaginaries; suppose them to be
2i = a + lb and Z2 = a — ib.
Let the third root be 23 = c, where c is real [(1), (288)]. Then we
show directly that (21 —22)^ is negative, and that (22— 23)^(^3— «i)^
is positive, hence the left member of the above' equation is nega-
tive; therefore G® + 4 if^ must be positive.
The quantity G® + 4 if^ is called the discriminant of the cubic
2? + 3ff2 + G = 0.
291] QUARTIC EQUATIONS 267
When all the roots are real, i.e., CP+^H^ <0,r and « are con-
jugate complex quantities; let them be
r^A+iB; s^A-iB.
In this case \^ and y/s cannot be evaluated algebraically. The
roots may then be obtained in trigonometric form. Let
A ^ ucoav; B ^ u sin v.
Then
r =s w (cos v + i sin v) ; s = t* (cos t; — i sin v).
Hence
^r ir I t; + 2 fcx , . . t; + 2 kir\
Vr = Vwlcos « h^sm ^ — j,
V« = Vw(cos — -^ ism 5 — I; fc = 0, 1, 2.
Here Vw denotes the real cube root of u.
We now find
2=</;:+7i = 27iicos?^^t|*?[;fc==o,l,2.
291. Ferrari's Solution of the Quartic Equation. — Write the
given quartic equation in the form
(1) x* + 2 ax3 + 6x2 + 2 ex + d = 0..
Add to both members (px + g)^:
(2) x4 + 2ax3 + (6 + p2)x2+2(c + pg)x + (d + g2) = (px + g)2.
The left naember will become a perfect trinomial square of the
form
(x2 + ox + A;)2
by putting
(3) p2=,fl2_6 + 2S; g2=«d + fc2. pg=:-c + a)fc.
Then equation (2) becomes
(x2 + ax + *)^ = (pa; + g)2j
or,
(4) x2 + ax + A; =±(px + (2').
Taking each sign in turn we have two quadratic equations
in X, which give the four roots of (1).
268 QUARTIC EQUATIONS [291
To obtain the values of p, q, and k in (4) we must solve equa-
tions (3) for these quantities in terms of the coefficients. On
equating the vdues of p^q^ frqm the product of the first two of
equations (3) and the square of the third equation we find a
cubic to determine k:
(5) 2 fc3 - 6ik2 + 2 (oc - d) fc +(6d - a^d - c2)= 0.
This is called the reducing cubic, and is to be solved for a real
value of k. Then p and q are obtained from (3).
Examjde. x* + 4 a;« - 3 a;2 - 16 a; + 5 = 0.
Here a = 2, 6 = - 3, c = - 8, d = 5.
Then (5) is 2 ifc» + 3 A^ - 42 A; - 99 = 0.
A real root is A; = — 3.
Then from (3), p = 1, g = 2; or, p = - 1, g =» — 2.
With either set of values of p and q (4) becomes
(a;2 + 2a;-3) = ±*(x + 2).
Hence
a^ + a;-5«0, or, a;2 + 3a;-l=0.
Therefore
^ -1 ± V2i _ -3± Vi3
^ 2 ' ""^^ 2
Exercises. Solve the following equations:
1. x»-3a;2 + 4«0. 9. a^ + 2x« + 2x2 - 2a; - 3 = 0.
2. x«-3a;-2=0. 10. x4-j_6a:» + 3a;2-2x-3 = 0.
3. 4 x« - 3 a; - 1 = 0. 11. x* - 4 a:» - 9 a;2 + 2 x + 3 = 0.
4. x3-24x-48 = 0. 12. x^ + 4x3 - 16x + H = 0.
6. x« - 7 x2 + 4 X + 24 = 0. 13. x^ + 4 x« - 16 x - 16 = 0.
6. x» - 3 x2 - 6 X + 1 = 0. 14. x^ - 3 x3 - 7 x2 + 15 X + 18 = 0.
7. x8 - 7 X - 6 = 0. 16. x4 - 4 x3 - 8 X + 32 = 0.
8. x«-x2 + a;-l =0. 16. x^ + x* - 2x2 -3x -3 =0.
CHAPTER XX
Spherical Trigonometry
292. Spherical Geometry. — We devote this article to a review
of some facts concerning the geometry of the sphere.
(a) A plane section of a sphere is a circle. When the plane
passes through the center, the section is a great circle; otherwise a
small circle.
(b) Any two great circles intersect in two diametrically opposite
points and bisect each other.
(c) The two points on the sphere each equally distant from all
the points of a circle on the sphere are callpd the poles of the
circle. A great circle is 90® distant from each of its poles.
(d) A spherical triangle is a figure bounded by three circular
arcs on a sphere. In this chapter we consider only triangles whose
sides are arcs of great circles. Any such triangle may therefore
be considered as cut from the spherical surface by the faces of a
triedral angle whose vertex is at the center. The face angles of
this triedral angle measure the sides of the triangle, and its diedral
angles the angles of the triangle.
(e) If a triangle be constructed by striking arcs with the vertices
of a given triangle as poles, the new triangle is called the polar
triangle of the given one.
Let the sides of the given triangle be a, fe, c; its angles A, B, C;
let the sides of the polar triangle be a', V, c' and its angles A', JS', C;
we assume that A is the pole of a', B of 6', and C of c'; then
a' = 180 - A ; A' = 180 - a ;
and similarly for the other sides and angles. That is, any part of
the polar triangle is the supplement of the part opposite in the given
triangle,
(f) The difference between the sum of the angles of a spherical
triangle and 180® is called its spherical excess.
.The area of a spherical triangle is to the area of the sphere as its
spherical excess, in degrees, is to 720®. That is, if E be the spherical
269
270
SPHERICAL RIGHT TRIANGLES
[293,294
excess in degrees and K the area, and B the radius of the sphere,
then
^ X4irB2.
K^
720
293. Spherical Right Triangles. — Let be the center of a
sphere and ABC a triangle on its surface having C = 90^. The
triangle shown in the figure has each
part, except C, less than 90°. The re-
sults below are true in any case, as may
be shown by drawing other figures, or by
assuming the right triangle as a special
case of the oblique triangle.
Cut the triedral angle 0—ABC by a
plane J_ OB, forming the plane right A
A'B'C, with C'=90^ Then also As OB'C and OB' A' are
right-atigled at B'. Further, Z A' B'C measures Z B (292, (d)).
Then
A'C
(a)
sin^ = sinA'B'C' =
(b)
cos B = COS A' B'C =
(c)
tan^ = iBXi A' B'C =
A'C
OA'
sin&
A'B'
A'B'
OA'
B'C
sine
B'C
OB'
tana
A'B'
A'B'
OB'
A'C
tan c
A'C
OC
tan6
B'C
B'C
sina
OC
Dividing (a) by (b) and comparing with (c) we have
(d) cos c = cos a cos b.
By combining these equations we may obtain others by which
any part of the triangle may be expressed directly in terms of
any two given parts, the right angle excluded. These formulas
are all contained in two simple rules.
294. Napier's Rules o^^ Circular Parts. — Let co-x denote the
complement of any part x ^f the triangle. Take the complements
295]
SPHERICAL RIGHT TRIANGLES
271
of c,Af By and arrange the five parts, a, &, co-A, co-c, co-B, called
circular parts in the order m which they occur in the triangle as
in the adjacent figures. Then if any one of the five be taken as
the middle part, of the other four parts two will be adjacent and
reiHA
iMR
the other two opposite to this part. Thus, if co-c be taken as the
middle part, co-B and co-A are adjacent, a and b opposite.
Rvles:
Prod/uct of tangents of adjacevvt partSy
or
Product of cosines of opposite parts.
Sine of Middle Part =
Exercise. Taking each part in turn as the middle part write out a com-
plete list of formulas relating to the spherical right triangle. Derive these
formulas from those given above.
296. Solution of Right Triangles.
Example. Given a = 35° 42'; B = 60;* 25'.
The diagram of circular parts is shown in
the figure. Taking (1), (2), (3) in turn as
middle part we have
(1) sin 35° 42' = tan 29° 35' tan b;
(2) sin 29° 35' = tan 35° 42' tan (co-c) ;
(3) sin (co-A) = cos 29° 35' cos 35° 42'.
Find &, c, A.
Hence,
tan&
sin 35° 42'
cote =
sin 29° 35'
tan 29° 35" tan 35° 42"
cos A = cos 29° 35' cos 35° 42'.
Check. The computed parts must satisfy the relation
sin (co-il) = tan b tan (co-c), or cos A == tan b cot c.
272
SPHERICAL OBLIQUE TRIANGLES
[296,297
CompuicUions.
log log log
sin 35° 42' = 9.7660 sin 29° 35' = 9.6934 cos 29° 35'
tan 29° 35' = 9.7541 tan 35° 42' = 9.8564 cos 35° 42'
tan & = 0.0119 cot c = 9.8370 cos A
b = 45° 17' c = 55° 30' A
Check. log cos A = log tan b + log cot c.
' 9.8490 = 0.0119 + 9.8370.
9.9394
9.9096
9.8490
45° 4'
Ambigiums Case, When the given
parts are an angle (not the right angle)
and its opposite side, two solutions
are possible, because the other parts
are then calculated from their sines.
The two triangles together form a lune, as A A' in the figure,
where -4, a are supposed to be the given parts.
296. Quadrantal Triangles. — A quadrantal triangle is one
having a side equal to a quadrant or 90°. Its polar triangle will
be a right triangle, which may be solved by Napier's Rules. The
parts of the given quadrantal triangle then become known by
(e) of (292).
Exercises. £oIve the following triangles,
C being the right angle:
1. o = 45** 10',
4. 6 = 100**,
7. 5 = 145** 53',
B = 70** 20'.
a = 40^
c = llO** 20'.
2. 6 = 65** 15',
5. il = 120** 42',
8. 6 = 132*' 16',
A = 25** 50'.
c = 56** 50'.
5 = 65** 46'.
8. c = 33** 18',
e. A^ 40^
9. c = 170** 4',
h = 30** 37'.
a = 30^
a = 175** 17'.
Solve the following quadrantal triangles:
10. a = 90^
11. A = 65** 15',
12. A = 122*' 10',
h = 50^
b = 90%
B = 70*^ 22',
c = 4o^
c = 50** 25'.
c = 90%
297. Oblique Triangles. Two Fundamental Formulas. — We
consider only triangles in which no part exceeds 180°.
I. Law of Sines. — Let ABC be a spherical triangle. Draw
CD ±ABy forming two right triangles (figiu'e).
In A ACDy sin p = sin 6 sin A.
In A BCDf sin p = sin a sin B.
298]
SPHERICAL OBLIQUE TRIANGLES
273
Therefore,
sin & sin il = sin a sin By or
sin a sin^
(1)
amb ninB
That is, the sines of the sides are
proportional to the sines of the
opposite angles.
Exercise. Discuss the case in which D falls on AB produced.
n. Law of Cosines. — In the figure above let ilD = m, so that
BD = c-m. Then in right A BCD
cos a = cos (c — m) cosp, . . . (d), (293)
= cos c cos m cos p + sin c sin m cos p.
But m A ACD
cosmcosp = COS&
and sinmcosp = sinCsinft X . ^ = sinftcosA.
sinC
Hence
(2) cos a = cos & cos c -f sin & sin c cos A.
That is, the cosine of any side equals the product of the cosines of
the other two sides phis the product of their sines by the cosine of
their included angle.
Exercise. Discuss the case where D falls on AB produced.
From the fundamental formulas (1) and (2) we shall derive a
series of other formulas adapted to the solution of triangles.
298. Principle of Duality. — By means of (e) of (292) any
formula relatmg to the spherical triangle can be made to yield a
second formula. Thus, let A A'B'C be polar to A ABC. Then
from (1) and (2)
sino' sin -4'
sin 6' sin B
t f
But
o' =
b' =
c' =
cos a' = cos b' cos c' + sin b' sin c' cos ^1'.
180 -A, il' = 180 - o,
180 - B', B' = 180 - 6,
180 - C, C = 180 - c.
^
274 SPHERICAL OBLIQUE TRIANGLES [299
Substituting and reducing, we have
sinA _ sing
sin B "" sin 6
(3) cos ^ = — cos B cos C + sinB sin Ccosa.
The first of these is simply the law of sines; the second is a new
formula.
299. Formulas for the Half Angle. — Solving (2) for cos il, we
have
cos a — cos & cos c
cosA =
sin & sin c
Then
• ^ A 4/1 — cos A /^„ , , . /l — cosA„\
sm^A = y 2 \^Why not ± y ^ 7
__ cos o — cos & cos c
sin & sin c
-v/_
▼ 2si
si n&sinc — coso + cos&cosc
2 sin h sin c
— c) — cos a
2 sin h sin c
«. a + ft — c. a — & + C
2 sm ;^ sm x
2 sin 6 sin c
Now let
(4) 2« = a + 6 + c;
then
= 9 — ana s = « — 1>;
therefore, '
(5) sini^ = J sin (* - 6) sin (. - :^.
2 T sm&smc
Similarly,
//sx 1^4 /sin « sin (« — a)
(6) cos- A = V ^—^^ '*
By dividing.
(7) tani^ = J^('-^)sf^(^-^) .
2 y Bins sm (s — a)
300,301] SPHERICAL OBLIQUE TRIANGLES 275
Given the three sides, one of these formulas, preferably the last,
will determine the angles. When all three angles are desired, let
(8) tanr- \/ ^ (s-a)sin(s- b) sin {s - c) .
sin^
then
(9) tan-^ =
2 sin (« — a)
/irk\ J. 1 ^ tanr
(10) tan-^=-r-7 rT>
2 sm (« — 6)
/tt\ X 1^ tanr
(11) tan-C =
2 sin (« — c)
300. Formulas for the Half Sides. — Proceeding as above with
(3) of (298), or by applying the principle of duality to formulas
(5) to (11) we have, on putting
(12) 28=^A + B + C
and
.-«v . ^ ^ / — cos 8
(16) xanM - V^Qg(^ _ ^)cos(S - B)cos{S - C)'
(14) sm-a = y-
2 V sin^sinC
n^\ --^- Jco^(s-B)cos{8-C)
(16) cos- a = V ■ — =— -: — -^ >
/,/sx X 1 . / — cos8 cos (;Sf — ^)
(16) tan-a = v Tz =:? — ^r- — ^>
^ '^ 2 V cos (S - jB) cos (;» - C)
(17) tan|a = tan^cos(S - ^),
(18) tan^ft = tan-Bcos (>8f - B)j
(19) tan|c = tan JB cos (;» - C).
301. Napier's Analogies. — Dividing tani-4 by tanJB and
reducing, we have
tan^il _ sin (s — b)
tan ^ B sin (s — a)
By composition and division,
tan iA+ tan j B _ sin (s — b)+ sin (s — a)
tan i A — tan J 5 ~ sin (s — 6) — sin (s — a)
276 SPHERICAL OBLIQUE TRIANGLES [301
Reducing tangents to sines and cosines and simplifying the result-
ing complex fraction, applying the formulas for sin (x ± y) on the
left and for sin u± sin r on the right, we have
(20)
or,
sin|(^ + ^) tan|c
sin|(^ — ^) tan|(a — 6)
^1. . sin|U-^)^ 1
^20'^ tan- (a — h) = . f , -tan-c.
^^^ »^ ^ sin|U + -B) ^
Multiplying tan i A by tan \ B and reducing,
tan i A tan i B sin (s — c)
1 sine
By composition and division, and reduction as above,
cos l(A + B) tan|c
(^^^ cos| U - ^) ~ tan|(a + &)'
or
(210 *^5^-- + ")=' cosiu + B) *^5--
These formulas determine the other two sides when two angles
and their included side are ^ven.
Proceeding as above with tan J a and tan J 6, or by the principle
of duality applied to formulas (20.) to (21'), we obtain
sin|(a + 5) _ cot|c
8in|(a-&) ~ tan| (A-sy
(22)
or.
1 , , sin I (a — &) 1
(220 ^5(-^--^) = sm (a + .) ^^^i^>
(23)
or,
cos I (g + &) _ cot I C
cos J (a - &) ~ tan K^-JTb)*
>
. 1 / . «v cos 5 ( a — &) 1
(230 tan- U + ^ = 1) ( cot- C.
^ ^ ^ cos 5 (a + &) ^
These formulas determine the other two angles when two sides
and their included angle are given.
302,303] SPHERICAL OBLIQUE TRIANGLES 277
302. Area of a Spherical Triangle. — This may be calculated by
(f ) of (312), namely,
g = ^ ^"iy ^^ X 4 T 222, or, X = i57 (radians) X 222.
To obtain JB, we may first calculate the angles. E may also be
obtained by one of the following formulas, which we add without
proofs.
1 tan I a tan s & sin C
tan--E = ^-T =-7 ;
» 1 + tan I a tan I & cos C
l-V^'
tanrE = V/ tan-tan—- — tan ^ tan
303. Solution of Spherical Oblique Triangles. — Six cases
arise, according to the nature of the three given parts.
I. Given two sides and an opposite angle.
Denote the given parts by a, 6, A, Calculate B by (1), then
C by (22) or (23), and c by (20) or (21).
Check* sin^^sinB
sm c sm C
which involves the compvied parts.
Ambiguous Case, Formula (1) will give two (supplementary)
values for B. Two solutions are obtained when both values of
B lead to values of C. Otherwise one or both values of B must
be rejected.
Rule, Retain values of B which make A — B and a — b of like
sign. Otherwise (20) and (22) take the impossible form + = — .
n. Given two angles and an opposite side.
Denote the given parts by A, S, a. Calculate b by (1), then
proceed as in I.
Ambiguous Case, Formula (1) gives two values of b. Retain
the valvs or values which make A — B and a — b of like sign,
m. Given the three sides.
Calculate the angles by (9), (10), (11).
4
^r 7.. sin A _ sin B _ sin C
sm a sm o sm c
278 SPHERICAL OBLIQUE TIOANGLES [304
IV. Given the three angles.
Calculate the sides by (17), (18), (19).
Check: As in HI.
V. Given two sides and their included angle.
Denote the given parts by a, 6, C. Calculate ^(A + B) by (23'),
i (A — B) by (22'); then A and B by addition and subtraction;
obtain c by the law of sines. Check by (20) or (21).
VI. Given two angles and their included side.
Denote the given parts by A, B, c. Calculate i (a + b) from
(21 0> i (^ ~ &) from (20') ; hence get a and b; obtain C by the law
of sines. Check by (22) or (23).
304. Example. Given a = 100^37', b = 62° 25', A = 120^48'.
Formvlaa.
. n sin b . 4
sm B = - — sm A ,
sma
cotiC = ?!5i(^tanH^-B),
^ sm J (a — 6) ^ ^ ^'
, , smi {A+ B) . 1 . ,x
tan * c = - — T-j-A — ST tan f (a — o).
^ sm J (A — B) ' ^ ^
Check:
sin b sin £
sin c sin C'
log sin 6 = 9.9476 a = 100° 37' A = 120° 48'
log sin A = 9.9340 6 = 62° 25 ' B = 50° 46 ^
colog sin a = 0.0075 a + & = 162° 62' A + B = 170° 94'
log sm B = 9.8891 a- 6= 38° 12' A-B= 70° 2'
B= 50°46'.5 i(a + &)= 81°31'H^+5)= 85°47'
or 129° 13'.5 J (a - 6) = 19° 6' H^ - 5) = 35° 1'
Reject the larger value of B by the rule in I.
log tan i (A - B) = 9.8455 log tan J (a - 6) = 9.5395
log sin i{a + b) = 9.9952 log sin ^ (A + B) = 9.9989
colog sin I (a - 6) = 0.4852 colog sin i (A - JS) = 0.2412
log cot i C = 0.3259 log tan J c = 9.7796
JC= 64°43'.5 ic = 31°3'
C = 129° 27' c = 62° 6'
Check: log sin 6 = 9.9476 log sin B = 9.8891
sin c = 9.9463 sin C = 9.8877
0.0013 0.0014
305,306] TERRESTRIAL SPHERE 279
Note. In the solutions of triangles, a complete form should he pre-
pared in advance, so that only numerical values need be inserted
when the tables are opened.
305. Exercises. Solve the triangles whose given parts are:
1.
2.
8.
4.
a =
53** 18'.3,
o= 42**15'.3,
a = 84** 14' 30",
A = 116° 8'.5,
6 =
36^ 5'.6,
6= 33**18'.8,
h = 44** 13' 46",
5= 35°46'.6,
c =
50** 24'.9.
c= 60°32'.l.
c=51** 6' 20".
C= 46**33'.7.
6.
6.
7.
8.
A^
97** 53',
il = 53** 42' 34",
= 89*' 0',
a = 70** 20',
B =
67** 59'.7,
5= 62** 24' 26",
h = 47** 30',
6 = 38** 28',
C =
84*'46'.7.
C = 155** 43' 12".
C = 36*' 0'.
C = 52** 30'.
9.
10.
11.
12.
6 =
19** 24',
a= 88** 24' 3",
a = 102** 22',
6= 76° 40' 48",
c =
41** 36',
c = 120** 10' 55",
B= 84** 30',
A= 84° 30' 20",
A =
84** 10'.
B= 49** 27' 50".
C = 125** 28'.
C = 130° 51' 33".
18.
14.
15.
16.
c =
104** 13'.4,
c = 108** 39' 10",
6= 54**18''16",
a = 88° 27' 50",
A =
63** 48'.6,
A= 64** 48' 52",
A = 127** 22' 7",
h = 107° 19' 52",
B^
51** 46'.2.
B = 40° 23' 17".
C= 72** 26' 40".
C = 116°15' 0".
17.
18.
19.
20.
6 =
83** 5' 36",
6 = 68** 45',
a= 56^37',
a = 48°,
c =
64** 3' 20",
B=58*' 5',
A = 123° 54',
b = 67°,
A =
57** 50' 0".
C = 50** 51'.
5= 57** 47'.
A = 42°.
21.
22.
23.
24.
6 =
8r,
a = 69** 34'.9,
a = 69** 11'.8,
a = 151° 01' 5",
A =
72%
c = 70** 20'.3,
b « 56** 38'.5,
b = 134° 10' 52",
B =
119^
C = 50**30'.l.
A = 68** 40'.
A = 144° 20' 45".
25.
26.
27.
28.
o =
40** 8' 28",
a= 88**12'.3,
c = 100** 49' 30",
A = 45°,
6 =
118** 20' 8",
A^ 63**15'.2,
B = 95** 38' 11",
a = 10°,
il =
29** 45' 32".
B = 132** 18'.
C= 97** 26' 28".
6=60°.
306. Applications to the Terrestrial Sphere. — We shall con-
sider the earth as a sphere with a radius of 3960 miles. Longi-
tudes are to be reckoned from Greenwich westward through
360** or 24**. We shall denote longitude by X, latitude by (f>.
Problem 1. Given the latitudes and longitudes of two stations,
to find the distance between them.
Let P be the earth's north pole, G Greenwich, Ai and A 2 the
two stations (figure). Let the positions of the two stations be
Xi, ^1 and X2, 02 respectively.
280
CELESTIAL SPHERE
[307
Then in AAiPA2y PAi =
90° - 01, PA2 = 90** - 02, and
Z A1PA2 = X2 — Xi- Hence
in AA1PA2 two sides and
their included angle are known,
and A1A2 (in degrees) may be
calculated as in V of (303).
Problem 2. A ship is to sail
from ill to il2 by the shortest
path (great circle). On what
course (at what angle with the
meridian) will she depart from
Ai\ on what course will she
arrive at -42?
Assuming the positions of
A\ and A2 given, we have two
sides and the included angle of the triangle A\PA2- We must
calculate angles A\ tod A2- This comes under V of (303).
Exercises.
1. Calculate the sides (in miles), the angles, and the area (In square miles)
of the triangle whose vertices are:
h m 8
New York ;i = 4 55 64,
San Francisco 8 9 43,
Mexico City 6 36 27,
= 40° 45' N.
37** 47' N.
19° 26' N.
2. A vessel sails on a great circle from San Francisco, ^ = S** 9" 43',
<l> = 37° 47' N. to Sydney, X = 13" 55- ICT, 4> = 33° 52' S. Find the courses of
departure and arrival and the distance sailed.
8. If the vessel in exercise 2 makes 12 knots an hour, what is her position
{X and 4) and on what course is she sailing 5 days after leaving San Fran-
cisco? (1 knot ~ 1 nautical mile =" 1' on a great circle.)
307. Applications to the Celestial Sphere. — For the purpose
of this article we assume the celestial sphere to be an indefinitely
large sphere concentric with that of the earth. On it as a back-
ground we see all celestial objects.
The projections on the celestial sphere of the earth's poles,
equator, meridians and parallels of latitude are named respectively
the celestial poles (P, P' in the figure), the celestial equaJtar or
simply equator (QrvQ^e), hour circles (as PSE), and parallels of
declination (as MSM').
307]
CELESTIAL SPHERE
281
An observer at on the earth's surface will have his zenith at
Z, where the plumb line at 0, if produced, would meet the celestial
sphere; his horizon is the
great circle swne, whose ^
pole is Z; his meridian is
the great circle nPZQs,
meeting the horizon in
the north and south
points.
Let S be a point on the
celestial sphere, as the
sun's center, or a star.
Because of the rotation
of the earth, S will appear
to describe the parallel
e'MSw'M'e'y rising at e'
and setting at w'. When
S has the position shown in the figure, HS is its altitude, denoted
by h (height above horizon) ; Z sZH (measured by arc sH) is its
azimuthy denoted by A ; ZS, or 90° — A, is the zenith distance of S
and denoted by 2. Thus h and A, or z and A, completely define
the position of S with reference to horizon and zenith.
With reference to the equator and pole,
ES is called the declination of S, denoted
by d, and Z QPE (angle which hour
circle PS of S makes with meridian PQ)
is called its hour angle, denoted by t; PS
or 90° — 5 is the polar distance of S, and
denoted by p. Thus the position of S is defined by d and t, or by
p and t
A PZS is called the astronomical triangle; its parts, except the
angle at S which we shall not need, are:
PZ = 90° - nP = 90° - 0; {<f> = latitude of 0.)
PS = p = 90° - d; ZS = 2 = 90° - h;
Z ZPS = t; Z PZS = 180° - A.
Problem 1. Given the latitude of 0, and the declination and
altitude of S, calculate the hour angle and azimuth of S.
282
CELESTIAL SPHERE
(307
Here the three sides of A PZS are known, and it is only neces-
sary to calculate the angles at P and Z (HI, 303).
Problem 2. In a given latitude, and for a given declination of
the sun, find the sun's hour angle at sunset and the length of day
(sunrise to sunset).
Here S is on the horizon and PZS a quadrantal triangle. We
obtain t by solving the polar right triangle for 180 — t. The length
of day will be 2 L
Problem 3. Given the sun's declination and its hour angle
when it bears due west (A = 90°), find the latitude.
Here PZS is a right triangle, with the right angle at Z; p and t
are known, and PZ may be calculated by use of Napier's Rules.
Problem 4. Find the
hoiu" angle and azimuth of
Polaris when at greatest
elongation, given the dec-
lination of the star and the
latitude of the station of
observation.
Let MSM' be the star's
diurnal path about the pole
(figure). When the star is
at greatest elongation, the
great circle ZS is tangent to the small circle MSM\ of which PS
is a radius. Hence A PZS is right-angled at S; PZ and PS are
known, and the angles at P and Z may be found by aid of Napier's
Rules.
Exercises.
1. In latitude 40^*49' the sun's altitude is observed to be 20** 20'; its
declination is 15° 12'; find its azimuth and hour angle.
2. With latitude and declination as in exercise 1, find the sun's hour angle
when it is due west; when it sets; find its azimuth at sunset; find the length
of day.
3. With latitude and declination as in exercise 1, find the sun's altitude
and azimuth when its hour angle is 45**.
4. The sun, in declination 12** 22', is observed to have an altitude of 30**
when due west. What is the latitude of the station?
6. The declination of Polaris being 8S** 49', find his azimuth and hour angle
at greatest elongation at a station in latitude 40** 49^.
307] CELESTIAL SPHERE 283
6. As in exercise 5 for the star 51 Cephei, d = 87° 11', and for d Ursse
Minoris, ^ = 86° 37'.
7. The stylus of a horizontal sundial consists of a rod pointing to the
north celestial pole. Hence its shadow falls due north when the sun is on the
meridian, that is, at apparent noon. What angle does its shadow make with
the meridian one hour after apparent noon, at a place in latitude 40°?
{Suggestion, In the first figure of this article let nP — 40° and Z ZPS =
l** or 16°. The stylus lies in the line P'Pj and its shadow, cast by the sun /S,
must lie in the plane <SP'P, and hence wiU faU on the plane of the dial, svmey
along the line of intersection of these two planes. This line will be deter-
mined by the center of the sphere and the point where arc SP produced wiU
meet arc ne. Call this point S\ Then JBirc nS' measures the required angle,
and may be found by solving right A nPS\ in which nP = 40° and Z nPS' = 15°) .
8. What angle does the shadow of a horizontal simdial make with its
noon position t hours after noon in latitude 4> ? {Ans» tan x — tan t sin <^,
X being the required angle.)
9. Calculate the angles which the hour lines of a horizontal sundial make
with the noon-line in an assumed latitude.
ANSWERS
(Answers are given only for the odd-numbered exercises.)
Article 10
1. i a - i. 8. .05 o2 - 3 at - 4.625 ac, 5. 63 x - 2 y - 4 2. 7. o^ft^c -
i a^c* + ^ a^ccP - 2 a^c. 9. 1.2 a^bcM^ - l.S cuAP + .3 a^(M^ - 3 ac^d^. 11.
7fi-5x^-\-3a^-\-6x^-7x+2, 18. x^ - 9 a:^^/^ .j. 7 a^ 4- 13 a^ _ 19 a;2y5 .j.
Sxj/^-y'r, 16. io2 + A. 17. x* - a2x2 - 62x2 -I- 0252. 19. 9o2_9a + 6.
A4 o o o« «n o Ar -07 (x* - 2 x2 + 1) „ 10 X , 25 3/ ,
21. -3m2p. 23. -fjxy2. 25. — ^ ^ ^-^. 27. — 5^ + 24^"^
4^-o^- 29. 3o2a;-4aa;2-|-x3. 81. x2 + 5 a:y + 3 2/2. 83. io8-ia26
iz x^ z x^y
+ }a62. 86. x3-|x2-2x + J. 37. 2 x2 + X2/ - 3 y2. 89. (a + 6)». 41.
Jx2-32/. 43. a6-|-c. 46. a^+l^+c^-\-(fi -2{db-ac-\-ad-hd-\-bc-{-cd),
Article 12
^ ^("^"4^2-^)- 3. (x-l)(3x-l). 6. (3x-2/)(2x + 7y).
7. X (2 X - 3 y) (4 x2 + 6 xy + 9 2/2). 9. (x + 2) (x - 2) (x + 3) (x - 3).
11. (x - 11) (x + 10). 13. (x - 9 o2) (x - a2). 16. {xy - 5 «) (xy + 2 2).
17. (x - 1) (x - 8) (x + 8). 19. (x + 1) (x2 - X + 1) (x - 1) (x2 -fx + 1).
21. -3x2/(x + 2/). 23. (ac + 6) (oc + d). 26. xj/ (x -|- y) (x - y)2.
27. (x^y - 22) (x22/ + 5). 29. (x + 2) (x2 + 7 x + 2).
Article 16
1. 3 (x + 1). 3. 4 (x2 + 2/2). 6. ax (a - x)2. 7. 3 o (2 a -|- 3 6 - 4 c).
9. (2X-3 2/). 11. (3 X- 2a). 13. (x2 + 7). 16.(5x2-1). 17. (x + 2/).
19. (a2 - a6 + 62). 21. 24 a^bx^y^. 23. (a + 6) (a - 6)2. 26. (x - 4)
(x+i)(x+3). 27. (3x-2)(2x+3)(2x-3). 29. (3x - 2a) (4x - 3a)
(3 X + 4 a). 31. (m + n) (m - n) (m + 2 n) (m - 2 n). 33. (x + 1) (x + 2)
(x + 3). 36. (x - 1) (x + 1) (x + 2) (x + 3). 37. (x - 1) (x + 1) (x2 + 1)
(x2-x + l). 39. (a-6)(a + 6)(a2 + a6 + 62)(a2~a6+62)(a6 + a363+6«).
,284
ANSWERS 285
Article 19
1- 4ax. 8. -^3^. 6. ^ 7. 1. 9- ^r+T"'
11. ^y 13. ^-(^. 16. -5- 17. 0.+^. 19. i^4p3. 21. 0.
«, 2 ~(3a:^-2a;8 + 3a;-4) nj ^ __^
*' (aj-l)(a;-2)(aj-3)' 2(aj + 1) (a;2 + 1) (a;2 -x + 1)* o« o2
_i_«^ «^ oo "^^/i 343 g8 \ ,.
(5a3 -9c3) (45c3 - 4963) (9c2 -6a3)
14,175_a8c«
Article 21
5- 33??i- 7. a^b^, 9. ^•
Article 33
1- *^^- 3- ^-^^i^'- 5- p^,- 7. ^^3;^. 9. ^27, V4. 11. ^-yie,
^^27^13. ^^, ^v^le, 727. 15. y/I' y/i^ y/Ji. n.
\/w* \/fl- "• ^' ^^' ^^" 21. ^'^^ ^<^^, Wo. 28. y/g^,
v4' \/?"' ^^- 75^' "^"^^^' 7S- ^^- 3 <^- ^^- ^ V2-.
31. 3i '\/5. 83. (3 + & - a) VS. 86. a \/x, 87. 1 + V3. 89. § (V2 +
VS). 41. i (V6 - Vli). 43. V2 + Vs. 46. V16 - V3. 47. 1.
49. 6 V2-3 Vi5+8 V3-6 VlO. 61. 8-8Vi2 + Vl8. _63. Vw2 - n.
66. a. 67. \/p2^. 69. 2. , 61. 4, 63. 3. 66. 3. .67. i/^- 69. ~~^\
7L w^". 78. a^. 76. a'. 77. a^. 79. (aj + y)^. 81.,^. 83. a^.
'^ a(l+VS) ft- 4a3 + 12V53 + 9 ^^ ll+2Vii 01 oV&-cV5 .
00. ^ • 07. ^i — 5 jr • OW. = • Vl. rr ^3-
1 — a 4 a3 — 9 5 a^b — c^d
6
98. o*. 96. 2o-2-7a*+6a-* + 7o-*-llo"*-2a-* + 7o-i-6. 97. 2o»
4 8 8 1
-5a^+10a*- 7a* + 6a^. 99. 4a-26"* - 12a"'6~5* + 9a"*6"'. 101.
a;^ - 3 xy* + 3 a;*y* - 2/2. 103. m"* (l + 4 m"^ + 6 m-3 + 4 rn"^ + m-e).
106. a* + a' + a^ + 2 (a^^ «a?l ^^^J) . 107. a*+ 4 6* + 9c + 16 d* + 2 ( -2oH*
H-3a*c* -4a*rf* -6 6*c* + 8 6*d* -12 c*dO. HI. a;* - x* + a;^ - a;* + 1. 118.
a* H- o^ft''^ + 6*. 116. a - Va. 117. 3 V^; 5 V^; 9 V^. 119. 3 Vi-
i
286 ANSWERS
121. 2 \/i. 128. 9 V=l. 126. m V^. 127. 47 - 1. 129. 4t V§ - 2.
181.-1. 188.1. 186. ^^^+i^^?T^- 137. -§. 139. 2aj=3. 141,
ax + & » c4. 148. 4x » 5. 146. x » 5. 147. x » 10. 149. x » 4.
Article 38
L 0. 8. -3,-4,-6,-7. 5. -3,-4. 7. 2, a. 9. 7, .3. 18. (p + g)
(-s)
Article 41
8. 2.9196, 0.9196, 9.9196 - 10, 8.9196 - 10. 6. 3.667. 7. 1.655; 11.695.
9. 52.22; 29.34. IL 0.1829. 18. log^- 16. log ^'Z^ . n. log ^' ^^ "*" ^ .
Article 46
L 0.975. 8. 88.444. 6. 0.99965.
Article 61
1. X= p; • 9%X^T' 0. • 7. 00. 9. £ T-' II, ; ,
2 g — n 6— a— 1 m-\-n'-a
00. 18. 1. 16. i.
Article 60
1. 6. 8. -^- 6. — ^. 7. 3tol. 9. lidays. 11. ^^
days. 18. 5^^ min. past 10; 21]^x ^i^* P<^t 10. 16. 1} hrs. ,
Article 64
1. 3, 5. 8. 32, - 17. 6. 9, 8. 7. 2, 3. 9. Inconsistent. 11. 0, 4. 18.
1, - 1. 16. Dependent. 17. 6, 12. 19. 12, 5.
Article 69
1. 6, 12. 3' 6, 12. 6. 9, 7. 7. 4, 3. 9. x = tSIIh; y = ^^'^^
1.0163725' ^ 1.0163725
5a — 5 6
2
11. Not independent. 18. 5, 6. 16. i, i. 17. 4, 7. 19. 7, f 21.
a + & 00 Q^ ofec/ OK -r - ^^< + n'p9v . _ ga^-magy ^_
2 og — aj og —of mqr + p« p« + wgr
_n^q_^ nmq ^^^ j^^ solution. 31. No solution. 88. 20, 17, 5. 36.
mq — np mq ^ np
3,2,1. 87. 3,4,5. 39. i, §, 00. 41. 1,2,3,4. 43. 1, .8, .2, .6. 46. 16Ahrs.,
7Hhrs. 47. $4000; 4§%. 49.36,9. 61.89,35. 68.13,17,20. 66. l,lf,l§.
67. 2, 3, 6 hrs. 69. $9150, $8600, $7550.
ANSWERS 287
Article To y
1. 2, - 6. 3. 2, - 8. 6. - 2, 7. 7. 3, - 4}^ 9. 5, li. 11. {, - A-
13. - 2i, 6. 16. 3, - 4^ 17. 26, - 6. 19. 2, c.
Article 8)S
1.6. 3. 0or3. 6.1. 7.13. 9.4. 11. VlSl- -29 ^^ ^ 16. 3or-§.
17. IJ. 19. 15. 21. ± t V|. 23. 4 o/- i. 26. 3. 27. ± ^/'^^mp. 29.
±6VS^^. 31. ±ia. 33. ±mV^. 36. ^ (^ :^2 V«) . 37^ feiVP'^^.
4 -Vi or - 8. 47. 4 or - 9. 49. 27 or_64. 61. or 9. 63. 14, 16, 18; or,
-14, -16, -18. 66. 30X60. 67. iVST^T. 69. b- a±y/<^+hf-6 ab^
in
61. i. 63. ^ , , n < i 66. 20; 60. 67. x > - 1 and < - 9; - 1 and
on — 1' " '
-9;x< -1 and > -9. 71. J (Vl7 - l); J (Vl7 + l).
Article 93
1. ±i^/2;±J^/5. 3. i^/5;-J^^. 6, ~6:b VTl . 3 ±2 Vll ^ ^ ^
or H; 3 or ji. 9. 20 > 20 U. w = ± 2.
Article 96
Vl3 Vl3 Vl3 Vl3 145
- 162 ± 2 V- 1559 ^ A±\r^3 , - 9 ± V^23 .. .^ 1 ^/r
146 ^- 13 ' 13 11. ±iV5.
Article 97
10101 a 65j:Vi29 . 1±Vi29 7±4V^ . 2T4V^^
1. 0, 1, 0, 1. 3. 32 ' 16 ^' 9 ' 3 • ^*
1± V5; ^'^2 • 9. -4±2>/3; -7±4V3. 11.-1.
288 ANSWERS
Article 99
i.i.-f;o.-f. 8.jV6;i. 5. i«±^Hi; ^li^VEB. 7.
f
Article 106
1. a; = ± } ^^, ±1 V5, y - ± § V2, T W5. 3. a; = 0, 3, ± A Vi3;
y = 2, 0, ^ A Vi3. 6. a: = 0, 9, V; V = 0, - 6, V- .
Article 106
1. ± W29±V41, ±iV7TV41. 3. ±AV-5; ±V^^. 6. ±V3,
±AV57, 0, TA V57.
Article 107
1. ±V3;±1. 3. ±-|-;±2- 6. ± -^; ±y.
Article 111
1. x=±25; y-±6. 8. ±5; ±4. 6. i^^; i^^. 7. }, J; §, }.
9- 7, III; 8, 9§ii 11. 13; 7. 13. ± 13; ± 7; two solutions. 16. 7, - |;
- 3, 17i. 17. 37 A, 4; 43 A, 7. 19. 4, 6; 4, 3; two answers. 21. 14?}, 5;
16H, 2. 23. 8, 9; 9, 8. 26. ±2, «; ±1,«. 27. ^^^^ ; ^^-i^^;four
solutions. 29. 3 ± V6, "" ^ "^"/^ ; 3 T V6, IL^L^yEJl. zi. - 2, oo;
0, «. 33. 7, 2; 2, 7. 36. 0, 5; 5, 0. 37. 5, - 6; 11, - 12; four answers.
39. 2,3, -3±V§; 3,2, -3=fV3. 41. 12,3, -8±2V7; 3,12, -8T2V7-
43. 18, |. n2^^^; 3g, -63T^3N^9 . ^ 4,3/_±:^.3,^
7^^^ 47. 4,7, "-^^ ; 7,4 /^^^^^^ . 49. 9,7; 7, 9; two
answers. 61. ± 2, ± t?s ^/516; ± 1 T -Tii- 63. x = 7, - 2, 7w, -2w,
7u;«, -2t£j«; y « 2, -7, 2t£j, -7w;, 2u;2, -7m;2; ws — :L::^:^^^ — ?. 55,
m = ll, -9, llti?, -9w;, 11 u^*, -9u;2;n«=9, -11, 9!(?, -llio, 9w;2^ -lltc^.
ANSWERS 289
57. X = 243, 32; y = 64, 729; two answers. 69. x » 3, - 1, + 1, - 3; y « 1,
- 3, 3, - 1. 61. ± Vl ± J V3; ± Vl T J Vs. Use both upper or both
lower signs under radicals; outside of radicals use all combinations. 68.
±V^"+4n2+m ±Vw2+4n2 - m ^ , .. «- 6*±aV2^-o2.
— ^^^ 2 > — ^ ^ \ *^o solutions. 65. ^ ;
-^— — ; two solutions. 67. -[±^—^-^ + l); -
( ^ V -3obm " ^/ ' *^^ solutions. 69. « « ± V-S + 1» y ** ± V--5 -1;
x = ±V^+l, y = ±V--T-l;^our solutions. 71. M«±iV±32V2-27+l;
» = ± i V ± 32 V2 — 27 — 1 ; four solutions. Use all possible combinations of
signs in u and in ». 78. ± ab^2a^ — 62 — a62 j _|_ ^5 ^/2a^ — b^ + a62; two
solutions. 75. ,, ^ : ,, ^ 77. 6, 3, 4 ±V-33; 3, 5, 4 =FV-33.
^p2 + g2 y p2 -^ g2
79. X = ± V— 3, y -T V— 3, 2 = 2; two solutions. 8L x = 2 or 00 ; y =»
-} or -1; 2-lorO. 88. x = g^ (p9 ~^ ± V(pg -r)2 - 4g«); y = ^^
(p5 - r T V(P3 - r)2 ~ 4g3); 2 = -; two solutions. 85. ± y,T V, ± V ;take
all upper or all lower signs. 87.x = i(a-6-c-2itV(2+6+c-a)2+4»(2+c));
y'«jq7^;2 = j^- 89. a^^^iorf; 2/ =± i V^, or ± }; «=±jV^,
or ± J.
Problems
1. 8, 6. 3. 48, 36. 5. x = 15, - 12; y = ll, - 16; two answers. 7. x= 19,
- 20; y = 17, ~18; four answers. 9. 33, 56. 11. 19,23. 18. 28, 20 ft. sec.
15. 13^; 45 days. Assume each man's pay proportional to amount of work he
does. 17. 42. 19. A. 21. 3,5yds. 28. «i = 15.4; 11.7 ft. sec.; «j=6.8; 12.2
ft. sec.
Article 114
1. 3. 8. - 3. 6. 0. 7. - 3. 9. 1. 11. - 4. 18. - f. 16. t- 17.
M3 . 19. J2g^. 21. 1,-3. 28. -V. 26. -6.
log y log a-'o* ' *
Article 122
26. 1. 27. V600,000. 29. 6i in.
Article 148
1. 7 + nT;2nir-5, (2n + l)ir + 5 ; ± 5 ± 2nT;2nT. 8. 2 nx - 41" 48',
4 O Do
(2 n + 1) T + 41* 48'; (2 n + 1) ir ± 70** 32'; 63** 26' + nx; 2 nr + IV 32';
(2 n + 1) X - 11** 32'. 6. 68** 12' + wt; 2 nx - 16** 35'; (2 n + 1) t + 16**
35'; 2 nx ± 5** 44'.
290
ANSWERS
Article 160
Bin
cos
tan
CSC
cot
1.
-1
±iV3
^^
-2
^^
1
8.
dbt
±1
1
±i
±i
i
6.
±1
* 2
1
V3
±2
V3
V3
7.
±A
-H
±A
±V
-ti
±v
9.
.6
n
m
h
±i
±i
1
m
n
1
h
1 "•
±i '
^ ^
, Vw»2 — n2
\m^ — n2
IL
-L m
. 1
-^ n
IS.
4.Vl-/t2
±\1 —h?
"■ Vl - A2
Vl-A*
A
4 m
cfi-b^
2ab
o2 + &2
o2-62
02+62
"^02- 62
o« + 6*
2ab
2ab
'*'a2-62
15.
■^a2 + 62
Article 161
- , tan a; -.2 ^cac^x — 1 . ^ , ^ r; 5-r
1. ± -===•• 8. — - — 5 6. cos ^ ± VI — cos2 $.
Vl+tan^x csc2a;
Article 169
1. i;0. 8. «!;«!. 6. ±>^^i^"; iV5^. 7. ±
[(12 db 63) ± (18 ± 14) V3]. 9. JSHJ. 11. J!- 18. ± iUh ± iXIJ,
i ill J; iJf. 16. ~ 17. §. 19. sin 202i^ = § V2 - \^; cos 202J'> =
-iV2 + V2; tan 202}*» = V3 - 2 V5, sin7§** = i^2yj2 - V3 - 1; cos 7J**
= i V2 V5 + V3 + 1; tan 7§" = Vl5 + 8 V3 - 10 ^^ - 6 V&.
111
Article 160 .
1. 4** 40', 3** 20*. 3. 8; 5.
Article 166
1. 2 nx ± 60^ 8. (n + i) x. 6. nx + 45%- nx + 71** 34'. 7. 2 tit + 36** 52'.
9. nx. IL nx;wxdb7. 13. nx;ilx±^. 16. (2 71 + 1)5. 17. ^/^ t \ ^-
4 6 '§ 2(p±g)
19. :^ ; -^. 21. nx; 135** + nx. 23. 2 nx - 60**; 2 nx - 120^ 26. 2 nx
r — s*r -{-s'
+30**; (2n + l)x - 30^ 27. ^; nx +|. 29. y ; (2n + 1)| db 30*.
ANSWERS 291
Article 168
1. r = ± 6, ^ == tan-if 8. r = ± 41, d = tan-i V. 6. r=±-y,^ =
tan-i 1. 7. r = ± 3 VS, ^ = tan-i (- 3). 9. r = 5 >^, <^ = tan-i i, ^ =
tan-i 1. 11. aJ2 + y2 = ^.2. 13. ^ 4. ^ = 1. 15. a;2 + y2 + ^2 = 1.
Article 179
L Area = 4828, A = 97** 48', B = 18° 21'. 8, C = 63° 50'. 2. 8. Area =
1445.7, A = 34° 24', B = 73° 15', C = 72° 21'. 6. 6 = 290.9, c = 289.0,
B « 72° 6'. 7. 6 = 5340, c = 6535, A = 81° 52'. 9. a = 9548, c = 10804,
C = 105° 59'. 11. No solution. 13. c = 3120, c' = 402.2, B = 26° 52', B'
= 153° 8', C = 131° 47', C = 5° 31'. 16. h = .5458, 6' = . 1814, A = 39°
37', A' = 140° 23', B = 117° 51', B' = 17° 5'. 17. c = .7105, A = 76° 20', B
= 44° 53', Area = .2024. 19. a = 13.534, B = 15° 9'. 4, C = 131° 19'. 6,
Area = 32.564. 21. A = 149° 49', B = 3° 2', C = 27° 9'. 23. B = 51° 9', B'
= 128° 51', C = 87° 38', C = 9° 56', c = 116. 82, c' = "20. 172. 26. 6 = 71760,
B = 146° 43', C = 14° 4'. 27. A = 57° 53', B = 70° 17', C = 51° 50'. 29. c =
38088, B = 48° 34'. 7, C = 49° 38'. 3. 31. A = 18° 12', B = 135° 51', C=25°
57'. 33. c = 748. 1, A = 42° 51', B = 64° 9'. 36. h = .000331, B = 83° 33',
C = 32° 36'. 37. c = 2406, c' = 227.6, B = 31° 58', B' = 148° 2'. C = 120°
44', C = 4° 40'. 39. c = 369.27, A = 39° 39'.6, C = 90°. 63. 7; Vi29;
20 V3. 66.6824. 67. 45°, 60°, 75°; 612.5ft.; 683ft. 69.698.3ft. 71. 121ft.;
390.ft. 73.1145 ft. 76.8640 ft. 77.62.00 ft. 79.969.2 ft. 81.19955 m.
88. 59.1; 513. 86. 25, 33J, 41§. 87. ^^|^. 89. 18.76 chains; 7.578 acres.
91. 3.620 acres, south of dividing line. 98. 10.802 chains east of A. 96. i =
tan-i^. 97. 20° 7'. 99. 12° 32'.
Article 183
1. 55; 403. 3. 14; 200. 6. 28; 364. 7. p - V ?; 20p - 95 g. 9. I = 150;
d = 3. 11. a = 9; d = 2. 18. a = 18; d = 5. 16. a = 17; I = 97. 17. a = }-;
I = Ap. W- n = 16, l=- 69. 21. n = 14; a = 12. 23. n = 103, a = 1281.
26. 8925. 27. 10 sec. 29. 29700 ft.
Article 187
1. Z = 256; S = 508. 3. J = 4096; S = 5461. 6. ' = "2^44; -S - -^^i'
7. Z = o (1 +x)i; S = iil±£)ill2. 9. j. 4g; 288, ± 1728. 11. ± 12, 4,
± I, J, ± 2^^. 18. 12, 3, i, A- 16. a = 2, 5 = 254. 17. a = 6, S =
19. n = 6, 5 = 126. 21. r = 3, n = 7. 23. r = J, n = 6. 26. n « 9,
I = 19683. 27- a = 6, 2 = 320.
292 ANSWERS
Article 189
L 3}. 3. V. 6- i 7. 8 sec.
Article 191
1. a « 115 or 1; d = -10 or +2. 3. o - -11 or ^f^; d = 4 or - ^\
6. First number V; com. diff. A V2989 or j^ V - 1779. 7. Middle num-
ber = 6; com. diflf. = ± y J (5 &2) ± i/qM + ^- 9. 65^ 60% 65^ 11. a,
ar*, or, ar', .... 13. ± 1, ± 10, ± 40, ± 160. 16. 10.11 inches. 17.
$1845X1010. 19. 2 a; 4 a VS.
Article 194
1. $2975+. 3. $1489+. 5. 20. 7. $497.80. 9.
rCl+r)*"-!
Article 203
1. Convergent. 3. Conv. if | a: | < 1. Div. if | a: I ^ 1. 6. CJonv. if 1 x |< ^ •
7. Conv. if 1 < X < 10. 9. Convergent. 11. Convergent for all values of x,
13. Conv. for all values of x. 16. Conv. when — 1 < x S 1.
Article 206
1. .41. 3. 1,261. 6. .0589+. 7. .0053+.
Article 209
1. |x2. 3.3x2-1. 6. ± — %=' 7. ± , ^ * 9. ± ^^
2 V-^ VaJ* - 1 Vic^ - 1
Article 214
1. 12x« + 15x2. 8.^ + -^- 6. -^-— .• 7. 2xca^. 9. -sinx
4x* 9x* 2x^ 3x*
6 X
+ 8ecxtanx. 11. t^—z — =-• 13. — sin x tan x + cos x log cos x. 16. sec^x.
3 x* — 1
17. sec X CSC X.
Article 216
1. 3 x2 + 2 X. 3. coi^ X — sin2 x. 6. c*. 7. 12 Z, where I = length of edge.
9. Area of base. 11. ??^^. 13. 6^ + 3; 603; 9; 3. 16. ^ =
6 — c cos A da c — 6 cos A da he sin A
., -— . s ,
a dc a dA a
ANSWERS 293
Article 222
^- "'^ 3 ^ 15 + 315 + • ^- "^ 3 ^ 45 315 + " ^'
l+2x-\-2^ + ^-f+ .... 7.2 + ^+^ + ^+ • •• . 9. 1-
5 + A-^s+ • • • •."• 1+--|-|+ • • • • 13. l + i.-i.^ +
^, x8 + .... 16. 1 -ja; + f«2- ^+ • • • . 17. l+2x+2aJ2 +
2 x^H . 19. 3* 1 TT Tp + • • • . 21. 2' = To
2.3^ 4.3^ 8.3^ 7.2' 49.2^
- -^^ + . . . . 23. irV27^ +6 V3^ + -F= - "74=. + ' ' 'l
343.2^ 6^L V3x V27a;» J
1 4x* , 14a;* 140x2 ,
* (16a)* (IGa)*"' (16a)* (16 a)^'
Article 229
1. i w8 + i n2 + i n. 3. i n3 + i n2 + J n. 6. nf a + ^^-^ d ^ .
Article 231
3. .0314; .0204. 6. 5'^16»»05«.59; 18" 48' lO'M.
Article 232
1. Sixth entry should be . 364. 3. Sixth entry should be 3' 30".
Article 234
1. an = an-2- Ctn-l] W>1, Z. l+3l^+X^-\-2x^+''', 6. fx —
, • , . . 1 _^ . T 2 17 , 83 X 383 x2 ,
ix2-ix3 + §x4+ .... 7.3^-^+^--gj-+---.
Article 239
. 3 1 ^2 2,1 . 2\/3 + 3
8(3x4-1) 8(x + 3) X x+2'x-2* 6(x-2~ V3)
2%/3-3 - __1 1 1__ Q 3 5-3x
6(x-2+V3) 4(x-l) 4(x + l) 2(x2 + l) •^•'x^x2+4
11. 1 - ;^- + o f o^.^os . 13. - ? + -^ - ^-^,. 16. ^
3 X ' 3 (x2 + 3) X ' X - 2 (x - 2)2 -"• 3 (x + 1)
^-2 . 17. 1_^+ 1_^^_ 3_^. ^3^ 4 • ]
3(x2-x + l) 2(x-l) ' 6(x-2) ' 10(x+3) x-2 x-1
12 2
21. — f^ =-s +
x+1 x+2'x-2
294 ANSWERS
Article 241
3. X = it, y = i> 2 = A- 5. a; = if, y = 1, « =• ft- 7« Not independent.
Article 249
1. 0. 8. 0. 6. 8. 7. 398. 9. 832. 11. 016263^4. 33. w = A*5, t' = - i,
t0 = i|. 36. Inconsistent.
Article 267
1. V5, - 45*; 5, 36** 52'; Vi46, 114'* 27'; 2, 90°; 2, 0°; 2, 0*»; 6, 30*»; 36,
- 60**; 4, 90^
Article 269
8. ±3; ±3i. 5. a;i = 2; X2 « 2 (cos 72** + 1 sin 72**)]^ a;s = 2 (cos 144**
+ i sin 144**); etc. 7. a;i = V3; iP2 - ^^^^ xs = ~^/|+^^ ; etc.
Article 260
1. Scos^dsin^-sin'^. 3. cos* d - 6 cos^ ^ sin^ ^ + sin* ^. 6. 6cos«i^8infl
- 20 cos8 ^ sin3 d 4- 6 cos ^ sin6 ^.
Article 263
1. 24. 8. 240.
Article 264
1. 20. 8. 120. 7. 190.
Article 266
1. 1260. 8. 360. 6. 6^4 X 14C7 + 6^5 X uC^ + cCe X wCs = 71500. 7. 73.
9. 4; there will be three different throws. 11. 36; there will be 21 different
throws.
Article 268
1. A. 8. A. 5. If. 7. aJg. 9. ^^ji. IL j^ihz'
Article 270
1. A. 8. A; 4- 6. 6. 7. A. 9. }.
Article 275
1. a;8-6x2 + lla;-6 = 0. 8. x*-2x»-4jc» + 8a:=x0. 6. 6x*-
5a;S-5a;2 + 5a; -1=0.
Article 280
1. - 1, 2, 2. 8. 3, 3,-2, - 2. 6. 3, 3, - 1, - 2. 7. 1, 1, 1, - 2.
9. 3, 3, ± J.
ANSWERS 296'
Article 286
La:» + 2a;2-4a;-8 = 0. 3. a;S-12a;-14=iO. 6. aJ2 + 2a; + l=0.
7. a;2 - 2 X - 2 = 0. 9. a;3 + a;* - 9 = 0. 11. x« - 9 x2 + 24 x - 16 = 0.
18. a;* + 6a:8 + x2 -24aj + 16 =0. 16. A = 1; a:^ - Sx = 0. 17.^ = 1;
a:8-9x-7=0. 28. ± V^, 2. 25. 2, ± 2 V2, - 1 ± V^.
Article 291
1. 2, 2, - 1. 8. 1, - i, - i. 6. 3, 2 ± 2 y/S. 7. - 1, - 2, 3, 9. 1, - 1,
- 1 ± V^=^. IW (- 1 ± VS), i (5 ± V37). 13. 2,-2, - 2, - 2.
16. 4, 2, - 1 ± V^.
Article 306
1. A = 79° 30'.8, B = 46** 15'.3, C = 70° 55'.6. 8. A = 130° 6'.4,
B = 32° 26M, C = 36° 45'.8. 6. o = 96° 24'.5, 6 = 68° 27'.4, c = 87° 31'.6.
7. c = 50° 6', A = 129° 58', B = 34° 30'. 9. a = 43° 18', B = 28° 48',
C = 74° 22'. 11. b = 78° 17', c = 126° 46', A = 96° 46'. 18. a = 76° 25',
6=58° 19', C = 116° 31'. 16. a = 124°12'31", c= 97° 12' 25", B= 51° 18' 11".
17. a = 58° 8' 19", B = 98° 20' 0", C = 63° 40' 0. 19. b = 75° 29', c = 108°
14', C = 46° 52'. 21. No solution. 28. c = 84° 30', B = 56° 20', C - 97° 19'.
26. B = 42° 37' 18", 137° 22' 42", C = 160° 1' 24", 50° 18' 55", c = 153°
38' 42"; 90° 5' 41". 27. a = 64° 23' 2(f", 6 = 99° 48' 50", A = 65° 33' 10".
Article 306
1. N.Y. - S.F. 2568 mi. N.Y. - M.C. 2090 mi. S.F.-M.C. 1889 mi.
Angles: N.Y. 48° 58', S.F. 55^48', M.C. 82° 40'. Area: 2025300 sq. mi.
8. X = 9* 34'» 15», <^ = 22° 6' N; course, S 44° 28' W.
Article 307
1. A = ± 92° 50'; « = ± 5* 4'» 12«. 3. A = 43° 27'; A = 70° 3'. 6. N 1**
33'.6 E or W; < =• ± 5* 55'» 54». 7. 9° 46'.4.
I
INDEX
PAQB
Abscissa 42
Addition 4
Altitude 281
Alternating series 173
Annuities 169
Antecedent 88
Approximations 199
to the roots of an equation. . 260
Area of plane A 148
of spherical A 277
Arithmetic progression 161
mean 162
Azimuth 281
Base of logarithms 189
Binomial series 196
convergence 196-7
Binomial theorem 33, 195
Celestial poles 280
sphere 280
equator 280
Chance 244
Circle 67
Circular measure 113
Circular parts, Napier's rules of 270
Co-factor 223
Combinations 242
Complex numbers 21, 233
Comparison test 174
Complementary function ... 97, 100
Computation 199
of logarithms 201-2
Conic sections 72
Conjugate complex numbers ... 21
Consequent 88
Convergence of series 171
of binomial series 196-7
PAGB
Coordinates . . . *. 42
polar 231
Cosecant 95, 100
Coisine 95, 100
Cotangent 95, 100
Coversed sine Ill
Cubic equation 264
Declination 281
Degree of a term 64
of a polynomial 64
De Moivre's theorem. . .' 235
Derivatives 184
higher 192
formulas 186
Determinants, general definition 220
of second order 217
of third order 218
properties 222
use in solving equations 226
Differences 203
Difference quotient 180
Discriminant of quadratic equa-
tion 57
of cubic equation 266
Division 5
synthetic 255
Ellipse 68
Equations, cubic 264
exponential 86
linear 37-53
of nth degree 249-63
quadratic 54-86
quartic 267
trigonometric 137
Equator, celestial 280
Evolution 18
297
298
INDEX
PAGB
Exponent, irrational 20
laws 17-21
negative 17
positive integral 7
rational 19
zero 18
Exponential equations 86
values of sin x and cos z 239
Extremes .' 88
Factor, highest common 11
theorem 10, 249
Factoring 9
Fractions 13
partial 213
Functions 90, 179
continuous 180
hyperbolic 240
trigonometric 103
inverse trigonometric 134
Geometric mean 164
progression 163
infinite progression 165
series 173
Graphic solution of linear equa-
tions 39-50
of quadratic equations. . 65-80
Graph of straight line 41, 43
of trigonometric functions. . . 105
Harmonic progression 167
mean 167
Highest common factor 11
Horizon 281
Hour angle 281
Hyperbola 70
rectangular 71
Hyperbolic functions 240
Imaginary number 21
Infinite series 171
solution of linear equations. 38
Infinity 6
Initial line 99,231
Integral expression 14
PAGK
Interest 168
Interpolation 206
Inverse ratio 88
trigonometric functions 134
variation 91
Involution 17
Irrational expression 20
exponent 20
number 19
Joint variation 92
Law of sines 144
of cosines 145
of tangents. 146
Least common multiple 11
Limit 171
Linear equations 37
graphic solution 39-50
simultaneous 46
Logarithms 28, 30
computation of 201-2
laws of 30
modulus of 203
natural or Naperian : . . 189
Maclaurin's series 193
Mean arithmetic 162
geometric 164
harmonic 167
proportional 89
Means, in a proportion 88
Meridian 281
Modulus of common logarithms 203
Multiplication 4
Naperian logarithms 189
Napier's rules of circular parts 270
analogies 275
Natural logarithms 189
Numbers, complex 21
conjugate complex . 21
imaginary 21
irrational 19
principal root of 22
rational 5
INDEX
299
PAGB
Numbers, real 20
surd 20
Ordinate 42
Parabola 59, 69
Partial fractions 213
Permutations 242
Polar coordinates 231
triangle 269
Pole 231
Power 8
Present worth 169
Principal value of an inverse
trigonometric function . . 135
of a root 22
Progressions, arithmetic 161
geometric 163
infinite geometric 165
harmonic 167
Proportion 88
Quadratic equations 54-86
formula 56
simultaneous 64
Quartic equation 267
Radian 143
measure 143
Radius vector 231
Ratio 88
inverse 88
Rational expression 5
exponent 19
number 5
Real number 20
Root of an equation 56
principal 22
Roo'.^ of unity 237
Secant 95, 100
Series, alternating 173
binomial 196
geometric 173
infinite 171
PAGB
Series, Maclaurin's 193
power 173
ratio test 176
Sine 95, 100
Slope 181
Sphere, celestial 280
terrestrial 279
Spherical excess 269
triangles 269-79
Straight line 41, 43
Subtraction 4
Surd expression 20
number 20
Synthetic division 255
Tangent, trigonometric ... 95, 100
to a curve 181
Terminal line 99
Terrestrial sphere 279
Triangles, plane right 98
plane oblique 144-155
spherical right 270-1
spherical oblique 272-8
Trigonometric equations 197
Trigonometric functions . . . 94-140
defined 95, 100
discontinuities 104
graphs 105
inverse 134
line values 101
periodicity. 106
signs 101
variation 103
Undetermined coefficients 211
Variable 90
Variation 90
direct 91
• joint 92
inverse 91
Versed sine Ill
Zero 5
exponent 18
APPENDIX A
The Greek Alphabet
f
Lettera.
Name.
Letters.
Name.
Letters.
Name.
A, a,
Alpha
i,h
Iota
P,P,
Rho
B,)8,
Beta
K,K,
Kappa
S, cr,
Sigma
r,7,
Gamma
A,X,
Lambda
T,r,
Tau
A,d,
Delta
M, M,
Mu
Y, u,
Upsilon
E,«,
Epsilon
N,v,
Nu
*,<^,
Phi
z, f ,
Zeta
B,{,
Xi
X, X,
Chi
H, u,
Eta
0,0,
Omicron
^,^,
Psi
e, e, 6,
Theta
n,^,
Pi
fi, CO,
Omega
List of Formulas
Factors of a** ± 6*, n being a positive integer (9).
a** — 6** is divisible by (o — b) and by (o + b) when n is even,
a** — 6** is divisible by (a — 6), not by (a + &), when n is odd.
o** + 6** is divisible by (a + b), not by (o — 6), when n is odd.
o'* + 6** is not divisible by (o + 6) or by (a — 6) when n is even.
Special Cases.
(x2 — i)2 _ (^ _[_ 5) (a — 6). a^ + &^ has no real factors.
a3-63 =(a-6) (a^ + ab + b^).
a3 + 63 =(a + 6) (a2-a6 + 62)^
a4 - 64 = (^2 _[. 52) (^2 _ 52), ^4 _[. 54 ^as no real factors.
a« - 6« =(a - 6) (a* + aSft + a^b^ + ab^ + b^).
a^ + b^=(a + b) (a^ - a% + a%^ - ab^ + b^).
Factor Theorem. — If / (x) reduces to zero when x = a, f (x)
contains the factor (x — a). (11), (272).
301
Wl FORMULAS
(90; to 'Ji,.
or Conqiler Hnmben. (2S.)
isV^; i2 = -l; i^=-i; i*==+l, etc.
x + iy = r (cosO + ism ^) = re*.
Sords. — If a+V6 = c+V5, where Vb and Vd are surds,
then a ^ e and b == d. (29.)
Logaritfams. (37), (39), (226).
If (^ ^m, thea x ~ logi m.
logafnn = logiin + login. logi— = l<^m — logan.
loga a = 1. 1<^ 1 = 0. logo = — 00, if a > 1.
Change of Base, logam = log^m X logab.
If a = 10 and b = e, then log. b = logio c = 3f . (Table V.)
Hence logio m = M log^ m.
Bhiomial Theorem. (42), (220-1).
^ n(n-l)(n~2) . . .(n-r+l) ^,.,^, ^ ^ ^ ^
/I I ^Nn_i . ^^ I n yn-l) g n(n-l)(n-2) o
FORMULAS 303
Quadratic Equation, ax^ + &» + c = 0. (74), (76), (78).
Roots real and unequal if 6^ — 4 ac > 0.
X =
-6±V62-4ac
Roots real and equal if 6^ — 4 ac = 0.
Roots imaginary if 6^ — 4 or < 0.
h c
Sum of roots = — . Product of roots = —
a a
Graph ot y = ax^ + 6a; + c is a parabola.
Standard Equations of Conic Sections.
Circle: x^ + y^ = r^. Parabola: y^ = 4:ax; x^ == iay.
Ellipse: -^ + p = 1. Hyperbola: "2 "" f2 ~ i !•
Rectangular Hyperbola: xy =±_k^.
Ratio, Proportion, Variation.
If, a:b = c:d,
then,
(1) a + b:b = c + d:d;
(2) a — b:b = c — d:d;
(3) o + 6 : a — 6 = c + d : c — d;
(4) • a'*:6'» = c'*:(r.
If ai : 6i = 02 : 62 = ^3 * 63 = • • ' >
then any of these ratios = ^ — r^r — ; — ^^ — ; —
pbi +qb2 + rbz+ • • •
where p, q, r are any multipliers;
also any of these ratios = y t^ , ,\ ■ i>\ ■ '
▼ bi + 02 + 03 + • • •
If y cc X then y = kx;
If 2/ « - then y = -, or xy = k.
Arithmetic Progression. (180.)
a = first term; d = common diff.; n = number of terms;
I = last or nth term; S = sum of n terms.
nth term = I = a + (n — I) d.
Arithmetic mean of o and 6 = — tz — •
• .
9
804 FORMULAS
Geometric Progression, (184.)
r = the ratio; o, n, I, Sj as above.
nth term = Z = af^"^.
o 1 — r** a — rl
1 — r 1 — r
Geometric mean of a and 6 ^^ab.
Smn of infinite geom. progr. = ^ __ , if | r | < 1.
Infinite Series. — Tests for convergence or divergence.
Series, U1 + U2 + wa + • • • + Wn-i + Un+ - - - .
Converges when the terms are alternately + and — , and
steadily decrease toward zero (199).
Converges when the ratio — ^ becomes and remains numeri-
Un-l
cally less than 1 for all values of n, provided always that
Km 1^ = 0. (202.) •
Diverges when the ratio — ^ becomes and remains greater than
'^— 1
1, or approaches 1 from the upper side. (202.)
Converges when its terms are numerically less than the corre-
sponding terms of a series known to converge absolutely. (201.)
Diverges when its terms are all of like sign and are numerically
greater than the corresponding terms of a known divergent series.
Test Series.
i4.a.4.a.2-4-x34- . . . ^conv.when|x|<l;
l + x + x -ha;-+ <j.^^ whenlxl^l.
jl^ J_ J^ , ( conv. when p>l;
p"'"2p"^3^''^ ' ' ' ^div. whenp=l.
Derivatives. (210.)
-. ^dy^y ^ ^ = slope of tangent to curve y = /(a;).
*^ "~ dx ""ax-o Aa; ( — rate of change of y relative to x.
FORMULAS 806
Formulas for Differentiation, (211-2.)
^ _ ^ ^ ^ — n <^(cy) _ ^
dz dy dx dx ' dx " dx
U(u + v + w + • • _ du dv dw
• • •
dx dx dx dx
d (uv) dv , du \v/ dx dx
-^^ = "^+''^- -E ? — .
dy _^ du
dx du dx
when y is SL function of u, and u a function of au
dx** - ^ , d loe a; 1 da'
, ■ — fUU
dx
dx
x dx - " '"* "•
d sin a;
— J — = cos X.
dx
d COS X
—3 — = — sm X.
dx
d tan X 9
— 3 — = sec-^ X.
dx
d cot X
^ — =— csc^x.
dx
d sec a; .
— 1 — - = sec X tan a;
dx
dx.
d CSC X
= — cscx cotx.
dx
d sin"^ X 1
dcos"^x —1
dx Vl — x^ dx Vl — x^
dtan~^x 1 dcot"^x '—1
dx 1 +x^ dx 1 + x^
ds^c"^x 1 dcsc~^x —1
dx a; Vx2 - 1 dx ^ Vx^ - 1
Maclaurin's Series. (218.)
/(x) = /(0) + xr (0)+ ^/"(0)+ S/'" (0)+ • • • .
Some Standard Series.
6*=l+x + T;r + r5+'«« . Always convcrgent.
Lf 15
sinx = x — T^ + l-T— ••• . Always convergent.
306
FORMULAS
co8a; = l-|2 + |4- • • •
^2 ^fA yA
loge (1 + x) = x - 2" + 3" "■ "4 +
• • •
Always convergent.
Convergent only if
- 1< a; = 1.
Theorem of Undetennined Coefficients. (233-4.)
If, for all values of x from a; = to a; = A where h is any number
other than zero, we have
oo + aix + a2X^ + • • • + dnX^ + • • • =0,
then Oo = 0, ai = 0, a2 = • • • an = 0, • • • .
If, for values of x as above, we have
Oo + aix + 020:^ + . . . = 6o + hix + b2X^ + . . . ,
then Oo = 6o> ai = &!> ^2 = &2> etc.
Partial Fractions. (235-8.) — The partial fractions may be
determined according to the factors of the denominator of the
given fraction by the following rules:
Form of factor:
(ax + 6),
{ax + 6)^
(orf + hx-{- c).
Corresponding fraction or fractions:
A
ax+ b
Ai
+
2
+
ax + b (ax + b)^
• • •
+
(ax + 6)~
ax^ + bx + c
(a»Hto+c)", .^j^+?', + .,d'f J:f^, +- • • + ^"^+^
dx^+bx+c (ax^+bx+c)^
(ax^+bx+c)
m
Determinants. (240-9.)
ai bi
0,2 62
ai 6i ci
0,2 &2 C2
as
bz C3
= ai&2 "" P^i*
— a\A\ — 61S1 + c\C\
= 0162C3 + 0263^1 + 036102
— 0362^1 — 0261 C3 — ai&3p2«
FORMULAS
807
Here A\, Bij Ci, are the minors of ai, 61, ci, respectively,
oi 61 ci di
02 ^2 C2 ^2
= aiili -61B1 +C1C1 -diDi,
03 O3 Cz CLz
(I4 04 C4 d4
where ili, Bi, Ci, Di are the minors of ai, 61, Ci, di, respectively.
Similarly for a determinant of any order.
Differences and Interpolation. (227-32.)
Let 1^0, wi, W2, • • • be a given sequence, and let Aii^o, A2i^o,
Aat^o, • • • be the first terms of the successive difference columns.
Also let nCi, nC2, nCa, • • • be the binomial coeflScients, i.e.,
r ^ r - ^ (^ - ^) r _ n(n-l) (n-2)
nCi = n, n^2 = 12 ' n^3 = Tq > CtC.
Let Un be the nth term of the sequence and «„ the sum of its
first n terms. Then
Un = Uo + nClAlUo + nC2^2Uo + nCsAsUo + ' ' • ;
Sn •= nClUo + nC2AlUo + nCzA2Uo + nC4A4iio + • ' • •
If uo^fixo), wi =/(xo + A), W2 =/(a^+ 2A), Us = /(xo + 3 A),
. . . , then
f(Xo + nh)=f(Xo) + nCiAif(Xo)+nC2A2f{Xo)+nC3Asf(Xo)+ • • • .
Here n need not be an integqf.
Useful Approximations. (224.)
When X, yyU,v, . . . are small (near 0) we have, approximately,
(l+x)(l+y) = l+x + y. _1
l+x
= 1 — X.
(l+x)(l-y) = l+x-y.
(1 - x) (1 -- 2/) = 1 - a; - 2/.
l~x
l+x.
l+x . , (l+a;)(l+2/) ..... . .
THf-'+^^-y- (i+i^)(i+S... '^^+^+^+'''"^""^""'''-
(1 + a;)** = 1 + na;.
Vl + x =l + ix.
1 . 1
' / =1 X.
VI + x 2
(l+a;)2 = l + 2x.
As special cases of this:
Vl -x = 1 -ix.
1 , . 1
f = 1 + - X.
Vl-x 2
(l-x)2 = l-2x.
e'^l+x. log, (1 + x) = X. logio (1 + x) = .43 x.
sin X = tan x = x (radians). cos x » 1.
308 FORMULAS
More accurately,
7? Q? a?
sinx = a;--y cosa; = l — y tanx = x + -^-
De Moivre's Theorem. (266.)
(cos B + i sin oy = cos nd + i sin nS.
. 2^ = r'* (cos nB + i sin nB),
The nth Roots of Unity. (259.)
Xk = cos h I sm ; fc = 0, 1, 2, . . . , n — 1
Expansions of cos n6 and sin n6. (260.)
cos riB = cos** B — ^ ,1" ^ cos'*-^ B sin^ ^
If
. n (n - 1) (n - 2) (n - 3) „ . ^ . . ,
+ — ^^ -^-Ta — — cos**--* B sin^d -- • '
sinnd = ncos'^-i ^sin ^ - n{n- 1) (n - 2) ^^^„_3 ^ ^.^3 ^ _^
Exponential Values of sin x and cos x. (261.)
sma; = ^r-, cosa; = 77
2t 2
Hyperbolic Functions. (262.)
smhx = -^, ^ + J3 + [5 +
coshx = " '^" =14- + +
2
e' + e-'
2
sinh X
coshx
1
• • •
• • •
, , «*x*^ ^ . , cosh X
tanh a; = — i coth x — -r-r —
smh X
sech X = — ^i: — • csch x = -^r-r —
cosh X sinhx
Permutations and Combinations. (263-4.)
„P, = n(n - 1) (n - 2) . . . (n-r+1). ^Pn = \n.
p _nPT _ n (n - 1) . . . (n-r + 1) _ ^
Plane Trigonometry
Definitions. (124, 132.) — In right triangle ABC, whose sides
are a, 6, c [figure of (124)],
sinA=-> cosA=-> tanA=r>
ceo
cscil=-> secA=r> cotA=--
aba
vers A = 1 — cosil. covers A = 1 — sin A.
More generally, if x be an angle of any magnitude, as XOP in
the figure of (132),
ordinate abscissa , ordinate
sinx = -jr— f cosx = j7-T y tanx = -r: — : — >
distance distance abscissa
distance distance . abscissa
CSC X = T^ —f sec X = -r — : y cot X = — -t: t" '
ordinate abscissa ordinate
Relations between the Functions of an Angle. Formulas,
Group A. (137.)
1 . 1 o X 1 ex COSX
1. smx = 3. tana; = — :— • 5. cota; = -:
CSC a; cotx smx
1 sin X ^- sin^ x + cos^ x = l,
2. cosx = • d. tana; = ^ ^ . x 9 o
sec X cos X 7. 1 + tan-^ x = sec^ x.
8. 1 + cot^ X = csc^ X.
Rules for expressing any function of any angle in terms of a
function of an acute angle. (139.)
Any function of any angle x is numerically equal to the
same function - . , j. . . i_ j i_ { even ,, .
. ^. of a; mcreased or diminished by any < , , multi-
co-function ( odd
pie of 90°.
The sign of the result must be determined according to the
quadrant of x.
309
810 FORMULAS
of + aj and - 05. (140.)
f(+x) = /(— x)j when / = cosine or secant/
f{+x) =— /(— a;), when/ = sine, cosecant, tangent, cotangent.
Angles Corresponding to a Given Function. (146.)
Let denote the smallest positive angle having a given func-
tion equal to a given number a. Then all angles such that
C sm X ** CL
I. ] Bie x^2nv + e and (2n + l)T — d;
II. \
cscx = a
cos a; = a
sec x » a
are a: = 2 nx ± ^;
___ ( tan X = a , .
III. < X are a; = nir + ^.
( cot X '^ a
Formulas, Group B. (166.)
9. sin (x + y) = sin xcosy + cos x sin y.
10. cos (x + y) — cos X cos 2/ — sin x sin y.
11. sin (x — y) = sin a; cos j^ — cos a; sin y.
12. cos (x — y) = cos x cos y + sin x sin y.
tan X + tan y
13. tan (x + 2/) =
14. cot (x + y) —
1 — tan X tan y
cot X cot 2/ — 1
cot X + cot 2/
^ - ^ , . tan X — tan y
16. tan(x-2/) = i + tanxtan2,'
. / N cot X cot 2/ + 1
16. cot(x-y)= eoty-cotx -
Formulas, Group C. (167.)
Double Angle. Half -Angle.
14. sin 2 X = 2 sin x cos x. 17. sin J x = ± y ""^^^^
16. cos 2 X = cos^x - sin2 x, 18. cos J x = ± y L±|^^.
= 1 - 2 sin2 X, iQ +n« 1 ^ - ^ 4 /I - cos x
19. tan
ia:=±y|
+ COS X
= 2cos2x-l. 1-cosx
=5 ; >
sm X
16. tan2x= ,^\^^^ > = sl5^_>;
1 - tan2 X 1 — cos X
f
FORMULAS 811
FormulaSi Group D. (168.)
on I- n . U + V U — V
20. sm w + sm v = 2 sin — x — cos ^ *
21. smu — smv = 2cos — ^ — siii"~;5 — *
22. cos u + cost; = 2cos — -^ — cos — ^ —
«.« _.w + i;.w — t;
23. cos w — cosv = — 2 sm — :r — sm — -^ — •
Solution of Plane Triangles
Right Triangles. — By means of the definitions of the trigo-
nometric functions write an equation involving the two given
paries and a required part; solve this for the required part.
Oblique Plane Triangles. (169-172.)
Law of Sines: 1. a:6: c = sin A : sinS: sin C (169)
Law of Cosines: 2. a^ ^l^ + <? —2 6c cos A. (170)
La. of Tar^erus: 3. ^ = ^f^ • ("1)
Half-Angles. (172.)
Let 8 = H« + 6+c) and r = y/ii^l°Hl^^Iii^ .
^. AnhA^^^^MH. -6. tanM=v/^^^^P^^-
^ oc ▼ s Cs — a)
M=v/'^'"'^^- . .^..1 ._ r
5. cosM=V ^\ ^ ' 7. tanM =
6c * s — a
Solution of Oblique Plane Triangles. (173-8.)
Case I. Given two angles and a side. (174)
Use law of sines.
Case II. Given two sides and the included angle. (176)
Use law of tangents^ then law of sines.
^12 FORMULAS
Case III. Given two sides and an opposite angle. (176)
Use law of sines. Ambiguous case.
Case IV. Given the three sides. (177)
Use one of the formulas (4), (5), (6), or (7) above,
preferably the last one.
Area = J a6 sin C = Vs(s-o) (s - 6) (s - c). (178)
Spherical Trigonometry
Spherical Right Triangle. (313-6.) — Let ^1 , ^B, C be the angles,
and a, b, c the sides. Arrange tjie five parts a, 6, co-B, co-c, co-A
in circular order. These parts are then connected by Napier's
Rules:
J? • jji X ^ product of cosines of opposite parts :
sine of middle part = ijxrx x r j- x ^
( product of tangents of adjacent parts.
To solve a spherical right triangle use Napier's Rules to write
a formula involving the two given parts and a required part.
To solve a quadrantal triangle, solve its polar right triangle.
Spherical Oblique Triangles. (317-22.)
Law of Sines : sin a : sin b : sin c == sin ^4 : sin B : sin C.
Law of Cosines: cos a = cos b cos c + sin 6 sin c cos A.
Half-Angles.
1 /^ I 1. I -\. 4. -_ 4 /sin (s — a) sin (s — b) sin (s — c)
s = o (^ + ^ + c) J *^^ ^
2 ▼ sin s
4. sin^^^v/^^ "^^'-"^'"'^
2 ▼ sm sm c
_ 1 J 4 /sin s sin (s — a)
5. COSHil=V ; — i—^^ -'
2 ▼ Sin sin c
6. tanlA = v/?i2iM^4EiiZ^).
2 ▼ sin s sin (s — a)
Q X 1 yi tan r
8. tSJlTzA = -r
2 sin (s — a)
FORMULAS 313
Half-Sides.
2 / cos S
5 = 2U + B + C); ^^^R=\/ ,osiS-A)cos(S-B)cos(S-C)
. 1 4 /— COS aS COS (S — A)
13. sm20 = y
2 V sin -B sin C
1 , /cos (S - B) cos (S - C)
14. cos2a = y-
2 V sin B sin C
^ 1 .1 — COSaS cos(S — il)
15. tan;^a = \/
2"" V cos (S - B) cos (S - C)
16. tan ^a = tan R cos (S — ^4).
Napier's Analogies.
iA X 1 / r\ sinH^ — B) , 1
19 tanj5 (a - 6) = . i /^ . px tan^c.
2 sin^ (il +B) 2
rt/^ X 1 / . r\ cos i (il — JS) , 1
20. tan^Ca + ft) = _^^^^^-^ tan ^ c.
21. tani(A-fi)^?!4feScotic.
2 sm f (a + o) 2
22. tanj (^ + B) = £5ii(5L^ cot | C.
2^ cosj(a + o) 2.
Spherical Excess.
B=(4+B + C)--180^
^„ X 1 ET tan^a tan^ft sinC
Jto. tan z:£j =
/
2 1 + tan J a tan \ h cos C
24. tan t ^ = Vtan i s tan i (s — a) tan J (« "" ^) ta^^^ (« — c).
Area = ^ ^^^f^^^^ X^irR^ = E (radians) X ii!^.
Solution of Spherical Oblique Triangle. (323.)
I. Given two sides and an opposite angle.
Use law of sines, then Napier's Analogies. Two solu-
tions possible.
II. Given two angles and an opposite side.
As in I.
814 FORMULAS
III. Given the three sides.
Use formulas for the half-angles.
IV. Given the three angles.
Use formulas for the half-sides.
V. Given two sides and their included angle.
Use Napier's Analogies^ then law of sines.
VI. Given two angles and their included side.
As in V.
APPENDIX B
Explanation of the Tables and Their Use
TABLE I
This table gives the decimal part; or mantissa, of the logarithm
of every positive number containing hot more than three sig-
nificant figures. The mantissas of the logarithms, of numbers
containing more than three significant figures are to be obtained
by interpolation (36). The integral part, or characteristic^ of the
logarithm must be supplied by the computer, according to the
position of the decimal point in the number.
Rules for Characteristics.
(a) When the number has n significant figures to the left of
the decimal point, the characteristic of its logarithm is n — 1.
(b) When the number is a decimal with n ciphers between the
decimal point and the first digit which is not zero, the characteris-
tic of its logarithm is 9 — n, and — 10 must be supplied to com-
plete the logarithm.
The reason for these rules will become evident when we consider
an example.
Example, Let us find log 302. In the table find 30 in the
left-hand column and run across the page horizontally to the
column headed 2. There we find that
mantissa of log 302 = .4800.
Now 302 lies between 100 and 1000, i.e. between 10^ and 10^.
Hence, by the definition of a logarithm, log 302 must lie between
2 and 3. Therefore the characteristic is 2, and
log 30i^ = 2.4800.
This is of course not the exad logarithm of 302, but only its value
to four decimal places.
Writing the last equation in exponential form, we have
302 = 102-*8oo.
315
816 EXPLANATION OF TABLES
Multipljdng both sides by 10,
3020 = 10 X 102-4800 = 103.4800, Hence, log 3020= 3.4800.
Multipljring again by 10,
30200 = 10 X 103-4800 ^ 104.4800. Hence, log 30200 = 4.4800.
Therefore, where a number is multiplied by 10, the character-
istic of its logarithm is increased by 1; the mantissa remains
unchanged.
Dividing the above equation successively by 10, we obtain
30.2 = 102-4800 ^ 10 = 101-4800^
3.02 == 101-4800 4- 10 = 100.4800^
.302 = 100-4800 ^ 10 = 100.4800-1^
.0302 = 100-4800-1 ^ 10 = 100.4800-2^
.00302 = 100-4800-2 ^ 10 = 100.4800-3^
and so on. As logarithmic equations these are:
log 30.2 = 1.4800,
log 3.02 = 0.4800,
log .302 = 0.4800 - 1 = 9.4800 - 10,
log .0302 = 0.4800 - 2 = 8.4800 - 10,
log .00302 = 0.4800 - 3 = 7.4800 - 40,
and so on. The second form in the last three equations is used
for convenience in computations; it is in accordance with rule (b).
To discuss rules (a) and (b) more generally, let m be any number.
Then by the definition of a logarithm, when
m lies between
log
m lies between
(1)
1 and 10,
and 1,
(2)
10 and 100,
1 and 2,
(3)
100 and 1000,
2 and 3,
(4)
1000 and 10000,
3 and 4,
and so on. Therefore, when m has
(1) 1 digit to the left of the point, log m = 0.+
(2) 2 digits to the left of the point, log m = 1.+
(3) 3 digits to the left of the point, log m = 2.+
(4) 4 digits to the left of the point, log w = 3. +
and so on. Hence rule (a).
EXPLANATION OF TABLES 817
In the case of decimal numbers,
when m lies between log m lies between
(1) 1 and 0.1, and - 1,
(2) 0.1 and 0.01, - 1 and - 2,
(3) 0.01 and 0.001, - 2 and - 3, ,
(4) 0.001 and 0.0001, - 3 and - 4,
and so on. That is, when m is a decimal number in which
(1) no cipher follows the point, log m = 9.+ • • • — 10
(2) 1 cipher follows the point, log m = 8.+ • • • — 10
(3) 2 ciphers follow the point, log m = 7.+ • • • — 10
(4) 3 ciphers follow the point, log m = 6.+ • • • —10; -^ v
and so on. Hence rule (b).
Interpolation. — Example, Find log 3024.
From the table,
mantissa of log 302 =.4800; difference = 0014
mantissa of log 303 = .4814;
Assuming that the increase in the logarithm is proportional
to the increase in the number, we have
mantissa of log 3024 =.4800 + 4 X.0014 =.4806.
The result is here given to the nearest unit in the fourth decimal
place, .4 X .0014 being taken equal to .0006 in place of .00056.
Proportional Parts. — For convenience in interpolation, the
tabular diflferences greater than 20 are subdivided into tenths and
tabulated under the heading " Prop. Parts." When the difference
is less than 20, the interpolation is best made mentally. If it is
degired, the table of proportional parts may be used when d < 20
by taking half the proportional part corresponding to double the
difference.
Examples.
1. log 164.3 = ?
Mantissa of log 164 = .2148; d = 27,
Correction for .3 = 8
log 164.3 = 2.2156
2. log (164.3)3 =?
log (164,3)3 = f log 164.3,
= i (2.2156) = 1.4771.
818 EXPLANATION OF TABLES
8. log .01047 - 7
Mantissa of log 104 » .0170; d <- 42,
C!orrection for .7 = 29
log .01047 =85199-10
log \^(.01047)* = ?
\^.01047« = (.01047)*,
log ^(.01047)* = i log (.01047),
= i (8.0199 - 10).
4 (8.0199 - 10) = 32.0796 - 40 =" 22.0796 - 30.
§ (22.0796 - 30) = 7.3599 - 10.
Note, When a logarithm which is followed by —10 is to be divided by a
number, add and subtract a multiple of ten so that the quotient will come
out in a form followed by —10. Thus:
i (8.2448 - 10) = i (38.2448 - 40) = 9.5612- 10.
Anti-logarithm. — The number whose logarithm is x is called
the anti-logarithm of x.
Thus, if a: = log m, then m = anti-log x.
Given a logarithm, to obtain the corresponding number (anti-logor
rithm).
Examples,
1. log m = 0.4806. m = ?
The given logarithm lies between the tabular logarithms .4800 and .4814,
to which correspond the numbers 302 and 303 respectively. Thus we have
Number. Mantissa of log.
302 .4800 ) J
m .4806 \ [U
303 .4814 )
Hence, without regard to the decimal point, m = 302 + A = 3024 -|-.
Pointing off properly,
m = anti-log 0.4806 = 3,024+.
2. logm « 7.0959 - 10. m = ?
mantissa of log 124 = .0934 I ok )
mantissa of log m « .0959 ) [35
mantissa of log 125 = .0969 ^
Hence m has the sequence of figures
124 + U= 1247 +.
Pointing off properly, ^
m = anti-log (7.0959 - 10) =.001247+.
Note, The value of the quotient §5 may be obtained from the column of
Prop. Parts by finding the number of tenths of 35 required to equal 25. We
have from this column,
.7 X 35 = 24.5 and .8 X 35 = 28.0.
EXPLANATION OF TABLES 3i9
Hence we see that to make 25 we need a little more than . 7 X 35. A close
approximation would be .71+, making m = .0012471 -|-.
When the tabular difference is large, it is possible to obtain correctly more
than four significant figures of a number when its four-place logarithm is given.
Cologarithm. — The cologarithm of a number is the logarithm
of the reciprocal of the number.
Thus: colog m = log— = log 1 — log m = — log m.
In practice we usually write it in the form
colog m = — log )n = (10 — log m) — 10.
Rule. To form the cologarithm of a number, subtract its
logarithm from 10 and write — 10 after the result.
Examples,
1. colog 302 = (10 - log 302) - 10
= (10 - 2.4800) - 10 =- 7.520a- 10.
2. colog .003024 = (10 - log .003024) - 10
= (10 - [7.4806 - 101) - 10 = 2. 5194.
Use of the Cologarithm.
Example. Calculate the value of gj.i v 0R9k '
Let m be the value of the given fraction. Then without the use
of cologarithms the calculation is as follows.
log m = log 302 + log .415 - log 541 - log .0828.
log 302 = 2.4800 log 541 = 2.7332 '
log .415 = 9.6180 ~ 10 log .0828 = 8.9180 - 10
12.0980 - 10 ' 11.6512 - 10
11.6512 - 10
log m = 0.4468, m = 2.7975.
•
To use cologarithms, we write
m = 302 X .415 X v^ X
541 ^^ .0828
log m = log 302 + log. 415 + colog 541 + colog .0828
log 302= 2.4800
log .415 = 9.6180 - 10
colog 541 = 7.2668 - 10
colog .0828 = 1.0820
log m = 20.4468 - 20
m = 2.7975.
\
820 EXPLANATION OF TABLES
As a last example, we calculate the value of the quantity,
^ _ ^ /(.00812)i X (- 47L2)3
-v^g
(- 522.3)3 X (.01242)*
To take account of the signs, which must be done independ-
ently of the logarithmic calculation, we note that the cube of a
negative quantity occurs on both sides of the fraction; hence the
sign of the fraction is plus.
We now write
log m = i [log (.00812)* + log (471.2)3 + colog (522.3)3
+ colog (.01242)*].
log .00812 = 7.9096 - 10 log (.00812)* = 8.6064 - 10
log 471.2 = 2.6732 log (471.2)3 =8.0196
log 522.3 = 2.7179 log (522.3)3 = 8.1537
log .01242 = 8.0941 - 10 log (.01242)* = 8.5706 - 10
Hence
log (.00812) i =
log (471.2)3 =
colog (522.3)3 =
colog (.01242)i =
8.6064-
8.0196
1.8463-
1.4294
10
•10
2
19.9017 -
-20
log TO = .
m =
9.9508 -
.8929.
-10
Exercises. Verify the following equations:
1. log 7 = 0.8451. 10. log jU = 7.1158 - 10.
2. log 253 = 2.4031. 11. log (.0022)8 = 2.0272 - 10.
3. log 253.5 = 2.4040. 12. log ^JOO^ = 9.1141 - 10.
4. log .0253 = 8.4031 - 10. 13. log (.01401)* = 8.5171 - 10.
6. log .002533 = 7.4036 - 10 14. log (.0003684) J = 7.9820 - 20.
6. log 6544 = 3.8158. 16. colog 200 = 7.6990 - 10.
7. log 4.007 = 0.6028. 16. colog .7 = 0.1549.
8. log .9995 = 9.9998 - 10. 17. colog .0448 = 1.3487.
9. log V766 = 1.4421. 18. colog V5475 = 8.1308 - 10.
J
EXPLANATION OF TABLES 821
19. colog (.0003684)' = 12.0180. 26. \/-.0822 = - .4348.
20. antilog 1.2222 = 16 68. ^^ (_ ^213)? = 2.076.
21. antilog 3.6675 = 4650. /
22. antilog 0.4000 = 2.5118. 28 / ^ ^ • = - .11858.
23. antilog (8.3250 - 10) = .021135. * V- (.00475)
24. antilog (6.9525 - 10) = .0008964. 1
26. (.748)3 = .4185. ^^' (72.32)1 " '^^^^^'
TABLE II.
This table gives the logarithma of the sine, cosine, tangent and
cotangent of angles from 0° to 90°, at intervals of 10'.
When the angle is taken from the left-hand column of the page,
the name of the function must be sought at the top of the page;
when the angle is taken from the right-hand column of the page, the
name of the function must be sought at the foot of the page.
When the function is numerically less than 1, —10 must be
written after its tabular logarithm. This is the case with the
sines and cosines of all angles between 0° and 90°, with tangents ^
of angles oetween 0° and 45°, and with cotangents between 45°
and 90°.
For convenience in interpolation the diflferences of the tabular
logarithms are given, and these differences are subdivided into
tenths in the column of proportional parts. Hence this column
contains the corrections to the tabular logarithms for each minute
of angle from 1' to 9' inclusive. These corrections are to be
added when the logarithm increases with the angle, and they
are to be subtracted when the logarithm decreases as the angle
increases.
When the logarithm of a function of an angle greater than 90°
is required, change to the equivalent function of an angle less than
90° (139). Algebraic signs must be adjusted independently of
the logarithmic calculation, as in the use of Table I.
Seconds of arc must be reduced to the equivalent fractions of a
minute of arc.
To obtain log sec x, take from the table colog cos x\ for log
CSC X use colog sin x,
Examples.
1. log sin 20^ 13' = ?
log sin 20^ 10' = 9. 5375; d - 34.
d for 3' (Prop. Parts) = 10.2
log sin 20** 13' = 9. 5385 - 10.
822 EXPLANATION OF TABLES
2. log cos 20*' 13' = ?
log cos 20M0' = 9. 9725; d » 4.
d for 3' = 4 X .3 1.2
log cos 20** 13' = 9. 9724 - 10.
8. log tan 29** 47' = ?
log tan 29^ 40' = 9. 7556; d = 29.
d for 7' (Prop. Parts)' = 20.3
log tan 29^ 47' = 9. 7576 - 10
The same result may also be obtained by starting with log tan 29^ 50', thus:
log tan 29** 50' = 9. 7585; d = 29.
d for 3' = 8.7
log tan 29** 47' = 9. 7576 - 10.
As a rule, in interpolating start from the nearest tabular number.
4. log cot 29** 47' - ?
log cot 29** 50' = 0. 2415; d - 29.
d for 3' = 8.7
log cfjt 29** 47' = 0. 2424.
5. log sin 58** 44' = ?
log sin 58** 40' = 9. 9315; d = 8.
d for 4' = 3^
log sin 58** 44' = 9. 9318 - 10.
6. log tan 67** 23'.5 « ?
log tan 67** 20' = 0. 3792; d = 36.
d for 3'.5 = 10.8 + 1.8 = 12.6
log tan 67** 23'.5 = 0.3805.
Here we obtain d for 3'.5 from d for 3' + d for 0',5. Note that d for
0.5 is simply one-tenth of d for 5'.
7. log cos 105** 51'.6 = ?
cos 105** 51'.6 = - sin 15** 51'.6.
Neglecting the algebraic sign we have
log sin 15** 50' = 9.4359; d = 44.
d for 1'.6 = 7X)
log sin 15** 51'.6 = 9.4366 - 10 = log cos 105** 51'.6.
8. log tan 250** 34' .3 = ?
tan 250** 34'.3 = tan 70** 34'.3.
log tan 70** 30' - 0.4509; d = 40.
d for 4' .3 = 17.2
log tan 70** 34'.3 = 0.4526 = log tan 250** 34'.3.
EXPLANATION OF TABLES 828
Angles near 0"^ or near wf.
When an angle, x, lies near 0°, sin x, tan x, and cot x vary too
rapidly with x to permit of accurate interpolation of their loga-
rithms from the table. The same is true of cos x, tan x, and cot x,
when X lies near 90°. We will show how accurate values of these
logarithms may be obtained.
Let S = log and T = log -^ — i
X ^ X
X being expressed in minutes of arc.
Then log sin x = log x' + S,
and log tan x = log x' + T.
When x is small the quantities 8 and T vary quite slowly with x.
The values of S and T are given in the last column of the first
page of Table II, x ranging from 0° to 5°; —10 is to be added to
the tabular numbers there given.
To get log sin x, reduce x to minutes of arc and take log a:' from
Table I; to this logarithm add S,
To get log tan x, add T to log x\
To get log cot X, first get log tan x and form the cologarithm of
the result.
For, log cot X = colog tan x.
To obtain log cos x, log tan x or log cot x, when x lies between
85° and 90°, calculate the co-function of the complementary angle
by the method given above.
To find f he angle from log sin x, log tan x or log cot x, when x
lies near 0°, we use the relations
log x' = log sin X — S;
log x' = log tan X — T;
log x' = — log cot X — r.
The necessary values of S and T can be obtained after finding
an approximate value of x from Table 11.
To find X from log cos x, log tan x, or log cot x, when x lies near
90°, replace
log cos X by log sin (90° — x) ;
log tan X by log cot (90° — x) ;
log cot X by log tan (90° — x).
824 EXPLANATION OF TABLES
Then 90° — x can be obtained by the method given above for
angles near 0°. Hence x is determined.
Examples.
1. Find log sill a;, log tan x and log cot x when a; = 1® 22' 12".
a; = 1** 22' 12" = 82' .2. log x' = log 82.2 = 1.0149.
logx = 1.9149 log a; = 1.9149
S = 6.4637 -10 T = 6.4638 - 10
log sin aj^= 8. 3786 - 10 log tan x = 8. 3787 - 10
log cot x = colog tan a; = 1 .6213.
2. Find log cos x, log tan x and log cot x when x = 89° 5' 50".
Let y •= 90** - a; = 54' 10" = 54'.17.
Then log cos x, log tan x, log cot x are equal respectively to log sin y, log cot y,
log tan 2/, which may be found as in example 1.
3. log sin X = 8.2142; a; = ?
From Table II, a; = 50' + ; hence S = 6.4637 - 10.
log sin X = 8.2142 - 10
S = 6.4637 - 10
loga;' = 1.7505; x = 56'.30 = 56' 18".
4. log tan a; = 8.0804 - 10; x = ?
Prom Table II, x = 40'+ ; hence T = 6.4638
log tan X = 8.0804 - 10
T = 6.4638 - 10
log x' = 1.6166; X = 41 '.36 = 41' 21".6.
6. log cot X = 8.6276 - 10; x = ?
Let y = 90** - X.
Then log tan y = log cot x = 8.6276 - 10.
From Table II, y = 2** 20'+ ; hence T = 6.4640.
log tan 2^ = 8.6276 - 10
T = 6.4640 - 10
log 2/' = 2.1636; y = 145'.73 = 2** 25' 44".
Hence x = 90° - y = 87** 34' 16".
Let the student obtain the results required in' the last five
examples by direct interpolation from Table II.
Exercises. Verify the following equations:
1. log sin 20** 40' = 9.5477 - 10. 10. log cos 81** 29' = 9.1706 - 10.
2. log cos 66** 30' = 9.6007 - 10. 11. log cos 81** 31' = 9.1689 - 10.
3. log tan 29** 35' = 9.7541 - 10. 12. log cot 9** 6' = 0.7954.
4. log cot 37** 25' = 0.1163. 13. log sin 152** 27' = 9.6651 - 10.
6. log sec 55^ 50' = 0.2506. 14. log sin 2** 10' 10" = 8.5781 - 10.
6. log CSC 44** 50' = 0.1518. 16. log tan 1** 34' 20" = 8.4385 - 10.
7. log tan 63** 27' = 0.3013. 16. log cot 0** 10' 22" = 2.5206.
8. log sin 81** 29' = 9.9952. 17. log cos 89** 28' 44 " = 7.9588 - 10.
9. log sin 81** 31' =9.9952. 18. log tan 88** 46' 14"= 1.6683.
EXPLANATION OF TABLES
325
19.
log sin X
= 0.7926,
! ^
= 38** 20'.
20.
log sin X
= 9.3548;
X
= 13° 5'.
2L
log sin X
-9.8867
; X
= 50° 23'.
22.
log cos X
= 9.6030
; X
= 66° 22'.
23.
log tan X
= 0.6278
\ ^
= 77° 44'.5.
24.
log cot X
= 0.0906
1 ^
= 39° 4'.
26.
log cot X
= 0.6648;
X
= 12° 12'. 5.
26.
log sec X
= 0.1374;
; X
= 43° 13'.
27.
log CSC X
= 0.2890-
; X
= 30° 56'.
28.
log sec X
= 0.6680,
; ^
= 77°35'.S.
29.
log sin X
= 8.3698;
X
= 1° 20' 34".
30.
log tan X
= 8.7659;
) ^
= 3° 20' 18".
31.
log cot X
= 1.2952;
X
= 2° 54' 3".
32.
log cos X
= 8.5387;
X
= 88° 1' 8".
33.
log cot X
= 7.9485;
X
= 89° 29' 28".
34.
log CSC X
= 2.3549
; X
= 0° 15' 11".
36.
log sec X
= 1.5102;
\ ^
= 88° 13' 48".
TABLE m
This table gives the numerical values of the six trigonometric
functions of angles from 0° to 90° at intervals of 10'. The func-
tions of intermediate angles are to be obtained by interpolation.
By using the tables inversely, an angle may be found, usually
to the nearest minute, when a function of the angle is known to
four decimal places.
TABLE IV
This is a conversion table for changing from sexagesimal to
radian measure, and conversely. The entries are given to five
decimal places in radians, corresponding nearly to 2" in sexagesi-
mal measure.
Examples,
1. Express 200° 44' 36" in radian measure.
200° = 3 X 60° + 20°
3 X 60° = 3 X 1.04720 = 3.14160 radians.
20° = 0.34907
44' = 0.01280
36" = 0.00017
200° 44' 36" = 3.50364 radians.
2. Express 3.50364 radians in sexagesimal measure.
3.0 radians = 171° 53' 14"
0.5 " = 28° 38' 52"
0.003 " = 10' 19"
0.0006 " = 2' 4"
0.00004 " = 8"
3.50364 radians = 200° 44' 37"
326 EXPLANATION OF TABLES
TABLE V
This table contains the values of a number of mathematical
constants, generally to fifteen places of decimals.
TABLE VI
This table gives the values of the natural or Naperian loga-
rithm of a;, and of the ascending and descending exponential
functions e* and e"*, from x = 0toa: = 5at intervals of 0.05.
As a rule the tabular entries are given to three decimal places.
TABLE Vn
This table gives the values of n^, v?, Vn, and Vn, for values of
n from 1 to 100.
The direct use of the table requires no explanation. A s an
example of its inverse use we find the approximate value of V320.
We have
(6.8)3 = 314.432 (n = 68),
(6.9)3 ^ 328.509 (n = 69).
Hence, interpolating linearly,
(6.840)3 = 320 approx., or V320 = 6.840+.
TABLES
328
TABLE I. LOGARITHMS OF NUMBERS
Ho.
10
11-
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
60
51
52
53
54
Vo.
0000
0414
0792
1139
1461
1761
2041
2304
2553
2788
3010
3222
3424
3617
3802
3979
4150
4314
4472
4624
4771
4914
5051
5185
5315
5441
5563
5682
5798
5911
6021
6128
6232
6335
6435
6532
6628
6721
6812
6902
6990
7076
7160
7243
7324
0043
045a
0828
1173
1492
1790
2068
2330
2577
2810
3032
3243
3444
3636
3820
3997
4166
4330
4487
4639
4786
4928
5065
5198
5328
5453
5575
5694
5809
5922
6031
6138
6243
6345
6444
6542
6637
6730
6821
6911
6998
7084
7168
7251
7332
0086
049*2
0864
1206
1523
1818
2095
2355
2601
2833
3054
3263
3464
3655
3838
4014
4183
4346
4502
4654
4800
4942
5079
5211
5340
5465
5587
5705
5821
5933
6042
6149
6253
6355
6454
6551
6646
6739
6830
6920
7007
7093
7177
7259
7340
8
0128
0531
0899
1239
1553
1847
2122
2380
2625
2856
3075
3284
3483
3674
3856
4031
4200
4362
4518
4669
4814
4955
5092
5224
5353
5478
5599
5717
5832
5944
6053
6160
6263
6365
6464
6561
6656
6749
6839
6928
7016
7101
7185
7267
7348
3
0170
0569
0934
1271
1584
1875
2148
2405
2648
2878
3096
3304
3502
3692
3874
4048
4216
4378
4533
4683
4829
4969
5105
5237
5366
5490
5611
5729
5843
5955
6064
6170
6274
6375
6474
6571
6665
6758
6848
6937
7024
7110
7193
7275
7356
0212
0607
0969
1303
1614
1903
2175
24Q0
2672
2900
3118
3324
3522
3711
3892
4065
4232
4393
4548
4698
4843
4983
5119
5250
5378
5502
5623
5740
5855
5966
6075
6180
6284
6385
6484
6580
6675
6767
6857
6946
7033
7118
7202
7284
7364
6
0253
0645
1004
1335
1644
1931
2201
2455
2695
2923
3139
3345
3541
3729
3909
4082
4249
4409
4564
4713
4857
4997
5132
5263
5391
5514
5635
5752
5866
5977
6085
6191
6294
6395
6493
6590
6684
6776
6866
6955
7042
7126
7210
7292
7372
6
0294
0682
1038
1367
1673
1959
2227
2480
271«
2945
3160
3365
13560
3747
3927
4099
4265
4425
4579
4728
4871
5011
5145
5276
5403
5527
5647
5763
5877
5988
6096
6201
6304
6405
6503
6599
6693
6785
6875
6964
7050
7135
7218
7300
7380
8
0334
0719
1072
1399
1703
1987
2253
2504
2742
2967
3181
3385
3579
3766
3945
4116
4281
4440
4594
4742
4886
5024
5159
5289
5416
5539
5658
5775
5888
5999
6107
6212
6314
6415
6513
6609
6702
6794
6884
6972
7059
7143
7226
7308
7388
8
9
0374
0755
1106
1430
1732
2014
2279
2529
2765
2989
3201
3404
3598
3784
3962
4133
4298
4456
4609
4757
4900
5038
5172
5302
5428
5551'
5670
5786
589%
6010
6117
6222
6325
6425
6522
6618
6712
6803
6893
6981
7067
7152
7235
7316
7396
9
Prop. Part!
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
1
2
3
6
5
6
7
8
9
43
4.3
8.6
12.9
17.2
21.5
25.8
30.1
34.4
38.7
41
4.1
8.
12.
16
20.
24.6
28.7
32.8
36.9
.2
.3
.4
.5
39
3.9
7.8
11.7
15.6
19.5
23.4
27 3
31.2
35.1
37
3.7
7.4
11.1
14.8
18.5
22.2
25.9
29.6
33.3
36
3.5
7.0
10.5
14.0
17.5
21
24
28
31
33
3.3
6.6
9.9
13.2
16.5
19.8
23.1
26.4
29.7
42
4.2
8.4
12.6
16.8
21.0
25.2
29.4
33.6
37.8
40
4.0
8.0
12.0
16.0
20.0
24.0
28.0
32.0
36.0
38
3.8
7.6
11.4
15.2
19.0
22.8
26.6
30.4
34.2
36
3.6
7.2
10.8
14
18
21
25.2
28.8
32.4
84
3.4
6.8
10.2
13.0
17.0
20.4
23.8
27.2
30.0
32
3.2
6.4
9.6
12.8
16.0.
19.2
22.4
25.6
28.8
Prop. Parta
LOGARITHMS OF NUMBERS. TABLE I
329
I'll
No.
55
1
2
7419
3
4
6
6
7
8
9
7474
Prop. Porta
Wioii
7404
7412
7427
7435
7443
7451
7459
7466
31
30
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
1
3.1
6.2
9.3
3.0
6.0
9.0
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
2
3
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
4
12.4
4 Cf tf
12.0
4 tf A
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
5
6
15.5
18.6
15.0
18.0
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
7
8
21.7
24.8
21.0
24.0
61
62
7853
7924
7860
7931
7868
7938
7875
7945
7882
7952
7889
7959
7896
7966
7903
7973
7910
7980
7917
7987
9
27.9
27.0
tf%j%
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
1
29
2.9
28
2.8
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
2
5.8
O 99
5.6
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
3
4
8.7
11.6
8.4
11.2
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
5
14.5
IT t
14.0
16.8
19.6
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
7
17.4
20.3
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
8
23.2
22.4
69
70
8388
8451
8395
8457
8401
8463
8407
8470
8414
8476
8420
8482
8426
8488
8432
8494
8439
8500
8445
8506
9
26.1
27
25.2
26
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
1
2
2.7
5 4
2.6 1
5.2 1
72
8573
8579
858^
8591
8597
8603
8609
8615
8621
8627
St
3
^0 • ^
8.1
7.8 ]
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
4
5
10.8
13.5
•10.4 . 1
13.0 !
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
6
16.2
10 A
15.6 1
10 A 1
^
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
7
8
18.9
21.6
18.2 1
20.8 1
76
77
8808
8865
8814
8871
8820
8876
8825
8882
8831
8887
8837
d893
8842
8899
8848
8904
8854
8910
8859
8915
9
24.3
23.4 1
9»
A^ M
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
4
26
24
n A 1
»?
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
1
2
2.5
5.0
2.4 I
4.8 1
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
3
4
7.6
10.0
7.2 j
9.6 !
•
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
5
A
12.5
15.0
12.0 I
14 4 j
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
7
17.5
16.8 j
83
9191
9196
9201
9253
9206
9258
9212
9217
9269
9222
9274
9227
9279
9232
9284
9238
9289
8
9
20.0
22.5
19.2 1
21.6 I
84
9243
9248
9263
■
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
23
- 22 1
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
1
2.3
A A
2.2 1
A A
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
2
3
4.6
6.9
4.4
6.6
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
4
9.2
4 4 ar
8.8 i
4 4 t^ 1
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
5
6
11.5
13.8
11.0 1
13.2 j
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
7
8
16.1
18.4
15.4 1
17.6
91
92
9590
9638
9595
9643
9600
9647
9605
9652
9609
9657
9614
9661
9619
9666
9624
9671
9628
9675
9633
9680
9
20.7
19.8 j
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
1
2
21
2 1
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
4.2
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
3
4
5
6.3
A 4
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
o. *
10.5
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
6
7
12.6
14.7
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
8
16.8
99
9956
9961
9965
9969
9974
9978
9983
9987
7
9991
9996
9
18.9
No.
1
2
3
4
6
6
8
1
9
Prop. Parta
I
i 330 TABLE II. LOGARITHMIC SINES, COSINES,
TANGENTS AND COTANGENTS. TABLE II
331
— (
I
i;
►
30'
40'
50'
10'
20'
30'
40'
50'
9° 0'
10'
20'
30'
40'
50'
10° 0'
10'
20'
30'
40'
50'
11** C
10'
20'
30'
40'
50'
12° C
10'
20'
30'
40'
50'
13° C
10'
20'
30'
40'
50'
14° 0'
10'
20'
30'
40'
50'
16° C
log lin
9.1157
.1252
.1345
9.1436
.1525
.1612
.1697
.1781
. 1863
9.1943
.2022
.2100
.2176
.2251
.2324
9 . 2397
.2468
.2538
.2606
.2674
.2740
9 . 2806
.2870
.2934
.2997
.3058
.3119
9.3179
.3238
.3296
.3353
.3410
.3466
9.3521
.3575
.3629
.3682
.3734
.3786
9 . 3837
.3887
.3937
.3986
.4035
.4083
9.4130
log cos
95
93
91
89
87
85
84
82
80
79
78
76
75
73
73
71
70
68
68
66
66
64
64
63
61
61
60
59
58
57
57
56
55
54
54
53
52
52
51
50
50
49
49
48
47
log cos
9.9963
.9961
.9959
9.9958
.9956
.9954
.9952
.9950
.9948
9.9946
.9944
.9942
.9940
.9938
.9936
9.9934
.9931
.9929
.9927
.9924
.9922
9.9919
.9917
.9914
.9912
.9909
.9907
9.9904
.9901
.9899
.9896
.9893
.9890
9.9887
.9884
.9881
.9878
.9875
.9872
9.9869
.9866
.9863
.9859
.9856
.9853
9 . 9849
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
3
2
2
3
2
3
2
3
2
3
2
3
3
2
3
3
3
3
3
3
3
3
3
3
3
3
4
3
3
4
log tan
log sin d
9.1194
.1291
.1385
9.1478
.1569
.1658
.1745
.1831
.1915
9.1997
.2078
.2158
.2236
.2313
.2389
9.2463
.2536
.2609
.2680
.2750
.2819
».2887
.2953
.3020
.3085
.3149
.3212
9.3275
.3336
.3397
.3458
.3517
.3576
9.3634
.3691
.3748
.3804
.3859
.3914
9.3968
.4021
.4074
.4127
.4178
.4230
9.4281
log cot
97
94
93
91
89
87
86
84
82
81
80
78
77
76
74
73
73
71
70
69
68
66
67
65
64
63
63
61
61
61
59
59
58
57
57
56
55
55
54
53
53
53
51
52
51
log oot
. 8806
.8709
.8615
0.8522
.8431
.8342
.8255
.8169
.8085
0.8003
.7922
.7842
.7764
.7687
.7611
. 7537
.7464
.7391
.7320
.7250
.7181
0.7113
.7047
.6980
.6915
.6851
.6788
. 6725
.6664
.6603
.6542
.6483
.6424
0.6366
.6309
.6252
.6196
.6141
.6086
0.6032
.5979
.5926
.5873
.5822
.5770
0.5719
log tan
30'
20'
10'
82° C
50'
40'
30'
20'
10'
81° C
50'
40'
30'
20'
lO'
80° 0'
50'
40'
30'
20'
10'
79° 0'
50'
40'
30'
20'
10'
78° 0'
50'
40'
30'
20'
10'
77° 0'
50'
40'
30'
20'
10'
76° 0'
50'
40'
30'
20'
10'
76° 0'
Prop. Parts
73
7.3
14.6
21.9
4t29.2
36.6
43.8
51.1
58.4
65.7
71
7.1
67
6.7
13.4
20
26
33.5
40.2
46.9
53.6
60.3
14
21
28
35
42.6
49.7
56.8
63.9
61
6.1
12.2
18.3
24.4
30.5
36.6
42.7
48.8
54.9
66
5.6
11.2
16.8
22.4
28.0
33.6
39.2
44.8
50.4
61
5.1
10.2
15.3
20
25
30
35
40
45.9
66
6.6
13.2
19.8
26.4
33.0
39.6
46.2
52.8
59.4
70
7.0
14.0
21.0
28.0
35.0
42.0
49.0
56.0
63.0
69
6.9
13.8
20.
27.6
34.
41.
48.
55.
68
6.8
13.6
62.1
60
6.0
12
18
24
30
36.0
42.0
48.0
54.0
66
5.5
11.0
16.5
22.0
27.5
33.0
38.5
44.0
49.5
60
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.5
66
6.5
13.0
19
26
32
39
45.5
52.0
58.5
69
5.9
11.8
17.7
23.6
29.5
35.4
41.3
47.2
53.1
M
5.4
10.8
16.2
21.6
27.0
32.4
37.8
43.2
48.6
48
4.8
9.6
14.4
19.2
24.0
28.8
33.6
38.4
43.2
64
6.4
12.8
19.2
25.6
32.0
38.4
44.8
51.2
57.6
20
27,
34.0
40.8
47.6
54.
61.
63
6 3
12.6
18.9
25.2
31.5
37.8
44.1
50.4
56.7
68
5.8
11.6
17.4
23.2
29.0
34.8
40.6
46.
52.
63
5.3
10.6
15.9
21.2
26
31
37
42
47
47
4.7
9.
14.
18.8
23.5
28.2
32.9
37.6
42.3
67
5.7
11.4
17.1
22.8
28.5
34.
39.9
45.6
51.
62
5.2
10.4
15.6
20.8
26.0
31.2
36.4
41.6
46.8
97
94
93
91
89
87
86
86
84
82
81
79
78
77
76
76
74
9.7
9.4
9.3
9.1
8.9
8.7
8.6
8.5
8.4
8.2
8.1
7.9
7.8
7.7
7.6
7.5
7.4
19.4
18.8
18.6
18.2
17.8
17.4
17.2
17.0
16.8
16.4
16.2
15.8
15.6
15.4
15.2
15.0
14.8
29.1
282
27.9
27.3
26.7
26.1
25.8
25.5
25.2
24.6
24.3
23.7
23.4
23.1
22.8
22.5
22.2
38.8
37.6
37.2
36.4
35.6
34.8
34.4
34.0
33.6
32.8
32.4
31.6
31.2
30.8
30.4
30.0
29.6
48.5
47.0
46.5
45.5
44.5
43.5
43.0
42.5
42.0
41.0
40.5
39.5
39.0
38.5
38.0
37.5
37.0
6S.2
56.4
55.8
54.6
53.4
52.2
51.6
51.0
50.4
49.2
48.6
47.4
46.8
46.2
45.6
45.0
44.4
67.9
65.8
65.1
63.7
62.3
60.9
60.2
59.5
588
57.4
56.7
55 3
54.6
53.9
53 2
52.5
51.8
77.6
75.2
74.4
72.8
71.2
69.6
68.8
68.0
67.2
65.6
64.8
63.2
62.4
61.6
60.8
60.0
59.2
87.3
84.6 83.7 181.9 180.1 178.3
77.4 76.5 1 75.6173.8172.9
71.1
70.2
69.3
68.4 67.5166.6
Log.trig.
fnnct's.
0** -'15°
90** - 75°
332 TABLE II. LOGARITHMIC SINES, COSINES,
r
log sinL d
0'
10'
20'
30'
40'
60'
^6° 0'
10'
20'
\
30'
40'
50'
.%7«> 0'
10'
20'
30'
40'
50'
18° 0'
10'
20'
30'
40'
50'
'19« 0'
10'
20'
30'
40'
50'
T0« 0'
10'
20'
30'
40'
50'
:l« 0'
10'
20'
30'
40
50'
?;.=' 0'
10'
20'
30'
9.4130
.4177
.4223
.4269
.4314
.4359
9.4403
.4447
.4491
.4533
.4576
.4618
9.4659
.4700
.4741
.4781
.4821
.4861
9.4900
.4939
.4977
.5015
.5052
.5090
9.5126
.5163
.5199
.5235
.5270
.5306
9.5341
.5375
.5409
.5443
.5477
.5510
9.5543
.5576
.5609
.5641
.5673
.5704
9.5736
.5767
.5798
.5828
log cos
47
46
46
45
45
44
44
44
42
43
42
41
41
41
40
40
40
39
39
38
38
37
38
36
37
36
36
35
36
35
34
34
34
34
33
33
33
33
32
32
31
32
31
31
30
log COS
9.9849
.9846
.9843
.9839
.9836
.9832
9.9828
.9825
.9821
.9817
.9814
.9810
9.9806
.9802
.9798
.9794
.9790
.9786
9.9782
.9778
.9774
.9770
.9765
.9761
9.9757
.9752
.9748
.9743
.9739
.9734
9.9730
.9725
.9721
.9716
.9711
.9706
9.9702
.9697
.9692
.9687
.9682
.9677
9.9672
.9667
.9661
.9656
log sin
3
3
4
3
4
4
3
4
4
3
4
4
4
4
4
4
4
4
4
4
4
5
4
4
5
4
5
4
5
4
5
4
5
5
5
4
5
5
5
5
5
5
5
6
5
log tan
9.4281
.4331
.4381
.4430
.4479
.4527
9.4575
.4622
.4669
.4716
.4762
.4808
9.4853
.4898
.4943
.4987
.5031
.5075
9.5118
.5161
.5203
.5245
.5287
.5329
9.5370
.5411
.5451
.5491
.5531
.5571
9.5611
.5650
.5689
.5727
.5766
.5804
9 . 5842
.5879
.5917
.5954
.5991
.6028
9 . 6064
.6100
.6136
.6172
log cot
50
50
49
49
48
48
47
47
47
46
46
45
45
45
44
44
44
43
43
42
42
42
42
41
41
40
40
40
40
40
39
39
38
39
38
38
37
38
37
37
37
36
36
36
36
log cot
6719
.5669
.5619
.5570
.5521
.5473
1.5425
.5378
.5331
.5284
.5238
.5192
1.5147
.5102
.5057
.5013
.4969
.4925
4882
4839
4797
,4755
4713
,4671
,4630
.4589
.4549
.4509
.4469
.4429
.4389
.4350
.4311
.4273
.4234
.4196
.4158
.4121
.4083
.4046
.4009
.3972
.3936
.3900
.3864
.3828
log tan
0,
Prop. Parts
76°
50
40
30
20
10
74^0
50
40
30
20
10
73°
50
40
30
20
10
72°
50
40
30
20
10
71°
50
40
30
20
10
70°
50
40
30
20
10
69° C
50
40
30
20
10
68°
50
40
30
60
5.0
10.
16.
20.
25.
30.
35.
40.
45.
46
4.6
9
13
18
23
27.6
32.2
36.8
41.4
42
4.2
8.4
12.6
16.8
21.0
25.2
29.4
33.6
37.8
38
3.8
7.6
11.4
15.2
19.0
22.8
26.6
30.4
34.2
34
3.4
6.8
10.2
13.6
17.0
20.4
23.8
27.2
30.6
49
4.9
9.8
14
19
24
29.
34.
39.
44.
48
4.8
9.6
14.4
19.2
24.0
28.8
33.6
38.4
43.2
47
4.7
9.4
14.1
18.8
23.5
28.2
32.9
37.6
42.3
46
4
9
13
18
22.5
27.0
31.5
36.0
40.5
.5
.0
.5
.0
44
4.4
8.8
13.
17.6
22.0
26.
30.8
35.2
39.6
43
4.3
8.6
12.9
17.2
21.5
25.8
30.1
34.4
38.7
41
4.1
8.2
12.3
16.4
20.5
24.6
28.7
32.8
36.9
40
4.0
8.0
12.0
16.0
20.0
24.0
28.0
32.0
36.0
39
3.9
7.
11.
15.6
19
23
27
31.2
35.1
37
3.7
7.
11.
14.8
18.
22.
25.9
29.6
33.
36
3.6
7.2
10.8
14
18
21
25.2
28.8
32.4
36
3.5
7.0
10.5
14.0
17.5
21.0
24.5
28.0
31.5
33
3.3
6.6
9.9
13.2
16.5
19.8
23.1
26.4
29.7
32
3.J
6.-!
9.6
12.
16.
19.2
22.4
25.6
28.8
31
3.1
6.2
9.3
12.4
15.5
18.6
21.7
24.8
27.9
Prop. Parts
TANGENTS AND COTANGENTS. TABLE H
X
logiln
d la
««»
d
logUn
d
logoot
Prop.PMU 1
30'
9.5S28
31 "
30
30
9656
9.6172
36
35
36
0.3828
30'
36
36
34
40'
.5859
9651
.6208
.3792
20'
60'
.5389
9646
.6243
.3757
10'
\
11
J;
'A
23= C
9.5919
29 «
30
29
9640
9.6279
35
34
35
0.3721
87= C
10'
.5948
9635
.6314
.3886
50'
30'
.5978
9629
.6348
.3652
40'
7
as!!
1
238
30'
.6007
29
29
28
9624
.6383
34
35
34
.3617
30'
g
308
M'
.6036
9618
.6417
.3583
20'
50'
.6065
9613
.6452
.3548
10'
w <y
0.6093
28 »
28
28
9607
9.6486
34
33
34
0.3514
se° c
10'
.6121
9602
.6520
.3480
50'
83
"M
~3i
20'
.6149
9596
.6553
.3447
10'
\
s«
3.;
\\
30'
.6177
28
27
27
9590
.6587
33
w
.3413
30'
ill
ll
12:4
10'
.6205
9584
.6620
.3380
20'
A
SO'
.6232
9579
.6654
.3346
10'
b!
^■'
18,«
26° V
9.6269
27 8
27
27
9573
9.6687
33
32
33
0.3313
66° C
g
't,A
¥
10'
.6286
9567
.6730
.3280
50'
e
28.8
27. B
20'
.6313
9561
.6752
.3248
40'
30'
.6340
26
26
26
9555
.6785
33
33
32
.3215
30'
40'
.6366
9549
.6817
.3183
20'
SO
^F
~K
50'
.6392
9543
.6850
,3150
10'
1
11
ae= ff
9.641S
26 8
26
25
9537
9.6882
32
32
31
0.311S
64° 0-
2
B.
i:
\\
10'
.6444
9530
.6914
.3086
50'
20'
.6470
9524
.6946
.3054
40'
s
l-i
1!:
IBS
30'
.6495
26
25
24
9518
.6977
32
31
32
.3023
30'
7
!7!0
20,
2«:
19 t
25,2
«'
.6521
9512
.7009
.2991
20'
so-
.6546
9505
.7040
.2960
10'
ar ff
9.6570
2S •
25
24
9499
9.7072
31
31
31
0.2928
6S= C
10'
20'
.6595
.6620
9492
9486
.7103
.7134
.2897
.2866
50'
40'
27
"M
"26"
30'
.6644
24
24
24
9479
.7165
31
30
31
.2836
30'
b.
J
40'
.6668
9473
.7196
.2804
20'
!:
I
10^0
2,S
60'
.6692
9466
.7226
.2774
10'
28- C
9.6716
24 «
23
24
9459
9.7257
30
30
31
0,2743
63° V
\
1
5.0
10'
.6740
9453
.7287
.2713
60'
a
I.
200
20'
.6763
9446
.7317
.2683
M'
a
u.
23:
22.6
30'
.6787
23
23
23
9439
.7348
30
30
30
.2662
30*
40'
.6810
9432
.7378
.2622
20'
50'
.6833
9425
7
.7408
.2592
10'
—
24
"M
"22
KP V
9.6856
8'
22
9418
9.7438
29
30
29
0.2562
81° (r
10'
.6878
9411
.7467
.2533
50'
6.
6.S
20'
.6901
9404
.7497
.2503
40'
*
30'
.6933
23
22
9397
.7526
30
29
.2474
30'
e
\.
13^2
40'
.6946
9390
.7556
.2444
20'
fl.
ifl
50'
.6968
9383
.7585
.2415
10'
334 TABLE n. LOGARITHMIC SINES, CX)SINES,
TANGENTS AND COTANGENTS. TABLE II
i
336
TABLE in. NATURAL FUNCTIONS
X
lin X
eoi X
tan X
OOt X
■00 X
00000 jr
Qo 0'
10'
20'
.00000
.00291
.00582
1.0000
1.0000
1.0000
.00000
.00291
.00582
00
343.77
171.88
1.0000
1.0000
1.0000
00
343.78
171.89
50'
40'
30'
40'
50'
.00873
.01164
.01454
1.0000
.9999
.9999
.00873
.01164
.01455
114.59
85.940
68.750
1.0000
1.0001
1.0001
114.69
85.946
68.757
30'
20'
10'
10'
20'
.01745
.02036
.02327
.9998
.9998
.9997
.01746
.02036
.02328
57.290
49.104
42.964
1.0002
1.0002
1.0003
57.299
49.114
42.976
89'' C
50'
40'
30'
40'
50'
.02618
.02908
.03199
.9997
.9996
.9995
.02619
.02910
.03201
38.188
34.368
31.242
1.0003
1.0004
1.0005
38.202
34.382
31.258
30'
20'
10'
go 0'
10'
20'
.03490
.03781
.04071
.9994
.9993
.9992
.03492
. 03783
.04075
28.6363
26.4316
24.5418
1.0006
1.0007
1.0008
28.654
26,451
24.562
88° C
50'
40'
30'
40'
50'
.04362
.04653
.04943
.9990
.9989
.9988
.04366
.04658
.04949
22.9038
21.4704
20.2056
1.0010
1.0011
1.0012
22.926
21.494
20.230
30'
20'
10'
8° 0'
10'
20'
.05234
.05524
.05814
.9986
.9985
.9983
.05241
. 05533
.05824
19.0811
18.0750
17.1693
1.0014
1.0015
1.0017
19.107
18.103
17.198
87° 0'
50'
40'
30'
40'
50'
.06105
.06395
.06685
.9981
.9980
.9978
.06116
.06408
.06700
16.3499
15.6048
14.9244
1.0019
1.0021
1.0022
16.380
15.637
14.958
30'
20'
10'
4° (T
10'
20'
.06976
.07266
.07656
.9976
.9974
.9971
.06993
.07285
.07578
14.3007
13.7267
13.1969
1.0024
1.0027
1.0029
14.336
13-.763
13.235
86** (K
50'
40'
30'
40'
50'
.07846
.08136
.08426
.9969
.9967
.9964
.07870
.08163
.08456
12.7062
12.2505
11.8262
1.0031
1.0033
1.0036
12.746
12.291
11.868
30'
20'
10'
5'> 0'
10'
20'
.08716
.09005
.09295
.9962
.9959
.9957
.08749
.09042
.09335
11.4301
11.0594
10.7119
1.0038
1.0041
1.0044
11.474
11.105
10 . 758
85° C
50'
40'
30'
40'
50'
.09585
.09874
. 10164
.9954
.9951
.9948
. 09629
.09923
.10216
10 . 3854
10.0780
9 . 7882
1.0046
1.0049
1.0052
10.433
10.128
9.839
30'
20'
10'
6«> C
10'
20'
. 10453
. 10742
.11031
.9945
.9942
.9939
. 10510
. 10805
.11099
9.5144
9.2553
9.0098
1.0055
1.0058
1.0061
9.5668
9.3092
9.0652
84° (K
50'
40'
30'
40'
50'
.11320
.11609
.11898
.9936
.9932
.9929
.11394
.11688
.11983
8.7769
8 . 5555
8.3450
1.0065
1.0068
1.0072
8.8337
8.6138
• 8.4647
30'
20'
10'
T 0'
10'
20'
.12187
. 12476
. 12764
.9925
.9922
.9918
. 12278
. 12574
. 12869
8.1443
7.9530
7.7704
1.0075
1.0079
1.0083
8.2055
8.0157
7.8344
83° 0'
50'
40'
30'
.13053
.9914
.13165
7.6958
1.0086
7.6613
30'
001 X
iin X
oot X
tan X
ooioe X
iOO X
X
NATURAL FUNCTIONS. TABLE III
337
X
■mjr
eo8 jr
tan X
OOt X
lae X
eofeo X
30'
40'
50'
.1305
.1334
.1363
.9914
.9911
.9907
.1317
.1346
.1376
7 . 5958
7.4287
7.2687
1.0086
1.0090
1.0094
7.6613
7.4957
7.3372
30'
20'
10'
8°
0'
10'
20'
.1392
.1421
.1449
.9903
.9899
.9894
.1405
.1435
.1465
7.1154
6.9682
6.8269
1.0098
1.0102
1.0107
7.1853
7.0396
6.8998
82° 0'
50'
40'
30'
40'
50'
.1478
.1507
.1536
.9890
.9886
.9881
.1495
.1524
.1554
6.6912
6.5606
6.4348
1.0111
1.0116
1.0120
6.7655
^7^363
•6.5121
30'
20'
9°
0'
10'
20'
.1564
.1593
.1622
.9877
.9872
.9868
.1584
.1614
.1644
6.3138
6.1970
6.0844
1.0125
1.0129
1.0134
6.3925
6.2772
6.1661
81° C
50'
40'
30'
40'
50'
.1650
.1679
.1708
.9863
.9858
.9853
.1673
.1703
.1733
^5.9758
5.8708
5.7694
1.0139
1.0144
1.0149
6.0589
5.9554
5.8554
30'
20'
10'
10°
0'
10'
20'
.1736
.1765
.1794
.9848
.9843
.9838
.1763
.1793
.1823
5.6713
5.5764
5.4845
1.0154
1.0160
1.0165
5.7588
5.6653
5.5749
80° 0'
50'
40'
30'
40'
50'
.1822
.1851
.1880
.9833
.9827
.9822
.1853
.1883
.1914
5.3955
5.3093
5.2257
1.0170
1.0176
1.0182
5.4874
5.4026
5.3205
30'
20'
10'
11°
0'
10'
20'
.1908
.1937
.1965
.9816
.9811
.9805
.1944
.1974
.2004
5.1446
5.0658
4.9894
1.0187
1.0193
1.0199
5.2408
5.1636
5.0886
79° 0'
50'
40'
30'
40'
50'
.1994
.2022
.2051
.9799
.9793
.9787
.2035
.2065
.2095
4.9152
4.8430
4.7729
1.0205
1.0211
1.0217
5.0159
4.9452
4.8765
30'
20'
10'
12°
0'
10'
20'
.2079
.2108
.2136
.9781
.9775
.9769
.2126
.2156
.2186
4.7046
4.6382
4.5736
1.0223
1.0230
1.0236
4.8097
4.7448
4.6817
78° 0'
50'
40'
30'
40'
50'
.2164
.2193 •
.2221
.9763
.9757
.9750
.2217
.2247
.2278
4.5107
4.4494
4.3897
1.0243
1.0249
1.0256
4.6202
4.5604
4.5022
30'
20'
10'
13°
0'
10'
20'
.2250
.2278
.2306
.9744
.9737
.9730
.2309
.2339
.2370
4.3315
4.2747
4.2193
1.0263
1.0270
1.0277
4.4454
4.3901
4.3362
77° 0'
, 50'
40'
30'
40'
50'
.2334
.2363
.2391
.9724
.9717
.9710
.2401
.2432
.2462
4.1653
4.1126
4.0611
1.0284
1.0291
1.0299
4.2837
4.2324
4.1824
30'
20'
10'
14°
0'
10'
20'
.2419
.2447
.2476
.9703
.9696
.9689
.2493
.2524
.2555
4.0108
3.9617
3.9136
1.0306
1.0314
1.0321
4.1336
4.0859
4.0394
76° 0'
50'
40'
30'
40'
50'.
.2504
.2532
.2560
.9681
.9674
.9667
.2586
.2617
.2648
3 . 8667
3.8208
3 . 7760
1.0329
1.0337
1.0345
3.9939
3.9495
3.9061
30'
20'
10'
16°
0'
.2588
.9659
.2679
3.7321
1.0353
3.8637
76° 0'
CMX
linx
cot X
tanX
eofeo X
•60 X
X
Nat. trig,
fiuot's.
0° - 15°
90° - 75°
338
TABLE in. NATURAL FUNCTIONS
/■-N
^
X
lin X
001 X
tan X
OOt X
•00 X
00000 jr
15° 0'
10'
20'
.2588
.2616
.2644
.9659
.9652
.9644
.2679
.2711
.2742
3.7321
3.6891
3.6470
1.0353
1.0361
1.0369
3.8637
3.8222
3.7817
75° 0'
50'
40'
30'
40'
50'
.2672
.2700
.2728
.9636
.9628
.9621
.2773
.2805
.2836
3.6059
3.5656
3.5261
1.0377
1.0386
1.0394
3.7420
3.7032
3.6652
30'
20'
.10'
16° 0'
10'
20'
.2766
.2784'
.2812 '
.9613
.9605
.9596
.2867
.2899
.2931
3.4874
3.4495
3.4124
1.0403
1.0412
1.0421
3.6280
3.5915
3.5559
•
74° C
50'
40'
30'
40'
50'
.2840
.2868
.2896
.9588
.9580
.9572
.2962
.2994
.3026
3.3759
3 . 3402
3 . 3052
1 . 0430
1.0439
1.0448
3 . 5209
3.4867
3.4532
30'
20'
10'
17° C
10'
20'
.2924
.2952
.2979
.9563
.9555
.9546
.3057
.3089
.3121
3.2709
3.2371
3.2041
1.0457
1.0466
1.0476
3.4203
3.3881
3.3565
78° 0'
50'
40'
30'
40'
50'
.3007
.3035
.3062
.9537
.9528
.9520
.3153
.3185
.3217
3.1716
3.1397
3.1084
1.0485
1.0495
1.0505
3.3255
3 .2951
3.2653
30'
20'
10'
18° 0'
10'
20'
.3090
.3118
.3145
.9511
.9502
.9492
.3249
.3281
.3314
3.0777
3.0475
3.0178
1.0515
1.0525
1.0535
3.2361
3.2074
3.1792
72° C
50'
40'
30'
40'
50'
.3173
.3201
.3228
.9483
.9474
.9465
.3346
.3378
.3411
2.9887
2.9600
2.9319
1.0545
1.0555
1.0566
3.1516
3.1244
3.0977
30'
20'
10'
19° C
10'
20'
.3256
.3283
.3311
.9455
.9446
.9436
.3443
.3476
.3508
2.9042
2.8770
2.8502
1.0576
1.0587
1.0598
3.0716
3.0458
S.0206
71° C
50'
40'
30'
40'
50'
.3338
.3365
.3393
.9426
.9417
.9407
.3541
.3574
.3607
2.8239
2.7980
2.7725
1.0609
1.0620
1.0631
2.9957
2.9714
2.9474
30'
20'
10'
20° C
10'
20'
.3420
.3448
.3475
.9397
.9387
.9377
.3640
.3673
.3706
2.7475
2.7228
2 . 6985
1.0642
1.0653
1.0665
2.9238
' 2.9006
2.8779
70° 0'
50'
40'
30'
40'
50'
.3502
.3529
.3557
.9367
.9356
.9346
.3739
.3772
.3805
2.6746
2.6511
2.6279
1.0676
1.0688
1.0700
2.8555
2.8334
2.8118
30'
20'
10'
21° C
10'
20'
.3584
.3611
.3638
.9336
.9325
.9315
.3839
.3872
.3906
2.6051
2.5826
2.5605
1.0712
1.0724
1.0736
2.7904
2.7695
2.7488
6d°' 0'
50'
40'
30^
40'
50'
.3665
.3692
.3719
.9304
.9293
.9283
.3939
.3973
.4006
2 . 5386
2.5172
2.4960
1.0748
1.0760
1.0773
2.7285
2.7085
2.6888
30'
20'
10'
22° 0'
10'
20'
.3746
.3773
.3800
.9272
.9261
.9250
.4040
.4074
.4108
2.4751
2.4545
2.4342
1.0785
1.0798
1.0811
2 . 6695
2.6504
2.6316
68° 0'
50'
40'
30'
.3827
.9239
.4142
2.4142
1.0824
2.6131
30'
001 X
sin X
cot X
tan X
OOMO X
iOO X
X
NATURAL FUNCTIONS. TABLE III
339
X
simr
COS jr
tan X
cot X
800 X
00800 X
30'
40'
50'
.3827
.3854
.3881
.9239
.9228
.9216
.4142
.4176
.4210
2.4142
2.3945
2.3750
1.0824
1.0837
1.0850
2.6131
2.5949
2.5770
30'
20'
10'
23° C
10'
20'
.3907
.3934
.3961
.9205
.9194
.9182
.4245
.4279
.4314
2.3559
2.3369
2.3183
1.0864
1.0877
1.0891
2.5593
2.5419
2.5247
67° 0'
50'
40'
30'
40'
50'
.3987
.4014
. .4041
.9171
.9159
.9147
.4348
.4383
.4417
2.2998
2.2817
2.2637
1.0904
1.0918
1.0932
2.5078
2.4912
2.4748
30'
20'
10'
24° 0'
10'
20'
.4067
.4094
.4120
.9135
.9124
.9112
.4452
.4487
.4522
2.2460
2.2286
2.2113
1.0946
1.0961
1.0975
2.4586
2.4426
2.4269
66° 0'
50'
40'
30'
40'
50'
.4147
\4173
.4200
.9100
.9088
.9075
.4557
.4592
.4628
2.1943
2.1775
2.1609
1.0990
1 . 1004
1.1019
2.4114
2.3961
2.3811
30'
20'
10'
26° C
10'
20'
.4226
.4253
.4279
.9063
.9051
.9038
.4663
.4699
.4734
2.1445
2.1283
2.1123
1 . 1034
1 . 1049
1 . 1064
2.3662
2.3515
2.3371
66° 0'
50'
40'
30'
40'
50'
.4305
.4331
.4358
.9026
.9013
.9001
.4770
.4806
.4841
2.0965
2.0809
2.0655
1 . 1079
1 . 1095
1.1110
2.3228
2.3088
2.2949
30'
20'
10'
26° 0'
10'
20'
.4384
.4410
.4436
.8988
.8975
.8962
.4877
.4913
.4950
2.0503
2.0353
2.0204
1.1126
1.1142
1.1158
2.2812
2.2677
2.2543
64° 0'
50'
40'
. 30'
40'
50'
.4462
.4488
.45lf
.8949
.8936
.8923
.4986
.5022
.5059
2.0057
1.9912
1.9768
1.1174
1.1190
1 . 1207
2.2412
2.2282
2.2154
30/
20'
10'
27° 0'
10'
20'
.4540
.4566
.4592
.8910
.8897
.8884
.5095
.5132
.5169
1 . 9626
1 . 9486
1 . 9347
1 . 1223
1 . 1240
1.1257
2.2027
2.1902
2.1779
63° 0'
50'
40'
30'
40'
50'
.4617
.4643
.4669
.8870
.8857
.8843
.5206
.5243
.5280
1,9210
1.9074
1 . 8940
1.1274
1.1291
1.1308
2.1657
2.1537
2.1418
30'
20'
10'
28° 0'
10'
20'
.4695
.4720
.4746
.8829
.8816
.8802
.5317
.5354
.5392
1.8807
1.8676
1.8546
1.1326
1.1343
1.1361
2.1301
2.1185
2.1070
62° 0'
. 50'
40'
30'
40'
50'
.4772
.4797
.4823
.8788
.8774
.8760
.5430
.5467
.5505
1.8418
1 . 8291
1.8165
1.1379
1.1397
1.1415
2.0957
2.0846
2.0736
30'
20'
10'
29° 0'
10'
20'
.4848
.4874
.4899
.8746
.8732
.8718
.5543
.5581
.5619
1 . 8040
1.7917
1.7796
1 . 1434
1.1452
1.1471
2.0627
2.0519
2.0413
61° 0'
50'
40'
30'
40'
50'
.4924
.4950
.4975
.8704
.8689
.8675
.5658
.5696
.5735
1.7675
1.7556
1 . 7437
1 . 1490
1 . 1509
1.1528
2.0308
2.0204
2.0101
30'
20'
10'
30° C
.5000
.8660
.5774
1.7321
1.1547
2.0000
60° 0'
001 X
8in;ir
cot X
tan X
eoioo X
800 X
X
15°
75°
-SO*
-60°
340
TABLE III. NATURAL FUNCTIONS
X
lin X
CM X
tanjT
eot X
see JT
eoMe jr
80° 0'
10'
20'
.50Q0
.5025
.5050
.8660
.8646
.8631
.5774
.5812
.5851
1.7321
1 . 7205
1 . 7090
1.1547
1 . 1567
1 . 1586
2.0000
1.9900
1.9801
60° 0'
50'
. 40'
30'
40'
50'
.5075
.5100
.5125
.8616
.8601
.8587
.5890
.5930
.5969
1 . 6977
1 . 6864
1.6753
1.1606
1.1626
1.1646
1.9703
1.9606
1.9511
30'
20'
10'
81° 0'
10'
20'
.5150
.5175
.5200
.8572
.8557
.8542
.6009
.6048
.6088
1 . 6643
1.6534
1 . 6426
1.1666
1.1687
1.1708
1.9416
1.9323
1.9230
69° 0'
60'
40'
30'
40'
50'
.5225
.5250
.6275
.8526
.8511
.8496
.6128
.6168
.6208
1.6319
1.6212
1.6107
1.1728
1 . 1749
1 . 1770
1.9139
1.9049
1.8959
30'
20'
10'
82° 0'
10'
20'
.5299
.5324
.5348
.8480
.8465
.8450
.6249
.6289
.6330
1 . 6003
1 . 5900
1.5798
1.1792
1.1813
1 . 1835
1.8871
1.8783
1.8699
68° 0'
50'
40'
30'
40'
50'
.5373
.5398
.5422
.8434
.8418
.8403
.6371
.6412
.6453
1 . 5697
1 . 5597
1 . 5497
1.1857
1 . 1879
1.1901
1.8612
1.8527
1.8444
30'
20'
10'
38° 0'
10'
20'
.5446
.5471
.5495
,8387
.8371
.8355
.6494
.6536
.6577
1 . 5^99
1 . 5301
1.5204
1 . 1924
1.1946
1 . 1969
1.8361
1.8279
1.8198
67° 0*»
60'
40'
30'
40'
50'
.5519
.5544
.5568
.8339
.8323
.8307
.6619
.6661
.6703
1.5108
1.5013
1.4919
1 . 1992
1.2015
1.2039
1.8118
1.8039
1.7960
30'
20'
10'
84° 0'
10'
20'
.5592
.5616
.5640
.8290
.8274
.8258
.6745
.6787
.6830
1.4826
1.4733
1.4641
1.2062
1.2086
1.2110
1.7883
1.7806
1 . 7730
66° 0'
60'
40'
30'
40'
50'
.5664
.5688
.5712
.8241
.8225
.8208
.6873
.6916
.6959
1.4550
1.4460
1.4370
1.2134
1.2158
1.2183
1.7655
1.7581
1 . 7507
30'
20'
10'
86° 0'
10'
20'
.5736
.5760
.5783
.8192
.8175
.8158
.7002
.7046
.7089
1.4281
1.4193
1.4106
1.2208
1.2233
1.2258
1.7435
1 . 7362
1.7291
66° CK
60'
40'
30'
40'
50'
.5807
.5831
.5854
.8141
.8124
.8107
.7133
.7177
.7221
1.4019
1.3934
1.3848
1.2283
1.2309
1.2335
1 . 7221
1.7151
1.7082
30'
20'
10'
86° 0^
10'
20'
.5878
.5901
.5925
.8090
.8073
.8056
.7265
.7310
.7355
1 . 3764
1.3680
1.3597
1.2361
1.2387
1.2413
1.7013
1.6945
1.6878
64° C
60'
40'
30'
40'
50'
.5948
.5972
.5995
.8039
.8021
.8004
.7400
.7445
.7490
1.3514
1.3432
1.3351
1.2440
1 . 2467
1.2494
1.6812
1 . 6746
1.6681
30'
20'
10'
87° 0'
10'
20'
.6018
.6041
.6065
.7986
.7969
.7951
.7536
.7581
.7627
1.3270
1.3190
1.3111
1.2521
1.2549
1.2577
1.6616
1 . 6553
1.6489
68° 0'
50'
40'
30'
.6088
.7934
.7673
1.3032
1.2605
1 . 6427
30'
eoBX
linX
OOtX
tanX
COMO X
i
leoX
X
I
I
- (.
NATURAL FUNCTIONS.
tabt;
E III
341
X
■imr
col JT
taa jr
cot X
■OCX
OOMOX
1
30'
40'
50'
.6088
.6111
.6134
.7934
.7916
.7898
.7673
.7720
.7766
1.3032
1.2954
1.2876
1.2605
1.2633
1.2662
1.6427
1.6365
1.6304
30'
20'
10'
38° C
10'
20'
.6157
.6180
.6202
.7880
.7862
.7844
.7813
.7860
.7907
1.2799
1.2723
1.2647
1.2690
1.2719
1.2748
1.6243
1.6183
1.6123
52° C
50'
40'
30'
40'
50'
.6225
.6248
.6271
.7826
.7808
.7790
.7954
.8002
.8050
1.2572
1.2497
1.2423
1.2779
1.2808
1.2837
1.6064
1.6005
1.5948
30'
20'
10'
39° (K
10'
20'
.6293
.6316
.6338
.7771
.7753
.7735
.8098
.8146
.8195
1.2349
1 . 2276
1.2203
1.2868
1.2898
1.2929
1.5890
1.5833
1.5777
61° C
50'
40'
30'
40'
50'
.6361
.6383
.6406
.7716
.7698
.7679
.8243
.8292
i3«
1.2131
1.2059
1 . 1988
1.2960
1.2991
1 . 3022
1.5721
1.5666
1.5611
30'
20'
10'
40° C
10'
20'
.6428
.6450
.6472
.7660
.7642*
.7623
fc*B391-
.8441
.8491
1.1918
1 . 1847
1.1778
1.3054
1.3086
1.3118
1.5557
1.5504
1.5450
60° 0'
50'
40^
30'
40'
50'
41° 0'
10'
20'
.6494
.6517
.6539
.6561
.6583
.6604
.7604
.7585
.7566
.7547
.7528
.7509
.8541
.8591
.8642
.8693
.8744
.8796
1.1708
1.1640
1.1571
1.1504
1 . 1436
1.1369
1 3151
1.3184
1.3217
1.3250
1.3284
1.3318
1.5398
1 . 5346
1.5294
1.5243
1.5192
1.5142
30'
20'
10'
49° 0'
50'
40'
30° - 45°
60° -45°
30^
40'
50'
.6626
.6648
.6670
.7490
.7470
.7451
.8847
.8899
.8952
1.1303
1 . 1237
1.1171
1.3352
1.3386
1.3421
1.5092
1.5042
1.4993
30'
20'
10'
42° 0'
10'
20'
.6691
.6713
.6734
.7431
.7412
.7392
.9004
.9057
.9110
1.1106
1.1041
1.0977
1.3456
1.3492
1.3527
1.4945
1.4897
1.4849
48° 0'
50'
40'
30'
40'
50'
.6756
.6777
.6799
.7373
.7353
.7333-
.9163
.9217
.9271
1.0913
1.0850
1.0786
1.3563
1 . 3600
1.3636
1.4802
1.4755
1.4709
30'
20'
10'
48° 0'
10'
20'
.6820
.6841
.6862
.7314
.7294
.7274
.9325
.9380
.9435
1.0724
1.0661
1.0599
1.3673
1.3711
1.3748
1.4663
1.4617
1.4572
4r 0'
50'
40'
i^
: '«>•
;§
30'
40'
50'
.6884
.6905
.6926
.7254
.7234
.7214
.9490
.9545
.9601
1.0538
1.0477
1.0416
1.3786
1 . 3824
1 . 3863
1 . 4527
1 . 4483
1.4439
30'
20'
10'
ii2
44° (K
10'
20'
.6947
.6967
.6988,
.7193
.7173
.7153
.9657
.9713
.9770
1.0355
1.0295
1.0235
1 . 3902
1.3941
1.3980
1.4396
1.4352
1.4310
46° 0'
50'
40'
30'
40'
50'
.7009
.7030
.7050
.7133
.7112
.7092
.9827
.9884
.9942
1.0176
1.0117
1.0058
1.4020
1.4061
1.4101
1.4267
1.4225
1.4184
30'
20'
10'
46° 0'
.7071
.7071
1 . 0000
1.0000
1.4142
1.4142
46° 0'
.
COSX
i^X
cot A^
tanX
COMC A*
fee X
X
'
342 TABLE IV. DEGREES TO RADIANS AND CX)NVERSELY
0m
Ade^eM
mninntM
it seooxuU
M
a radiant into
MM
into radiani
n
de^ee meaforo
0.00000
0.00000
0.00000
1
0.01745
0.00029
0.00000
0.00001
0**
0' 02"
2
0.03491
0.00058
0.00001
0.00002
04
3
0.05236
0.00087
0.00001
0.00003
06
4
0.06981
0.00116
0.00002
0.00004
08
6
0.08727
0.00145
0.00002
0.00005
O**
0' 10"
6
0.10472
0.00175
0.00003
0.00006
12
7
0.12217
0.00204
0.00003
0.00007
14
8
0.13963
0.00233
0.00004
0.00008
17
9
0.16708
0.00262
0.00004
0.00009
19
10
0.17463
0.00291
0.00005
11
0.19199
0.00320
0.00005
0.0001
O''
0' 21"
12
0.20944
0.00349
0.00006
0.0002
41"
13
0.22689
0.00378
0.00006
0.0003
1 02
14
0.24435
0.00407
0.00007
0.0004
1 23
15
0.26180
0.00436
0.00007
0.0005
0«
1' 43"
16
0.27925
0.00465
0.00008
0.0006
2 04
17
0.29671
0.00495
0.00008
0.0007
2 24
18
0.31416
0.00524
0.00009
0.0008
2 45
19
0.33161
0.00553
0.00009
0.0009
3 06
20
0.34907
0.00582
0.00010
21
0.36652
0.00611
0.00010
0.001
0«
03' 26"
22
0.38397
0.00640
0.00011
0.002
06 53
23
0.40143
0.00669
0.00011
0.003
10 19
24
0.41888
0.00698
0.00012
0.004
13 45
25
0.43633
0.00727
0.00012
0.005
0**
17' 11"
26
0.45379
0.00756
0.00013
0.006
20 38
27
0.47124
0.00785
0.00013
0.007
24 04
28
0.48869
0.00814
0.00014
0.008
27 30
29
0.50615
0.00844
0.00014
0.009
30 56
SO
0.52360
0.00873
0.00015
.
31
0.54105
0.00902
0.00015
0.01
O*'
34' 23"
32
0.55851
0.00931
0.00016
0.02
1
08 45
33
0.57596
0.00960
0.00016
0.03
1
43 08
34
0.59341
0.00989
0.00016
0.04
2
17 31
36
0.61087
0.01018
0.00017
0.05
2°
51' 53"
36
0.62832
0.01047
0.00017
0.06
3
26 16
37
0.64577
0.01076
0.00018
0.07
4
00 39
38
0.66323
0.01105
0.00018
0.08
4
35 01
39
0.68068
0.01134
0.00019
0.09
5
09 24
40
0.69813
0.01164
0.00019
41
0.71558
0.01193
0.00020
0.1
5°
43 46"
42
' 0.73304
0.01222
0.00020
0.2
11
27 33
43
0.75049
0.01251
0.00021
0.3
17
11 19
44
0.76794
0.01280
0.00021
0.4
22
55 6
DEGREES TO RADIANS AND CONVERSELY. TABLE IV 343
A degrees
n minutes
n seconds
0m
a
into radians
MM
degree measure
45
0.78540
0.01309
0.00022
0.5
28° 38' 52"
46
0.80285
0.01338
0.00022
0.6
34 22 39
47
0.82030
0.01367
0.00023
0.7
40 06 25
48
0.83776
0.01396
0.00023
0.8
45 50 12
49
0.85521
0.01425
0.00024
0.9
51 33 58
60
0.87266
0.01454
0.00024
51
0.89012
0.01484
0.00025
i:o
570 17. 45,,
52
0.90767
0.01513
0.00025
2.0
114 35 30
53
0.92502
0.01542
0.00026
3.0
171 53 14
54
0.94248
0.01571
0.00026
4.0
229 10 59
66
0.95993
0.01600
0.00027
6.0
286*^ 28' 44"
56
0.97738
0.01629
0.00027
6.0
343 46 29
57
0.99484
0.01658
0.00028
7.0
401 04 14
58
1.01229
0.01687
0.00028
8.0
458 21 58
59
1.02974
0.01716
0.00029,
9.0
515 39 43
60
1.04720
0.01745
0.00029
10.0
572^ 57' 28"
TABLE V. MATHEMATICAL CONSTANTS
1
T» 3.14159 26635 89793.
««» 9.86960 44010 89359.
ifl » 31.00627 66802 99820.
Vx - 1.77245 38509 05516.
- = 0.31830 98861 83791.
^ = 0.10132 11836 42338.
i - 0.03225 15344 33199.
1
0.56418 95835 47756.
180°
1 radian = -^ - 57°.29577 95131,
IT
^"°"" =3437'.74677 07849,
648000"
« 206264".80624 70964,
radians.
1° « 0.01745 32925 19943.
(1°)2 - 0.00030 46174 19787.
(1°)» - 0.00000 53165 76934.
radians.
1' =" 0.00029 08882 08666.
(l')2-= 0.00000 00846 15950
(l')»= 0.00000 00000 24614
1" » 0.00000 48481 36811-
(1")2 = 0.00000 00000 23504.
sin 1° = 0.01745 24064 37284.
sin 1' » 0..00029 08882 04563.
sin 1" = 0.00000 48481 36811.
e = Naperian base = 1 + 1-5 + 1-5 + . . .
2.71828 18284 59045.
Af » 0.43429 44819 03252; logion » M Iog« n.
^-2.30258 60929 94046; log« n « j^ logio n.
Radians
to
degrees
and con-
versely.
Math,
const's.
344 TABLE VI. VALUES OF LOGeX, e AND e
— «
X
log«x
«•
e-»
X
lOgeJf
e»
e-»
0.00
— 00
1.000
1.000
2.50
0.916
12.18
0.082
0.05
-2.996
1.051
0.951
2.55
0.936
12.81
0.078
0.10
-2.303
1.105
0.905
2.60
0.956
13.46
0.074
0.15
-1.897
1.162
0.861
2.65
0.975
14.15
0.071
0.20
-1.610
1.221
0.819
2.70
0.993
14.88
0.067
0.25
-1.386
1.284
0.779
2.75
1.012
15.64
0.064
0.30
-1.204
1.350
0.741
2.80
1.030
16.44
0.061
0.35
-1.050
1.419
0.705
2.85
1.047
17.29
0.058
0.40
-0.916
1.492
0.670
2.90
1.065
18.17
0.055
0.45
-0.799
1.568
0.638
2.95
1.082
19.11
0.052
0.50
-0.693
1.649
0.607
3.00
1.099
20.09
0.050
0.55
-0.598
1.733
0.577
3.05
1.115
21.12
0.047
0.60
-0.511
1.822
0.549
3.10
1.131
22.20
0.045
0.65
-0.431
1.916
0.522
3.15
1.147
23.34
0.043
0.70
-0.357
2.014
0.497
3.20
1.163
24.53
0.041
0.75
-0.288
2.117
0.472
3.25
1.179
25.79
0.039
0.80
-0.223
2.226
0.449
3.30
1.194
27.11
0.037
0.85
-0.163
2.340
0.427
3.35
1.209
28.60
0.035
0.90
-0.105
2.460
0.407
3.40
1.224
29.96
0.033
0.95
-0.051
2.586
0.387
3.45
1.238
31.50
0.032
1.00
0.000
2.718
0.368
3.50
1.253
33.12
0.030
1.05
+ 0.049
2.858
0.350
3.55
1.267
34.81
0.029
1.10
0.095
3.004
0.333
3.60
1.281
36.60
0.027
1.15
0.140
3.158
0.317
3.65
1.295
38.47
0.026
1.20
0.182
3.320
0.301
3.70
1.308
40.45
0.025
1.25
0.223
3.490
0.287
3.75
1.322
42.52
0.024
1.30
0.262
3.669
0.273
3.80
1.335
44.70
0.022
1.35
0.300
3.857
0.259
3.85
1.348
46.99
0.021
1.40
0.337
4.055
0.247
3.90
1.361
49.40
0.020
1.45
0.372
4.263
0.235
3.95
1.374
51.94
0.019
1.50
0.406
4.482
0.223
4.00
1.386
54.60
0.018
1.55
0.438
4.711
0.212
4.05
1.399
57.40
0.017
1.60
0.470
4.953
0.202
4.10
1.411
60.34
0.017
1.65
0.501
5.207
0.192
4.15
1.423
63.43
0.016
1.70
0.531
5.474
0.183
4.20
1.435
66.69
0.015
1.75
0.560
5.755
0.174
4.25
1 .447
70.11
0.014
1.80
0.588
6.050
0.165
4.30
1.459
73.70
0.014
1.85
0.615
6.360
0.157
4.35
1.470
77.48
0.013
1.90
0.642
6.686
0.150
4.40
1.482
81.45
0.012
1.95
0.668
7.029
0.142
4.45
1.493
85.63
0.012
2.00
0.693
7.389
0.135
4.50
1.504
90.02
0.011
2.05
0.718
7.768
0.129
4.55
1.515
94.63
0.011
2.10
0.742
8.166
0.122
4.60
1.526
99.48
0.010
2.15
0.766
8.585
0.116
4.65
1.537
104.58
0.010
2.20
0.789
9.025
0.111
4.70
1.548
109.95
0.009
2.25
0.811
9.488
0.105
4.75
1.558
115.58
009
2.30
0.833
9.974
0.100
4.80
1.569
121.51
0.008
2.35
0.854
10.486
0.095
4.85
1.579
127.74
0.008
2.40
0.876
11.023
0.091
4.90
1.589
134.29
0.007
2.45
0.896
11.588
0.086
4.95
1.599
141.17
0.007
2.50
0.916
12.182
0.082
5.00
1.609
148.41
0.007
TABLE VII. SQUARES, CUBES, SQUARE AND CUBE ROOTS 346
n
n2
ns
Vw
\/n
n2
n»
Vw
^n
1
1
1
1
1
61
2601
132651
7.141
3.708
2
4
8
1.414
1.260
52
2704
140608
7.211
3.733
3
9
27
1.732
1.442
53
2809
148877
7.280
3.756
4
16
64
2.000
1.587
54
2916
157464
7.348
3.780
5
25
125
2.236
1.710
55
3025
166375
7.416
3.803
6
36
216
2.449
1.817
56
3136
175616
7.483
3.826
7
49
343
2.646
1.913
57
3249
185193
7.550
3.849
8
64
512
2.828
2.000
58
3364
195112
7.616
3.871
9
81
729
3.000
2.080
59
3481
205379
7.681
3.893
10
100
loeo
3.162
2.154
60
3600
216000
7.746
3.915
11
121
1331
3.317
2.224
61
3721
226981
7.810
3.936
12
144
1728
3.464
2.289
62
3844
238328
7.874
3.958
13
169
2197
3.606
2.351
63
3969
250047
7.937
3.979
14
196
2744
3.742
2.410
64
4096
262144
8.000
4.000
15
225
3375
3.873
2.466
65
4225
274625
8.062
4.021
16
256
4096
4.000
2.520
66
4356
287496
8.124
4.041
17
289
4913
4.123
2.571
67
4489
300763
8.185
4.062
18
324
5832
4.243
2.621
68
4624
314432
8.246
4.082
19
361
6859
4.359
2.668
69
4761
328509
8.307
4.102
20
400
8000
4.472
2.714
70
4900
343000
8.367
4.121
21
441
9261
4.583
2.759
71
5041
357911
8.426
4.141
22
484
10648
4.690
2.802
72
5184
373248
8.485
4.160
23
529
12167
4.796
2.844
73
5329
389017
8.544
4.179
24
576
13824
4.899
2.884
74
5476
405224
8.602
4.198
25
625
15625
5.000
2.924
75
5625
421875
8.660
4.217
26
676
17576
5.099
2.962
76
5776
438976
8.718
4.236
27
729
19683
5.196
3.000
77
5929
456533
8.775
4.254
28
784
21952
5.291
3.037
78
6084
474552
8.832
4.273
29
841
24389
5.385
3.072
79
6241
493039
8.888
4.291
30
900
27000
5.477
3.107
80
6400
512000
8.944
4.309
31
961
29791
5.568
3.141
81
6561
531441
9.000
4.327
32
1024
32768
5.657
3.175
82
6724
551368
9.055
4.344
33
1089
35937
5.745
3.208
83
6889
571787
9.110
4.362
34
1156
39304
5.831
3.240
84
7056
592704
9.165
4.380
35
1225
42875
5.916
3.271
85
7225
614125
9.220
4.397
36
1296
46656
6.000
3.302
86
7396
636056
9.274
4.414
37
1369
50653
6.083
3.332
87
7569
658503
9.327
4.431
38
1444
54872
6.164
3.362
88
7744
681472
9.381
4.448
39
1521
59319
6.245
3.391
89
7921
704969
9.434
4.465
40
. 1600
64000
6.325
3.420
90
8100
729000
9.487
4.481
41
1681
1764
68921
6.403
3.448
91
8281
753571
9.539
4.498
42
74088
6.481
3.476
92
8464
778688
9.592
4.514
43
1849
79507
6.557
3.503
93
8649
804357
9.644
4.531
44
1936
85184
6.633
3.530
94
8836
830584
9.695
4.547
45
2025
91125
6.708
3.557
95
9025
857375
9.747
4.563
46
2116
97336 .
6.782
3.583
96
9216
884736
9.798
4.579
47
2209
103823
6.856
3.609
97
9409
912673
9.849
4.595
48
2304
110592
6.928
3.634
98
9604
941192
9.899
4.610
49
2401
117649
7.000
3.659
99
9801
970299
9.950
4.626
50
2500
125000
7.071
3.684
100
10000
1000000
10.000
4.642
n
n2
n'
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n
n2
n«
Vw
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