-NRLF A TEXT-BOOK ON ADVANCED ALGEBRA AND TRIGONOMETRY WITH TABLES BY WILLIAM CHARLES BRENKE, Ph.D. ASSOCIATE PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF NEBRASKA /^ NEW YORK THE CENTURY CO. 1910 Copyright, 1910, by THE CENTURY COMPANY Published, August, 1910 Stanbopc iprcaa F. H. GILSON COMPANY BOSTON, U.S.A. Q .^■'-v TABLE OF CONTENTS CHAPTER I. Page The Operations of 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 Common Multiple. 15. Exercises. 16. Fractions. 19. Exercises. CHAPTER II. Involution; Evolution; Theory of Exponents; Surds and Im- AGINARIES 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. Logarithms; 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. 69. Exercises and Problems. iii iv - TABLE OF CONTENTS CHAPTER V. Page 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. Standard 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, Proportion, Variation 88 115. Definitions. 116. Laws of Proportion. Exercises. 118. Variation. 119. Direct Variation. 120. Inverse Variation. 121. Joint Variation. 122. Exercises. CHAPTER VII. The Trigonometric Functions 94 124. Trigonometric Functions. Exercises. 128. Functions of Com- plementary Angles. Cofunctions. 129. Application 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. 134. Line Values. 135. Variation of the Trigonometric Functions. Graphs of the Trigonometric Functions. Exercises. 136. Periodicity of the Trigonometric Functions. 137. Relations between the Functions. 138. 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 of Several Angles 121 1.52. Functions of (x ± 2/). Exercises. 156. Functions of 2 x. Exer- cises. 157. Functions of ^x. Exercises. 158. Addition Theorems. Exercises. 159. Exercises. Paob TABLE OF CONTENTS CHAPTER IX. ^ sin x tan .x ^ ^ „ Ratios — — and — = — Inverse Functions. Trigonometric A' -* 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 XL 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. Page Undetermined Coefficients. Partial Fractions 210 234. Theorem of Undetermined Coefficients. Exercises. 235. Par- tial Fractions. 239. Exercises. CHAPTER XVI. Determinants 217 240. Determinants 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 XVIL Polar Coordinates. Complex Numbers. De Moivre'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 ?i0 and cos nd. Exercises. 261. Exponential Values of sin x and cos x. Exercises. CHAPTER XVIII. Permutations. Combinations. Chance 242 263. Permutations. Exercises. 264. Combinations. Exercises. 266. Exercises. 267. Probability or Chance. Exercises. 270. Exer- cises. CHAPTER XIX. Theory of 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 of the Roots. 291. Ferrari's Solution of the Quartic Equation. Exer- cises. TABLE OF CONTENTS vii CHAPTER XX. Paob Spherical Trigonometry 2G9 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. 299. Formulas for the Half Angle. 300. Formulas for the Half Sides. 30L 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 of Formulas 301 Appendix B. Tables I to VH 315 Protractor Inside of back cover y 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 mysterious realm. For a similar reason the basic 'idea of the DilTerential 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 ALGEBRA AND TEIGONOMETRY CHAPTER I The Operations of Algebra 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 I feet, of a weight of w pounds, or of a velocity of v feet per second. Here the letter used, I, 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 h, h, h, • . - , In, or by l^^\ l^-\ l'^^\ ' • ' , l'^^^- Three or four different lengths may be indi- cated by accents (called " primes "), as V, I", V", .... Fixed or known quantities are usually designated by the first letters of the alphabet, as by a, 6, c, . . . ; unknown quantities which are to be determined from given data are represented by the last letters of the alphabet, as hy x, y, z, . . . . \i x denote a quantity of a certain kind, other quantities of the same kind are indicated by Xy, X2, X3, . . . (read, "x sub-one, x sub-two, X sub-three, etc."), or by x^^\ x''-\ x^^\ . . . (read "x 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.; 5^, read " is not equal to "; , read ' ' is greater than "; <, read ' ' is not less than "; >, read ' ' is not greater than = , read ' ' is identical with "; = , read " approaches." V 4 FUNDAMENTAL OPERATIONS [3 Signs of Aggregation. — When several quantities are to be treated as a single one, they are enclosed by parentheses ( ), brackets [ ], or braces \ j, or a line is drawn over them, called a vinculum, . Signs of Quality. — These are + , positive; — , negative; | |, absolute value. The first two simply indicate opposite qualities; thus, if -{-v, or simply V, denote a velocity in one direction, then —v denotes an equal velocity in the opposite direction; if -\-t 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 with 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, -r-, 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 = h -\- d. 2. If a = 6 and c 7^ d, then a + c 7^ b -\- d. 3. a -\- b = b -{- a. (Commutative law.) 4. (a -\- b) + c = a -{- (b -^ c). (Associative law.) Laws of Subtraction. — (Subtraction defined by (a— 6)4-6= a.) 1. If a = 6 and c = d, then a - -c = b-d. 2. (a-c)-]-b = (a + b) - c. 3. a + (6 - c) = (a + 6) - c. 4. (a-hc) -b = (a-b)+ c. 5. (a - c) -b = (a-b)- c. 6. a - (6 + c) = (a-b)- c. 7. a-(b-c) = (a - 6) + c. Laws of Multiplication. 1. If a = 6 and c = d, then ac = bd. 2. If ti = 6 and c 7^ d, then ac 7^ bd. 3. aXb = b X a .. (Commutative law.) 4. aX (b X c) = (aXb) X c. (Associative law.) 5. (a-\-b-c)Xd = aXd + bXd-cXd. (Distributive law.) 4,5] RATIONALITY. ZERO 5 Laws of Division. — (Division defined bj' (a -^ 6) X 6 = a.) 1. U a = b and c = d, then a -^ c = b -^ d, provided c, d j^ 0. 2. {a ^ b) X c = {aX c) ^ b, provided b 9^ 0. 3. a X {b ^ c) = (a X b) -^ c, provided c 7^ 0. 4. (a ^ 6) -f- c = (a H- 0) -^ 6, provided 6, c 7^ 0. 5. a -^ (6 -^ c) = (a -^ 6) X c, provided b, c 9^ 0. Some Working Rules. — Tiie 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 even 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, b, . . . X, y, . . . . Any expression which can be built up from these quantities by means of the four fundamental operations is called a rational expression (or function) in terms of the quan- tities involved. 5. Zero. — Zero is defined as that number ichich may be added to any quantity without changing the value 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 -r- 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 + 5^ a. Now consider the quotient r , and suppose a to be fixed, and 6 to be taken smaller and smaller. As b tends toward zero, the quotient r increases without limit and becomes larger a than any assignable number. But as b approaches zero, t takes the form - and at the same time increases without limit so that no value can be assigned to this form. Example. Let z = 1. Then x = a? (1 + X) (1 - X). and 1-X = 1-X2=( Dividing by 1 — X, we have 1 = 1 + X. Therefore 1=2, since x We are led to this fallacy by dividing by zero in the form of 1 — X. Since we assumed a; = 1, therefore 1 — rr = 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 b = ("b approaches zero "), if a is any fixed number, r increases without limit. Such quantities, which are larger than any assignable number, are all indicated by the same symbol, oo (read "infinity"). As an example, consider the law of gases, pressure times volume is constant, or c pv = c, or p = -• 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 v = 0. This is often indicated by lim i> = « r = (read " the limit of p is infinity, when v approaches zero ")• We are thus led to write the equation, — = QC, when « 9^ 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 limit can be made as large as we please after being increased or diminished, multiplied or divided by any number," we have oo-|-a=co, CO— a=co; ooXa=oo, co-^a = oo. ' 8. Powers. — For brevity we put a X a = a~, a X a X a = a^, and a X aX 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= a? -2 ah + 62. 3. a-- 62 = (a + 6) (a - 6). 4. a3+ 6-^ = (a + 6) {a^ - ah + 62). 5. a3-63 = (a - 6) (a2 + a6 + 62). 6. (a + 6 + c)2 = a2 + ^2 _^ c2 + 2 {ah -^ ac + he). 7. The square of any polynomial 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" + 6" is divisible by (a + 6), not by (a — 6), when n is odd. 11. a'' + 6" is not divisible by (a + 6) or (a — 6), when n is even. 8 . EXERCISES [10 10. Exercises. — Simplify, by removal of parentheses and col- lection of like terms: 1. ila-h) + a-ia). 2. il a% - I ab^) + (I a% + | ab^). 3. (0.8 a2 - 3.47 ab - 17.25 ac) - (f a^ - 0.47 ab - 12t ac). 4. (I x2 + 3ax - I a2) -(2a? -ax- la^). 5. Mx+ {[482/ - (6z +Sy- 7x) +4z]- [48y-8x+2z - (4x + y)]}. 6. 6a - {4a -[86 -2a + 6] + (36 -4a)}. Perform the operations indicated in the following exercises and simplify the results when possible: 7. -I a^c (I 62 - 4 c^ + I a(P - 3). 8. 3 xy"^ (x^ -3a?y + 3 xy^ - 2). 9. 0.6 ac^d'^ (2 arb - 3cd^ + hac^ - 5). 10. 3i a26c (6 a^ - 4 62 + 2 a63 - 3 c2). 11. (x^ -2x + l){x^ -Sx+2). 12. (3 a^b - 2 a%~ + 06^) ( 2 a^ - a6 - 5 62). 13. {x^ -lxY + ^ xy^ - y^) {x^ -2xif + y% 14. (a + bf + (a - bf. 15. {\a-^lf-{ha-l)\ 16. (x2 + 1 - 2/2 + 22/) (^2 + 1 + 2/2 - 22/). 17. [ar + (a + 6) a; + a6] [x- - (a + 6) .r + a6]. 18. [{x + o)2 - a.r] [{x - af + ax]. 19. a (a + 1) (a + 2) - (a - 1) (a - 2) (a - 3) 20. [x{y-\)-y{x- 1)] [(x + yf - (x - yf]. 21. 31-^ ?/i^np5 _=. _ io| mhip^. 22. a26c7 ^ A a462c8. 23. ifxV-^ -fia;V- 24. 3 a2 (6 + x)3 ^ 6 a^ (6 + x)^. 25. 1.75 x^ (a;2 - 1)" 4- 25 x^ (1 - a;2)2. 26. (8 a^6 - 24 0*6^ + 16 a768) -- - 8 a%. 27. (8 xhj - ^ xy' - i 2/' + 2 2/') - - t :^y^^ 28. x^ {or + 62) - 2 x* (a? + b^f ^ x^ (a2 + 62). 29. (6 ah - 17 a2x2 + 14 ax^ - 3 x^) - (2 a - 3 x). 30. (4 2/" -18 2/^ + 22 2/2 -7 2/ + 5) -^ {2y-5). 31. [2 x» +'7 xhj - 9 2/- (x + 2/)] ^ (2 X - 3 2/). 32. (-Jd^ + id*-nrf''+^) -(-f^+2d). 33. (r\ a'-ia'b + U ci%~ + i 06^) ^ (| a + -J 6). 34. (-/j m^ + 2^ to2^ _ 2 5 ^^2 ^ 1^4 „3) ^ (1 ,„ _ 7 ,j) 35. (x'^ - §3 X* + U x^ - I x2 - \V X + t) H- (x2 - i X + 5). 36. (2«3 -.i6a + 6) ^ (a + 3). 37. (4 x" - x22/2 + 6 X2/^ - 9 2/") ^ (2 .r - X2/ + 3 2/2). 38. (x" + 4 x22/2 - 32 y') ^ (x - 2 2/). 39. («^ - 5 «'''6'- - 5 0=6^ + 6->) -r {a~ - 3 a6 + 62). 40. (x='-8 2/^) -~{x-2y). 41. (A^'-O//) -^(i.r+3 2/). 11] FACTORING 9 42. (27 a^b^ + 64 xV) ^ (3 ab + ixy). 43. (a^b^ + c») H- (aV _ abc +^). 44. (m=^ -32y^) ^ (ii-2v). ^ 45. (a - 6 + c - d)-. \ v> 46. (x - i ?/ - 2 it + 'f')-. 11. Factoring. — To factor an expression is to find two or more quantities wliose 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 ivhose terms contains a common factor. Example. I xhfz' + \ ^'iz - A xVz' = i x^-z (h xz^ + J - } x^)- (6) Expressions whose terms can be grouped, so that each group contains the same factor. Example, x' - 7 x^y + Uxif -Sy^ = (x^ -Sy^) - (7 x-y - 14 xif) = {x-2ij){jr + 2xy + 4i-) -7xy{x-2y) =.(x-2 2/)(x2-oxy/ + 4 2/2) = {x-2y){x-y){x-4y). (c) Trinomials of the form ax^ -{-bac -^ c. Let h,khe 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 h. If this is true in the first scheme, the factors are {hx -\- n) {kx -\- m) ; in the second, the factors are (Jix + m) {kx + n). Example. 12 x' — 7 x — 10. Here h, k may be one of the pcairs 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, 5, or +2, -5. By trial we find that h, k must be 3, 4, and ni, n must be 2, —5. The factors are therefore (3 x + 2) (4 x — 5). To find the factors of 12 x^— 7 xij — 10 y-, we would proceed as above and obtain {3 x -\- 2 y) {4 x — 5 ij). (d) Expressions ivhich can he written as the differetice of the squares of two quantities. The factors are the sum and the difference of the two quantities respectively. Example. a* + a%- -\-b^ = a^ + 2 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. 15. Exercises. — Find the H. C. F. of 1. 6 (x + If and 9 (x^ - 1). 2. a" - 6" and a'' - 6". 3. 12 (x2 + 2/2)2 and 8 (.r" - 1/). 4. u^ — v^ and vr — v". 5. {c?x — ax"f and ax {a? — x^f. 6. 27 (a" - h^) and 18 (a + hf. 7. (24 a^ _i- 36 ab - 48 ac) and (30 a? + 45 a% - 60 a\). 8. 125 x^ - 1 and 35 x^ - 7 a: + 5 ax - tt. 9. 4x2- 12x1/ +9?/2 and 4x2 -9?/2. 10. x2 + 2x - 120 and x2 - 2x - 80. 11. 12x2 -17ax + 6a2 and 9x2 + 6ax -8a2. 12. x3 + 4 x2 - 5 X and x3 - 6 X + 5. 13. x^ + 3 x2 + 7 X + 21 and 2 x" + 19 .r + 35. 14. 0^ + 703+702 - 15o and n^ -2a^ - 13a + 110. 15. 20x'' + x2 - 1 and 75 x" + 15 x^ - 3 x - 3. 16. x" - ox' - a2x2 - o^x - 2 o" and 3 x^ - 7 ax- + 3 a'x - 7 a^ . 17. x^ - y\ x^ + ?/, and x^ + y^. 18. x2 - 2 a2 - ax, x2 - 6 a2 + ax, and x2 - 8 a2 + 2 ax. 19. o^ + 02^2 + b\ a* + ob^, and a% + ?>^. 20. 3x3 - 7x2^/ + 5x1/2 - y3^ ^2y _|_ 3x2,2 _ 3x3 - yS^ and 3x3 + 5 x22/ + X2/2 - 2/3. 16,17] FRACTIONS 13 FindthcL.C. M. of: ■ 21. S a-. r-y^ and 12 ahx'y-. 22. 4ax^y', Gn^xy^, and ISa^x-y. 23. a^ - b^ and (a - b)~. 24. cv^bx — abhj and abx + Iry. 25. .r- - 3 X - 4 and x' - x - 12. 26. x2 - 1 and x2 + 4x+3. 27. 6x2 + 5 X - 6 and 6 x^ - 13 x + 6. 28. 12x2 + 5x - 3 and 6x=' + x^ - x. 29. 12 x2 - 17 ax + 6 a^ and 9 x^ + 6 ax - 8 a^. 30. a3-9a2+23a-15 and a2_8a + 7. 31. ??i^ + 2 m^n — y^m^ — 2 n^ and ?«^ — 2 ??r-/t — ?/;/r + 2 n'. 32. x^ — ir, (x — ?/)2, and x -\-y. 33. x^ + 3 X + 2, x^ + 4 X + 3, and x" + 5 x + 6. 34. X" + 5x + 10, x^ - 19x - 30, and x^ - 15 x - 50. 35. x^ + 2 X - 3, x^ + 3 x2 - X - 3, and x^ + 4 x- + x - 6. 36. 6x- - 13x + 6, 6x2 + 5.^ _ 6, and 9x2-4. 37. x2 - 1, x2 + 1, and .t3 + 1. 38. x2 + 1, x"" - 1, and x" - 1. 39. a^ - 6^ a9 - b^, and a^ - b^. 40. x2 - i/2, x^ + y\ x3 - y^, and x^ + 7/. 16. Fractions. — An algebraic fraction is the indicated quotient of two algebraic expressions. It is written in the form y,, N being called the numerator and D the denominator. When A'' and D have a common factor F, so that we may put A^ = NiF and D = D^F, then the fraction may be simplified as follows: ^ N ^NiF ^N, D DxF Di' When all factors common to A^ 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 . oi N and D, and remove it as above. 17. Sign of a Fraction. — By the rules for division we have, N ^_ ^^ ^_ Ji ^ ZL^ 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 fraction can he 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 D[^D,^ ^"^ D A mixed expression is reduced by the formula N PD + N ^'^ D~ 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' W~ D^N' N'D ' D' Examples. 1. Simplify X y X y First reduce each fraction to a simple fraction, thus: 2 2 2X2/ 1 1 y -X y-x X y xy y y xy 2/ x-y x-y X X 19] EXERCISES 15 __x ^ _ X ^ _ xy ^ xy 1 _ 5 ?/^_^ y - X X - y y y Reducing to the common denominator x — y, we liave ^^ + _^ + ^^ = Ill^^Jl£^±f2/ = (provided X ^ 2/). y -x ' x -y x -y x-y o , x2 , x2 , x2 2. X3 ; = X3 1 - X4 ,1 - X4 X (1 - X4) X X a:2 , x2 , , 1 x4 + 1 X3 r = X3 + X — X (1 + X2) — X3 X X 19. Exercises. — Reduce to simple fractions or to integral expressions: • W^±f_«^^\(a2_:,2) n. x^+y_2^ 3x2+3y2 ^ ^- U - ^ « + -^i ■^^- x2 - 2/2 • x + y 2 "'~^% "'~^'. 1„ 45 (x - 2/) . 27(x-y)2. a3 + 63 2 a6 12. -^ a4_-64 x2J^xy_+y_2 x3 - 7/3 ^ a2 + 62 • 13. ^- (M)(^^)- /x5 _ 7/\ /x _ 2/\ xi2 + yi2 _ x4 + 2/4 14. 16. xi2 - 2/12 x^ — 2/8 32(3+2/) • 1286(z+2/)2 a2-4x2 a2 + 4 ax rt2 _ 2 ax ■ ax + 4 x2 a2 + a6 . a2 + 62 • ah (a + 6)2 a* - 64 t|3 - 1,3 Ulv2 - ui m2 4- liv + ti2 p2 + 3 p + 9 . p3 _ 27 x + 1 , y + 1 ,,, x6 - 2/^ ^ x2 + x?/ + 2 /2 (X - 2/)2 • X X - 2/ x3 - 2/3 _1_ ^_JT-/ J7_ 2/ ' (a; - 2/)2 X - 2/ 11 X y -o a: + 2/ x3 + 2/^ 9. 1+^- ' x + 2/ X2+2/2 ' x+- X -y x^ -yi X -4-+-^- 19. — ^ — a: y_ 1 I 1 +y x- y x + y 3-2/ 1.2.1 20 21. X (X - 1) ' 1 - X2 ' X (X + 1) 1 2,1 x2-5x + 6 x2 -4x + 3 ' x2-3x + 2 16 EXERCISES [19 22. " I ^ I ^^' ' U +V U — V U' +v^ - 1 2 (x - 2) , x-3 23. 24. a;2-5x+6 x2 - 4x + 3 ' a;2 - 3x +2 7+3x2 5-2x2 3 -2x+x2 4-x2 4+4x+x2 4-4x+x2' l-2x 2x-3 , 1 3 (x2 - X + 1) 2 (x2 + 1) ' 6 (x + 1) 6c ac a6 a/ \ a + b + cj 27. /^! + «^-?-« + l 28. \a X 6 ?// Va S ^ 2// 29 (l- ^^"1 /7x 49x2 343x3 \ 29- 1,^ ir^j ini/ + 121 2/2 + 1331 W (ah 3bc\ /5ac 7abW3b _ ah \ V3c2 5a2J \762 902^ Uo2 3c2^ M x3j/2 _ 3x2j/3 2xy4 _ ?/\ /2x2y _ 3xy2 _ 3y3\ ^ i 5a3 2a26 +3a62 h^)\Za-i 5 ah 2h'^l' CHAPTER II Involution. Evolution. Theory of Exponents. Surds 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 n factors (8), n being a positive integer. Hence, if m be a second positive integer, we have by cancellation, (1) — = a"-'" when n > m; (2) — = ^— when n -262H + (c,„_i+c,„)a" + i-"'6'" + (c,„+c,„ + 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- ^''"^2^~ -^ (l+ci); • . . {Cm + Cm+i) = ^i-l-l ^^^- 1 + ^"'^^ These equations all become identities on substituting from (2). Therefore all three laws are true for the expansion of (a + 6)"+^ provided that they are true for the expansion of (a + 6)". But they are true for (a + 6)^, hence for (a + b)-^, hence for (a + b)^, 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) (w - 2) ci=n; C2= ^ ^ 2 ' ^^ = TTTs ' * * * _ n (n - 1) (n - 2) . . . (n - ?n + 1) ^'"~ 1 . 2 . 3 . . . . w where c^ is the coefficient of the (m + l)th term. In place of 1 • 2 • 3 • . . . w we use the symbol [w or m! (in either case, read "factorial m"). Then equation (1) becomes , , 7i(n-l)(n-2) . • • (u-m + 1) ,._„,,„,, I'm When o = 1 and & = a; we have, \2 [3 44. The expansion of (a + b)" may be reduced to that of (1 + x)" thus: (a + br = a"(l + ^y = fl4l + 'i^ + • • •]• In place of Cm to denote the coefficient of the (m + l)th term of the expansion of (a + 6) , the symbols nCm or {^J are often used. These are called the binomial coefficients. 45] BINOMIAL THEOREM 35 Table of Binomial Coefficients w = 1 n = 1 1 1 n = 2 12 1 71 = 3 13 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 Example 1. Expand (a* -2 6^)*. [{ai)+{-2lr)Y = (a*)' '+4C a^n-2l/)+Q(a^)H-2l^r + 4C ai)(-2by + (-2b'y = a'- 8a¥ + 24 a6^ - 32 aH^ + 16 6^ Example 2. Find the fifth term in the expansion of(x-i-hy'r. This term will be 8-7-6 1 .2-3 ^(- ^K- -^^r=f-^^^- 45. Exercises. Expand: 1. ix-y)^. 2. (2a -3 6)6. 3. (a-i +6-2)4. 4. (x-i-rr. 6. (x^+rr. 12. (1 + a^)7. 13. (a^ + 62/)6. 14. (a^+2/+a^-2/)5. 15. (x«-' - 2/"»')6. 6. {2p--3q'-y. 16. {x' - yyy. 7. (ax + 6!/)8. 17. (^/'.x + ^yY. 8. (^u-2+2!;2)7. 18. (x^^« -«^^)^ 9. {^2~x - 3 52. Graphic Solution of Linear Equations. — Suppose that a given equation has been reduced to the standard form, 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 = aoc + b. Then we want that value of x for which y = 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 x - 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 Ues between and 1. 53. Graph of the Equation y = 2 x — l. — 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 ,, ,t ,j length (say one-fourth of an inch, or ~' ''^^' ' one side of a square of the cross- section paper), mark on XX a series of points whose distances from shall equal the assumed values of x. When x is positive, the distance is laid off 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 .r = ±1, ±2, ±3, H-2^ and -If. Now to represent the value of y corresponding to a given value of X, mark the representative point of x on A^Y, and at this point lay off a segment perpendicular to XX and having a length equal T)!] GRAPH OF LINEAR EQUATION 41 >■ ■ n p^ y- 3 r, fl Y 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 (52), we obtain the figure below. We thus get a series of points. Pi, P2, . . . , Pg, whose distances from the line XX are the values of the binomial 2 a; — 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 +5; also that y = between a: = 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 gra-ph of the equation y = 2 x — 1. The points Pi, P2 . . . are said to lie on this graph. 54. Graph of »/ = ax + b. — 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 collinear. Let xi, X2, and xs be any three values of x; let yi, 2/2, and 2/3 be the corresponding values of y. Lay off the corresponding and (xs, 2/3) so that (see figure) X, =OMu yy =MiPi, 2/2 = M2P2, X3 ■■ But since by putting x = Xi in y = ax -\- b, and similarly for ?/2 and 7/3, we have axi-{-b 2/2 - 2/1 = a{x2 - X,), ys - 2/2 = a(x3- X2). values (xi, yi), {xo, 2/2) P, K OM2, OM3, 1J3 = M3P3. 7/1 is the value of y obtained M, Af> M, 7/2 = ax2 + b ys = axs-\-b Therefore, 2/ 2 - .V l ^ 2/3 - ?/2 X2 Xi X3 - X2 i=a). 42 GRAPH OF LINEAR EQUATION [55,56 But y2 - yi = M2P2 - Ml Pi = M.Po - M.H = HP2; ?/3 - 1/2 = il/sPs - M2P2 = KPs; X2-X1 = OM2 - OMx = M1M2 = PiH; and 0:3 - 3:2 = OM3 - OM2 = M2M3 = P2K. Substituting these in the two fractions above we obtain HP2^KPs PiH P2K Therefore A P1HP2 is similar to A P.KP^. Hence the points P\ , P2, P3 he on a straight hne. Theorem: The graph of the equation y ^ ax -{-b is a straight line. Corollary: To construct the graph of the equation y = ax -\- h, construct two points on it and draw a straight line through them. 55. Exercises. Draw the graphs of the equations (each set to the same reference Hues): 1. y = X + 1, 2 2/ = 2 .T + 2, 5 X = 5 X + 5, -^ 2/ = i X + |. 2. 2/ = 3 X - 4, 2 ?/ = 6 X - 8, % = 3 Lc - 4 /t. 3. 2/ = x + l, 2/=x+2, ?/=x+3, ?/=x-l. 4. 2/ = 3 X - 4, 2/ = 3 X - 2, 7/ = 3 X, 2/ = 3 X + 1. 5. 2/=:c + l, 2/ = 2x + l, 2/ = 3x + l, 2/ = ix 4jr*. 6. 2/=3x-4, 2/ = 6x-4, 2/ = 9x-4, 2/ = ix --4. 7. 2/=-^ + l,2/=-3x-4. 8. 2/ = -c - 1, 2/ = 3x + 4. Explain the effect on the graph of 2/ = ax + 6, of : 9. Multiplying 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 YY as before, and let P be any point in the II , I plane (see figure), obtained by laying off a [/' pair of corresponding values of x and y. \}i The position of P is completely determined ^ ^ I ^ — \ — ^ as soon as x and y are given. Therefore P^ x and y are called the coordinates of P, X being called the abscissa, and y the III IV ordinate. A point whose coordinates are x and y is referred to as the 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 quadrant, 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 (53)) that the signs of the coordinates in the four quadrants will be as in the following table: Quadrant Abscissa Ordinate I + + II - + III IV + - 57. 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, as y = 2 x or y = x -j- 1, or 2 X — 3 y -\- 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 + Bij + C = is called the general linear equation in two variables. Theorem: The graph of the linear equation Ax -\- By -\- C — is a straight line. ■ j^ (J Proof: If 5 7^ 0, we can write ?/ = — 77 a: — 75 , which has the t> li form y = ax -\-b. Therefore the graph is a straight line when 5 7^ 0. (7 If 5 = 0, the equation reduces to Ax -{- C = 0, or x = — ^' unless A = 0. But if A = and B = Q, then C = and the equation vanishes identically. Excluding this, we may reduce Ax -\- C = C toa:= — -j)Ora: = a constant. But this is a straight line parallel to the 7/-axis. Therefore the given linear equation represents a straight line. (Hence the term " linear " equation.) Exercises. 4 C 1. Show that the equations Ax -\- By + C = and y = — '.,x — - have the same graph. 2. Show that the equations Ax + By + C = and kAx + kBy + kC = have the same graph, k being any constant. 3. How is the graph oi Ax -{■ By + C = aff(;ctcd by a cliangc in C? inBfm Af 44 USE OF THE GRAPH [58 58. 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 quantities as the coordinates of a point. Example 1. A man starts 5 miles south of A and walks due north at the rate of 3 miles an hour. How far is he from A at the end of x hours? Solution. Let y be the required distance. Also let y be negative to the south of A, positive to the north. Then the relation between y and x is 3x -5. y Sx The graph is shown in the figure. Here one square on 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 respec- tively. Compare with the values obtained from the equation. In this example negative values of x and the corre- sponding values of y may be interpreted as follows: Let the time 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 positive, and before it, negative. Thus, 3 hours before this moment would be represented by x = — 3. The corresponding value of 2/ 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 = UF - 32). Draw the graph. We shall retain the letters 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 about 1°. Exercise. Read ofT the values of C corresponding to F = — 40°, F = 0°, F = 57° 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 V = I'O (1 + 273 0, 59,60] EXERCISES AND PROBLEMS 45 w , ZOOcu^ ^ ^ lOOai-ft. -soo-' ' -zoo' -lOO' o' ' ' ' UOO" *200' where vq = volume at temperature 0°, and V = volume at temperature t°. Represent graphically the rela- tion between volume and tem- perature for a quantity of gas whose volume at 0° is 100 cu. ft. Replacing 273 l:)y its approxi- mate value .0037, and I'o by 100, the equation becomes V = .37 < + 100. The graph is given in the adjacent figure. 59. Exercises. 1. From the figure, read off 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 vo ft. per second, acquires in t seconds a velocity given by t; = gt -{- vo, in which g = 32.2. As- sume a value of vo and draw the graph of the equation. 5. 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 + X = 15. 8. 3 X = 27. 9. 2 (x - 1) 10. ix-FJx 60. Problems. 11. 12. x-2 3x-5 ?x-l __ _8 ^x 3' 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. 5. 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 Hquids, 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 V 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? 11. 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? 15. 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 -\- hy -\- c = 0, a'x + h'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 in the last step of the solution. Example. Solve for x and y : ^^+x = 15 and ?^ + y = 6. Clearing and simplifying: 4 X -f 2/ = 45 and x -h 4 ?/ = 30. 62] SIMULTANEOUS LINEAR EQUATIONS 47 From the first of these, Substituting in the second Hence Then X - y Check. 2/ = 45 - 4 x. z + 4 (45 - 4 x) = 30. 15 X = 150 or X = 10. ?/ = 45 - 4 X = 5. 10 + y + 5 = 1+ 5 Second Method — By Elimination. — Multiply the first equa- tion by a', the second by —a, and add the resulting equations together. This eliminates x, 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 h' , the second by -6, and adding. The proper multipliers for the two eliminations are conveniently indicated thus: h' a' ax-hby + c = 0, - h — a a'x + h'y + c' = 0. Check. Substitute the values of x and y in either of the given equations. Example. Solve for x and y : 8 X - 15 y + 30 = and 2 X + 3 !/ - 15 Indicating the multipliers: 3 2 8a:-15y + 30 = 15 -8 2x-\- 3 i/- 15 = 0. < - 54 ?/ +- 180 = 0, or y = V, 5ix - 135 = 0, or a: = i. Check. 8 X § - 15 X V + 30 = 20 - 50 -f- 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 = W, and c = k'c', 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 x, substituting in one of the equations and solving for y, 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 = ka' and h = 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 = kc', which is not true. In this case there is no finite solution possible. For if we assume X = xi and y = yi to he a, solution of either equation, the other equation will not be satisfied by these values because c 7^ kc'. 63. Graphic Solution of Two Simultaneous Linear Equations. Let the equations be (1) ax -{- by + c =0, (2) a'x + h'y + c' = 0. The graph of each equation is a straight line. Suppose Li 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 = kb', but c 7^ 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. 2x + y = 11. 3x — 2/ = 4. 2. 3a; + 8y = 19. 3x-y = 1. 3. 2X+7J =47. X + y = 15. 4. 3x+42/ = 85. 5a; + 4?/ = 107. 5x + 7y = 101. 7 X -y = 55.1 2x -y - 1 =0. Qx -32/ -3=0. 15 X -Ty = 9. 92/ -Ix = 13. 2x -72/ = 8. 42/ -9x = 19. 65,66] SIMULTANEOUS LINEAR EQUATIONS 49 9. x-2y-^2 =0. 3a; -62/ + 2 = 0. 10. Bx + Zy = 3. 12a; + 9?/ = 3. 11. X + J 2/ - 3 = 0. 12. 3y-4x-l=0. 18-3x = 42/. 13. 2x = ll+92/. 3x - 15 = V2y. 14. 2x4-72/ = 52. 3x -52/ = 16. 15. 3x-f42/-5 =0. ^x + 5 2/-i =0. 16. 5 2/ -2x =21. 13x-42/ = 120. 17. -+^ = 7 2^3 ^• 2x + 3 2/ = 48. 18. ?^+^=34. ¥+8^-^- 19. 4+. = ^^. .-s = V^. 20. f = IO-|. 4|2/ = 5x - 7. Simultaneous Linear Equations in More than Two Unknowns 65. Three Equations in Three Unknowns. Let the given equations be, (1) ax + hy + cz + d^ 0, (2) a'x + h'y + c'z + d' = 0, (3) a"x + h"y + c"z + d" = 0. Eliminate one of the variables, say z, 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 x, y, and z so found in either of the equations not used in the last step of the solution. 66. Exceptional Cases. 1. The given equations 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 infinitij of solutions, 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. ^' 4 + 6 " ^^"- ^- 3 + 2 " 3 3 8 2" 2"'"3 a' 69] EXERCISES 51 , x+y y -X ^- ~Y + 2 X x + y = 5. 4. £±I+L^t'.5. ^-±y-^^.ro. 5. V+'^^-- 2x + =-V^^=21. 6. i^^=.-.. 2^-»+2» = i. 16 2 (5 - 11 X) 11 -7 ?y "• 11 (x - 1) "^ 3 - 2/ 7 + 2x _ 125^ 144?/ ^ 3 — X 30 (y -|- 5) 7-6x 4-3x 17. 18. 19. 107/ -19 5y-ll 6x-10y-17 _^ 4x-Uy -.5 . 3x-5y + 2 ~2x-72/ + 12 1 ^^ ^2 l-x + ^ x + y-1 3' 1 ^3 1 4' 1 I -x+y 1 1 x + 7. .25 X + 3 ?/ = 10. 4.5 X - 4 7/ = 6. 8. 4.2 !/ + 4 X = 33. 0.77 7/ - 0.3 X =2.95. 9. 0.2525 x + 0.33 7/ = 280. 3.122 x + 0.055 7/ = 3096. 10. 0.2 7/ + 0.25 x = 2{y -x) 0.8 X - 3.7 7/ = - 15.3. 11. 0.1 7/+ 0.3 X =0.3. 0.05?/ + 0.15 X =0.15. 12. |x-0.6 7/ = 5 (x - 1) 0. 5 2 (2 7/ + 3) 6 13. x^y 30 1 1 _ 1 X 7/ 30* 14. X 7/ 15. 1 '> ~ + - = 10. x^y ^(-^■- 20. ax — by = m. cx+dy = n. 21. x + y = 3a-26. x-7/ = 2a-36. 22. ^^^■ X 23. ^"-f=c. a b ^^«■ 24. ^^''■ "+?- 25. 7HX ,vy_i 71 ^ q ''-'M = v. n s 26. X - 7/ + 1 7/ - X + 1 ^ -^ = 7?in. X - y + 1 52 EXERCISES = 0. = 0. + 3=0. = +1=0. 0. 35. 2 X + 3 2/ = 12. 3x + 22 = 11. 3 2/ + 4 z = 10. li X + n y = 10. 38. x + 2 ?/ + 3 z = 32. 25 X + 2i 2 = 20. 2 X + 3 y + 2 = 42. 3i 2/ + 3i 2 = 30. 3 X + 2/ + 2 2 = 40. 39. U^=2. 40. -^ = 1- y z X + 2/ 5 ' + '=*• ^-s- X z X + z 6 1 , 1 ^g _y^ ^1. X 2/ ■ 2/^27 41. X + 2 2/ = 5. 42. 2/ + 2 + w = 2. 43. 3 x + 2/ + z = 4. 2/ + 22=8. 2 + M+x = 3. x+42/ + 3m = 6. 2 + 2u = ll. m+x+2/ = 4. 6x+z + 3u = 8. M + 2x = 6. x+2/+z = 5. . 82/ + 32 + 5U = 10. 44. Find two numbers whose sum is 1735 and difference 555. 45. 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 \\% 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 corn for $252; 60 bushels of wheat and 200 bushels of corn 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. 50. 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. G9] EXERCISES 53 How long will it take both pipes to fill the tank, and what is the discharge per minute of each pipe? 51. How many pounds of copper, and how many of zinc, are contained in 124 pounds of brass (alloy of 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.? 52. 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 xV of its weight? 53. 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. 54. Find three numbers which are in the ratio of 2 : 3 : 4 and whose sum is 999. 55. 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? 56. 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. 57. 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 1| hrs. How long will it take each pipe alone to fill the vessel? 58. 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? 59. 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. I^et X = one of the numbers; then 3 — a; = the other number, and x{5 — x)= Q or x~ — 5x-{-Q = 0. To determine x we must solve this equation. Definition. An equation of the form rtic^ -\-bx -\- c = 0, where a: is a variable and a, b, c are constants, is called the gen- eral equation of the second degree in 07ie variable, or, a quadratic equation in x. Methods for Solving the Equation ax^ + 6x + c = 0. 72. 1. By Factoring. When the trinomial ax^ -\- bx -\- c can readil}^ be factored, then each factor, equated to zero, gives a value of X. Example. x2 - 5 x + 6 = 0, . or (X - 2) (x - 3) = 0. X - 2 = or X - 3 = 0. Hence x = 2 or x = 3. 73. 2. By Completing the Square. (a) The equation is reduced to the form (x + /i)2 = k, whence x + h =± Vk, and x =- h ±Vk. This reduction is effected as follows : Given ax- + fex + c = 0. Transpose c : ax- + bx = — c. 54 73] • QUADRATIC EQUATIONS 65 . , b c Divide by a: ^*' "^ a^ ^ " a' Add l^j to both members: h , / b\- c , / h or, a \2a) a \2a ^Ll 62- 4 4ac a2 W^ — 4ac 4a2 •=± -b±' v/6«- 4ac -+2^-±V^T^=±2^^'''^'-*- a Hence, (b) The equation is reduced to the form (2 ax + h)- = k, whence 2 ax -f /i = -t Vk, and x = ^ z a, Given ax" + hx-}-c = 0. Transpose c: ax^ + bx=- c. Multiply by 4 a: 4 a2a;2 + 4 a6a; = - 4 ac. Add 62: 4 , ^2^2 _^ 4 (j^l)j^ _|_ ^2 = 52 _ 4 ^f.^ or, (2 ax + 6)2 = 62-4 ac. 2 ax + 6 = ± ^62 - 4 ac. - 6 ± V62 - 4 ac Hence, 2a Example. 2 2;2 + X - G = 0. (a) Transpose -6: 2 x2 + X = 6. Divide by 2: x2 + i X = 3. Add (1)2: x2 + ^x + a)2 = 3+a)2, or (a; + \y = n- x+\ =±h and X =±l -I = I or -2. (b) Transpose -6: 2 x2 + X = G. Multiply by 8: IG x2 + 8 X = 48. Add 12or 1: IG x2 + 8 X + 1 = 49, or, (4 X + 1)2 = 49. 4 X + 1 = ± 7. Hence X = i or — 2, as before. 56 QUADRATIC EQUATIONS [74-76 74. 3. By Formula. In (72), by completing the square accord- ing to either method, we obtained — b ± Vb^ — ^ac Of. = _ 2 a Any quadratic equation in x may be solved directly by means of this formula, by merely inserting for a, h, and c their values from the given equation. The formula should be carefully com- mitted to memory. Example. 2 x2 + x - 6 = 0. - l±Vl-4X2X( (-< ^ = - 1 ± 7 3 4 2 or — 2. 11. 5x (X - 2) + i = 1 -3 X. 12. (1 -3x) (x-6) = = 2(x + 2). 13. {X + l)(2x + 3) = 4x2 -22. 14. Ix 2 - 48 = 2 X (x + 7). 15. 13 x2 - 30 = 6 (1 -X)2 + 63. 16. ax + b = X2. 17. bx - 2 62 + x2 = ; 2 6x. 18. X' + mn = — (m • ■fn)x 19. ex -2c-x2 =- -2x. 20. X2 ax a2 2 ~ 2' 4 75. Exercises. Solve for x: 1. x2 + 4 X - 12 = 0. 2. x2 -8x= -7. 3. x2+6x = 16. 4. x2 + 12 =7x. 5. 14 = x2 -5x. 6. 5x2 _3x-2 =0. 7. 3x2+5x -42 =0. 8. 3x2 _50 =25x. 9. 2x2 -13x =-15. 10. 3x2 _7x -6 = 0. 76. Definition. A root 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 ajc"^ + 6ac + c = 0. — The values of x obtained by the formula X = h ± Vb^ -4ac 2a will be 1. real and unequal if b^ ~ 4 ac > 0; 2. real and equal if b^ — 4 ac = 0] 3. imaginary if &^ - 4 ac < 0. 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 x reduce to the common value — 6 -i- 2 a; in the third case the radicand is negative, hence both square roots are imaginary. The expression 6^ — 4 ac, on whose value depends the nature of the roots, is called the discrhninant of the equation ax- -{-bx-{-c = 0. When the discriminant vanishes, the roots are equal; ax^ + 6x + c is then a perfect square. 77. Exercises. — Without solving the equations, determine the nature of the roots of: 1. Exercises 1-10 of (74). 6. ^x^-hx-i^O. 2. 4x2+4x + l=0. 7. 0.1x2+0.5x+0.8 =0. 3. x^+x + l =0. 8. 1^ x2 - 6J x + 8} = 0. 4. 6x2 +2X-1 =0. . 9. ^^2 -ix + ?j =0. 5. 9 x2 + 12 X + 4 = 0. 10. 0.06 x2 + 0.22 x + 0.08 = 0. For what values of the Hteral quantity involved in the following equations will the roots be real and unequal, equal, or imaginary respectively : 11. x2 +2x+c = 0. 18. 2x2+4/ix -/i2 =0. 12. 4 x2 + 4 X + ;i = 0. 19. 2x2 + 4 ax - a = 0. 13. 3x2 - 2x - A; = 0. • 20. ax2 + 2x + 1 = a. 14. ^ x2 - J X + 4 a = 0. 21. a2x2 + ax + 5 = 0. 15. x2 + 2 ?jx + 4 = 0. 22. 2 cx2 + 3 X - c2 = 0. 16. 3x2 - 4/cx + 5 = 0. 23. (l + k) x^ + x - k = 0. 17. 6 x2 + - X - 3 = 0. 24. - + ^ X + — ^ = 0. 7n n 2 n + 1 78. Relations between the Coefficients and Roots of a Quadratic Equation. — The roots of the equation ax^ -f- 6x + c = Xi = Hence _ 5 _^ V62 - 4 ac - 6 - V6-' - -4ac 2 a ' ^- 2 a Xi + X2 = ' and X1X2 = -• a a 58 QUADRATIC EQUATIONS [79-81 That is, if the equation he divided by the coefficient of x^, 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 Trmomial wx^ + &x + c. — If xi and X2 be the roots of the equation ax- + bx + c = 0, the trinomial is divis- ible by X — xi and x — X2. But (x — xi) {x — X2) = X- — {xi -\- X2) x + a;ia;2 . , 6 , c = x~ -\- - X -\ a a Therefore a{x — xi){x — Xo) = 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 a; p, q; m + n, m — n. 4. V^^, - V^T; 1 + V^^, 1 - V^^; a + b V^, a-b \f^^. 5-14. Factor the left members of the equations in exercises 1-10 of (74). 15-24. Same for exercises 1-10 of (76). 25. 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 = Q, we must find the values of x which reduce the trinomial ax^ + bx -\- c to zero. When a, b, 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 x2 + a; - 6 = 0. S2,S3] QUADRATIC EQUATIONS 59 0, + 1, + 2, + 3, G, -3, + 4, + 15 Designate the trinomial on the left by y, so that y = 2x^ + x-6. Now form a table showing the values of y corresponding to a series of assumed values of x : x= . . . -3, -2, -1, y= . . . +9, 0,-5, We see that y = when x = — 2, which gives one root exactly. Also, y must be zero again 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 through them (figure). Scaling oflf the values of X for which y = 0, we have x = — 2 and a: = 1.5 approximately. 82. Parabola. — The curve in the figure is 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 X-axis, are called the ic-intercepts. 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 equation r o X 1 V / zj = 2x' + X \ Y 5 / A O A X \ \ •^ I Parabola y = ax.^ -\- bx -\- c with its axis parallel to the ?/-axis. (We is always a parabola assume a ?^ 0.) » The parabola will cut the x-axis in two distinct points, or be tangent to the a:-axis, or will not cut the a:-axis at all according as the equation ax'^ -\- hx -{- c = Q 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 rr, and in the third for no real value of x. These three cases are illustrated in the figures below. IS Y r) \ 1 \ \ 1 \ \ y X ._! 9 -2 S = x* — 4x + 3 y = x' — 4x + 4 V-AaOO 6^-4ac = y = x^ — 4x + 5 V- 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?/ = x2+2x, y = x'^-\-2x-\-\, and 2/ = x2 + 2 x + 2. 21. Same as in 20 for ?/ = -x2-2x, y= -x2-2x-l, and ?/=-x2-2x-2. 22. What is the effect on the graph of y = ax2 + 6x + c when c is increased or diminished? 23. What is the efifect on the graph of changing the signs of all terms of the trinomial? 85. Equations Reducible to Quadratics. Example 1. 2 x^ - 7 x2 + 6 = 0. Solve for x^ as the unknown quantity. ^,_7±V49-48^2or | X = ± V2 or ± i/|- Example 2. x-^ -8x-* =9. Solve for x" " * as the unknown. ,-J.8±>0,9„, I. 1 X = ;,}^ or - 1. 729 86] QUADRATIC EQUATIONS 61 Example 3. (2 x2 + 5 x)2 - 6 = 2 x2 + 5 x. Solve for 2 x2 + 5 X as the unknown. (2 x2 + 5 x)2 - (2 x2 + 5 x) - 6 = 0. 2 x2 + 5 X = ^-^ = 3 or - 2. 2x2 + 5x=3 or 2x2 + 5x= -2. X = 1 or — 3; or, X = — I or — 2. Exercise. Verify the answers in the above examples by substitution. Example 4. X + V2 x2 + 1 = 1. Transpose: V2 x2 + 1 = 1 - J Square and collect terms: x2 + 2 X = 0. Therefore X = or Example 5. X- V2x2 + 1 = 1. Transpose: - V2 x2 + 1 = 1 - ; Square, etc.: x2 + 2 X = 0. Therefore X = or Exercise. Ve rify the answers in examples 4 ar 2, as in example 4. On substituting the values found in examples 4 and 5 in the given equations, we find that the first equation is satisfied by both values 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 = '\/2x- -\-l. 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 greater than 1, V2 a;^ is certainly greater than x, and therefore V2 x^ + 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. ). Exercises and Problems. "5 = VSx + l. 3. V2x2 _5x + l = ^x + l. V3(5x + 7). 4. 4 X - 1 = V7x2 -2x + 4. 62 QUADRATIC EQUATIONS. EXERCISES 6. ^Jx + 3 + Va;+8 = 5 V^ 6. V2x + l + V7x-27 = V3a;+4. 7. Vx+^ + V3x -3 = 10. 8. Va; + 17 + Va; -4 = I V2x. 9. V2a; + 1 + Va: - 3 = 2 V^- 10. V12+X = V7a;+8 - 2. 4 4 15 16. 17. 2/ + V4 5 y - ^/4 -y^ 5 X + \/x^ + 5 X - Va;2 + 5 4^=^ = 1+^(757-1). VS 2 + 1 2 11. 12. 13. 14. 12_ ' 7 Vx2 + X + 3 ^3_ V2x2+5x-3 ~4 V3x2+x + 5 ^3_ V4x2 -x + 1 2 V9x2+6x + l ^3_ Vl8x2-3x-2 2 V9 x2 + 6 X + 1 ^ _ 3. Vl8x2-3x-2 2 18. ^ / ^ + 3 V2 y + 1 = 7 V«^. 19. V2t^ + 1 V3s + V5s = V3s + 1. 20. V7iT4 +iUiii = 7 V4^ 21. V4< V3 x2 + 1 - V2 x2 + 1 _1 V3 x2 + 1 + V2 x2 + 1 7 (Or, by composition and division rationalize the denominator.) 22 V27 x2 + 4 + V9 x2 + 5 ^^ ' V27 x2 + 4 - V9 x2 + 5 „« %/5x ■4 + X _ Vix + 1 VSx - 4 - VS - X V4 X - 1 24. V2 V2 X Vx + X2 1 + \/x2 - 16 + Vx + 3 = 14. 7 86] QUADRATIC EQUATIONS. EXERCISES 63 30. X + m _p — X X — m p + X n — X _ X + p n -\- X X — p a^ X X X ft2 a2 _ ft2 ab — X _ b — ex b — ax be — X 36. 31. ?i±i^+^iziA^ = 2x4. 1 ^• l2 - X2 X a X b X - 7n* y^ — m^ 10 n Va. 2/ ■ I y/a -m y + m . + X - Va - X = . m - - V2 my - y2 m + V2 /ny - r/2 ;. Vx + V2a-x =~ \Jx 37, y/x + '\/b Va - X + VT Vj — Vfe \la — x — \lb — : . a/t + Va - V^ + a;2 = Va- X + V — («- + ax) \Ja — 40. + Va Vo — X — V^ — b /I la — X 41. 2 X Va: - 3 ^« a - X 42. — F= X -6 = 20. V:t 43. l/^ + iA-+^ = . \' 6 + X y o — X 44. v/F^-v/^^ = yb-x ya — X 45. x'^' -16x^ =512. 46. x2"> + 2 ax'" = 8 a2. 47. x2 4- V5x + x2 = 42 - 5 z. 48. Vx -5Vx2= - 18. 49. 7 V^^ + V.^2 = _ 12. 50. x2+24= 7x- Vx2-7x+18. 51. ^(TT^^- ^V(i^^ = ^/^^^ 62. Find three consecutive integers, the sum of whose squares is 1202. 53. Find three consecutive even integers, the sum of whose squares is 776. 54. The sum of the squares of three consecutive integral multiples of 4 is 3104. Find the numbers. 55. A rectangle, twice as long as it is wide, has an area of 1800 square feet. Find its dimensions. 56. 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 ? 57. As in 56, except that the original dimensions are aXb inches and the remaining area A square inches. 58. 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 59. As in 58, when the original dimensions are a X b inches. 60. State some values of a and b for which exercise 59 is impossible. 61. Find the radius of a cylinder whose height is 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. 63. As in 61, except that total surface equals n times the volume. For what values of n is the problem impossible ? 64. What number exceeds twice its square root by 3 ? 65. The sum of the ages of a father and his son is 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 of x is x^ + 10 x + 9 positive ? zero ? negative ? (Graph.) 68. 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 a 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%™y" is of degree m\n.x,nmy,m + n in x and y, m -\- n -\- y m 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 + qy -\- r = 0, (2) aa;2 + by^ -\- cxy -\- dx -\- ey -{- f = 0. Solution. 1. Solve (1) for one of the variables in terms of the other. Thus : px + r y = — ' 2. Substitute this value in (2), obtaining a quadratic equation 89-91] SIMULTANEOUS QUADRATICS 65 3. Solve this quadratic for x, and let its roots be Xy and X2. 4. The corresponding values of y are now found by substituting these values for x in the first step. Thus: Tpxi + r , vx2 + r Vi = — ^-— — — ' and 7/9 = — ' ■• Example. (a) x + y = 1, (b) a:^ + 2/2 = 4. From (a), y = I — x. Substituting in (b) : x'- + {1 - xy = 4 or 2 x^ - 2 x - 3 = 0. Hence xi = 5 + i V7; xa = 5 — i V"- Then ^1 = ^ - i V7; 2/2 = I + W7. Reducing to decimals, we have approximately (xi, 2/1) = ( + 1.8, -0.8) and (x2, 2/2) = (-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 he real and un- equal, real and equal, or imaginary. 90. These three cases may be illustrated by means of the equations, (1) (2) Then x^ + {k - xY Hence xi = h {k + V yi = i{k- VS-F) and 2/2 = U^' + Vs"^^/^). These solutions will be real and unequal if A- < 8; real and equal if A;- =8; imaginary if k^ > 8. 91. Graphic Solution of the equations (1) x-hy=l, (2) x^ + y^ = 4. x + y=k, x' + t = 4. = 4, or 2x? ■ -2kx-{- (k- X2 = h(k- --4) V8- = 0. '8-k'') and k-n. 66 SIMULTANEOUS QUADRATICS [92 r / K ~~^ N, / X X \ V \ y ' Straight line x + y Circle x^ + if' 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 2/ = ± V4 - X". Assuming a set of values for x, and calculating the corresponding values of y, we have a: = 0, i, 1, U, 2, 2^, . . . ; y = ±2,±h Vi5, ± V3, ± ?; 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 (:ri, 2/0 = (1.8, -0.8) and (0:2, 2/2) = (-1-8, +0.8), as in (88). 92. Graphic Illustration of the Three Cases of (90). — In (1) of (90), let us put successively A; = 1, 2 V2, and 4, so that li- < 8, = 8, and > 8 respectively. We have then the equations, (1) a; + y = 1; a: + 2/ = 2 V2; X + 2/ = 4, (2) a;2 + i/^ = 4; x2 + 2/2 =4; x^ + if = A. The three straight lines and the circle are shown in the adjacent figure. When k = 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 x-+ y- r\ \ \ \ y \ --^ \ \ 1 \ \ \ \ \1 \ y 93,94] SIMULTANEOUS QUADRATICS 67 93. Standard Equation of the Circle. — The equation x- + if = r' 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 called the standard equation of the circle. x' + y- = 7^ Exercises. Solve for x and y, and check care- Circle, radius r, center at fully by graphs. ,x2 + 2/2 (x-y =0. ^' U-y = 2. 4. (X2+2/2 \2x + y 7. ( X2 + J/2 = 4, origin X2 + 2/2 =9, \3x + 4y = 12. [x2+2/2 =9, ! 4 X - 5 2/ = 20. ;4a;2 + 42/2 = 1, |3x-2/ = l. (x2+2/2 = 1, (x2 + 2/2 = 16, * (x-2/ = ^A2. (2x -3?/ = 4. 10. Determine A; so that the line x + y = k shall be tangent to the circle a;2 + 2/2 = 4. 11. Determine m so that the line y = mx + 5 shall touch the circle x2 + 2/- = 5. 12. As in 11, for the line y = mx + 2 and the circle x- -\- y- = f. 94. Consider the equations x-y = 1, 9^4 Proceeding as in (88), we obtain Xi 2/1 + 12 V3 ^ 13 -4 + I2V3 13 2.3 12%/^ 1.3 13 -4 0.9 12 V3 13 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 y=±\ V36-4z^'. Whena:=-3, -2, -1, 0, +1, + 2, + 3, y= 0, ± 3 Vo, ± 5 V'2, ± 2, ± ^ V2, ± § Va, 0. 68 SIMULTANEOUS QUADRATICS [95,96 / B J / ^ "^ X^' ( / X A \ " / A ^s ^^ / ^ / X,Y., k 32+2^ 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 = Q and B'B — 4; OA = 3 and OB = 2. Scaling off the coordinates of the points of intersection of the two Ellipse Straight line x — y = 1 graphs, we have as our graphic solution (xi,yi) = (2.3, 1.3); (x2,y^^ = (-0.9, 95. Standard Equation of the Ellipse. — Every equation of the form represents an ellipse, whose major axis is 2 a, minor axis 2 h, center at the origin. It is calle'd the standard equation of the ellipse. Exercises. Solve and check by graphs : 1, .1 rx2 , 2/2 '9+4 [x + 2/ = 0. 9 + 4 ^' [x + y = 5. 9^4 ^' 4x2 + X + 2/ X2 y2 = 36, -Vl3. Ellipse, semi-axes a and b respectively (9x2 + 162/^=25, ^' l2x-3y = 6. ^ +y^ = 1, x + 2y = 2. x2 + 4 2/2 = 4, X - y = 3. 12x2 + 3^2 = |3x + 2/ = 2. ,9x2+42/2 = [x-y = 1. 12, [x + y = VI3. 10. Determine k so that the line x — a;2 + 4 2/2 = 4. 11. Determine m so that the line 4x2 + 92/2 =36. k shall be tangent to the ellipse TOX + 3 shall touch the ellipse 96. Consider the equations x-y = 2, y- ^ ix. 97' SIMULTANEOUS QUADRATICS 69 Y / v^ ^ ■^ V ^ ■^ / / / / / / f / X / \ / ^^ ■ ^^ •^ _ 1 ^ ■ Solving as in (88), we find xi = 4 + 2 V3, a^o = 4 - 2 V3, ?/i = 2 + 2 V3, 7/2 = 2-2 V3. The graphs are shown in the figure, that of the equation y- = 4 x being a parabola, whose vertex is at the origin and whose axis is the x-axis. Exercise 1. Compare the graphic solution with tliat obtained by formula. Exercise 2. For what value of k will the Parabola, y^ == 4 x line X — y = k be tangent to the parabola U- = 4 X? Why arc there not two values of k as in the exercise of (95) ? 97. Standard Equations of the Parabola. — The equation 2/' = 4 ax 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 a is positive, to the left when a is negative. The equation ic^ = 4 « J/ always represents a parabola, whose vertex is at the origin and whose axis is the 7/-axis. The curve extends upward when a is positive, downward when a is negative. Parabolas \ Y .r= 4 ay \ "/ \ / \ 1 i \ x= = -4 ay y- = — 4 ax y- = 4 ax Exercises. Solve and check l^y graphs: [y = X. (X + y = \. ;4y2=x, |2x-2/ = 4. 70 SIMULTANEOUS QUADRATICS [ 98, 99 \Zx + y = 3. \y = 2x. ( .t2 = 4 y, \x+2y- : 2 X + 5 2/ = 10. ;x2 = -4?/, !2/-2x = l. 10. Determine k so that the line 3 x + y = A; shall touch the parabola 2/2 +4x =0. 11. Determine m so that the line y = mx + 2 shall touch the parabola t/2 = 8 x. 98. Consider the equations x-2y = Z, Hyperbola, '^^ - ^j = 1 Straight Line, x— 2y = Z The graphs are shown in the figure. The graph of the second equation is an h3rperbola, 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 ^3 ^.2 b" = 1 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 b, parallel to the coordinate axes,- with its center at the origin. The equation ^ _y^_ ^ _ ^ a^ b' represents an hyperbola whose axis coincides with the y-axis. 100,101] SIMULTANEOUS QUADRATICS 71 Hyperbola, -^ — vj = 1 a' h Exercises, Solve and check by graphs: 4. ^ 9 ~ 4 5x + 2/ 1. 2. \2x -y- 32/ 2/" (X2-2/2 ^ =36, 6. 2/ = 2. erbola, a? ¥ 7. 2 x2 - 3 y= = 6, 3x + y = 6. 8. (.r2-2/2=_i, l2/-3x = l. 9. (4 x2 - 9 2/2 = - 122/ -x=0. 36, fc shall be tangent to the 5. ;4x2 -9 2/2 =36, |4x + 2/ = 2. [x2 -4?/2 =4, 12/ = 2x-6. 10. Determine k so that the hne x — 2y = hyperbola 4 x2 - 9 2/2 = 36. 11. Determine m so that the line y = mx - 2 shall touch the hyperbola x2 - 2/2 = 1. 100. Rectangular Hjrperbola. — 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. \ Reclanqular hyper Red. Hyp. xy = ¥ Red. Hyp. xy = - 1^ 101. The general equation of the second degree, «x- + 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 eases 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 x2 ^ y2 = 0. The graph may also be imaginary, that is, the equation cannot be satisfied by any real values of x and y, as a;2 -f- i/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 coefficients a, 6, 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 + 2/2 = 9. 11. x2 = 4 y. 2. 4 x2 + 4 2/2 = 16. 12. 4 x2 = 2/. 3. 3x2+3 2/2 = 15. 4. 4 x2 + 2/2 = 4. 5. x2 + 4 2/2 = 4. 6. 16x2+25 2/2 = 400. 7. 25x2 + 16 2/2 =400. » 8. 2x2+4 2/2 = 9. 9. 2/2=4 X. 10. 4 2/2 = X. Construct the graphs of the preceding £ Construct the graphs of the equations: 21. x2 + 2/2 - 6 X - 8 2/ = 0. 22. (X - 2/)2 = 1. 23. 3x2+2x2/ + 32/2-162/ + 23=0. 24. x2 - 5 xy + 6 2/2 = 0. 25. 3 x2 + 2 2/2 - 2 X + 2/ - 1 = 0. Solve graphically and by formula several of the preceding equations with the equation (a) x-2/ = 1. (b) 2X + 3 2/ = 6. (c) x + 2/ = 0. (d) 2x-2/ =2. 13. 2/2= -4x. 14. - 4 2/2 = X. 15. x2 = - 4 y. 16. 4x2 =_y. 17. 16 x2 - 25 2/2 = 400. 18. 16x2 -25 2/2 =-400. 19. 25x2-16 2/2 = 400. 20. 25 .x2 - 16 2/2 = - 400. qual tions on cross-section papei 26. x2 + 2 X2/ + 2/2 = 0. 27. 5 x2 + 2 X2/ + 5 2/2 = 0. 28. 4.r2/+6x -S2/ + I =0. 29. 2/2 - X2/ - 5 X + 5 2/ = 0. 30. xy -2/2 = 1. 103-105] SIMULTANEOUS QUADRATICS 73 103. Solution of Two Simultaneous Quadratics. — When both quadratics are of tlie general form, as ax~ + by- -\- cxy -\- dx -\- cy -\- f = 0, a'x^ + b'y'~ + c'xy + d'x + c'y +/' = 0, they cannot usually be solved by elementary methods. For, if we solve one equation fOr 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. Whe7i the graphs intersect in less than four points (tangency is counted 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. 105. Case 1. Two quadratics, one of which is factorable. Ride: Factor the equation, put each factor equal to zero, and solve each of the resulting linear equations with the other quad- ratic. Rule 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 two values of y are of the form y = ax -{- b and y = a'x + 6. The required factors are then (y - ax -b) (y - a'x - b'). Graphically, 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 1 + ^ = 1 E_ + y = 1 9 4 9^4 Four real solutions, Four real solutions, all distinct. two being equal. t + t = l 9^4 (x - iy+ if = y Two real distinct solutions, two imaginary. Two real and equal solutions, two imaginary. X' - 7/ = 1 X - y =Q Two solutions, both infinite. &Q f-|' = l;i(x-6)^ + r = f All four solutions imaginary. xy = 1 xy = - I Four solutions, all infinite. The student is urged to draw, or to picture to himself mentally as far as possible, graphs corresponding to all equations considered. He should be able to recognize at a glance the standard forms of equation of the conic sections. 105] SIMULTANEOUS (QUADRATICS 75 all). Hence the four solutions may be all real and distinct, or equal in pairs, or imaginary in pairs. x2-2.Ti/-3i/2 =0, x2 - 4 (/2 - 4 = 0. Example 1. The factors of the first equation are, by inspection, {x + 7j)ix -3y) = 0. X + y = or X - 3 2/ = 0. \ Y ^. \ ./ >s \ ^> \ N o_. ^ f X r \ \ ^ ^ \ xj ^ \ \ \ Hence we have to solve Hyperbola, x^ — Ay- — 4 =0 Straight lines, x^ — 2xy — 3 ?/' = or X -V y = and x — 3 y = U + 2/ = 0, ( x2 - 4 y2 _ 4 0, Solving the first pair, we have {^h y\) h) and (^2, 2/2 3 2/ =0, -42/2-4 4 4 These are imaginary. The line x + ?/ = does not cut the hyperbola (figure). Solving the second pair, fe, 2/3) V5 Vo/' ' V Vs These solutions are real, and the approximate values may be scaled off from the figure. Note. An equation of the form Ajc' + Bxij + Cj/"' = cayi always he factored. Divide by the square of one of the variables, and solve for the ratio - or -• X ij The factors will be imaginary if 5- — 4 ^C < 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 5^ — 4 AC > 0, and coincident if .B- — 4 ^C = 0. Example 2. Factor 2 x2 - 2 xi/ + ?/2 = 0. Divide by x2: /^ V _ 2 ^| j + 2 = 0. ^ = 1 + V^n[ or 1 - V^l. X Hence the factors are [2/ - (1 + V^T) x] [2/ - (1 - V^^) x] = 0. 76 Example 3. SIMULTANEOUS QUADRATICS 0. [106 i2x2-2/2 \x^ -Ay - xy + 3 2/ - 2 0. \ \ Y 20 / \ / V 10 /. / \ ■' / 1/ v^ / ^ K !■ y A i \IO Solving the first equation for x in terms of we have Parabola, a;^ - 4 y = C Straight lines, 2x'-y--xij + 3y-2 or x — y + 1 = and 2x + y-2==0 ?/± V9?/2-24y + 16 ^ 2/± (3y-4) 4 4 Hence, X -y + 1 =0 or 2x+?/-2=0. Solving the first of these with the second equation above, we have ixi,yi) = (2+2V2, 3+2V2); (0:2,2/2) = (2-2V2,3-2V2). From the second equation we obtain (x3,2/3) =(-4 + 2V6, 10-4V6); (.T2, 2/2) = ( - 4 - 2 V6, 10 + 4 Ve). Exercises. Solve for x and y, and check graphically: (x2 + 2/2 = 1, \x^ + yx -2 2/2=0. (3:2 + 2/2 =4, ^^2- 2/2=0. ( 4 x2 + 9 2/2 = 36, (2x2 + 5x2/ +32/2 =6x + 62/. (X2-2/2 = 1, |x2/-22/+x =2. ( 2/2 - 4 X = 0, |6x2+x2/- 12 2/2 = 4 2/2 106. 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 Ax^ + Bxij + Ci/ = D, (1) D'. (2) AV + 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) A"a;2 + B"xy + C'Y~ = 0, which may be solved with either of the given equations by factoring, as in Case 1. 106] 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 factors 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 coincident 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 roots we may represent by vi and v-z. Then y = v\X and y = vox. Substituting these values in turn in either of the given equations, we have two quadratic equations in X alone. Example 1. 2x2 4x2/ 3 x;/ + 4 =0, 5 7/2-3=0. Transposing the constant terms we have 2 x2 - 3 x?/ = - 4. 4 x?/ - 5 2/2 = 3. Multiplying the first equation by 3, the second by 4, and adding, 6 x2 + 7 xy - 20 2/2 = or (3 X - 4 2/) (2 X + 5 2/) = 0. — — - "f - /I- ~ / > V ^. / '-. ^^ '/ (Z) 'TzT "" ~~- ,' ' ■■- X / " ~- /. >^ \ " ^ ^ „ ,, i2x'-3x2/ + 4 = 3 X - .4 y = Straight lines, \2x + 5y Equating each of these factors to zero, and solving with one of the given equations, we have, from the first factor, (xi, y\) from the second factor, N, 2/3) = (2 V^ - (4,3); (x2,2/2) = (-4, -3); J V^); (x4,2/4) = (- h V-5, I V-5). Hence two solutions are real and two imaginary. The figure shows the graphs of the given equations and of the factors of the auxiliary equation. 78 SIMULTANEOUS QUADRATICS [107 To solve by the second method, transpose the constant term as before, then put y = vx. 4 !;x2 - 5 !;2x2 = 3. 2-3?; 4 4 V — 5 1^2 Clearing, etc., 20v'^ -7 v - & = Q. Hence, ?; = f or — f . Therefore y = Ix or y = — Ix. (These are the linear factors of the auxiliary equation found above.) Substituting these values of y in either of the given equations, we find X as before. Example 2. 9 x2 + a;!/ + 2 ?/2 = 60, 8 x2 _ 3 a;?/ - 2/2 = 40. The auxiliary equation is 6 x2 - 11 x?/ - 7 2/2 = 0, or (2 X + ?/) (3 X - 7 2/) = 0. Solving each factor with one of the given equations we obtain (xi, 2/i) = (2, -4); (X2, 2/2) = (-2,4); Ellipse, 9 x= + xy + 2 2/2 = 60 Hyperbola, Sx^- Zxy -y- = 40 Straight lines, {2x + y) {3x - 7 y) = The graphs are given in the figure. Exercises. Solve for x and y: (^^'^^^=(^'^) = (x2 + 2/2 =9, I x2 - X2/ = 10. JX2-2/2 = 1, (x^ — xy + y^ = ( 4 x2 - 9 2/2 \y^ + xy = {xi, 2/4) 7__ V2' _3 2V2 }■ (x2+2x2/ = 2, 1. *• (2x2/ -2/' =6. ( x2 + X2/ + 2/2 = 3, 1 2 x2 - 3 2/- = 6. i2.r2 + X2/-3 2/2 = 2, \x'--xy + 2y^ = l. 107. Case 3. The given equations are of the forms ax- + by^ = c, Rule. Consider a;^ and \f- as the unknowns, and solve by the method of Hnear equations. lOS] SIMULTANEOUS QUADRATICS 79 Graphically, we have two conies in standard form. The four solutions may all be real, or equal or imaginary in pairs. Example, x- — 4 y^ = 4, 9 x2 + 16 2/2 = 144. By elimination we obtain, Hence x = ± 4 Vi§; 2/ = ± 3 VS. Taking eitlier value of x with either value of y, we obtain the four solutions. The approximate values may be scaled off from the Figure. r >^ ^-^ "~~~ V- ^ - ^^ \'^ ' ^ o 7J\ x I 2 \i X >^ ' ^ ^-^ --^ ^: Hyperbola, x^ — 4 t/^ = 4 Ellipse, 9x' + 16?/' = 144 Exercises. Solve for x and y, and check graphically: ^ (x2 + ;/2=4, 3_ j2x2 + 52/2 = 10, ^ 1x2-7/2=2. (X2- 2/2 = 1, 1x2 + 4 2/2 = 4. I 4 x2 + 2/- = 4. :x2 + 2/2 = 9, ' 4 x2 + 9 2/2 = 36. ( x2 - 2/2 = 9. i x2 + 2/- = 1, 1x2 + 2/2 =4. 108. Case 4. Symmetric and Skew-Symmetric Equations. — A symmetric equation is one which remains unchanged when the variables are interchanged. A skew-sijmmetric equation is one whose variable terms all change sign when the variables are interchanged. Thus x^ + y^ + X + y = 0, x'^-y^ + 2x-2y=\ 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 X = ^ (u -{- v) and ?/ = V (m — v). Note. Equations of higher degree than the second may often be solved by this methorl. Example. x^ + y' x~ + y :-y- = 9, ■vy = 3. Let u + V and y 80 SIMULTANEOUS QUADRATICS [109 "Substituting and reducing: u* + 14 w2y2 + ?;4 = 9, m2 + 3 y2 = 3. Let u2 = s and v"^ = t. Then s^ + U si + t^ = 9, s + 3 < = 3. Solving: (s, <) = (3, 0)or (f, I). (If s and / be considered as the coordinates of a point, the equations in s and t represent an ellipse and a straight line respectively.) Since w = ± Vs and v =±^l, we have (m, f) = ( ± Vs, o) or (±^' ±^ where the signs are to be taken in all possible ways. Then X = w+t; = VS, -V3, V3, -V3, 0, 0; y =u-v = -\JZ, -\l2,, 0, 0, V3, - V3. Here corresponding values of x and y appear in the same vertical line. 109. Case 5. 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, xy = 4. Squaring the first equation: x"^ -\- 2 xy + y"^ = 25. Subtracting four times the second : x2 — 2 x?/ + 2/2 = 9. Hence x - y = ± Z. . ' But X + 2/ = 5. X = 4 or 1 ; y = 1 or 4. Example 2. (1) x^ + xy + y^ = 6. (2) x2 - xy + 2/2 = 10. Subtract (2) from (1) : 2xy =- 4, or xy -' - 2. Add x2/=-2to(l): x2 + 2x2/ + 2/2 = 4, or x + y=±2. Subtract 3 x?/ = - 6 from (1) : x2 - 2 X2/ + 2/^ = 12, or x -y =±2 V3. Hence x = ± 1 ± Vs and 2/ = ± 1 T \/3. Simultaneous values of x and y are then obtained 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 u- = s and v^ = t. We shall give two more simple illustra- tions. 1 Example 1. i-.-15. xy If we let ^ = s X and 1 _ y ~ t, and we obtain, s + < = 2, St =- 15. These may be solved by the method of (109). Example 2. + x^y2 +2xy = 4, x2i/2 -2xy =0. Let X -{- y = s and xy = t. Then <2 - 2 < = 0. The last two equations are readily solved, and give s = + 2; - 2; 0. ^ = 0; 0; 2. The values of x and y may now be found by solving the pairs of equations, (x+2/ = 2, (x + y=-2, (x + i/ = 0, (xy =0. \xy =0. \xy = 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) x^ -xy = 18 y, (2) xy -y^ =2 x. Dividing (1) by (2), we have - = 9^ or (^y= 9 or x = ± 3 y. y X \yj 82 SIMULTANEOUS QUADRATICS [111 Substituting each of these values of x in either of the given equations, we can solve for y and so complete the solution. Example 2. (1) ix'^y-x = l, {x(x2/-l)=l, (2) ( a:%2 -x^- = 3; \ x'~ (x%2 _ 1) = 3. Divide (2) by (1): x (xy + 1) = B. Divide this equation by (1): ^ ^ ^ = 3. 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, y = 2. Examiple 3. (1) x4 + y" = 1, (2) X - y = 1. Raise (2) to the fourth power and subtract from (1): (3) 4 x3y - 6 x2y2 + 4 xy3 = 0. Square (2) and multiply the result by 4 xy: (4) 4 x3y - 8 x2y2 + 4 xy^ = 4 xy. Subtract (4) from (3) : 2 x2y2 = - 4 xy, or x2y2 + 2 xy = 0. Hence xy = 0, or xy = - 2. Solving each of the last two equations with (2) we have (x,y) = (l,0), (0,-l),(^^^ , V^^j\—^' 2 }' All four solutions also satisfy equation (1). 111. Summary 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. Equation (1) is factorable. Rule: Put each factor separately equal to zero and solve with (2) as in (a). Graph, two straight lines and a conic. Rule for factoring: Solve for y in terms of x (or x in terms of y) ; the quantity under the radical must be a perfect square. Ill] SIMULTANEOUS QUADRATICS 83 Case 2. (l) Ax' + Bxij -\- Cif= D; (2) A'x' -{- B'xy + CY = D\ Form the auxiliary equation, (1) X D' — (2) X Z> = 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' + BY = 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. x2 + y2 = 661. 8. x + y j<, 15. 5 X + 2 2/ = 29. X^ - y2 = 589. X -y 5 X2/ = - 105. 2. 2/2 - x2 = - 80. x^-y^= 48. 16. xy = 80. x'- + f' = 82. 9. 5 x2 + 2 2/2 = 373. x = 52/. 3. 3x2-^2 = 59. 2x + 5 2/ = 54. 17. 4x2-3 2/2 = -83. 2a;2 + 3 7/2 = 98. 10. x2 + 2/2 = 10. X - 2/ = 2. 3x + 22/ = 26. 4. x + 2/ = 12. X2/ = 35. 11. x2 - 2/2 = 120. X + 2/ = 20. 18. 3x2-2/2 = 83. X + 2/ = 15. 5. x + y = 1. xy= -I. 12. x2 - 2/2 = - i x + y = l 19. X2/ + X = 20. xy - y = 12. 6. x^ + y^ = 74. x + y = 12. 13. x2 + X2/ = 260. xy + y^ = 140. 20. 2 X + 3 2/ = 20. Sxy-y'~ =38. 7. x + y = l xy = h 14. x2 + 2/2 = 218. xy-y^= 42. 21. 5x2-42/2 = 109. 7 X - 5 2/ = 25. 22. x+xy + y = 4:7. 27. x2 + X2/ + 2/2 = 4. x + y = 12. x2 - xy '+2/- =2. 23. x2 + xy + 2/- = 217. 28. X2 + xy + 2/2=7. x + y = 17. X - f 2/ + xy = 5. 24. 1+1=1. 29. X2 xy + 2/- = 3. = 5 (X + 2/). 1+1=1. 30. .r3 + 2/3 = 9. x2 ^ 2/- -i^ X2/ = 4. 26. 2x2 -3x2/ + 2/2 = 3. 31. X2 -42/2=4. x2 +2x2/ -32/2 = = 5. X2 -2x2/ + 2x =42/. 26. x2 - X2/ + 2/2 = 37 . 32. 2x :2-2 ! 2/2 +3x2/ = -x-2j/. x2 - 2/2 = 40. X2 -4i y2 - X + 2 2/ = 0. 84 SIMULTANEOUS QUADRATICS 33. -u2 + y2 -I- uv = 67. u+v = 9. 34. p^ + pq+q'^ = 79. p2 _ p5 + g2 = 37. 35. r2+s2 +rs =25. r +s =5. r2 + -rs == 84. 2. 37. u + t' + u2 + i;2 = 162. ,, _ ,, 4- ^2 _ i,2 = _ 102. 38. p+g + p2 +q2 = ig. y _ p + g2 _ p2 = _ 1. 39. a;2 + 2/2 + a; + 1/ = 18. 2xy = 12. 40. ;i2 + fc2 _ A; + /i = 32. 2 /ifc = 30. 41. x2 + y2 + X + 2/ = 168. •Va;y ■= 6. 42. m2 + n2 - ?rt + n = 2400. ^/mn = 30. 43. 9tt2+t;2+3M + i' = 3042. Vl6w = 48. 44^ ^3 _ s3 = 1304. r - s = 8. 45. p* + ^ = 337. . p + 9 =7. 46. x" - 1/ = 609. X - 2/ = 3. 47. u4 + ?^^ = 2657. M+i; = 11- 48. m3 + n3 = 152. m2 - mn + n^ = 19. 49. p + <7 + V/H^ = 20. p3 + ^3 = 1072. 50. .t3 + 1/ = 280. x2 - X2/ + 2/2 = 28. 51. m2 + 3 ifl = 7. 7 u2 - 5 wy = 18. 52. p3 -)- gS = 152. p2q + pg- = 120. 53. x3 - 2/3 = 335. X2/2 - x22/ 70. 54. s3 + i3 = 855. St (s + = 840. 55. w3 - n3 = 602. mn{n — m) = — 198. 56. W2y4 + 1,2 = 17. W2;2 + y = 5. 57. xl -{-yi = 35. xi + 2/^ = 5. 58. x2|/2 -18x2/ + 72 =0. 6x2 -17x2/ + 12 2/2 =0. 69. x4+x22/2+2/4 =91. x2 - X2/ + 2/2 = 7. 60. x3 - 2/3 = 7 (x2 - 2/2). x2 + 2/2 = 10 (x + y). 61. s6 + <6 = 65. S4 + i4 = 17. 62. x2 + 2/2 = a. x2 - 2/2 = 6. 63. X - 2/ = wi. X2/ = n2. 64. 7^2 + g2 = a2. p + g = 6. 65. Vw + V^ = «• u + 1; = 62. 66. x2 + 2/2 = a (x - y). x2 + 2/2 = b (x + 2/). 67. ax-by = m. a3x3 - 632/3 = nx2/. 68. 6(x + 2/) =a(x-2/)- x2 + 2/2 = w2. 69. x* + 2/* = - 8- X - 2/ = 2. 70. p* + g^ = - 9- p -(/ = 3. 71. u* + t;4 = 175. M — V = 5. 72. 7-2 -|- rs + s2 = a. r3s + rs3 = 6. SIMULTANEOUS QUADRATICS 85 73. l-i=l. X y a 75. X3 + X2/2 = p. y3 -1- x2y = q_ Ui^=i^- 76. m^ — 7i3 = a (w — n). a;2 ^ y2 62 7h3+ 7t3 = 6 {)ii +n). 74, ti2 4- Mi; = m. 77. r5 + ,s5 = 3368. v^ + 7/y = n. r + s = 8. 78. x(.r + y-2) = 1. 80. xy = Sz. 82. X (x + 2/ - 2) = «• ?/ (x + ?/ - 2) = 2. xz = IS 2/ 2/ (x + 2/ — 2) =6. z{x + y-z) = 3. 2/2 = X. 2 (x + 2/ - 2) = c. 79. X + 2/ + z = 2. 81. x?/ 4- X = 1. 83. X + 2/ + 2 = p. x;/ =3. 2/2 +2/ = - 1. X2/ = (7. xyz = 6. xz -\- z = 3. xyz = r. 84. xix+y+z)^ a2. 87. X2/ + X = a. 2/ (x + 2/ + 0) = &2. 2/2 +2/ = ^• 2 (x + 2/ + z) = c2. xz -\-z = c. 85. (x +y)(x+z) = 4. 88. x- -\- y- = 17 z. (x + 2/)(y+2) = 1. 3(x + 2/) =52. (x + 2) (2/ + 2) = 16. X -2/ =2. 86. (x + 2/) (x + 2) = a. 89. X2 + 2/- + 22 = iU. (x +y}{y+z) = b. 2/-+.J^ = ^l. (x + 2) (2/ + 2) = c. 22+X = i|. Problems. Oo^)y ^^— 1. The hypothenuse of a right triangle ia [Too" ft. long. Find the other sides, if their ratio is 3 : 4. 2. The product of two numbers is 735, and their quotient ^ Find the numbers. 3. Find two factors of 1728 whose sura 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? 5. The product of two numl)ers 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 546. 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 yj 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 \% 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 original 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 t/-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: Vct^ =a2^-i; (m^+i)-^ = ^-2^-2; a^+i =62x-i^ 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" = fo", then w log a = t; log h. Example 1. Vo^ = a^^-^. This may be written a^ = a^-^-i. •■• 1=^^-' - -=!■ Example 2. (TO^ + 1)x = ;,j-2x-2. m^^ + x = 7,1-21-2. x2 + a;=-2x-2 or x2 + 3a; + 2 Hence X = - 2 or - 1. Example 3. 2^+1 = 32x-i_ Taking logarithms: (x + l)log2 = (2x - l)log3. ••• X (log:2 - 2 log 3) = - log 2 - log 3. or log 2 + log 3 log 6 21og3-log2 logl' Using common logarithms to four places, .= 0-7781 + _,,,,,^ 0.6532 + ^•^"^^'' ^• 114. Exercises. Solve: 1. 2^ = 8. 4. (i)^ = 2i2. 7. i0-x = i000. 2. 2^ = J. 5. (i)x = 1. 8. 1000^ = 100. 3. 4x = Jj. 6. {^UY = 253. 9. 3^ + 2 = 33. 10. \/a^^=a-'-2. 18. 42^-3 =7x-i. 11. -^p2x + 8 = pO. 19. a3-r+2 = 62X-1, 12. 42*-i =26^+8. 20. 3^'-x-6 = 1. 13. \frn^ =Vw^^ + 2. 21. 8^^+2-^=512. 1 3, • ?^ x''-! 1*- osi = Va6-i3x. 22. 5-^-2 =252'x + i'. 15. ''^■'-\/^=''"v'a"S. 23. (ax-2)3x + 4 = ax(3x + i,. 16. 3x = Vs. 24. (6x + 3)3x-i = 53x(x + i.. 17. 3^ + 1 = 522=. 25. that is, 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 5 the speed in miles per hour, then ,10 ,1 t = — or / cc - ■ s s 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 , .n 10 miles when s = 4.5, 7.8, and 15.6 miles st = IQ . ' ' 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 vanj ivith them jointly. Thus, if 2 oc xy, or z = kxy, 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 cyHnder 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. 6. 7. liyccx and x cc z, show that y <^ z. If a; cc 2 and y °^ z, show that ax + li x^ccy and z^ oc y, show that xz cos^l tan^ sec^ = cot^ = 125. 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°. 5. 15°. 8. 80°. 11. 57° 3. 60°. 6. 18°. 9. 10°. 12. 38° Check the results of the preceding exercises by means of the following table. f Of THf \ 96 TRIGONOMETRIC FUNCTIONS [126 Angle. Sin. Cos. 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 0.176 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. cot A = ; Scaling off the a-ngle with a protractor, we have A = 37°. By taking from the table the angle whose tangent is .75 we have A = 37° as before. 1. A 3 / opp. side t^"^ 4 1 adj. side, Lay off AC = = 4andrB =3, ±AC. Then AB = V42 + 32 = 5. Hence • ^ 3. .4. sin A = ^ , cos A = -^ , A 5 . 5. csc A = ;^ ; sec A = J , Sec A 3 hyp. ' 1 adj. side; 127, 128] TRIGONOMETRIC FUNCTIONS 97 Lay off AC = 1. With A as center and radius = 3, strike an arc to cut the -L 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. sin A = 2. sin A = 3. sin A = 4. cosA. = 5. cos A = 6. tan A = 2 • 7. tan A = 4. 8. tan A = V3. 9. cot A = 1. 10. cot A = 1.5. 11. sec A = 2. 12. esc A = 2 • 13. cos A = 0.2. 14. CSC A = 2.4. 15. tan A = 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 l.v^ 21. cos A lies between and 1. 22. sec A and esc A are always greater than 1. 23. tan A and cot A may have any value from to oo . 128. Functions of Complementary Angles. — Since the sum of the two acute angles of a right triangle is 90°, they are complementary. By definition, we have . „ opp. side h . A o c sm B = -^^ = - = cos A. hyp. c Also, cos B = sin A ; tan B = cot A ; esc J5 = sec ^4 ; sec .B = esc A ; and cot B = tan .4. Complementary Functions, 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 a^ -\- b^ = c^. 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 ABC, given A = 40°, C = 90°, and b = 60°. Find the other parts of the triangle. To get B, we have B = 90° - A = 50°. To get a, take t = tan A or a = b tan A. Finally, c is determined from - = cos A b ^. A or c = r — osecA. cos A From the table of (125), tan 40° = 0.839 and A eo 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 a ^ 50.340 „„.„ ^ = cos5 or 78:300 = 0-643. 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 (125).) TRIGONOMETRIC FUNCTIONS 99 1. A= 25°, a = 100. 6. B = 10°, a = 0.15. .2. A = 70°, b = 150. 7. A = 4' ', c = 0.045 3. A = 51°, c = 75. 8. B =S5°, c = 1.25. 4. fi = 38°, c = 50. 9. 5 = 57°, a = 16f . 5. B = 65°, 6 = 750. 10. A = 20°, 6 = ,K. 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. 15. Find the 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 XOP (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 OF^, OP 2, OP3, and OP4 successive- ly, thus generating the angles XOP\ , XOP2, XOP:i, and XOP4 respec- tively. OX is called the initial line, and OPx the terminal line of the angle XOPi, and similarly for any other angle. 100 TRIGONOMETRIC FUNCTIONS [132 An angle is positive when the generating hne 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 OPo, 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 H 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 quadra nt it may 1 . ,.^ „ ordmate (of P) sm xor = -jr- ; , ' ; distance (of I*) CSC XOP distance ordinate ' ,.^ ^. abscissa cos xor = -rr- ; distance sec XOP distance abscissa ' abscissa cot XOP abscissa 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 the Trigonometric ''unctions Quadr. sin. COS. tan. cot. sec. CSC. I + + + + + + II + — — — — + III — — + + — IV - + + - 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 Mpy sin XOP = MP cos XOP = 0M 1 / i-r) Hence approximate values of the sines and cosines of all angles may be read off directly from the figure. The other functions MP may be obtained by division, since tan XOP = ^^rj, etc. They may also be constructed graphically by a method explained in the next article. The lines OM and MP, whose lengths represent the sine and cosine of Z XOP, 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 t^e tangent at S in T. Then by similar triangles, MP ST Ts Y / Tz v-"^^ /7y T £. I \\ f \] T, /^V^^^^^ > / \ Ts tan XOP OM OS ST. In the same way, tanZOPi =STi; tan XOP2 = ST.. Hence when an angle is in the first quadrant, its tangent is measured by the segment of the tangent line from S 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 XOP3 = - S'T^; tan XOP4 = - ^'^4; tan XOP 5 ST, Here ST4 and ST5 are themselves 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. ^ W,e have further. sec XOP = OP^ OM OT ^^ = OT OS ^ 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 R' (figure below). Then cot XOP CSC XOP OM^ MP OP MP m-^^' OT OR = OT. 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 off by the terminal line of the angle produced; (b) 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. N 135. Variation of the Trigo- nometric Functions. — 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 MP, 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 from — 1 to 0. Similarly cos A'OP varies in the four quadrants successively from +1 to 0, to -1, -1 to 0, and Oto +1. Consider next tan XOP=^' When XOP = 0°, MP = and 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°= co, and say that the tangent varies from to ^ 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, CSC XOP Hence esc XOP varies from « to 1 in the sin XOP first quadrant, and from 1 to <» 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°. 104 TRIGONOMETRIC FUNCTIONS [135 This abrupt ci^iange in the cosecant when the angle passes through 180° is ex) pressed by saying that the cosecant has a dis- continuity at 180°; 5'^ec 180° may be either +oc or -oo, according to the side from which XOP approaches 180°. In the third quadrant esc XOP is negative and varies from — 00 to —1; in the fourth quadrant from —1 to — oo. There is another discontinuity at 360° or 0°, The variations of the six functions are shown in the following table. , . Quadr. sin. CSC. COS. sec. tan. cot. II III IV Oto, + 1 +1 toO Oto -1 -1 toO + C0 to +1 + 1 to +00 -co to -1 -1 to -oo + 1 too oto -1 -1 toO oto +1 + 1 to +00 -00 to -1 -1 to -00 + 00 tol to +00 -ootoO to +00 -ootoO +00 toO to-x 00 toO Oto-oo The range of values covered by each of the six functions is indi- cated in the diagram below. sec X and esc x sin X and cos x sec X and esc x tan X and cot 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 ; cscO° = -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 0° to 360°. ,5. By use of the results of exercise 4, trace the graph of the equation y=sm 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 squares. At successive points x on the horizontal axis erect perpendiculars equal to sin x, upward or downward according to the sign. See note at end of (143)]. 136] TRIGONOMETRIC FUNCTIONS 105 Graphs of the Trigonometric Functions Sine Curve *i (full line) Cosine Curve o (broken line) Tangent Curve -^i (full line) Cotangent o Curve (broken line) Secant Curve h (full line) Cosecant Curve o (broken line) I 4^0" [S^O" 106 TRIGONOMETRIC FUNCTIONS [136,137 6. Trace the graph of ?/ = cos x. (On same diagram as ?/ = sin x.) 7. Trace the graphs of y = tan x and y = cot x. 8. Trace the graphs of y = sec x and 7/ = esc x. '^ISG. Periodicity of the Trigonometric Functions. — Since the position of the terminal Hne 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 ± 7i.dQ0°, n being an integer. That is, fix) =/(x±n.360°), where / stands for any one of the trigonometric functions. Hence the trigonometric functions are periodic, with a period of 360°. (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, 1 1^1 sin -JO = ; cos x ^ ; tan oc = — —- . CSC X sec ic cot X ordinate ordinate distance sin .r . ^ cos x tan X = — j — ; = — r — r— - = , cot x = -: • abscissa abscissa cosic sin a? distance Whatever be the quadrant of angle XOP = x [figure of (132)], we have (ordinate)^ + (abscissa)'^ == (distance)'. Dividing this equation through in turr^ by (distance)-, (abscissa)-, and (ordinate)-, and expressing the resulting ratios as functions we have sin^ic+ cos^cc = 1, ■ 1 + tan' X = sec' x, 1 + COt^ X — CSC' X. 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 thorouglily committed to memory. They are collected below, as 138] TRIGONOMETRIC FUNCTIONS 107 Formulas, Group A. ,,^ . 1 ... . since (6) sin-x + cos-iT = 1. (l)sinx = (4)tanx = ^ CSC a; cos J) /7N . , * •• (7) 1 + tan-.r = sec'iT. /ox 1 /r\ X COS.r (2) COSX = (5) cot .*• = -: /o\ i _L ^^4-'i -^^^-J secx ^ sinx (8) 1 + cot'x = csc-u^. (3) tan.r= — ^ — cotoj We shall apply these formulas in two examples. Example 1. Prove that tan x + cot x = sec x esc x. sin X . cos a; sin2 x + cos^ x tan X + cot X = 1 — -. — = — -. — cos X sm X sm x cos x sm X cos X Example 2. Prove that esc X ;r j r- = cos X. tan X + cot X CSC X sec X. tan X + cot X sin .c , cos x sin2 x + cos^ x cos X sm X sm x cos x sm X cos X CSC X sm X cos x = cos x. In both examples all the steps taken are true for all values of x, since 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 idcntilies. 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. cot X = J sec X 2. -— = sin X. e / • 1 •> NO I cot X sec X 5. (sm^ X + cos- .r)- = 1. « . sec X _ cos 9 , „ - 3. tanx = 6. -^— rr — ^ = cot2». esc X sm 6 tan 7. (esc e — cot e) (esc + cot ») = 1. 8. (sec X — tan x) (sec x + tan x) = 1. 9. (sin 9 -f cos 5)2 = 1 + 2 sin d cos 0. 10. sin2 a + cos2 a = sec2 a — tan2 a. 11. (sina — cosa)2 = 1 — 2sinQ:cosa. 108 TRIGONOMETRIC FUNCTIONS [139 12. sin4 X — cos4 x = sin2 x — cos2 x. 13. (1 - sin2 x) csc2 x = cot2 x. ..-'^4. C0t2 e - C0S2 e = C0t2 C0S2 6. 15. tan + cot = sec d esc 9. 16. tan ^ sin ^ + cos = sec 9!>. 17. sin2 sec2^ = sec2 u ^ ^^^ tiple of 90°. As an equation, ^ ± f(x ± n-90°), neven; ) ± co-f(x ± »^.90°), nodd. The sign of the result must be determined by noting the quadrants of X and x ± n ' 90°. no TRIGONOMETRIC FUNCTIONS [140, 141 When the new angle, a; ± n • 90°, lies in the first quadrant, give to the result the sign of the given function of x, f (x). Examples. 1. sin 680° = sin (50° + 7 X 90°) = - cos 50°. 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 —x. — The figure shows two cases, x in the first quadrant and x in the second quadrant. In either case, sin a; = — sin {— x); esc a; = — CSC (— x); cos a; = cos ( — a;) ; sec X = sec {— x); tan X = — tan (— x); cot a; = — cot (— a;). 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 of the angle are equal to the negative of the corresponding functions of the negative angle. Or, f{x) = /(— x) when f stands for cos. or sec. f {pc>) = — / ( — ic) 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°. 5. 359°. 9. - 321°. 13. - 1060°. 2. 165°. 6. - 25°. 10. 742°. 14. - 401°. 3. 230°. 7. - 12.5°. 11. - 665°. 15. 525°. 4. 340°. 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°. 23. 790°. 18. - 80°. 20. 255°. 22. 600^ 24. - 510°. 142, 143) TRIGONOMETRIC FUNCTIONS 111 the values of the functions of: 25. 120°. 29. - 30°. 33. - 240°. 26. 135°. 30. - 45°. 34. 315°. 27. 150°. 31. - 60°. 35. 600°. 28. 300°. 32. - 120°. 36. - 510°. Find the i. 120°. 29. - 30°. 1. 135°. 30. - 45°. '. 150°. 31. - 60°. 1. 300°. 32. - 120°. 142. Versed Sine and Coversed Sine. — The expressions 1 — cos a: and 1 — sin a: occur often enough in the applications of trigonometry to warrant the use of special symbols for them. These are 1 — cos X = versed sine of oc = vers x; 1 — sin x = coversed sine of « = covers oc. Their line values are (figure), versa: = MN, covers a; = HK, X being in the first quadrant. Exercises. Find the values of the versed sine and coversed sine of: 1. 30°. 7. 150°. 2. 45°. 8. - 30°. 3. 60°. 9. - 120°. 4. 90°. 10. - 225°. 6. 120°. 11. - 300°. 6. 135°. 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 apgle 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 seco7ids. 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 radius of the circle, and is called a radian (figure, p. 112). Since the circumference contains the radius 2t times, the entire central angle of 360° contains 2 tt radians, i.e., 2 TT radians = 360°. 112 TRIGONOMETRIC FUNCTIONS [144 Hence, T radians = 180 radians = 90°; 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 the radian is implied. Thus, 2t = 360°, w = 180°, - = 60°, 2| = 2| radians, and so on. o Note. To get the true form of the graphs of the equations y = sinx, y = cosx, etc., take x in radians on thex-axis, thus: x = 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 2 t (radians) = 360°, 360° 180° _ 180° 2x ~ IT "3.1416- also, 1 degree = ^ (radians) = -— (radians) therefore, 1 radian 57°.29+; (radians) = .017 + (radians). 57.29 + Rule: To convert radians into degrees, multiply the number of radians by 180 or 57.29 + . To convert degrees into radians, multiply the number of or .017 + . ' 180 ^^57.29- By taking a sufficiently accurate value of tt, 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. y 145." Exercises. Reduce to degrees, minutes and seconds the angles whose circular measures are: ^ IT Sir 5ir 5ir 7 ir 8' T' T' T' T' 1 1 7 2. 1 2 i, -' 2 3 4 11 ^1 ^^1 2 7r+3 1.^ - + 1, 2 + 3' -6— 6 TT IT — 6 ir2 TT + 1 7r2 + ll-7r7r-l Reduce the following angles to circular measure: 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 l^ 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? 13. 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 tt radians per second, how many revolutions per minute does it make? 146. Angles Coiresponding 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 7nr has the terminal line OX (figure) coincident with the initial line, and angle (2 ?i + 1) tt has the x' cn-^-mr terminal line OX'. Suppose now we wish to write down all angles x such that sin a; = |. 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 tt - 30°. 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 2 riT + 30° and 2 nx + (tt - 30°) = (2 n + 1) TT - 30°. Hence all solutions of the equation "*■ sm a; = ^ are given by a; = 2 nvr + 30° or (2 n + 1> tt - 30° In general, if d denote the smallest positive angle whose sine is a, then all solutions of the equation (1) sin a; = a are x = 2 tit -h and (2 n + 1) tt — ^. Hence also, if denote the smallest positive angle whose cosecant is a, the solutions of the equation (2) CSC x = a are x = 2 tit -\- 6 and (2 n + 1) tt — ^. Consider next the equation COS X = h. The two simplest solutions are a: =+60° and a; =-60°. All possible solutions are given by X = 2 WTT + 60° and x = 2 nir - 60°, or X = 27nr ± 60°. In general, if 6 be the smallest positive angle whose cosine is a, all solutions of the equation (3) cos a; = a are x = 2 mr ± 0. Hence also, if be the smallest angle whose secant is a, all solutions of the equation (4) sec X = a are x = 2 ht ± 6. Finally consider the equation tan a; = 1. The two simplest solutions are X = 45° and a; = tt + 45°, 147] TRIGONOMETRIC FUNCTIONS 115 and all possible solutions are x = 2mr-\-45° and z = 2 rnr + (tt + 45°), the second set being the same as a: = (2 n + 1) w + 45°. Both sets are contained in the single equation X = TIT -{- 45°, the first set being obtained when n is even, the second set when n is odd. In general, if 6 be the smallest posi- tive angle whose tangent is a, all solutions of the equation (5) tan X = a are x = mr -{■ 0. Hence also, if be the smallest positive angle whose cotangent is a, all solutions of the equation (6) cot X = a , are x = nr -'r 0. Summary of equations (1) to (6). Let denote the smallest positive angle having a given function equal to a given number a. 11. III. ^ sin X = a I CSC Jc = a ( cos a? = a I sec X ^ a ( tan X = a ) cot J^ = a Then all solutions of the equation are £c = 2 jtir + 6 and (3n. + l)TT — 6; are x = 2 uir ± 6 ; are x. = » = .750. 149] TRIGONOMETRIC FUNCTIONS 117 149. Given one function of an angle, to find the other functions. Fiiui the Example 1. sinx ■ other functions. ' Take ordinate = 1 and distance = 2; then abscissa = db V3 (figure). Then Wo cos X = ± jL2. , tan x = ± 2 cotx = ±V3, 2 sec X = ± —7Z. , CSC X = 2. V3 V3 We have found two values for each function except csc x, which is the reciprocal of the given function. Similar results will be found in general. Exam-pie 2. tan X = -3 +3> +4 ^---^y The two possible positions of the ter- X minal line are shown in the figure. Hence, sin x = ± -> cosx = ± ~> 4 ,5 5 cot X = — ij > csc X = ± - > sec X = ± -7 • Example 3. Then (figure), ±2 +3 -^) Vl3 VT3 V3 Vl3 Example 4. sinx= -■ Ordinate = h; distance = k; hence abscissa = ± v A;^ — K^. 118 TRIGONOMETRIC FUNCTIONS • [150,151 Then cos a; = ± -^^ , tan x = ± —r- , etc. k ■\lk'i - h^ Exercise 1. Construct figures for the cases when , is (a) plus; (b) minus. Exercise 2. Is the problem possible for all values of h and A;? Example 5. tan x = fl -6 /_ - (fl -b) \ 2 \/ab [ - 2 \/ab ) Here ordinate = a — b, abscissa = 2 V"'?! or, ordinate = — (a — 6), abscissa = — 2 Va6. In either case, distance = + y{a — 6)'' + 4 a6 = | a + 6 1 . TT • , fl - ft ,2 -s/ab Hence, smx = i , r~r,' cos x = ± , 1-^-,' etc. |a + 6| \a + b\ Exercise 1. Calculate the values of the six functions when a = 2, b =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? 150. Exercises. Find the other functions, given that 1. sin X = — 5. 6. CSC X = — T L *^ « ^ ^ ^' 11. CSC^ = 2. cosx = i 7. secx =- |J-. 12. tan 9 = a. 3. tanx = ^. 8. cotx=-.75. ., ., , * 13. sm 9 = fl. 4. secx = 4. 9. sinx = .6. 14. cot0 = V^. 5. cotx = V3. ^^ b ^^ , a2 4-;^2 10. cos = - • 15. sec (/> = 2ab 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 _ cosx _ abscissa I distance Hence let abscissa = cos x and distance = 1. Then ordinate = ± Vdist.^ — absc- = ± Vl — cos^ x. loll TRIGONOMETRIC FUNCTIONS 119 The figure shows this graphically when cos x is positive. Taking into account both values of tlie ordinate, we have sina;= ± Vl-cos-a:; Vi — cos^ X tan X = ± cot .T = ± CSC X = ± COS a: COS a: V 1 — cos- X ' 1 V 1 — COS^ X sec a; 1 cos X Exercise 1. 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. cot X abcsissa cot a; = cot a; - 1 ordinate Hence let abscissa = cot x and ord.nate = 1 ; or let abscissa = — cot x and ordinate = — 1. In either case, distance = + Vl -|- cot^ x. (See figure, where we assume cot a; > 0.) Hence sin a; = ± vl +cot2a; cos X = ± cot a; Vl + cot- X etc. 120 TRIGONOMETRIC FUNCTIONS 151 By taking each of the functions in turn, and proceeding as above, we obtain the results shown in the fo lowing table. The given function and its reciprocal are uniquely determined; the other four functions are ambiguous in sign. sin X. cosx. tan X. cot X. sec X. cscx. tanx 1 ±Vsec2x-l 1 . ±Vl-COs2x ±VH-tan2x 1 ±Vl+C0t2x cotx secx 1 secx cscx Vcsc2x-i ± Vl-sin2x sinx ±Vl+tan2x ±Vl+C0t2x 1 cotx CSCX ± Vl — C0S2X 1 tanx ±Vsec2x-l 1 ±Vl-sin2x cosx cosx 1 tanx ±VCSC2X-1 itVl— sin^x cot a: ±VCSC2X-1 sinx 1 ±V1-C0S2X 1 cosx 1 ±Vsec2x-l ±Vl+C0t2x ±Vl+tan2x ±Vl-sin2x 1 sinx cotx secx ±VCSC2X-1 ±Vl+tan2x ±Vl+COt2x ±Vl-COs2x tanx ±Vsec2x-l Exercises. 1. Express sin x cos2 x + sin^ x in terms of tan x. 2. Express tan x sec x + sec2 x in terms of sin x. 3. Express cos2 x tan x + sin2 x cot x in terms of esc x. in terms of sec d. 4. Express :; — , , _ „ + 1 + sm ( _ ^ COS0 . 6. Express ;; ytztt, + 1 - sin sinff 1 - tan 1 - cot in terms of cos d. CHAPTER VIII Functions of Several Angles 152. Formulas for sin {oc + y) and cos (x + 2/). — 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. M X 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 _L OX and KP _L NQ. Then Z KQP = x, and in either figure. sin (x + y) NQ MP + KQ OQ OQ MP KQ OQ '^ OQ Hence (a) MP . OP. KQ PQ OP ' OQ.^ PQ^' OQ^ sin (ac -\- y) = sin oc cos y + cos x sin y. 121 122 FUNCTIONS OF SEVERAL ANGLES [153 Also, noting that ON in the second figure is a negative line, , , . ON OM-NM OM KP cosix + y) = ^= QQ =-OQ-OQ ^OM OP__KP PQ OP'OQ PQ'OQ' Hence (b) cos (.-r + y) = cos x cos y — sin x sin ij. 153. 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 (a') sin (A + 5) = sin A cos B + cos A sin R, (W) cos (A -\- B) = cos A cos B — 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 + w . 90°; B = y-^m- 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 for w and m. In particular, let n = 1 and w = 2. Then A=x-^ 90°; B = y+ 180°; A^B = x-\-y-^ 270°. Hence, if formulas (a') and (b') are true, ' sm{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 reduce to cos (x + y) = cos X cosy — sin x sin y, sin (x + ?/) = sin x cosy -\- cos x sin y. 154,155] FUNCTIONS OF SEVERAL ANGLES 123 But these are true since x and y are acute angles; hence also (a') and (b') are true. In exactly the same way the truth of these equations may be shown for any integral values of 7i 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 —y in (a) and (b), and noting that sin (—?/) = — sin y and cos {— y)= cos y, we have (c) sin (.K — y) = sin a? cos y — cos « sin y; (d) cos {x. — y)= cos oc cos y + sin oc sin y. Equations (a), (b), (c), (d) are usually called the addition and subtraction formulas of trigonometry. All the other working formulas are deduced from them. - 155. Dividing (a) by (b), we have , , , sin (x + v) sin x cos y -f cos x sin y tan {x -{-])) = 7 — ,-^ = ^ -. ^^ • cos {x + y) cos X cos y — smx sm y sin X cos 7/ cos x sin y cos X cosy ' cos X cosy sm X sm y cos X cos y ence, (e) . , . tan .X + tan y Similarly, (f) ^ , , , cot r cot V — 1 Also, from (c) and (d), by division, (g) tan .X — tan y (h) cot(x-,v) = £2l£lM^. Exercises. 1. If sin X = J and sin ?/ = f, calculate sin (x -\- y). (Four answers: ^ [± Vo ± 4^^].) 12^ FUNCTIONS OF SEVERAL ANGLES [156 2. If cos a; = I and cos y = |?, calculate cos (x + y). 3. If sin a = 3 and sin /3 = |, calculate cos (a — /3). Show that, 4. cos (60° +x) + cos (60° - x) = cos x. 5. sin (45° + 0) - sin (45° - 6) = \f2 sin 5. . « , . A cos (g - (f>) 6. cot e + tan = -^^ — „ ^^ , • sin cos 9 7. cos (A + 45°) + sin (A - 45°) = 0. 8. sin nd cos + cos nd sin = sin;(n + 1) 9. 9. tan^e-^) + cot(0+^) =0. 10. From the functions of 30° and 45° calculate 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. Formulas, Group B. (9) sin (« + ?/)= sin xcosij + cos oc sin p. (10) cos {x + 2/) = cos oc cos 2/ — sin ao sin y. (11) sin (« — 2/) = sin ic cos ?/ — cos ic sin 2/. (12) cos (x — 2/) = cos X cos 2/ + sin oc sin y. , , s tan iT + tan 2/ (13) tan(x + ,v)= ^_^^^^^^,y , cot a? cot 2/ — 1 (14) cot(x + 2/)= eotx + cot2/* tan £r — tan y (15) tan(a.-2/)=^^tanxtan2.* . cot X cot 2/ + 1 (16) cot(x-2/)= ,ot^_eota.' 156. Functions of 2 a?. — Putting ?/ = a; in (9), (10), and (13) of Group B, we have (14) sin 3 ic = 2 sin a? cos cc, (15) cos 3 a? = cos'* X — sin^aj, = 1 — 2 sin^ oc, = 2 cos^ a? — 1. 2 tana? (W) tan 2 a? = : — 5— * ^^^^ 1 - tan* X For cot 2 a: use - — ^ • tan2x 157 ] FUNCTIONS OF -SEVERAL ANGLES 125 Exercises. 1. Verify these formulas when x is 30°; 45°; 150°; -60°. Show that, 2. 2 CSC 2 X = sec x esc x. 1 - tan2 X 3. cos 2 X 1 + tan2 X. sin 2 X = tan X. 1 + cos 2 X 6. tan X + cot X = 2 esc 2 x. 6. Calculate the functions of 2 x when sin x = 1|. Ans. sin 2x = ±||§; cos2x = ^J^ ; etc. 7. Calculate the functions of 2 x when tan x = \. 157. Functions of | x. — The second and third values of cos 2 x in (15) are cos 2 a; = 1 — 2 sin^ x, cos 2 a; = 2cos2x — 1. cos 2; Solving these for sin x and cos x respectively, we have , /I — cos 2 X , . /l sin a: = ± y s » cos a: = ± y - Replacing a; by i x, these become (17) sin|x= ±y , (18) cos|x=±y ^ Dividing (17) by (18), ,^f.. . . , . /l — cos X 1— coscc (19) tan J a? = ± y - + cos X sin a? 1 + cos x Formulas, Group C. (14) sin 3 a; = 2 sin a? cos a?. (17) sin|a7 = ±y — cos X /I ox 1 , t A + cos X (18) cos |x = ±y (15) cos 2 a? = cos' a? — sin' a? '^ = 1 — 3 sin' as /im * i ,./l — cosx (19) tan^a? = ±Vrn = 2cos'x-l. Vi + cosx _ 1 — cos X sin X ,. ^. , „ 3 tan X. sin x (16)tan2x = - — J-. = — 1 — tan' X 1 + cos X 126 FUNCTIONS OF SEVERAL ANGLES [158 Exercises. 1. Calculate the functions of 15° from those of 30°. 2. Calculate the functions of 22|° from those of 45°. 3. Calculate the functions of 75°. 4. Calculate the values of tan (2 x - y), when sin x = I and cos y = \i. Show that, 5. sin 4 a; = 2 sin 2 a: cos 2 x. ^^ 1 + sec 9 ^ ^ cos2 \ 6. sec 2 - sec2 X 6. cos 2 X = ^^^2 X ' 13. sin/3cot J|3 = 1 +cos^. 2 X 1 — tan X 1 + sin 2 X . 1 + tan x cos" 9 - sin" 61 = cos 2 6. 15. cot /3 cos3 e - sin3 d 2 + sin 2 14. 1 + tan /3 tan 5 /3 = sec / 2 cot? COS — sin 2 fl 1 + tan ^ 10. cot X + CSC X = cot I X. 16. , ^"^^ = i. 1 - sm /3 1 _ tan ^ 11. (sin 1 + cos § 0)2 = 1 + sin 0. 2 158. Formulas for sin « ± sin r and for cos u ± cos f . — For- mulas (9) and (11) of Group B are sin (x + y) = sin xcosy -\- cos x sin y, sin {x — y) = sin a:; cos ?/ — cos a; sin y. Adding: sin (x + y) + sin (x - ?/) = 2 sin x cos 2/. Subtracting: sin (x -\- y) - sin (.r - ?/) = 2 cos x sin ?/. Let X -\- y = u, and x - ?/ = y ; ?^ + y , u — V then a: = — ^ and ?/ = ^ ■ Substituting in the two preceding equations, we liave . if + r n — r (20) sin u + sin *' = 3 sin — - — cos — - — u + r . 11 — r (21) sin n - sm v = 2 cos ^ sin ^ • 158] FUNCTIONS OF SEVERAL ANGLES 127 Proceeding similarly with formulas (10) and (12) of Group B, we obtain, (22) cos u + cos f = 2 cos — — - cos 3 (23) cos n — cos v = — 3 sm —7; — cm — The last four cciuations, called the addition theorems of trigo- nometry, we collect as the Formulas, Group D. (20) sm u + sm v = 3 sm — :; — cos — ^ — • (21) sm a — sm *' == 3 cos — - — sm — - — */ + V 11 — V (22) cos u + cos V = 2 cos — - — cos — -— • (23) cos u — cos V = — 2 sm — - — sm — - — • Example 1. Show that ^!"^+^!°^ = ^- ^ smx — smy X — y tan-^— _ . X + y X — y 2 sin — i^ cos sin X + sin 7/ 2 2 sin X — sin w _ x -\- y . x — 2 cos ' sm — ?r- . tan -^ , X + y .X — y 2 = tan — ^ cot _ = 2 2 ,_ .r-^ r. 7 r. C1L XI. i. cos 75 + cos 15° /^ Example 2. Show that ^^o"^ 1^6 = - V3. cos 75 — cos 15 cos 75° + cos 15° 2 cos 45° cos 30° *,-o .or>o y^ .^Fs-^ r?5 = — .-, ■ .go ■ or^o = - cot 4o cot 30 = - v3. cos 75 — cos 15 — 2 sm 45 sm 30 128 FUNCTIONS OF SEVERAL ANGLES [159 Exercises. Show that: 1. sin 3 a; + sin 5 a; = 2 sin 4 x cos x. 2. sin 10 + sin 6 5 = 2 sin 8 e cos 2 9. ■ 3. cos 2 x + cos 4 X = 2 cos 3 X cos x. 4. sin 7 a — sin 5 a = 2 cos 6 a sin a. 5. cos 4 e — cos 6 e = 2 sin 5 d sin 6. 3 X X 6. cos x + cos 2 X = 2 cos — cos ^ • 7. sin 30° + sin 60° = V2 cos 15°. 8. sin 70° - sin 10° = cos 40°. 9. sin 5 X cos 3 X = ^ (sin 8 x + sin 2 x). 10. 2 cos 10° sin 50° = sin 60° + sin 40°. . ^ sin A + sin B , A + B cos A + cosB 2 ^_ sin9 + sin30 , „. 12. — 1 5-;: = tan 2 d. cos + cos 3 13. 2 cos a cos /3 = cos {a — 13) + cos {a + p). 14. sin 4 sin e = I (cos 3 — cos 5 0). 15. cos 8 X — cos 4 X = — 4 sin 2 x sin 3 x cos 3 x. 16. sin (2 X + 3 2/) + sin (2 X - 3 y) = 2 sin 2 X cos 3 y. 159. Exercises involving the use of formulas (1) to (23). 1. If sin X = t and sin y = |, find the value of sin (x + y) and cos (x + y) 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. 3. If cos X = I and cos y = |?, calculate sin (x + y) and cos (x + 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 2/ = f , calculate all values of sin (x± y). 6. If sin a = i and sin /3 = §, calculate all values of cos (a ± 0). 7. If cos a = f and cos /3 = |, calculate all values of tan (a ± /3). 8. Calculate sin {x + y + z) when sin x = tV, sin y = ^j, sin 2 = 5^, and X, y, z 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 x = |, cos ?/ = if, cos z = H, and X, y, z all lie in the first quadrant. 11. As in exercise 10, when x, y, 2 all lie in the fourth quadrant. 12. Calculate tan (x + y) when tan x = 1 and cot y = V3. 13. Calculate all values of sin 2{x -y) and of tan {2x -y) when tan x = i and tan y = j\. 14. Calculate all values of cos (a + /?) when tan a = m and tan ^ = n. 15. Calculate cot (a — /3) when tan a == a + I and tan /3 = a — 1. 16. Calculate tan (a + /8) when tan a = — —r and tan (8 = ^ — xT" X -f- 1 Z X "T 1 17. If tan a = f and tan /3 = j\, calculate tan (2 a + /3). 159] FUNCTIONS OF SEVERAL ANGLES 129 18. Calculate sin 75°, cos 75°, and tan 75°, by use of the relation (a) 75° = -iy-; (b) 75° = 135° -60°. 19. Calculate the functions of 202^; of 7^°. Prove the following identities: 20. sin X sin (y — z) + sin y sin (z — x)+ sin z sin (x — y) = 0. 21. cos X sin (y — z)+ cos y sin (2 — x) + cos 2 sin (x — y)= 0. 22. cos (x + y) cos {x — y) + sin {y + 2) sin (y—z) — cos (x + 2) cos (x — 2) = 0. 23. cos (x — y + 2) = cos x cos y cos 2 + cos x sin y sin 2 — sin X cos 2/ sin 2 + sin x sin y cos z. 24. sin 3 X = 3 sin X — 4 sin^ x. t^ _. cos (a + /3) 26. cos 3 X = 4 cos^ x — 3 cos x. 3 tan X — tan^ x 1 - 3 tan2 X cot3 X — 3 cot X - - ^ o 3 tan X - tan3 x _. 26. tan 3 X = -^j 5- — s 31. 1—3 tan2 X 27. cotSx 28. tan 4 9 3 cot2 X - 1 4 tan g (1 - tan2 g) 1 -6tan2|9 + tan4 9' sin a cos /3 = cot a — tan p. cos (a - /3) COS a sin /3 = cot /3 + tan a. sin (a — 0) COS a cos /3 = tan a — tan 0. sin (x + y) tan X + tan y sin (x - ?/) tan X — tan y cos (x + y) _ cot X - tan 2/ 29. '-^5j£_±^ = tan« + tan^. 34. , , . ^. cos a cos /3 cos (x — y) cot x + tan y 35. sin (0 + (f)) sin {e — (p)= cos2 ^ - cos2 9. 36. cos (w 4- y) cos (w — v)= cos2 u — sin^ v. 37. sin (A - 45°) = — (sin A -cos A). V2 OQ ,/, ir\ cotA+1 on + /a 'r\ tanO-l 38. cot ^A - ^ j = -^-^^ . 39. tan (^ -^j = ^^^^^^-^. 40. tan(^ + .] = ^ 41. tan fa +^j + tanf a - ^j = tana cot C5f cot2 a - 3 ' . 5 7r 5 IT 49. \/2sin(0 + 45°) = sin9 + cos». sin Y^r- cos — r 42. ^ 1^ = 2V3. 50. sin2x ^tanx 51. sec2x 1 + tan2 X csc^x 43. tanQ+9l= j^ r- csc2x-2 52. cot - cot 2 fl = CSC 2 0. 53. sec2 « COS 2 a = 1 - tan2 6. 54. 1 + tan « tan 2 a = sec 2 e. 55. 1 — COS 2 X = tan x sin 2 x. cot2 + 1 56. sec2( 44. cosfe+^Wsin(0-^) = O. 45. cot (0 + ^ W tan (^ - ^) = 0. 46. cotfe-^j + tan(0 + ^) = O. """ ''"^ " " " cot2 - 1 „ sin2(? ^ „ 47. cot I - tan I = 2. 67. p^^^^ 2-0 = *^" ^• 48. 2cos^ = V2 + V2. 68. , ^'"^^.,, = cot0. 8 1 — cos 2 e 130 FUNCTIONS OF SEVERAL ANGLES [159 59. cot20-l =2cot0cot2&. _ 1 - cos 2 x 60. 2 - sec20 = sec''0cos2e. ^'^' ^an x - ^ _^ cos2a;' Ri ^°^ 2 '' 1 - tan g cos 3 (?, sin 3 9 ^ ^ ^ ^^' l+sin2g = r+^^- 64. --j-^+___-=2cot2e. 62. '-^^ = 2COS2X - 1. 65. ^^"^ + ^"^^ = sec2.. cos X cot 9 - tan 66. tan (45° + ) = 2 sin f ^ - o) cos (^ - (/.V V2. 101. 2 sin I a + , j sin | a — 7 1 = sin2 a — cos'^ . 102. sin I ^ + a j — sin I ^ — a I = V- sin a. V3-l_,,o 103. sin 40° - sin 10° V2 tn.A • r, . • . - n ../v.- Sin 75 + sin 15 17, 104. sin 3 X + sm x = 4 sm x cos2x. 105. ~. — ^^^^ -. — ^^^ = y3. sin 75 — sin 15 ^Qg^ COS x + cosy ^ _ ^^^ X +^ ^^^ X- y cos X — cos y 2 2 .„ sin 70° + sin 20° . ..„ sin 100° + sin 40° .., , _„ 107. -^o , p^TT^ = 1. 108. - ■ ,„^ ■ .r>o = V3 tan 70 cos70 +cos20 sin 100° — sm 40 109 (sin « + sin 0) (cos a: + cos 0) ^ _ ^^^2 (sin a — sin 0) (cos a — cos 0) j^^Q (sin g + sin 0) (cos a - cos 0) ^ _ ^^^g (sin a — sin /3) (cos a + cos 0) 111. (sin 75° + sin 15°) (cos 75° + cos 15°) (sin 75° — sin 15°) (cos 75° — cos 15°) ., ^ _ cos 2 X + cos 12 X cos 7 x — cos 3 x ^ oi.* ^ -^ _ /% cos 6 X + cos 8 X cos x — cos 3 x sin 2 x 113. sin X + sin 2 X + sin 3 X = 4 cos ^ x cos x sin | x. (Hint. Replace sin x + sin 3 x by 2 sin 2 x cos x and sin 2 x by 2 sin x 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 X + cos 2 X + cos 3 X = 4 cos ^ X cos x cos i x — 1. 115. sin 2 X + sin 4 X + sin 6 X = 4 cos x cos 2 x sin 3 x. ... sine +sin20 + sin3 ^ _„ 116. -^, K-^'r o-^ = tan 2 9. cos6» + cos2e + cos3 117. cos 20° + cos 100° + cos 140° = 0. 118. cos 9 + cos 3 + cos 5 + cos 7 = 4 cos 5 cos 2 cos 4 6. 119. sin + sin 3 + sin 5 + sin 7 = 16 sin cos2 cos2 2 0. 120. 4 sin2 f,6 cos2 0. Vl + a2 Let X = tan- la and 2/ = sin- Vl + 0.2 then tan x = a and sin y = Vl +a2 163, 164] TRIGONOMETRIC EQUATIONS 137 To prove that ^ = V, or that sin x = sin y. Now since x and y stand for principal values, and a is positive, both angles are in the first quadrant. Then from tan x = a we find (149) ffl sin X = , _ , Vl+a2 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: ,5,, ,1 TT 6. cos-ii+2sin-iA = 120°. 2. tan-i^ + tan-ig = ^. ^ 2 tan-i 3 = sin-i f. 3. 2tan-ii=tan-i|. g^ 3 sin- 1 .^ = sin- 1 -| • 4. tan-i3+^ = tan-i(-2). ^ o cot-i2 = csc-i |. 5. tan-i^ + csc-i VlO =^- 10. 4tan-ii = tan-i^^g + |- 2 11. tan- 1 ^ + tan- 1 1 + tan- 1 ( ^ ,4 , . , 8 , . ,13 TT 12. sm-i^ + sm-i^ + sm-ig^ = 2- 13. cos-i|i + 2tan-4 = sin-i|- 65 5 5 i>i o. ,2 ,5 . ,33 14. 2 tan- 1 - — esc- 1 - = sm- 1 — • 3 o 65 ')- 15. sin- 1 a = cos- 1 Vl — «^> if « > 0- 2w 16. 2 tan- 1 ?« = tan- 1 1 -m2 cot2 e - tan2 (9 1 17. 2 tan- 1 (cos 2 0)= tan 164. Trigonometric Equations. — A trigonometric equation is an equation which involves one or more trigonometric functions of one or more angles. Thus: sin^ x-\- COQX = 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 hy substitution. Examples. 1. sin2 X + cos X = 1. Expressing all terms by means of cos x, we have 1 — cos2 X + cos X = 1, or cos2 X — cos X = 0. cosx = 0, or cosx = 1. Hence x may be any odd multiple of | or any multiple of 2 7r; i.e., if n be any integer or zero, x=±(2n + l)^ or x=±2mr. Exercise. Check these answers by substitution. 2. tan0 +sec0 = 3. Expressing all terms by means of tan d, we have tanO ± Vl +tan2!? = 3, or ± Vr+tan2(? = 3 _ tanff. Squaring and reducing, tan = ^ ; hence = 53° 8' ± nir. o When n is odd, these values of e do not satisfy the given equation. Hence the solutions are = 53° 8' ± 2 nir. 3. cot a CSC a = 2. Then ± cot a Vl + cot2 « = 2, or cot^ a + cot2 a = 4. Hence cot a = ± V— | ± ^ 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 nx. The reason for the additional values in the last two examples is that in example 2 we really solved both the equations tan e ± sec 6 = Z, and in exam- ple 3, both the equations cot a esc a = ± 2. 165. 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 sin2 X — 3 sin x cos x = 1. Since 2 sin2 x = 1 — cos 2 x and 2 sin x cos x = sin 2 x, we have 3 9 1 — cos2x - -sin 2x = 1, or tan2x=— ". ^ 6 Hence 2x = tan-M - ^J = - 33° 41' ± titt. X = - 16°50'.5 ± n|- Exercise. Check these answers. Solve the given equation by expressing cos X in terms of sin x. 5. sin 3 v/ — sin 2 f/ = 0. By formula (21) of (158) this becomes 5.1^ 2 cos 2 ?/sm~?/ = 0. 5 1 Hence cos ^ ?/ = or sin - ?/ = 0. ^y = ± {'^ n ^r \)\^ ov ^y = ±nir. 2/ = ± (2n + l)^, or y = ±2 nir. 6. cos X + 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)2. or 2x=±^^±2mr. X = ± (2 n + 1) ^ or ± 5 ± n-n-. 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 equals the cotangent of an angle B, A + B must be an odd multiple of J Hence 4a4-5a= ± (2n + l)| a=±(2n + l)^- Here a is any odd multiple of 10°. Otherwise thus: tan 4 a — cot 5 a = 0; hence — -; : — r — ■ = 0: cos 4 a sm a sin 4 a sin 5 a — cos 4 a cos 5 a cos 9 or or — cos 4 a sin 5 a cos 4 a sin 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 x in example 4. 8. a sin + & cos B = c. We might reduce to sin e or cos e 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 in and M such that (o = 7/icosM, ^ (m=V^M=;6^ w • Ttf whence \ ,, . , o /6=msmAf; >M=tan-i-- The given equation then becomes m (sin cos M + cos 6 sin M) = C or sin {d + M) = — . Hence if we let sin- 1 x represent all angles whose sine is x, + M =sin-i — , or = sin-i M. m m Va2 +62 a Graphic Solution. As an example, we take the equation sin 2 + sin e + 2 = 0. We want the values of d which reduce the expression sin 2 ^ + sin + ^ to zero. Let 1/ = sin2 + sin^ + s- Calculate y for a series of values of ^, as = 0°, 10°, 20°, . . . , and plot the points (^, y) in rectangular coordinates. The result- ing curve will show the approximate values of Q for which y is zero. Any convenient scales may be used on the axes of Q and y. Let the student read off the required solutions from the graph below. 166] TRIGONOMETRIC EQUATIONS 141 Exercise. By means of this graph solve the equations (a) sin 20 + sin0= 0; (b) sin20 +sin(?= 1; (c) sin20 + sin0= \. 166. Exercises. Solve the following equations: 1. 2 sin2 X — 3 cos x = 0. 2. 4 sin2 a + 1 = 8 cos a. 3. sin a + cos a = V2. 4. tan e + coi9 = 2. 5. tan /3 + 3 cot /3 = 4. 6. 2 sin2 X + 3 = 5 sin X. 7. 2 (1 — sine) = COS0. 8. 5 sin e + 10 cos = 11. 9. cos 2 X = cos^ X. 10. 2 cos 2 X = 1 + 2 sin x. 11. 4 cot 20 = cot2 0-tan2 0. 12. cos = sin 2 d. 13. tan 2 X = 3 tan x. 14. sin 2 2/ = cos 3 y. 15. tan a = cot 3 a. 16. cot 8 ?!> = tan ^. Solve some of the above equations 12, 13, 14, 15, 26, 28, 29. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29 30. graph sec px = CSC qx. tan ?/ = cot 6 y. sin rd = sin sd. cot (30° - x) = tan (30° + 3 x). sin 4 a = cos 5 a. _ sin(60°-x)-sin(60°+x)=iV3. sin 2 + sin 4 = V2 cos d. sin(3O°+0)-cos(6O°+0) =-.VV3. sin 4 a = cos 3 a + sin 2 a. sin 3 /3 + sin /3 = cos /3 — cos 3 /3. sin X + sin 2 X + sin 3 X = 0. sin X + sin 3 X + sin 5 X = 0. cos X + cos 2 X = 008 i x. ically, in particular 1, 2, 4, 5, 7, 8, 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 all the angles deter- mined by the given function. Examples. 1. Solve for r and 6: r cos0 = x, r sin a = y. Squaring and adding, r^ = x^ + y^; hence r = ± Vx2+t^. Divide the second equation by the first, tan ^ = - ; hence 6 = tan- 1 - . a; ' . X 2. Solve for a and /3: a sin a + 6 sin (3 = c, d sin a + e sin (S = /. Solve for sin a and sin /3 as unknowns; hence get a and 0. Exercise. Carry out the solution of example 2. Is the solution possible for all values of a, 6, . . . ,/? (62.) 3. Solve for r and d: dr sin + &r cos = c, a'r sin 9 + Vr cos 6 = c'. Solve for y sin 6 and y cos as unknown; then proceed as in example 1. Exercise. Carry out the solution in example 3. 4. Solve for X and 2/ : y = sin x, y = sm2x. Subtracting, sin 2 x — sin a; = or 2 sin x cos x — sin x = 0. Hence sin x = or cos x = A. X = ± nw or ± 60° ± 2 mr. y = or ± i V3. Exercise. Solve example 4 graphically. 5. Solve for y and <: y = a sin (n< + &), y = a' sin {nl + b'). Equating the values of y, and expanding, a (sin ni cos b + cos nt sin b) = a' (sin nt cos b' + cos nt sin b'). Dividing by cos nt and solving for tan nt, a' sin b' — a sin b tall nt = r ; r, • a cos — o cos b' This determines a set of values of nt. Then y is obtained by substituting in either of the given equations. 168] TRIGONOMETRIC EQUATIONS 143 6. Solve for r, 9, and f/>: x = r cos 6 cos (f>, y = r cosO sin 0, z = r sin 6. Dividing the second equation by the first, we have - = tan 0: hence 6 = tan-i -• X ^' '^ X Squaring the first two equations and adding, x2 4- 2/2 = r2 cos2 0; hence r cos 5 = ±\Jx^ + ?/2. Combining this result with the third equation, as in example 1, we have tanO = ; hence = tan" ± Va;2 + 2/2 ' ± Va;2 + 2/2 r2 = X2 + 2/2 + 22. 168. Exercises. Solve for r and »: 1. r cos = 3, r sine = 4. 6. rsin(e + ^)=2, 2. r cos = 12, rcos[e-^)=l. r sin 9 = — 5. 7. r = sin^e + ^), 3. r cos (? = - 9, r sin 9 = - 40. 4.' r cos 9 + 2 7 sin = 3, 2 r = sin f e - ^] • 8. r = 2sin^2e-^j r = 3sin(e + 'f). rsinO = 1. 5. r (2 sin + 3 cos 6) = 1, r (sine + 4cose)= 1. Solve f or r, 6, and = — 1, r cos sin c/) = 4, r cose sin<^ = 1, r sin e = 5. r sin e = - 2. Eliminate from the following equations: 11. X = r cos 6; y = r sin e. 12. X = a cos e; 7/ = 6 sin e. 13. X = a3 cos-'' 9] y = b^s'm^ 9. 14. -cose + ^sine = 1; - sin e - |co3 e = - 1. a a 15. Eliminate 9 and ?!> from the equations X = r cos 9 cos ; y = r cos e sin ; 2 = r sin e. 16. The same for the equations X = a cos 9 cos (!>', y = b cos e sin ^; z = c sin e. 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, h = bsmA; from A BCD, h = asmB; equating the values of h, b sin A = asm B, or a : 6 = sin A : sin B. In the second figure, from A ACD, from A BCD, h = h sin (t — A)= h sin A ; h = a sm B; equating the values of h, 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:b.c = smA:sinB : sin C. 170. The Law of Cosines. — In any plane triangle, the square of any side equals the simi of the squares of the other two sides, minus twice their product by the cosine of their included angle. In the above figures let AD = m. Then First figure. ' - Second figure. in A AC D, m = b cos A; m = b cos(ir—A) = —h cos A; inABCD, a^ = h'^-{-{c-m)2 a^ = h^-\-{c-\-my^ = h^+c^-2cm+m^ =h^-\-c^-^2cm + m~ = 62+c2-2cm. =62+c2+2c?/^ Replacing m by its value above, we have in either case, (2) tr = ft' + c' — 2 be cos A. (2') Similarly, b^ = tr -{■ c' — 2 ac cos B. (2") jci = ff' -\-b'*-2 (lb cosC. 171. The Law of Tangents. — In any plane triangle, the dif- ference of two sides is to their sum as 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 ~ s'mB' By composition and division, and subsequent reduction we have, a — b sin A — sin B a + b s'mA + sinjB _ 2 cos H^ + B) sin i (A - B) 2 sin \ {A + B) cos \ {A - B) = cot \ {A + B) tan h {A - B). That is, (3) a-b tan \ {A - B) 146 OBLIQUE PLANE TRIANGLES [172 Similarly, (3') (3") «-c _ tan|U- C) a-j-c tan J (^ + C) ' &- c _ tan 3 (B - C) & + c~ tan|(^+ C) 3 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, tan^(A - B) = ^tan^(A +5). Similarly for the other two. 172. Functions of the Half -Angles. — 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 A 2 6c In practice this formula is not convenient unless a, 6, and c happen to be small numbers. Now 1^^^1-cosA, (^^^^^^^^1-^.) sm^ But 1 — cos A 1 - 2 6, 62 + c2 - a2 2 6c c - 62 - c2 + a^ 0? 2 6c - (& - C)2 (a 2 6c + 6 - c) (a b+c) 2 be Let 2s = a + 6 + c, or s = h (a-\-h -\- c). Then a + 6 - c = 2 (s - c), and a - 6 + c = 2 (s - 6). rp, , , 4 (g - 6) (s - c) Then 1 - cos A = -^ —r^ > 2 6c and (4) .„1. = V/Il^^lll^. 172] OBLIQUE PLANE TRIANGLES 147 Similarly, . 1 , /(s - a) {s - c) (4') sm-^ = \/ ^^ (4") sin 1 „ _ / (s - a) (s - />) 3 ^ " V ah Observe that the sides appearing explicitly under the radical include the angle to be calculated. To obtain cos j A, we have cos^A=y/^ , X 2 _1_ f,'. But 1+ cos A = 1 + 2 6c (b + c)2 - a2 2 6c (6 + c + g) (6 + c - g) 2 6c 4 g (g. — a) 2 he ' Hence 1 . / s(s - a) V 6c (5) cos-^ = V^^ Similarly, 2 V ac (50 cos^ 1 , /s (s — c) (5") "=5C = V^^- Dividing sine by cosine we have (6) tanl.^s/ il^ffj^- (6") tanlc = v/(l^^^i^. If _ J (s - 180' Exercises. solutions. Solve the following triangles, including graphic 1. A = 55° 2. A = 6.5° 25' 3. C = 34° 48' 4. B = 115° 10'. 5 5. J3 = 88° 20' B = 72° B = 78° 23' A = 100° 17' C = 40°22'.3 C = 105° 30' a = 1000. a = 4.245. 6 = 0. 5575. c = 0.00275. a = 10. 175] OBJ.IQUE PLANE TRIANGLES 151 175. Case II. Given two sides and the included angle, as a, b, C. To solve the triangle we calculate H^ + ^) as the comple- ment of AC; then h {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. Formulas. HA+B)=90°-hC > tan HA-B)= ~~ tan h (A -\- B), sin C ^sinC c = a-. — J = smA ^inJ5- Scheme for Logarithmic Solution. a = log (a - 6) = HA+B) = h = colog (a + 6) = HA-B) = a + b= log tan \ (A + 5) =^ A = a-b= log tan A {A - B) = B = log a = log 6 = log sin C = log sin C = colog sin A = colog sin B = logc = logc = 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. 152 OBLIQUE PLANE TRIANGLES [176 Logarithmic Solution. HA+B)=90°-hC = 90°- 13° 42' = 76° 18'. a = 20.63 log (a - 6) = 0.9074 HA-\-B)= 76° 18' b = 12.55 eolog (a + 6) = 8.4792 ^iA-B)= 44° 58'.4 a + 6 = 33.18 log tan H^ + g)= 0.6130 A =131° 16'.4 a-b= 8.08 log tan h{A-B)= 9.9996 B = 31° 19'.6 1.3145 9.6630 0.0682 log a = log sin C = cologsinA = logc = B) = 9.9996 log b = log sin C = cologsin5 = logc = A =131' B = 31' 1.0986 9.6630 0.2841 1.0457 1.0457 c = 11.11 c = 11.11 Graphic Solution. This is shown in the figure above, student scale off the known parts. Exercises. Solve the following triangles: Let the 1. a = 1500, 2. 6 = 15.25, 3. a = 1.002, 4. & = 6238, 5. 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'. 5 = 36° 54'. 176. Case III. Given two sides and an opposite angle, as a, 6, A, This is known as the ambiguous case. We begin by studying the graphic solution. Lay off angle A and on one of its sides take AC= b. With C as center and radius equal to a, strike an arc of a "circle. The figures show the various possibilities arising in the construction, the first three for A < 90°, the last three for A > 90°. 176] OBLIQUE PLANE TRIANGLES 153 In each case the perpendicular from C on the other side of angle A is equal to 6 sin A. Inspection of the figures then shows that when A < 90° and a < 6 sin ^, no triangle is possible; when A < 90° and a = b sin A, a right triangle results; when A < 90° and b > a > b sm A, two oblique triangles result; when A < 90° and a > b, one oblique triangle results; when A > 90° and a = 6, no solution is possible; when A > 90° and a > b, 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,b, A. f . „ b . . (C = 180°-(A + 5). sinB = -sniA. „ ' ^,. B' = 180°-5. sin C , sin C c = a- — 7 = b- — ^' smA sin is , sin C , sin C c = a -. — T- = 0-. — ^ sin A sin B Check. The agreement of the values of c and c' 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 B = 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, b Logarithmic Solution.* log6 = 2.88316 loga = colog a = 7.22019 log sin C = log sin A = 9.79217 colog sin A = log sin S = 9.89552 logc = B = 51° 50'.0 c = 5' = 138°10'.0 loga = A+B = 90° 7'.3 logsinC' = A+B' = 166°27'.3 colog sin ^ = C = 89° 52 '.7 log c' = C = 13° 33'.7 c' = = 764.1, A = 38° 17'.3. 2.77981 log6 = 2.88316 0.00000 log sin C = 0.00000 0.20783 colog sin B = 0. 10448 2.98764 logc = 2.98764 971.9 2.77981 logb = 2.88316 9.36960 log sin C = 9.36960 0.20783 colog sin 5' = 0.10448 2.35724 log c' = 2.35724 237.6 Graphic Solution. This is shown in the figure, from which the unknown parts may be scaled off. Exercises. Solve the triangles whose given parts are: 1. a = 29.95, 6 = 37.17, A = 42° 24'. 2. a = 1756, h = 745, A = 67° 30'. 3. h = . 728, c = . 542, B = 105° 44'. 4. h = 6.174, a = 2.614, B = 32° 22'. 177. Case IV. Given the three sides, a, b, 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. s = 2 (a + & + c) ; r = \/ (s — a) (s — b) (s — c) tan 2 A tan- S r 1 -3^; tan^C Check. HA-h B + C)=90° * The fifth 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 = Logarithmic Solution. a 428.6 h 806.2 c 542.4 428.6, h = 806.2, c = 542.4. cologs 7.0513 log (s - a) 2.6628 log(s - h) 1.9159 l_A 14°47'.7 A B 55° 51 '.5 \ C 19° 20'.5 2 s 1777.2 log (s - c) 2.5393 Check 89°59'.7 s 888.6 s-a 460.0 2|4.1693 log r 2.0846 A 29° 35'.4 S 111°43'.0 s - b 82.4 log tan *^ 9.4218 log tan ^B 0.1687 C 38° 41'.0 s - c 346.2 179° 59'.4 Check 1777.2 log tan iC 9.5453 Graphic Solution. This is shown in the figure, we find A = 29°, B = 112°, C = 38°. B By measuring ScaU Exercises. Solve the triangles whose given parts are: .1. a = 6192, b = 4223, c = 7415. 2. a = 156.21, b = 300.15, c = 410.32. 3. a = 0.00245, 6 = 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 = Wic = I ac sin li. (h = a sin b.) 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 h B. Then K = ac sin h B cos -> B = acy a)(s _£_) ./sji b) 156 OBLIQUE PLANE TRIANGLES [179 by (4') and (5') of (172). Hence, (9) JS: = Vs (s - a){s - b){s - 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. a = 183.9, b = 584.9, c = 166.6. 5. a = 183.7, A =36° 55'. 9, C =70° 58'. 2. 2. a = 1.925, b = 2.243, c = 7.25. 6. a =283.6, A =ir 15', B = 47° 12'. 3. a = 42.31, 6 = 71.70, c = 71.35. 7. ^ a = 783, B = 42° 27', C = 55° 41'. 4. a = .41409, 6 = . 49935, c = .18182. 8. c = 22.504, B = 55° 11', C = 45° 34'. 9. b = 3069, B = 15° 51', A = 58° 10'. 10. b = 100.2, B = 48° 59', C = 76° 3'. 11. a = 3186, b = 17156, A = 147° 12'. 12. a = .8712, b = .4812, A = 24° 31'. 13. a = 1523, b = 1891, A = 21° 21'. 14. A = 61° 16', a = 95.12, b = 127.52. 15. a = .39363, c = .23655, C = 22° 32'. 16. 6 = 147.26, c = 109.71, A = 41° 15'. 17. b = .5863, a = .8073, C = 58° 47'. 18. a = 10.374, c =9.998, J5 = 49° 50'. 19. b = 6.4082, c = 18.406, A = 33° 31'. 20. b = .8869,^ a = 3.0285, C = 128° 7'. 21. a = .8706, b = .0916, c = . 7902. 22. a = 20.71, b = 18.87, C = 55° 12' 3". 23. A = 41° 13', o = 77.04, b = 91.06. 24. a = 4663, & = 4075, C = 58°. 25. a = 43031, c = 31788, A = 19° 12'. 7. 26. a = 16082, c = 13542, C = 52° 24'. 3. 27. a = .00502, & = .00558, c = .00466. 28. b = 2584, c = 5726, A = 27° 13'. 29. b = 37403, a = 49369, A = 81° 47'. 30. a = 6148, c = 7512, A = 133° 30'. 31. a = .01520, fe = .03366, c = .02114. 32. b = 8204, c = 9098, A =62°9'.e 179] OBLIQUE PLANE TRIANGLES 167 33. 34. 35. 36. a = 532, a = 290, a = .000299, c = 7025, b = 704, c = 356, c = .000180, b = 8530, C = 73°. C = 41° 10'. A = 63° .50'. C = 40°. 37. 38. 39. 40. b = 1482, a = .2785, B = 50° 20' 54", C= 49° 47' 26 a = 12S4, b = .2275, a = 235.64, c = 725.52, .4 = 27° 18'. B = 65° 40'. b = 284.31. b =950.04. In any triangle ABC, whose sides, opposite angles A, B, C, respectively, are o, b, c, show that: 41. b{s-b) cos2 1 =a{s-a) cos2 ^ • 42. a = b cos C + c cos fi. 43. (a - 6) (1 + cos C) = c (cos B - cos A). a2+ 62 + c2 . . cos A , cos jS , cos C 44. r a 6 c 2o6c 45. (b +c - a) tan .4 {c +a -b) tan - ■ 46. {b + c) (1 - cos A) = a (cos B + cos C). 47. (a2 - 62 + c=) tan B = (n= + &" - C") tan C. 48. cot 2 + cot 7j + cot 2 .^ .5 .^ cot 2 cot 2 ^ot 2 ■ 49. The radius of the inscribed circk V— )(s-b)(s- c) 50. The diameter of the circumscribed circle is o 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 assumed set of values of the literal quantities: 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 b, their included angle C; show that the area is ab sin C. , 65. 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 side is 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 ground 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] OBLIQirE PLANE TRIANGLES 159 75. From a point GO ft. above sea level the angle between a distant ship and the sea horizon (the offing) is 20'; how far away is the ship ? [Consider 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 is 14° 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 ground 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.; ZACD = 30°25' 15"; Z ACB= S5° iO' 20"; Z BDC = 35° 14' 50"; Z BDA = 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, ■ Z A = 44° 36', Z A = 56° 32', Z C = 55° 30', ZC =40° 0'; ZC =50°20'; Z.& = 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 5 = 69° 55' 0"; Z C == 46° 48' 0"; Z fi = 73° 29' 10"; Z ^ = 45° 20' 40"; calculate EF. 82. In a chain of five consecutive triangles, each having a side in common 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 farther end of this line by the sides of the triangle. 85. A right triangle whose perimeter i.s 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 being equilateral, has sides 40, 20, 30 ft. and rests on the shortest side. Ans. —^ 88. The sides 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 ABCD 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 AD, and the area of the field in acres. (1 chain = 4 rods.) 90. The following measurements of a field ABODE are made: A to B, due east, 25.52 chains; J5 to C, E 40° 26' N, 22.25 chains; C to A N 48° 26' W, 33.75 chains; DioE,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 _ TT7 sin i + M cos i cos i — M sin i Calculate P when W = 1000 lbs., i = 30°, /x = 0.1. 95. In exercise 94, what is i if P = J T7 and ^ = 0.1 ? 96. If I be the length of a plane inclined to the horizontal at an angle i, fi 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 / ^-^ y g (sin i — m cos i) What is T when I = 25 ft., i = 20°, m = 0.1 ? 97. In exercise 96, find i when I = 100 ft., m =0.1, T=5 sec. W / 98. When light passes from a rarer to a denser medium, the index of refraction m is determined by the equation __ • _ sin i :-~ '^ ~ sin r ^-l^l--"" 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 mci- dence i, and the index of refraction m. CHAPTER XI The Progressions. Interest and Annuities 180. Arithmetic Progressions. — Let a,h,c, . . . , k, I 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, h, c . . . ,k, I are called the terms of the pro- gression, a and I the extremes, and h, 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 -f 2 d, the last or nth term is a -\-{n — 1) d; that is, (1) I = a+ {n - 1) d. Also S = a+(a + f/) + (a + 2d)+ • • • +(a + n - If/); S = l +{l -d)-\-(l -2f/)+ •••+(/ -7T=n[f/). Adding, 2.S=(a + + (a + 0+ • • • +(« + /)= n(a + 0. Hence (2) S = r^{a + l). Putting for I its value from (1), ■ 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 quantities, 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, l = a-\-{m-\-2- l)d; hence J I — a Then the required means are a + rf, a + 2d, . . . , a + md. When w = 1 we have only a single mean, called the arithmetic mean. It equals ^ (a + 0. 182. Examples. 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 S = ^ (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, S = - 27; to find n. From (2'), -27 = n U -""^ X3^, or n2 - 3 n - 18 = 0. Hence « = 6 or - 3. Since n must be positive we discard the second value. 3. Find four numbers in A.P., such that the sum of the first and last shall be 12 and the product of the middle two 32. Let the numbers he a — 3 d, a — d, a + d, and a + 3 d, with a common difference 2 d. Then a-2d+a + 3d = 12 and ^ (a-d)ia+d) = 32. Hence a = 6 and d = ± 2. Therefore the numbers are 0, 4, 8, 12, or 12, 8, 4, 0. 1S3, 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; 5. 63, 58, 53, . . . , to 8 terms; 2. 5,8,11. . . . , to 12 terms; g^ x,x+2y,x+ i y, . . . , tolOterms; 3. 2, 2i, 3, . . . , to 25 terms; 4. 1,1.1,1.2,.. . . ,to200terms; '^' V,V-hq,V-q,. .., to 20 terms. Find the other elements of the A. P., given that : 8. a = 10, n = 14, S = 1050; 16. n = 35, S = 2485, d = 3; 9. a = 3, n = 50, S = 3825; 17. n = 50, >S = 425, d = h 10. a = - 45, n = 31, 'S = 0; 18. n = 33, S = - 33, ri = - f; 11. Z = 21, n = 7, S = 105; 19. S = 624, a = 9, rZ = 4; 12. / = 49, n = 19, 5 = 503J; 20. S = 2877, a = 7, d = 3; 13. Z = 148, n =27, S = 2241; 21. S = 623, d = 5, I = 77; 14. Z =-143,n = 33, S=-2079; 22. S = 682.5, rf = 1.5, Z = 45; 15. n= 21, S = 1197, d = 4; 23. S = 95172, d = - 7, Z = 567. 24. Find the sum of the first 100 odd numbers. 25. 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 long 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 them one at a time to a basket standing beside the first apple ? 184. Geometric Progressions. — If the numbers a, b, c, . . . , k, 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," ''means," "extremes," 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, I, 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) 1 = ar''-^. Also S = a -{- ar -{- ar^ -\- • • • -\-ar"'-'^ and rS = ar -{- ar- -{- • • • + ar" - ^ -|- ar"*. Therefore rS — S = ar" — a, or (r-l)S= (r" - 1) a. Hence ■y" — 1 1 — *■"' (2) S = a- - = a- Substituting from (1) in (2) we have (20 s = ':^. 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, 185. Problem. To insert m geometric means between two given numbers a and I. The total number of terms being m + 2, we have, if r denote the ratio, m+in I = ar'^ + s-^ or r = V -. ▼ a The required geometric means are then ar, ar^, . . . , ar^. When m =^ 1, the resulting single mean between a and I is y/al. The square root of the product of two quantities is called their geometric mean. 186. Examples. 1. Find the sum of the first 10 terms of the G. P. 2, 22, 23, . . . . 210 — 1 Here a = 2, r = 2, w = 10; hence S =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. From S = a ^" ~} we have 63 = ^!! ~ ] ; 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) 02,-3 = 32. Multiply (1) by a and in the result replace a?r^ by 32. Then a^ + 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, ',, i',, . . . , to 10 terms. 3. 1, 4, 16, . . . , to 7 terms. 6. 8, 2, f , to 20 terms. 7. a, a (1 + x), a (1 + xY, ... to 8 terms. 8. rrfi,mn,m.-hi^, . . . , 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 /j. 12. Insert 6 geometric means between 3 and 49152. 13. Insert 4 geometric means between 48 and o\- 14. Insert 5 geometric means between 81 and V/- Calculate the unknown elements, given : 15. Z = 128, r = 2, n = 7. 22. a = \, Z =2401, S = 2801. 16. Z =78125, r = 5, n = 8. 23. a = 10, I = h, 5 = 191-^. 17. l=i^, r = \, n = 5. 24. a = 3125, I =b, 5 = 3905. 18. a=9, ^ = 2304, r = 2. 25. a = 3, r =3, 5 = 29523. 19. 0=2, Z = 64, r = 2 26. a=8, r =2, 5 = 4088. 20. rt = 3, / = 192V2, r=V2. 27. r = 2, n = 7, 5 = 635. 21. 0=2, Z = 1458, 5 = 2186. 28. Z = 1296, r =6, 5 = 1555. 188. Infinite Geometric Progressions. — Consider a line segment AB of unit length, and bisect it at Ai, then bisect Ai5 at A-z, A2B at .4.3 and so on (figure). The points of bisection A,, A., ^3, • • • ^ ^ ^'' '^^ ^^^ continually approach B and the sum of -- - - the segments AA\ 4-^1^12 + MA?, + • • • approaches AB or 1. But the sum of these segments is represented numericallj^ by the series ^ + i + §+' • •' ^^ 2 + 2^ + 23+' • •' 166 THE PROGRESSIONS [189 and hence by taking n large enough we can make the sum ^ =14-1-4-14- . . . -^1 ^n 2 ^ 22 "^ 23 "•" "^ 2" differ from 1 by as little as we please. Hence we take 2 + 4 + 8+ ■ ■ ■ toii^finity = 1. The sum Sn above is a geometric progression with r = I and a = |. Its sum to n terms is therefore 1 (hr - 1 As n increases, {\Y approaches 0, and Sn approaches the value - y = 1, as found above. 2 2 — 1 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 r'^ - \ 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 S = _ > provided | r | < 1. (When r = 1, or when r > 1, -S is infinite.) Exam-pie. 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, h 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 S = j^ = 180 ft. 189. Exercises. Sum the following infinite geometric progres- sions: 1. 8, 2, *, . . . . 3. 5, 3, § 5. 1, -h, +i -i, .... 2. \,\,-h, .... 4. 2,?, A, ... . 6. 3, -1, i -I, . . . . 190, 191 ] THE PROGRESSIONS 167 7. If in the example worked above the ball requires 4 seconds for the first 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,b, c, . . . , k, I are 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, b, c form a H. P., then b is called the harmonic mean between a and c. Let the student show that we then have , 2 ac b = — , — 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 1 165. 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. 5. 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, nr^, .... 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 h, the sum of their squares k. 15. 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 [192 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 _L is dropped on the hypotenuse, forming a new right triangle which is treated similarly, and so on indefinitely. Find the sum 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 sum 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 A, =P(l+r). 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 = Ai(l-\-r)=P(l+rr, and so on. Hence ii Anhe the amount of P dollars in n years, interest at rate r compounded annually, we have (1) A,, = P(l-\-rr. If interest is compounded every t years instead of annually, then, after n compoundings, the amount is (!') _ ^„ = P (1 -f rty\ 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 = t!o; t = \; n = 8. Then An = 100 (1 -f .04 X i)^ = $100 (1.01)^ = $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)"-i; the second n — 2 years, its amount ^ (1 + r)"-^; and so on, to the payment next before the last, which is at inter- est one year, its amount A {1 -\- r); the last payment amounts to A. The total amount S is therefore AS = A+A(14-*-)+A(l+r)2+ • • • +^(i-f-,.)n-i^ or (2^ ^-^^ 1 + ..-1 -^ — ; 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 ?i — 1 years? Let Q be the amount required, called the present worth of the annuity. Let Qi be the sum which with interest will yield in n — 1 years the amount of the first payment, or A (1 + r)"-^ Then Qi(l+r)«-i =^(l+r)"-i or Q^ = A. Let Q2 be the sum which with interest for w — 1 years will yield the amount of the second payment, or A (1 -f r)"-^. Then Q2(l+r)"-i =^(l+r)"-2 or Q2 = ^ 1+r Similarly if Qs, Q4, . . . Qn be the present worths of' the 3rd, 4th, . . . last payments of the annuity we have, Hence 170 INTEREST AND ANNUITIES [194 The sum in the parentheses is a G. P. with ratio ^ • Apply- ing the formula and reducing, (1 + rY - 1 (3) Q 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. 5. 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 ^1 dollars, compound interest r%, the first payment now due. (Q = Qi + Q2 + Qs + • • • ad inf.). 9. As in exercise S, except that the first payment falls due in m years. CHAPTER XII Infinite Series 195. Limit of a Variable Quantity. — When a variable quantity changes in such a way that it approaches a fixed numerical value, so that the difference between the variable and the fixed quantity becomes and remains less than any assignable magnitude, however small, then the fixed quantity is called the limit of the variable. For example, as x varies the variable quantity 1 -{- x can be made to differ 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 -f 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 -|- a;) = 1. (= is read "approaches.") 1 = Exercise . Show that : (a) Hm ^ = 1 ; (6) Hm (l + J) = 2; (c) lim log (1 + a:) = 0; (rf) lim fl - ^V 0; (e)lime"=l; (/) lim fl + -Y = 1. n = 10\ n/ x^O n = oo\ 11/ 196. Infinite Series. — A sequence or succession of terms, ui, U2, Us, . . . , iin, . . . , unlimited in number, is called an infinite series. The sum of the first n terms of a sequence we denote by 5„. 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 5„, 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. (a) 2 "^ 22 "^ 23 "^ ' ' ■ "^ 2" "*" ' 1.1, .11 i^r - 1 g. (188. ^^n 2 "^ 22 "^ ■ ■ ■ "^ 2'' 2 i - 1 lim *S„ = 1 = >S. The series converges to the value 1, 71 — 00 or, ^ + ^2+ • • • +^+ •••=!. [(188), figure.] (b) l + 2 + 3+---+n+.-.. *S„ =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 - I + 1 - 1 + • • • •. Here Si = 1; >S2 = 1 - 1 = 0; ^Ss = 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 be shown that Sn, the sum of its first n terms, 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 *S„, and let h 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^ + • • • + ar"'-^ + • • • 1— r'* Here *S„ = 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 = oo -i 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 -\r ar -{- «/■'+ • • • > converges when \r\ < 1, and diverges when \r\ =1. Putting a = 1 , we see that the smpZe pother series, I+cc+cc^tI- • • • ^ converges when \oc\ < 1 and diverges when | a? | = 1 . 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 W1+W2+ Ms + • • • +Wu+ • • • . Then Sn = ui-{-U2-\- • • • +m„ and Sn-i = ui-^U2-\- • • • +Wn-i. Hence Un = Sn — Sn-i- By taking n large enough, both *S„ and Sn-i 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, w„, can be made to differ from zero by less than any assignable small quantity. lim Un = 0. n = oo 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 iin 9^ 0. n = oo 199. Alternating Series. — A series whose terms are alternately + and — is called an alternating series. Theorem 2. An alternating series converges provided that (a) each term is numerically less than the preceding, and (b) the linvit of the nth term is zero as n increases indefinitely. Proof. Let the series be Wi — M2 + Ms — M4 + W5 — Me + • • • . Write this in the two forms, (mi - M2) + {Uz - U4) + (M5 - Uq) + • • • ; Ml - (M2 - M3) - (M4 - Ms) - • • • . Each set of parentheses incloses a positive quantity according to condition (a) of the theorem; hence assuming that mi, U2, M3, . . . 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 wi. Also, since hm w„ = 0, the sum n = 00 cannot oscillate. Hence the series converges to a value between and its first term. Exam-pie. The alternating series, l-h + \-\+ ' • • converges to a value between and 1. 200. Absolute Convergence. — A series is said to converge abso- lutely when it remains convergent if all its terms are taken positively. Thus if wi, M2, W3, • • • be in part negative and in part positive, the series Wi + W2 + W3 + • • • converges absolutely provided that the series I Wi 1+ I W2 I + I W3 I + • • • converges. Exercise. Show that the series \ ■\- X -\- x"^ -{- • • • and a -\- ax -^ ax^ + - • • both converge absolutely when \x\ < 1. 201. The Comparison Test. Let wi + W2 + W3 + • • • be a series known to converge absolutely or to diverge. . Let vi -\- V2 -\- v^ -{- ■ • • be a series to be tested for convergence or divergence. Then, (a) If the u-series converges absolutely and, for all values of n, v^ is numerically less than Un, the v-series also converges absolutely; (b) If the u-series diverges and Vn is numerically greater than Un, and if all the terms of the v-series have the same sign, the v-series also diverges. Proof. Let r/n = I Wl I + I W2 I + I W3 I + • • • + \Un\ and Vr,= \vx\ + \v2\ + \vz\+ • • • +|t'„|. Then by condition (a), Un approaches a limit, say C/, as n = oo , and also, 7„ < f/„. Hence, since F„ must increase steadily with standard Test Series. (For use i: (1) a -\- ax -{■ ax^ + • • . + aa:'^ + (2) l-\-x^x'+ . ■ ■+x'^+ ■ • (3) l+^ + |l+- • •.^-H.. (4) 1 + 1+1+.. . + ^--- (5) j-^ + ^^-^h^' ■-^- 201] INFINITE SERIES 175 n, but is always less than f/„, it must approach a limit 7, less than U. Hence the v-series converges. Under condition (b), [/„ increases without limit, and also, Vn > Un. Hence F„ also increases without limit and the t^-series diverges. , i Conv. when | a; | < 1 ; ) Div. when \ x \ =1. Convergent. Divergent. ^ Conv. whenp > 1; S 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 + ^ + ^)+(^+i^0+ • • • +tV)+ • • • . We can form in this way an infinite number of parentheses, each of which is > ^. 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: p + (^2^ + 3]^ j + (^4^ + • • * + 7^ j + (s^ + ■ • ■ + i5^j + • • • . Considering each group of terms as a single quantity, we see that this series is less, term for term, than the series 1+2+1+8 2'' 4^ 8^ 1.1.1,1. or 1 + 2?n + 4^rri + g^i + • • • • I 2P- fore the given series converges. 176 INFINITE SERIES [202 Examples. 1. The series 1+22 + 33+' ■■+^+''' converges; for it is less, term for term, than (3). 2. The series 1 +^ ?, +-, ^ + • • • + , „ ^ + . . . diverges; for logio2 logio3 logion it is greater, term for term, than (4). 202. The Ratio Test. — The series Wl+M2+^*3+ ■ • • +Mra+ • • • converges absolutely if, beginning at some point in the series, the ratio Uji -r- Un-i becomes and remains numerically less than a fixed positive number which is itself less than 1. Proof. Assume that ^^ < r < 1 for all values of n > iV, \Un-l I A^ being a fixed positive integer. Then | m„ | < r\ Un-\ \ when n> N. Hence putting n = A^ + 1, A^ + 2, . . . , we have \uN+i\+M + ' .. .,! + ... Un Un- n - 1 1 n , which approaches 1 from the lower s 4. 12 3 2 + 3+4 + • •■+nTl+- Un _ n Un-l n + 1 .71-1 n 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: r T-2 T?i 3. 1 + 2 X -I- 3 x2 4- • • • + (n + 1) X" -I- • • • . 4. cos X -f cos2 X 4- • • • 4- cos" x + • • • . 5. tan X 4- tan2 x 4- • • • 4- tan" x 4- • • • • 6. sin-ix 4- (sin-ix)2 4- • • • 4- (sin-i x)" 4- • • • . 7. logio X 4- (logio x)2 4- • • • 4- (logio x)" 4- • • • . „ 1 -2 , 2-2 , 3-4 178 INFINITE SERIES [203 9 J-+A_ . J_+. . . l-2^2-3^3-4^ 10.* \lx+\2x'i + \3x'+---+\nx^+---* 11. H-X + I+I+---. ^2- ^-IS + IS-JT+'-'- ^. -. 1 ,1-3 1-3.5,, 1.3-5-7, 1*- l-2^+2T4^-2-Tr:6^ + 2.4.6.8^ ' 15. Z - i l2 + 4 X3 - 1 X-l + • • • . * [n = 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 function of x. Thus y = X- -{-1, y = a^, y = sin {ax -\- h) are all functions of x. As an equation, we indicate that y is a function of x by writing y = f(x). When a body is dropped from rest, the space s (ft.) fallen through in the time t (seconds) is s = | gf. Here s is a function of t, or s=fit); f(t)^hgt'- When a train is running at 30 miles an hour, the space s (miles) covered in the time t (hours) is s = 30 t. Hence s=f(t); f(t)=^Qt. When the relation between y and x is given by an equation of the form y = f(x), y is called an explicit function of x. Suppose the relation between x and y to be given in the form, a;2 + ?/ = !. 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 1/2 = 4 x" sin (x -\- y)= 1; a"" -{- a'-' = b. 205. Variation of Functions. — Consider the relation y = x^. When x = a, then y = a-; when x = a -\- h, y ={,a-\- h)~. 179 180 DERIVATIVES [206 As X changes from a to a -{- h, y changes from a^ to (a + h)-. The total change in x is h, and the corresponding change in y is (a + hy - a2 or 2 ah + h^. Let us designate a change in x by Aa; (read " increment of x," or "delta x") so that in this example Ax = h; let the corre- sponding change in y be Ay, so that we have in this case Ay = 2ah-\-h^ = 2aAx + Ax^. In general, if y = f (x), then to the values x and a: + Aa; of the variable x correspond the values / (x) and f{x-{- Ax)* of y. Hence the change in y, corresponding to the change Ax in x, is Ay=f{x-\-Ax)-f{x). Continuous Function. — When Ay = 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 a; = 2 and Ax = 0.1. 2. As in exercise 1, when y = 's/x. 3. As in exercise 1, when y = x^. 4. As in exercise 1, when y = 10-^. 5. 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 = \ and Ax = 0.01. 206. Difference Quotient. — The fraction change in y Av change in ..c Ax is called the difference quotient of y relative to x. Thus, if y = x-, then Ay ={x + Ax)^ - x^ = 2xAx-\- Ax^. Hence the difference quotient is Ay^2xA^^fA^^2x + Aa:. A.T Ax We shall abbreviate Difference Quotient by writing D. Q. Exercises. Calculate the D.Q. in the exercises of (205). * / (x 4- Ax) stands for the result obtained by replacing x by x + Ax in /(x). 207,208] DERIVATIVES 181 207. The D.Q., -r^, geometrically. — Lot the curve in tlie fig- y ure represent a part of the graph of the equation y = f{x). Let P be a point on the curve hav- ing coordinates (x = OM, ij = MP), and P' a second point {x -{- Ax = OM', y-{-Aij = M'P'). Let the secant PP' make an angle 6' with the rc-axis. Draw PQ II OX. Then from A PQP', tan^' = ^. Ax Slope. — The tangent of the angle which a line makes with the rc-axis is called the slope of the line. Hence, the difference quotient, -v-;, is the slope of the secant drawn through the points (or, y) and (x + Ax, 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. i 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 = f{x), {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 d be the angle which the tangent to the curve at P = {x, y) makes with the x-axis, then ^ X (read, ''tangent of 6 equals the tan e =^ljm|^j jj^^ ^^ A| ^^ ^^ approaches 0." When — approaches a definite limit a tangent is thereby deter- Ax ^^ 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 y = X- tanO = 2x 209. Examples. 1. y = X-. y + Ay={x + Aa;)2 = x2 + 2 a; Ax + A?. Ay = 2x Ax + A? and ^ =2x+Ax. Ax Hence lim — = 2 x = tan 9. Ax^O ^^ Here the slope of the tangent at any point equals twice the abscissa. 2. y = ^\xK y + Ay = 2V (x + Ax)3 = 2V (x' + 3 x2 Ax + 3 X A? + Ai^). Ay = ii (3 x2 Ax + 3 X Ax2 + Ax^) 'and ^ = ^V(32:2 + 3xAx+ Ax^). Ax rt Hence lim v^ = na;2 = tan 0. y = 5V a;3 tan e = 1x2 and 3. 2/ =x2 -2x. y + A?/ = (x + Ax)2 - 2 (x + Ax) = x2+2xAx+Ax^-2x-2Ax = x2-2x+ (2x-2)Ax +Ax^. Ay = (2 X - 2) Ax + A? Ay = (2x-2)4-Ax. Ay y = x2 — 2 X ton = 2 X - 2 lim — ^ =2x -2 = tan». Az^oAx 4. y2 = X. Here y is an implicit func- tion of X. Solving, we have y = ±'s/x. 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 = V-c and y + Ay = Vx + Ax. A(/ = \Jx + Ax - V-C- Multiplying and dividing by Ay V-c + Ax+ Ax, we get ( Vx + Ax - Vx) ( Vx + Ax + Vx) \Jx + Ax + '^x Ax ton fl = ± Vx + Ax + Vx Hence ^ = -=L ^, Ax V^ + Ax + V^ lim --^ = ~ = tan 9. Ax-^oAx^ 2VaJ For the lower part of the curve, replace Vx by -sjx. 5. x2 + y2 = 100. Solving for y, we get 2Va; and ij=± VlOO - x2. Considering first only the upper half of the circle (figure) we have VlOO - (x + Ax)2 y = VlOO - x2 ; y + Ay .-. Ay Multiplying and dividing by the sum of the two radicals, VlOO - (x + Ax)2 - Vioo - x2. VlOO Ay ■2x Ax - Ax2 Hence and VlOO - (x + Ax)2 + VlOO - x2 Ay _ 2x + Ax Ax lim Ay VlOO - (x + Ax)2 + VlOO 2x X = tan 6. X2 Ax=^oAx 2 VlOO - x2 VlOO At any point on the lower half of the circle, tan d =-\ . • VlOO — x2 In all these examples the slope of the tangent at any given 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 = I x3. 4. 2/2 = 4 X. 7. x^ - y^ = 1. 2. y = 2 x2 - 3 X. 5. y2 = - 9 X. 8. 9 x^ + 16 y' = 144. 3. y = x3 - X. 6. x2 + y2 = 1. 9. 4 x2 - y2 = 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 ( -^j 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 a; as follows. Calculate f (x -{- Ax) — f (x) or Ay, divide by Ax, and pass to the limit by allowing Ax to approach zero. Call the new function of x so obtained f'{x), so that /'«=i™(i!)- This is called the ^rs^ derived function off{x) or thej^rs^ derivative off(x), and the expression sx^oXAxJ is called the first derivative of y with respect to x. It is usually written in one of the forms Hence the slope of the tangent to the curve y = f(x) at a point (x, y) is dv doc 211. Calculation of Derivatives. — We have already calcu- lated the derivative of y with respect to a; 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 calculation 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. J)^ (c) = 0, c being a constant. (1) For if c is a constant its change is 0, hence Ac = 0. (2) Therefore ~ = 0. Ax (3) Hence lim ^^ = or D, (c) = 0. Ax = Ax 211] DERIVATIVES 185 II. D^ {cij) = c D^i/, c being any constant. Proof. (1) The increment in y being Ay, the increment in cy will be c Ay, (2) Dividing by Aa:, the D. Q. of cy relative to x is c-r^- Ax Av (3) Let Aa: = 0. Then c does not change, while — ^ becomes Dx (y) . Hence DAcyy=Vimc^ = cD,y. Ar=o Aa; III. When ?/ is a sum of several functions of x, as y = u-{-v-\-w-\- • • • , where u, v, w, . . . are functions of x, then n^u = D^ti + n^v + D^w + • • . Proof. When x takes an increment Ax, let the corresponding changes in u, v, w, . . . he Au, Av, Aw, . . . respectively. The total change in y is, therefore, (1) Ay = Au -\- Av -]- Aiu -\- ■ ■ • . (2) Then ^- = f^ + ^ + p + . . . . Ax Aa; Aa: Aa; (3) Let Ax = 0. Then by definition (210), ^ approaches D^y, rr- approaches D^u, etc. Hence D^y = D^u + DxV + D^w + • • • , when y = u -{-v -\- w -\- ■ ■ • . IV. Let y be the product of two continuous functions of x, say u and v. y = u ' V. When X is changed to a; + Ax, let u change to u + Am and v to V -}- Av. Then 2/ + A?/ = (u + Au) {v + Av) = ?^y + M At' + y Af< + Am Ay. (1) Hence Ay = u Av -\- v Au -\- Au Av. (2) Then -^ = -u-_ + y-— 4-Au — • Ax Ax Aa; Ax 186 DERIVATIVES [212 (3) Let Ax = 0. Then t^' -;— ' -r~ approach D^y, D^u and D^v ^ ^ LX LX diX respectively. Also Aw = 0, since we assume m to be a continuous function of x (205). Hence (2) becomes D^y = u D^v + V Da^u, when y = u • v. V. Let y = -,u and v being continuous functions of x. u -\- Aw Then y + C.y = -^^^-^, u-\- Ml u V iya — u^v (1) and Ay = -^^ - - = ,2 + ,Ay " Am Ay V- U-— Av Ax Ax (2) Hence ^^ = „. + „a, A?/ t^D^ii — uD^t^ (3) and I>.2,=Jim-= ;, VL Let y be a function of u, where m is a function of x. Thus y = w2^2m; m = 2a;2 + l. When x changes to re + Ax, u changes to m + Aw and y to Ay _ Ay Aw Now A^-Aw'Ax' Hence I>^y - 1>„2/ • D^u. Collecting our formulas we have: (A) D^c = 0. (B) D^ {cy) = c I>^?/. (C) i>x (« + t' + ^t' + • • • ) = I>ocU + I>xi^ + DooW + • • • . (D) n^ (u ' v) = u D^v + ^ I>a>u. /ii\ vD^ii — uD^v (F) D^y =- D.,y ' n^u. 212. We next derive the following standard formulas: (G) i/=x"; DJ/ ^nx'^-K (H) i/-logx; !>.?/ = ^- 212] DERIVATIVES 187 (I) y = r/'; n^!/ - a* log a. (J) 2/ = sinx; D^y = cos a?. (K) y = cosir; D^y = — sinx. (G) y =^ x"; assume n to be a positive integer. (1) Hence A2/ = nx"-iAx + ''^^^^^^a:''-2Ai'+ • • • + aI-". (2) Then ^ = nx" - + "^^f^ ." -' A. + • • • +Ax"-'. ^ ^ Ax 1*2 (3) Let A.T = 0. All terms on the right of the last equation vanish except the first, and lim ^ = D,y = nx^-K The proof when n is not a positive integer will be given after formula (H) is derived. (H) y = logx; ?/ + A?/ = log {x + A.t). y- _J_ At / At \ (1) A2/ = log (x + Ax) - log X = log — ^- = log(^l+-^j- (3) Let Ax = 0. We must evaluate X ' I Ax lim log I 1 + -^ Ax = \ X Let z = — ; then 2 = 20 when Ax = 0, provided x ?^ 0. [x = Ax is excluded by our standing assumption of continuity (205).] We must now evaluate Let z = 1, 2, 3, . . . , n. The corresponding values of fl + -j are 2,2.25,2.37, ...,(! + -). As n 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, y-^n) ~^'^'\^ 1-2 n2+ 1.2-3 n^^ to (n + 1) terms _■,.,, }~n) V~7ilV~^il , ^to(n + l) ■^"^1-2^ 1.2-3 "^ \ terms. 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- 2" ~ ■*■ whose sum is less than 2. (-^■) 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 1 lim 1 + z when z 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 ( 1 + - ) = log e, and hence Z>x (log x) = ^ log e. 212] DERIVATIVES 189 Let us now take e as the base of our system of logarithms, so that log X shall mean loge x. Then loge = logeC = 1. Hence D^ (log 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 y = x"" we have logy = n log x. (Base e.) Hence Z)x (log y) = Dj. (n log x). Now in formula (F) replace y by log y and u by y. It becomes D,(log y) = D,(log y) • D,y = \^D,{y), from (H). Also Dx{n\ogx) = -> from (B) and (H). -D^y =- y X or Dxy = —- > where y = x". X Hence D^x"" = — = nx""-^. X (I) y = a^. Taking logarithms, logy = x log a. Hence D^ (log y) = D^ (x log a). But Dx (log y) = -Dxy (see above) and Dx (x log a) = log a. Hence - D^y = log a, or D^y = y log a, where y = a*. 190 DERIVATIVES [213 Therefore Z)jO'^ = a"" log a. (J) y = B\nx', y-\- Ay = sin {x + Ax). I Ax\ Ax (1) Ay = sin (x + Ax) — sin a: = 2 cos ix-{- — j sin -^ • (158.) / , Ax\ .Ax .Ax 2 cos kc + -^ sm -p^ - a x sm Ay -^--V-^T;-T / A.N-2 (2^ Ax- ^^ ^°'V +2; A^ 2 . (3) Let Ax = 0. Then . Ax sm — cos fa: + ~] = cos x, and — ^ — = 1. U60. Replace x by^-j lim --^ = 1>^ sin a? = cos a?. AX = Arr (K) y = cosx; ?/ + A^/ = cos (x + Ax). / Ax\ Ax (1) A?/ = cos (x + Ax) - cos x = - 2 sin f x + -^ j sin y • (158.) . Ax . sm -^ Ay . / , Ax\ 2 (2) Ax = -^^n"+T)-^- 2 (3) .'. lim -^ = J)^ 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 Dxi^x^ + ^x). Dx (4 x3 + 3 x) = Dx (4 x3) + Dx (3 x) (C) = 4 Dxx^ + 3 Dxx (B) = 12 x2 + 3. (G) 2. Calculate ^^(l+bgx) 214,215] DERIVATIVES 191 J. ( c^ \ _ (1 + log x) D:ce' - fi^D x (1 + log x) .„. "^'[l+logxl (1+logx)^ ^""^ (D, (C), (H), (1 + log x) e^ — e^ - ' (r+iogx)2 ^ xd+logx)-! X(l+l0gx)2 • 3. Calculate DxCSsin^x). Dx(3sin2x) =3Dxsin2x (B) = 6 sin X Dj; sin x (F) ; {u = sin x) = 6 sin X cos x. 214. Exercises. Calculate D^^y when : 1. y = 3 x4 + 5 x3. 10. ?y = log (x + 2). 2. y = 2-3 + 1. 11- 2/ = log(3x2 - 1). Q 1 J , 1 i 12. 7/ = f^ log X. » 3. t/ = ^x' +ix*. 4. y = x = - 2 x^. 13. 2/ = sin X log cos X. , 1 14. 2/ = esinx. 5. 2/ = -L+_L. Vx v^; 6. y = sinx + e^^. 7. w = e^. 16. y = cotx I c i / sin X 15. y = tan x = V cos X ,8. y = a^^ 17. y = log tan X. 18. y = sec x, 215. The Derivative as a Rate of Change. — The difference Aw 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 rate of change. Rule. To find the rate of change of one quantity relative to another, calculate the derivative of the first quantity with respect to the second. Examples. 1. y = x2. Then Dxij = 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 s = I gt^. Then Dis = gt = velocity at time i. 3. Find the rate of change of the volume of a sphere relative to the radius. F = |7rr3; DrV = 4irr2. That is, the volume of a sphere changes 4 irr2 times as fast as the radius. 216. Exercises. Calculate the rate of change of: 1. y relative to x, when y = x^ + x^. 2. y relative to x, when y = sin x. 3. y relative to x, when y = sin x cos x. 4. y relative to x, when y = sin2 x + cos^ x. 5. 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 t = and moves so that the space described in time t (seconds) is s = 16 <2+ 10. Find its velocity when t = 10; t = 5; t = 0. 13. The space-time equation being s = 2t^ + 3t — 5, find the velocity at any time /; what is it when t = 10; i = 1; i = ? 14. As in 13, when s = 10 sin ( 3 t + |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 given sides and to the given angle. 217. Higher Derivatives. — When i/ 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 Dly. The derivative of the second derivative is called the third derivative, written D^y, and so on. ' Exam-pies. 1. 2/ = x3. Dxy = 3x2; D|2/ = 6x; D^y = & ; Dly = 0. 2. y = smx. Dxy = cos x; Dly = — sin x; D^?/ = — cosx ; etc. 3. y = X". Dj2/ = nx«-i; D^y = n (n — l)x"-2 ; .... D> = ?i (n - 1) . . . 1 = |n. 218] MACLAURIN'S SERIES 193 218. Maclaurin's Series. — Suppose that a given function of X, f {x) , can be represented by a converging power series in x, thus : (1) f {x) = Co -\- cix + cox"^ -{- czx^ -{■ ' ■ ■ +c„a;"+ • • • . To find the values of the coefficients Cq, ci, C2 ■ ■ • . Put x = in (1) and we have Cq determined by /(0)=co. To get Ci, calculate DJ{x) or f'(x) from (1); (2) f'(x) = ci +2c2X + 3c3X- + • • • + /ic„x"-i + . . . . Put X = in (2) and we have ci determined by /'(O) = cu From (2) calculate DJ'(x) or f"{x); (3) /"(x)=2c2 + 2.3c3a;4- • • • +w(n- l)x"-2+ . . . . Put a: = in (3) and we have r(0)=2c2 or C2 = ^r(0). Calculating DJ"{x), or f"'{x), we have (4) /'"(a:) = 2 . 3 C3 + • • • + n (n - l)(n - 2) a;"-^ + • . . . 1 When X = 0, /' "(O)- = 2.3c3 ; C3 2 ~/'"(0) Similarly, C4 = = 2-. 1 3.4 r(o)= (0), ^ /^"H0)= ,^-/^"K0). n (n — 1) . . . 1 •' 1 7i Hence /(x) = /(0)+a:/'(0) + ^r(0) + ^'r'(0)+ • • • +fVn)(o)+ . . . . Here f"\0) is found by differentiating /(x) n times in succession and putting x = in the result. The above result is called Maclaurin's series for the function fix). In obtaining it we have tacitly assumed that, if f{x) be represented by a power series, the derivative f'{x) can be calcu- lated by differentiating the series term by term. 194 MACLAURIN'S SERIES [219 219. Examples. 1. Develop e^ in a power series in x. fix) = e^; nx) = e^', /"(x) = e^; . . . ; }^^\x) = e^. Putting X = 0, we have /(O) = 1 ; /'(O) = 1 ; /"(O) = 1 ; . . . ; /<"H0) = 1. Hence ->+-+|+i+---+|+ 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, 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) = sinx; /'(x)=cosx; /"(x) = -sinx; /'"(x) = - cos x, . . . . When X = 0, /(0) = 0; /'(0)=1; /"(0)=0; /'"(O) = - 1, etc. Hence X^ X** x^ sinx=x-|3+^-|y+- • • . 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 ^'''^^°^h>-\[v^'-^m[l^'- Note. In computing mth an alternating series {signs alternately + and — ), the error committed in rising only a few of the first terms of the series is always numerically less than the first term neglected. Thus the error in sin 10° as obtained from the three terms written above is less than A^ fiLV or less than .000 000 000 98. 5040 \ 18/ Hence the error is less than 1 unit in the ninth decimal place. Exercise. Show that cosx = l-|-2+||-^+ • • • . Calculate cos 10° to five places. 220] BINOMIAL THEOREM 195 3. Develop log (1 + x) in powers of x. /•(j)=log(l+x); /(O) = logl = ^'^^)=i+x: /'(0)=1. f"(T\ — •'■ /"(0) = -l. ^^^)- (1+X)2' t"'(^\ — " /"'(0)=2. ^ ^^)-(l+x)3' — 2-3 ^•^(^)=(l+x)-- /'^•(0) = -2.3. log(l + x)=x-|+|' -?+ ■'■•.■ 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 + x) 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 . In particular, when x = 1 we have log, 2 = l-i + 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+;r)" = l + na: + '^^^^^a:2+ • • • + x". We shall now derive the formula for expanding (1 + x)" in powers of x for any value of n, positive or negative, integral or non-integral. Let /(x)= (l+:r)". * Apply (199) to the neglected part of the given series. 196 BINOMIAL THEOREM [220 Then /'(x)=n(l+rc)'»-i; /'(0)=n. fix) = n (n - 1) (1 + a:)"-2; /"(O) = n (n - 1) J"'{x) =n{n- 1) (n - 2) (1 + a;)"-^; /'"(O) = n (n - 1) (n - 2). /""' {x)=n{n-l){n-2) . . . (n - m + 1) a;"-"*; /'"»> (0) = n (n - 1) (n - 2) . . . (n - w + 1). Hence by Maclaurin's series, , n(n — l) ^ , n(n — 1) (n — 2) „ , (1 + ic)" = 1 + na? + -J-^^' + 1.2.3 "^ + * " ' n (rt - 1) . ♦ ' (n - m + 1) «.,... "^ 1 . 2 .... m -T • • > provided that the serfes on the right, called the Binomial Series, converges. Convergence of the Binomial Series. — Denote the mth term of the series by u^, the (w + l)th term by w^+i- Then _ n (n - 1) (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 - 1) . m Um+l = — ' \ ^\ ^-Tin TV^. ^" Applying the ratio-test (202), we have u^ ^ n-m + 1 ^ ^ /n±l _ ^ ^_ u^ m \ m J The quantity in the last parenthesis is numericallij less than 1, when m is larger than w + 1 ; to secure this we simply start far enough out in the series to make m > n + 1. Then the ratio Um + \ -^ Um will be numerically less than x, and hence, if x he numerically less than 1, the series converges. When x is numeri- cally greater than 1, the series diverges. For the ratio m^+i -^ Um 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 numer- ically. Finally, when a; = ± 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 (I + x)" only when |x| 2 for all values of n. Hence 1 and therefore R„, < <2 (2m + l)A;^'"+i (2?^ + 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 (2 7W+ 1) (2 71+ l)-^'«+i 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 stop- ping with the mth term is less than 4 (2m + l)ll2'"+i* 4 Hence when m = 1, the error is less than ^ 3 ; 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 TT-TT^ , so that the first two terms will give log 5 correct to 5 places, O'lP 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-phice 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 6 = 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 n = y, or, n = e^ and n = 10^. Hence 10^ = e\ To solve for y, take logarithms of both members to the base 10. Then y = a:logioe, . which proves our statement. To find the value of logio e, we need only calculate loge 10 and take the reciprocal of the result. Exercises. 1. Calculate the modulus Af 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 uq, ui, U2, . . . , Un, . . . , and form the differences, Auq = ui — Uq, Aui = U2 — Ui, . . . , Aun-i = Un — Un-i, ■ • ■ , callcd 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 Aq. uo Ao Ai A2 As Mo Ml - Mo U\ Uo — Ml M2 - 2 Ml + Uo ^^ - 3 U2 + 3 ?/.i M2 M3 - M2 M3 — 2 U2 + Ml U3 Un-2 M„_i - Un-2 M„_i Un — Un-l Un — 2li„_i + Un-2 Un 204 DIFFERENCES . [228 We observe that the coefficients follow the binomial law. Let the student prove by induction that this law is followed in all the successive difference columns, 228. The nth term 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 kth difference column be denoted by A^Wo- Then we have Wo = Wo, AiWo = ui — Uo, A2W0 = U2 — 2ui+uo, A3U0 = W3 — 3 W2 + 3 wi — Uo, Solving successively ior uq, ui, U2, . ■ . , we have Wo = Wo, wi = Wo + AlWo, W2 = tio + 2 AlWo + Aotto, W3 = Wo + 3 AiWo + 3 A2W0 + A3?*o, Here the coefficients again follow the binomial law, and there is suggested the formula (1) W„ = Wq + nCiAiWo + nC2^2U0 + • ' • + A„Wo- F' Assuming the formula true for m„, we can show that it holds for Un+i. For apply formula (1) to the nth term of the first order of differences, which is Un+i — Un. We obtain Wn + l - ^ln = AlWo + nClAzWo + „C2A3Wo + ' • " + A„ + iWo. Adding equation (1) to this we get Wn + l = Wo +(„Ci + 1) AlWo +(„C2 + nC\) A2W0 + (nC3 + nCa) A3W0 + • • • + A„+iWo. But „Ci + l=n + lCi, „C2 + «Ci=n + iC2, n^S + n^. = n + l^g, • • • , as is easily verified by substituting in the values of the binomial coefficients. Hence Wn + l=Wo+n + lCiAiWo + n+lC2A2Wo+n + lC3A3Wo+ " ' " +A„+iWo. Hence, if (1) holds for m„, it also holds when n is replaced by n + 1, that is, for Un+\. But we have shown that it holds for W3; hence it holds for u^, hence for wg, 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 7i — 1 difference columns. From the equations just preceding formula (1) we have, by- addition, uo = Uq, iio + wi = 2^0 + AiWo, Wo + wi + 1^2 = 3 Wo + 3 AiWo + AoUo, 2^0 + Ml + W2 + W3 = 4 Wo + 6 AlWo + 4 A2W0 + AsllQ. The coefficients on the right are respectively those of the expan- sions of (1 + xy, (1 + x)-, (1 + x)^, and (1 + x)\ the first term of the expansion being omitted in each case. Let s„ denote the sum of the first n terms of the sequence; S„ = Wo+Wi+W2+ • • • +W„_i. Then by analogy with the preceding equations we assume that (2) S„ = „CiWo + nCo^ltk) + nC3A2Wo + „C4A3Wo H (- A„_iWo. We show by induction that (2) holds for all values of n. Adding (1) of (228) to (2) and noting that Sn+i = s„ + Un, we have 5n + l = (.Ci + l)Wo + UC2 + „C,)AiWo + UC3+„C2)A2Wo+---+A„Wo = „ + lCiWo + n + lC2AiWo+,i + lC3A2l<0+ ' * • + A^Wfi. Therefore (2) is true when n is replaced by w + 1. But we veri- fied above that (2) is true when n = 4. Hence it is true when n = 5, hence when 71 = 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^-iWo, and we have (3) w„ = Wo + „CiAiWo + „C2A2Wo + • • • + „Cr_iA,_iWo. (4) Sn =nClUo + nCsAiWo + nCaAgW,, + • • • + „CVA,_iWo. Example. Find the sum of the squares of n consecutive integers beginning with 10. Sn = 102 + 112 + 122 + . . . + (10 4- n - 1)2. 206 INTERPOLATION [230 Our difference table is as follows: Ao Ai A2 . A3 100 21 121 23 2 144 25 2 169 27 2 196 Hence r = 3. Then Sn = nCim + nC2AiMo + nC3A2Uo n (n n X 100 + 1-2 H2n3 + 57n2+541n) ^X21+^ l)(n-2) 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. 5. 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; 2 n2+ 2071 + 50. 230. Interpolation. — Suppose the terms of the sequence Uo,ui, U2, . . . to be the values of a function / (.r) for a series of equally values of x. Thus : Wo =fM, ui =f{xo -\rh), U2 =/(.ro + 2/i), Un =fi.xo -hnh). Y. !/= f(x) ' X \_^ ' o > < > ' > < > /X These values are shown graphically in the figure, as ordinates of the curve y = j{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) / (rro + nh) = / (xo) + n^^f (xo) + n{n- l)(n- 2) "^ 1.2.3 n (n — 1) 1 . 2 A3/(-io) + Ao/ M }^-^ ^3 230] INTERPOLATION 207 This formula has been derived when n is a positive integer. It is also true for fractipnal values of ??, 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 differences Ai/(.To), A2/(;ro), . . . become rapidly small, so that first differences are usually 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 (xq -\- 2 h) and f{xQ + 3h). 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 /(a;) = log x; xo = 100; h = I. To calculate log 100.7 we put n = .7. Our difference table is, /(x) ^i/(x) ^1 f ix) log 100 = 2.00000 101 = 2.00432 + .00432 428 - .00004 102 = 2.00860 103 = 2.01284 424 421 * 4 3 104 =2.01703 416 5 105 = 2.02119 412 4 ., lOG = 2.02531 407 5 107 = 2.02938 404 3 108 = 2.03342 401 3 109 = 2.03743 Then fixo + nh)^ = log 100.7 = log 100 + .7 X .00432 - .7(.7 - 1 X2 ^ X .00004 + (' = 2 + .00302 + .00000 = 2.00302. Here the second differences are so small that they can he neglected, and our result is that obtained by ordinary or linear interpolation. Graphically this amounts to replacing the curve y = f (x) by its chords. To calculate log 107.35, it is best to consider log 107 as the first term, or / (.To), and put n = .35. (We might take/(xo) = log 100 and put n = 7. 35. J We find log 107.35 = log 107 + .35 X .00404- '^^l'^~^^ X .00003 + ■ • • =2.03079. J X ^ Here also second differences arc 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 : A3/(X) - .0006 fix) sin 10° = 0. 1736 15° = .2588 20° = .3420 25° = .4226 Ai/(x) + .0852 832 806 ^2 fix) - .0020 26 32 30° = .5000 774 38 35° = .5736 736 44 40° = .6428 45° = .7071 692 643 49 Herexo = 10°; h = 5°; then 17 °20' = 22 xo + Y^h and ' Then OO 1 c I sin 17° 20' = sin 10° + ~ X .0852 - y^ ^ ' X .0020 22/22 ^^^^^ ' '^" 'X .0006+ • • • =.2979. 1X2X3 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. n "-^^^^ 4. Altitude. 10° 12° 14° 16° 18° 20° 22° 24° 26° Calculate the tabular number Calculate the refraction for alti- when n = 22; when n = 33.6. ' tudes 14° 40' and 21° 25'. Vn (« - 1) 10 0.0711 15 465 20 346 25 275 30 229 35 196 40 171 45 152 50 136 Refraction. 5' 13" .1 4' 22" .5 3' 45" .2 3' 16" .6 2' 54" .0 2' 35" .7 2' 20" .5 2' 7" .6 1' 56" .6 232] INTERPOLATION ' , 209 6. Greenwich Moon's Moon's mean time. right ascension. decUnation. h I h m a i\- ' 5 14 32.14 18° 47' 37". 7 2 ,". 7' 5 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 5 35 40.59 18° 50' 49". 4 10 5 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 determine 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: ' 3. Ai A2 .0300 279 - .0021 254 25 258 4 235 23 223 12 212 11 15° = .268 16° = .287 17° = .306 18" = .325 19° = .344 20° = .369 21° = .384 22° = .404 23° = .425 24° = .445 rfi 2.0 .250 2.2 .207 2.4 .174 2.6 .158 2.8 .127 3.0 .111 3.2 .098 3.4 .087 3.6 .077 Altitude. Refraction, 10° 5' 13" 11° 4' 46" 12° 4' 22" 13° 4' 2" 14° 3' 45" 15° 3' 34" 16° 3' 16" 17° 3' 4" 18° 2' 54" 19° 2' 35" CHAPTER XV Undetermined Coefficients. Partial Fractions 233. A useful method for expanding certain expressions in series depends on the following Theorem on Power Series. If the equation (1) ao + aix + a2X^ + • • • + a„a;" -f • • • = is true for all values of x from a: = to a; = a:o inclusive, where Xq ^ 0, then all the coefficients are zero, that is, ao = 0, ai = 0, a2 = 0, . . . , a„ = 0, . . . . Proof. Since (1) is true when a; = we have, putting for x, ao = 0. Then (1) reduces to aix + a2X^ + • • • + a^a;" + • • • =0, or (2) a;(ai+a2a:+ • • • +a"x"-i+ • • • ) = 0. This must be true for all values of x from to a;o. Choose for x a value £ between and a;o. Then £(ai+a2s+ • • • +a''£"-i+ . . . )=0. Then, since £ 5^ 0, we must have ai+a2£+ • • • +a,j£''-i+ . . . = 0, or, ai = - £ (as + age + • • • + a„£"-2 +...). The series in the last parenthesis converges, and therefore has a finite sum S. For, putting a: = £ in (1), and omitting the first two terms, we have left the convergent series a^e'- + a3£3 + . . . + a,,s^ + • • • , and this remains convergent after division by £2. Hence ai = — sS where S depends on e, but is finite for all values of £ between 210 234] UNDETERMINED COEFFICIENTS 211 and xq. Assume now that ui is not equal to 0; say ai = h. We can now take s so small that eS shall be numerically less than h; hence ai cannot equal h. .'. ai = 0. Then (1) reduces to aox- + a-sx^ + • • • + anX"" + • • • = 0, or, x^ (a2 + asa; + • • ■ + anx"-~ +...)= «• Choose for x a value e (not necessarily the same as e above) between and Xq. Then £2(«2 + a3^'+ • • • +an£"-2+ . . . )=0. Hence, since e 9^ 0, we have 02 + «3=- + • ■ • + ans""" + • • • = 0, or, 02 = - c (03 + • • • + a„e--' +•••)= 0. Here again the series in parentheses converges and has a finite sum. Hence by taking e sufficiently small we can show that 02 cannot equal any number h, however small. /. 02 = 0. Similarly we show that each coefficient must be zero. 234. Theorem of Undetermined Coefficients. — If two power series in x are equal to each other for all values of x from a: = to X = xo inclusive, then the coefficients of like powers of x in the two series must be equal. Hypothesis: (1) ao + aix + a2X" + ■ • • + anX" + • . • = 60 + bix + &2a:2+ . . . -f 6„a;" + • • • when ^ a; ^ a^o- Conclusion: ao = bo, ai = 61, a2 = &2, • • • , ct/i = &n, • • • • Proof. From (1), by transposition, we have ao-&o + (ai-&i)^+(«2-&2)a;2+ • • • +(a„-6„)a:"+ • • • = 0. Hence by the preceding theorem, ao - 60 = 0, ai - 6i = 0, a2 - 62 = 0, . . . , a„ - h„ = 0. Hence the conclusion stated above. Corollary. The theorem remains true when either or both of the infinite series reduce to polynomials. We consider a poly- nomial of 7n terms as an infinite series in which all coefficients after the mth are zero. 212 UNDETERMINED COEFFICIENTS [234 Assume 1 +x 1 -X2 + X - = ao + aix + 02x2 + asx^ + Y X'' Example 1. Develop t— : — ^^—^ into a power series. 1 - Clearing, and writing the coeflBcients of like powers of x in vertical columns, we have 1 -x2 + ai X + a2 x2 + as + ai + a2 - ao -ai X3 + Equating coefficients of like powers of x, we have ao = 1, or, oo = 1, oi + oo = 0, ai = — 1, 02 + ai — ao = — 1, 02 = 1, OS + 02 — oi = 0, as = — 2. Hence 1 - 1 +x = 1-X + X2-2X3+ • 1 + 2 X Example 2. Develop ^ , » . ^3 into a power D X OX ~\~ X If we put 1 +2x i X - 5 X2 + X3 Oo 4" OlX + a2x2 + 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 + x3 should start with — . To obtain the development we put 1 +2x 1 1 +2x 6x-5x2+x3 X 6-5xH-x2 Developing the last fraction as in example 1, 1 +2x Hence _lj_17 4_Ii 24_.293 3 6-5x + x2~6 + 36^^216'' "^1296^ "^ 1+2X _ 1 17 79 . j93^ ) X - 5 x2 + x3 6 X ^ 36 ^ 216 ^ 1296 ^ t Exercises. 1. In example 1, find an in terms of on -1 and on-2. ■ind the first four terms of the expansions of: 1+ X , X ^ 1 + X2 1 +X +X2" 1 -X 1 - X -X2' 2 - X + 3 x2 2 x^ + 3 X x2 + 2x + 2' 1 + 3 X + a;3 2 - 3 X + x2 3x + 4x2 -x3' 235] PARTIAL FRACTIONS 213 235. Partial Fractions. — It is sometimes desirable to resolve a given rational fraction into a sum 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 {ax + 6), no two factors being equal. Rule. The fraction can be resolved into a sum of simple frac- tions, of the form — r , equal in number to the factors of the ' ax -{-0 given denominator. Here A is a constant. 5a; -1 5x-l ■ . A . B Example. ^2 _ 6x + 5 " (x - l)(x - 5) " x - 1 + x - 5* Clearing: 5 x - 1 = A (x - 5)+ 5 (x - 1), or, 5x-l ={A+B)x -{5 A +B). Since the given fraction must be equal to its partial fractions for all values of X except x = 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 = A+B; -l=-i5A+B). Hence A=-l; 5=6. 5x-l ^ -1 . 6 x2-6x + 5 x-l^x-5' A shorter method for finding A and B is as follows: consider again the equation 5x-l=Aix-5)+B{x-l). Let x=5; 24 = 4 5; B = 6. Let x = l; 4 =-4 J.; A=-\. 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 ^ _1_ ^ x2-6x + 5 x-l'x-5 must hold except when x = 1 or x = 5. Hence 5x-l =A(x-5)+5(x-l) 214 PARTIAL FRACTIONS [236 is true for all values of x, except perhaps x = 1 and x = 5. It is therefore true when x = 1 + £, however small e may be ; that is, (1) 5(l + £)-l =^(l + £-5)+5(l+£-l). Suppose our equation is not true when x = 1 ; let the two members differ by a quantity h, so that 5X1-1 =A(l-5)+B(l-l)+/i, or, 4=-4A+/i. From (1) we have 4 + £ =-4^ + £A + £fi. From the last two equations, by subtraction, etc., Since A and B are fixed numbers, h can be made as small as we wish by taking £ small enough. Hence h cannot equal any number except 0. . 236. Case 2. — The denominator contains a linear factor repeated r times, as {ax + hy. Rule. Corresponding to the factor {ax + hf, take a set of par- tial fractions of the form Ai A2 _._ Ar {ax + 6) ^ {ax + 6)2 "^ ' "^ {ax + 6)'- This is the most general set of fractions having constant numer- ators and common denominator {ax + hy. Example. 3 x2 - X + 1 A B C D (X + 2)(x - 3)3 X + 2 "^ X - 3 "^ (X - 3)2 "^ (X - 3)3' Clearing : 3 x2 - x + 1 = A (x - 3)3 + B (x + 2)(x - 3)2 + C (x + 2)(x - 3) + D (X + 2). Let X = 3; then 25 = 5 D; D = 5. Let X = - 2; then 15 = - 125 ^; A =- is- 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 + 2D. Put x = l; 3 =- 8^ + 12B-6C + 3D. Substituting the values of A and D already found, and solving for B and C, we have Hence 3x2-x + l _ -3 3 ■ ^_12_^ . _J__. (x + 2)(x-3)3 ~ 25 (X + 2) "^ 25 (x - 3) "^ 5 (X - 3)2 "^ (x - 3)3 237] PARTIAL FRACTIONS 215 237. Case 3. — The denominator contains a quadratic factor, {ax~ + bx 4- c), which cannot be resolved into real linear factors. Rule. Corresponding to a quadratic factor {ax^ -^bx -{- c), take a partial fraction of the form Ax + B ax^ + 6x + c The reason for this assumption may be illustrated by a simple example. 2 X — 1 Example. Resolve _ -iw 2 -|-4 > ^^*° partial fractions. If i = V — 1. the factors of x2 + 4 are x + 2 i and x — 2i. Suppose now we assume 2x- 1 A B C ~i ~ I o V "I" ! (x-l)(x2 + 4) x-1 ' x + 2i ' x-2i Combining the last two fractions into a single one, we have B C ^ {B + C)x + 2{C-B)i x + 2i x-2i x2 + 4 If now we introduce two new constants M, N in place of B, C, by the relations B = M + iN; C = M - iN, we have B + C = 2M; i{C -B) = -2v^N=2N. Hence in place of the fractions B , C x + 2i x-2i' where B and C involve i, we take the single fraction Mx+4Ar x2 + 4 ' where M and N are real. Then, using B in place of M and C in place of 4 N, let 2x-l A Bx + C (x - l)(x2 +4) X - 1 "'" x2 + 4 ■ Clearing: 2 x - 1 = A(x2 + 4) + (Bx + C)(x -1). Put x = l; then 1=5 A; A = I. Put X = 0; then - 1 = 4 A - C; C = |. Equate coefficients of x*; then = A + B; B =- A =-|. Hence 2x - 1 1 , - X + 9 (X - 1) (x2 + 4) 5 (x - 1) ' 5 (x2 + 4) 216 PARTIAL FRACTIONS [238,239 238. Case 4. — The denominator contains a repeated quadratic factor, {ax^ -\- hx -}- cY. Rule. Corresponding to a repeated quadratic factor {ax^ + 6a; + cY, take the partial fractions, + 7-i^fTe^Tv>+ • • • + {ax^ + bx + c)'^ (aa;2 -{-bx + c)^^ ' {ax' -}-bx + cY Example. 10 x3 + 7 X + 4 A Bx + C Dx + E (X - 2) (x2 + 3)2 X - 2 "^ x2 + 3 "^ (x2 + 3)2 Clearing: 10x3 + 7x ,+ 4 = A (x2 + 3)2 + (Sx + C) (x - 2) (x2 +3)+ (Dx + E){x - 2). Put X = 2; 98 = 49 A; A = 2. Equate coeflBcients of x^, x^, x2, and x": = A + 5, 10 = C - 2 5, = 6.4+3B-2C + D, 4 = 9A-6C-2J5;. Hence, B = - 2, C = 6, D = 6, £; = - 11. Therefore, 10 x3 + 7 X + 4 2 -2x + 6 6x - 11 (X - 2) (x2 + 3)2 X - 2 ^ x2 + 3 "^ (x2 + 3)2 239. Exercises. Resolve into partial fractions: -^- x6 + x" - 8 x3 - 4x 5x + 12 3 x2 + 10 X + 3 3x - 1 8 x2 + X - 6 x2 + 6 X - 8 x3 - 4x 9. 1+X2 10. X -X3 X 11. x2-4x + l X4 12. x3 + 2x2-x -2 in 3x2 -2x ■• x3 + 4 X , 1 . '• (X2-1)2 X3 - 1 '• X3 + 3 X , .3 + 1 .^. • 20.^' + ^^+-^ X (X - 1)3 x3 - 3 X2 + 2 X x2 + 3 X + 4 x3 + 2 x2 + x' '""• x4 + 3 x2 + 2 13. X - 8 x3 - 4 x2 + 4 X 8 14. 1 X4 + x3 + X2 + X 15. 1 X3 + 1 16. X2 - 1 x2 - 4 17. x2 -3 x3 - 7 X + 6 18. x5 - 2 X + 1 X" + 2 X3 + X2 21. X2 + 8 x + 4 X3 + X2 - -4x -4 9.?.. X2 - ■ 2x - 1 CHAPTER XVI Determinants 240. Determinants of the Second Order. taneous linear equations aiX-\-biy = ci, aox + 62^ = C2, are solved for x and y, we find When two simul- 62C1 — b\C2 y = aiCo — (I2C1 aib2 — a2bi ' aibo — a2&i To express these results it is convenient to use the notation lai fei 02 fe^ (0162 — «2fcl), 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, 02, 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 constitute the diagonals of the array and giving these products the signs indicated in the scheme below: M' This process is called " expanding the determinant." The above values of x and y may now be written in the forms, c\ bi a I C] X = Co bo ' y = 02 C2 ai 61 ai 61 a2 62 a2 &2 217 218 DETERMINANTS [241 Exercises. 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 = I, 3. 4x-32/ = 5, 4. 8x + 5y-6 = 0, 2x + ?/ = 3. 2x + ?/ = l. 4x + y + 4=0. 5. 2x+?/ + l=0, 6. 2x + y + l=0, 6x + 3y + 2=0. 6x + 32/ + 3=0. 241. Determinants of the Third Order. — We shall now define a determinant of the third order in terms of determinants of the second order by the following equation: fli a2 as fli 62 &3 - 02 61 63 + ^3 &l &2 &1 62 &3 — C2 C3 Cl C3 Cl C2 C\ Co 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 a3&2Cl — a2&lC3 — ai&3C2- h \^^^^ \ aih2C3 + aobsCi + 036102 Cl \\\j 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, each product to be given the sign indicated. +^ We may now verify by direct calculation that the values of x, y, z, obtained by solving the linear equations aix -\- hiy -\- ciz = di, ' a2X + 62?/ + C2Z = ^2, asx + h^y + c^z = dz, 242] DETERMINANTS 219 di bi ci ai di ci ai 6i di d2 bo c-i ao do C2 a2 62 4 ds b;i Cs ' y = as d^i C3 ' z = «3 ^3 ^^3 ai 61 ci ai 61 ci ai 61 Ci 02 &2 C2 a2 62 C2 02 62 C2 ^3 h C3 aa &3 C3 as 63 C3 X + 2 2/ -It,3 2 = 2, 3x + 2z/+ 2 = 3. ^. 2x + 2y - 2 = 2, x+ 2/ -22 = 1, X - y + 2 = 4. X - ?/ + 2 = 2, 2x+ y + 3z = l, 2x-2y + 2z = 4. 2x- y + 2z=2, x-2?/ + 42=3, 3x -3?/ +62 = 1. Exercises. 1. Verify the last statement. 2. State a rule for solving three equations of the form just considered. Solve the following systems of equations: X- y+ z = \, 5. 5x + 62/-32 = 4, 7. 4x-5?/ + 22 = 3, 2x-Zy+ 2 = 1. 5. 3x-6j/ + 9z = 2, 8. x+ y+ 2 = 1, x-2;/ + 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 column be multiplied by a factor k, the determinant is multiplied by k. 242. Inconsistent or Non-independent Linear Equations. — Consider the equations aix + bxy = Ci and kaix + kbiy = Co. These are inconsistent if C2 7^ kci; they are dependent if Co = kci, since in this case the second equation is k times the first. In either case the determinant of the coefficients of x and y is 0. On solving by the determinant method, we find X = 00 and y = , when the equations are inconsistent; a; = ^ and y when the equations are dependent. That is, the inconsistent equations have no (finite) solution, while the solution is indeterminate in case of dependent equations. Geometrically, the equations represent two straight lines which are parallel, and distinct if C2 7^ c\k; they coincide if c^ = 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 number of solutions, and hence x and y appear above as indeterminate forms. [See exercises 5 and 6 of (240).] Exercises. 1. Consider the equations ' aix + h\y + ciz = di, kaix + khiy + kciz = 6,2, asx + bsy + czz = ds. The first two are inconsistent if ^2 9^ kd\, and dependent when di = kdi. Show that in the first case the only possible sokitions of the three equations are infinite, and in the second case there is an infinite number of solutions. 2. Show that the equations aix + h\y = and a2X + h^y = have one solution (0, 0), or an infinite number of solutions, according as the determinant of the coefficients is different from or equal to 0. Discuss also geometrically. 3. Show that the equations aix + h\y + c\z = 0, a2X + 62?/ + C2Z = 0, asx + hy + c^z = 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. Eliminate z so as to get two equations in x and y and discuss these as in exercise 2.) 4. Show that the equations 2x-32/ + 52 = 0, x + y-2=0, 3x-7?/ + llz = are not independent. What is the relation between them? {Hint. To find the relation between the equations, find ^1 and k2 such that ki times the first trinomial plus /c2 times the second shall equal the third.) 243. General Definition of a Determinant. — The array of n rows and n columns, ai a2 as . . a, 61 62 &3 • . 6, Cl C2 Cg . . . c, l\ h h ■ ■ • is called a determinant of order w. 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 column, 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, b^, C3, . . . , l,, form the principal diagonal. The term aihoCs . . .In is called the principal term of the ex- pansion. 244. Every term of the expansion of the determinant can he 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 onl}' 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 A„ = S ±ai&2C3 ■ ■ ' In, (^ = 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 |^- Proof. We need only show that the number of possible arrange- ments of the subscripts 1, 2, 3, . . . n, is jw. 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 ... n, 2 1 3 ... w, 2 3 1 ... n, 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 n — \ 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—l){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 aj})2Czdi 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„ hy the rule that the sign shall he 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 fll 02 03 Cl4 bi 62 bs bi Ai = Cl C2 C3 Ci di do dz d\ 247. Properties of Determinants. 1. A determinant is unchanged in value when its rows and col- umns are interchanged. 247] DETERMINANTS 223 For the expansion remains 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 he interchanged, the sign of the determinant changes. 3. If all the elements of a row or column are 0, the determinant = 0. For each term of the expansion contains a zero factor. 4. When all the elements of a row, or column, contain a common factor, this may he taken out and 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 hy any factor, we need only multiyly the elements of any row or column hy this factor. 5. If two rows or columns are alike, the determinant = 0. For by interchanging them we would have A« = — An; .'. An = 0. 6. If the elements of two rows or columns differ only hy 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 A„ = ai a2 as . . . an hi b2 hs . . . fe„ il J2 J3 ■ ■ ■ jn U h h 3\Ji -h J2J2 + • • • jnJn Here Ji is called the cof actor of the element ji, and similarly for /2, . . . , Jn. 8. A determinant is unaltered in value when the elements of any row are increased hy a constant multiple of the corresponding elements 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 An' = ai + kb h Cl I, a2-\-kb2, . . . , ttn + kb 62 . . . b^ C2 ... C„ h I2 . . . . In AuA2, . , so that . . ,A„ be the cofactors of the elen An = (ai + A;6i) Ai+ (a2 + kb.) A2 + • • • + (a„ + kbn) A„ = {aiAi + a2^2 + • • • + anAn) + A; (61^1 + 62^2+ • • • +Mn). ai a2 . . Ctn 61 62 . . &n + k h h . . . Zn bi 62 fei 62 h h The first of these determinants is A„, 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 =0. 12 3 Subtract the second row from the first. The new form is 3 6 9 99 98 97 1 2 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 aiAi + 02^2 + + anAn. Replace the a's by the elements of any other row, as the second. The result is biAi + 62A2 + + Mn = 0. 10. If we strike out from A„ the jth row and kth column, the remain- ing determinant, of order n — 1, is designated by A/,fc, and is called the minor of the element standing at the intersection of the row and the column struck out. Thus the minors of ai, 02, and 03 in the determinant are, respectively, ai ao as 61 &2 &3 Cl C2 C3 &2 h 61 63 , and 61 62 C2 C3 Cl C3 Cl C2 m, the minor of any We shall consider a We shall now show that, except as to element equals the cofactor of that element determinant of third order, although the argument will apply to determinants of any order. We have aiAi + a2A2 + asAs, \ai 02 as A3 =61 62 63 Ici Co Cs where ^1, A2, A3 are the cofactors of ai, 02, a^, respectively. Then Ai = Ai.i. For, since ai^i contains all the terms of A3 which involve ai, and since the expansion of Ai,i contains all possible products of ele- 226 DETERMINANTS [ 248 ments, one from each column and each row except the first, there- fore Ai and Ai,i must be identical. Now interchange the first two columns, so that A3 becomes —A3. Then ao ai as 62 61 h C2 Ci C3 A3 = 62 61 63 = — 02^2 — fll^l — «3-43. The minor of a2 is unchanged, namely . The expansion Cl C3 of this multiplied by a2 gives all the terms of the expansion of — A3 containing 02- But these are also contained in —02^2- Hence Ai,2 = — A2, or ^2 = — Ai,2. In the same way, by moving the third column into first place by two successive interchanges, which does not alter the sign of the determinant, we find Ai,3 = ^3. Let Aj^k denote the cof actor of the element standing at the intersection of the jth row and kth column of A„; we can bring this element. to the intersection of the first row and column by j — 1 -\-k — 1 successive interchanges of rows and columns. Hence A„ will become (-iy+^-2 . A„ or (-iy+^A„, since (-l)-^ = 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 A„= aiAi,i- «2Ai,2+a3Ai,3- a4Ai,4+ • • • +(-l)""^Ai,„. 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 system of equations aiX-\-biy-}-CiZ = di, d'zX + 62Z/ + C2Z = d2, asx + hy + C32 = (Zs.' Let the determinant of the coefficients be denoted by A, so that tti 61 Ci A = a2 &2 C2 as &3 C3 Let the cofactors of ai, 02, as be Ai, A2, A3 respectively. Multiply the given equations in order by Ai, A2, A^, and add the results. We obtain (aiAi + 02^2 + 03^3) X +(61^1 + 62A2 + Ms) y + {ciAi +C2A2 + C3A3) 2 = diAi + ^2^2 + dsA-i. From (7) and (9) of (247) we see that the coefficient of x is A, and of y and z zero. Hence we get d]Ai -\-d2A2 + dsA: di &i Cl d2 62 C2 4 63 Cs a\ hx Cl a2 &2 C2 as 63 C3 Similarly by multiplying by the cofactors of 61, 62, 6s and adding we get y, and by multiplying by the cofactors of Ci, C2, cs 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., c?i = 0, ^2 = 0, c^s = . . .), and A 5^ 0, the only solution is a; = 0, y = 0, 2 = 0, . . . ; when A = 0, there exists an infinite number of solutions. 228 DETERMINANTS [249 249. Exercises. Evaluate the following determinants: 1. that a 1 3 a + 1 2 2 a + 2 3 1 4 2 10 3 2 4 6 2 3 3 1 1 1 5 2 -2 1 4-5 10, 2. a h g h g f 9 f c 1 1 1 1 3 4-3 1-1 4 2, S 4 4 1 € o a — a o -b -d — c —e h c d e J -J 12. Show, without expanding, 1 -7 10 5 3 -7 = 0. 11. 3. a h —a c — b —c 6 2 8 2 2 8 1 6 4 3 2 5 4 4 4 -1 -9 -1 9 -1 7 1-1 9 16 27 23 Ol as 63 o C3 O Qi bi a di 13. Show that 1 1 1 X y z x2 y 2 ^2 (y - x) (s - x) (3 - 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 15. Give two pairs of values of x and y which satisfy the equa- tion X y 1 3 1 1 1 -2 1 16. Give the coordinates of two , points on the line y 1 = 0. 17. Trace the graph of y 1 1 1 -2 1 = 0. 18. Give the coordinates of two points on the line 1 X y 1 I ai 61 1 = 0. \ a2 ^2 1 249] DETERMINANTS 229 19. Give three sets of values of x, y, z which satisfy the equation X y z I 3 1-21 1-221 •14 11 20. As in 19, for the equation y z \ Ci a2 Qi 1 62 bs C2 C3 21. cos (x + y) = 23. cos 2 X = 24. 26. Prove the following identities cos X sin X sin y cos y cos X sin X I I sin X cos X I a be sin X sin y sin 2 cos X cos 2/ cos z 6in2x cos^x 1 8in2 y cos2 ?/ 1 =0. 22. sin (x — y) = sin X cos X sin u cos w a sin {y — z) + b sin (2 — x) + c sin (x — y). sin2 z cos2 z cos X sin X cos x cos x (s cosy 27. + sin z) sin y cos y Cos ?/ (sin x + sin z) cos z sin z cos z cos 2 (sin x + sin y) sin X sin 2 x sin 3 x sin2 X sin2 2 x 8in2 3 x sin 2 X sin 4 X sin 6 x = 0. 2 sin X sin 2 X sin 3 x (sin 2 x X). 28. Show that a + a' b + b' c -\- c' d e f g h k 29. Show that the equations -4X + 2/ + 2 = are satisfied by I 1 1 X : y : z = \ a h r a' b' c' = dc f + d e f g h k g h k and 2y + -4 1 30. Show that the equations are satisfied by X :y : z = Ix + my + nz =0, [ I'x + m'y + n'z = n \ — \l 71 \ ' n'\ ' \l' n'\ I 7n I' m' 230 DETERMINANTS [249 31. Show that the equations \3x + 5y + Qz = 0, [4:x + 6y + 7 z = 0, are eatisfied hy x : y : z = 1 : — 3:2. Solve the following systems of equations: 32. 2 x + 3 y - 4 3 + 7 = 0, 34. - r + s + t + u = 4, 7a;-4y-l=0, r - s + t + u = S, 9x-4z + l=0. r + s - t + u = 2, 33. 20 u + 2 r - 7 = 0, r + s + t-u = l. 4f + 5w;-l=0, Z5. 2x - y -3z + w = I, . 4w-3w + 2 = 0. x + 2y + z-w = 2, Sx-Sy-z + 2w=-l, -x-y + 2z-Zw = 0. 36. k + l + m-2n = l, 2k -l + 2m - 4:n =2, -k + 2l + 3m-6 7i=-2, k -l + 4:m -8n =-1. CHAPTER XVII Polar Coordinates. Complex Numbers. DeMoivre's Theorem. Exponential Values of sin x and cos x. Hyperbolic Functions 250. 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 and 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 OF = r from 0, called the radius vector, and by the angle XOP = 6, called the vectorial angle. Then r, are called the polar coordinates of P, and the point is indi- cated by {r,d). Similarly Pi is the point (ri, ^i). The coordi- nate d is positive when measured counter-clockwise from OX; r is positive when measured from along the terminal side of 0; it is negative when measured from along the terminal side of d 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); ^- "^-, -ITT Z 6"' 3 Calculate the rectangular coordinates of each of these points, taking as origin and OX as the x-axis. 231 232 POLAR COORDINATES [251,252 251. 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 = Va;2 + 2/2, ^ X = r cos d, ly = rsin^; = tan 252. Curves in Polar Coordinates. — When r and 6 are unrestricted, the point (r, 6) may take any position in the plane. When r and 9 are connected by an equation, the point (r, 6) is in general restricted to a curve, the equation between r and 6 being called the polar equation of the curve. Example 1. Trace the curve whose polar equation is r = sin d. Assume a series of values for 6, calculate the corresponding values of r and plot the points whose coordinates are corresponding values of r and 6. 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 0, .5, .87, 1.0, .87, .5, 0, -.5, -.87, -1.0, The graph is shown in the figure. For values oi > 360°, and for negative angles, no new points are obtained. The curve is a circle, with radius = ^. Example 2. Trace the curve r =-2 0. Here is understood to be in radians. 300° -.87, 330°, -.5, 360°. 0. r = 0,-, 7r,-j,2^,.. For negative values of 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 6 respectively. 263 COMPLEX NUMBERS 233 Exercises. Trace the following curves: 1. r = 2 sin d. 6. r = sin- 1 e. 2. r = cose . 6. r = tan- 1 e. 3. r = tan d. 7. rO = 1. 4. r = sec 0. 8. r = 2*. 10. r = logio 6. 11. r = 4. 12. 9=^ 71/ 253. Complex Numbers. — Let a and h denote any two real numbers and i = V— 1. Then the quantity a + ih is called a complex yiumher. It may be considered as made up of a real units and h imaginary units, a X 1 + 6 X t. 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 -\- ih. When 5 = 0, P lies on the a;-axis, and the complex number reduces to a real number. Thus all points on the x-axis corre- spond to real numbers, and this line is called the axis of real numbers. Let P (figure) be a point {x, y) in the plane, and let z be the com- plex number represented by P. Then z = oc-^iy. 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, d). Then Hence X = r cos 6; y = r sin z = X -\- iy = r (cos 6 -^ i sin 6) . Here r is called the modulus and d the angle of the complex number z. 234 COMPLEX NUMBERS [ 254, 255 When r is fixed, and 6 is changed by integral multiples of 2 tt, we obtain a set of complex numbers of the form, z = r [cos {d+2mv) -\-i sin {d-\-2 mr)\) n = Q, ±1, ±2, . . . . All these numbers have the same representative point. 254. Addition of Complex Numbers. — The sum of two com- plex numbers, z = 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 z, z' 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 Q are {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 - ij'). Exercise. Give a geometric construction for the representative point of z — z' . 255. Multiplication of Complex Numbers. — The product of the two complex numbers, z = r (cos d -\-i sin 6) and z' = r' (cos 6' + i sin 6'), we define by the equation zz' = rr' (cos 6 -\- i sin 6) (cos 6' -{- i sin 6'). / 256,257] DE MOIVRE'S THEOREM 235 Multiplying out the product of the two binomials wc find zz' = rr' [cos d cos 6' — sin d sin 0' + i (sin d cos 6' -\- cos 6 sin d')\ = rr' [cos {e + 0') + i sin (9 + ^')]- Therefore the modulus of the product zz' equals the product of the 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" . . . = rr'r" • • • [cos {9 -\- 6' -\- 6" + • • - ) -j-ism{d-\-d' -{-d" + • • • )] for any finite number of factors z, z', z", .... When the factors are all equal this reduces to z" = r" (cos nQ + i sin uQ), n being a positive integer. Exercise. Show that the above definition of the product zz' is the same as zz' = xx' - yy' + i (xy' + x'y), where z = x -\-iy and 2' = x' + iy'- 256. De Moivre's Theorem. — When r = \, then z = cos 6 + i sin 6. Hence by the above result we have (cos 6 + i sin 6)" = cos nS + * sin «6. This equation contains what is kno\\Ti as De Moivre's Theorem. 257. Definition of z^\ — Let p be any real number, positive or negative, rational or irrational. Then by analogy with the result for 2'' when n is a positive integer, we define zp by the equation zP = rP (cos i>6 + i sin p^), where z = r (cos 6 -\- i sin 6). Then, if q also be real, we have zi = 7-9 (cos qd -\- i sin qO), and ^v^q ^ ;.p+g[cos {p-\-q)d + i sin (p + q)d]= z^^^. 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 cos 0; — 4 = r sin 0. r = V32 + 42 = 5; tan » = ^ . i-t)- The angle lies in the fourth quadrant. 2. Express 2 (cos 150° - i sin 150°) in the form X +iy. 2 (cos 150° ism 150°) =2( -^ V3 -^ 3. Find the value of (1 + i)^ (2-3 i). {l+iy = l+2i + i^ = 2i. (1 + i)M2 - 3 i) = 2 i (2 - 3 i) = 4 i - 6 i2 : V3-i. + 4 i. Exercises. 1. Find the modulus and angle of 1-i; 4 + 3i; -5 + Hi; 2i; 2; (l+i)(l-t); 3V3+3i; (3V3-3i)^ {l +i^3)(\/S +i). Give figure for each case. 2. Find the value of : (l+i)3; (l-i)i; (l+i)2(l+2i)2; (3-3i)2 (VS + O^ (l-i^/s)\ 258. Theorem. If P and Q are any real quantities and if pj^iQ= 0, then P=0 andQ = 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 + tQ = by the equivalent equations p = 0; Q = 0. 259] ROOTS OF UNITY 237 As a corollary we have, if P-\-iQ = P' + iQ', then P = P' and Q = Q'. For the given equation is equivalent to (P — P') -\-i{Q — Q') = 0. 259. The nth Roots of Umt^ — To solve the equation X" - 1 = 0, or X" = 1, replace 1 by its value cos 2 kw + i sin 2 kir, k being an integer. We obtain re" = cos 2kT -\- i sin 2 kir. Jaking the nth roots of both members we have, by putting P = - in (257), 2k7, , . . 2kir X = cos h 1 sm ?i n Here fc 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, fc = 0; Xi = cos + i sin 0=1, 7 1 27r , . . 27r fc = 1 ; Xo = cos h * sm — > n n TO 47r , . . 47r A; = 2: Xs = cos h ^ sm — > 2 (n - 1) TT , . . 2 ( n - 1) 7r fc = n — 1 ; a-^ = cos ^ 1- i sm 238 EXPANSION OF SIN nd AND COS ni [260 The representative points of Xi, X2, xs, . . . x„ 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 of its arithmetic nth root by the nth roots of unity. Example. Find the cube roots of unity. These are given by x = cos— 5 — hisin-i^; fc = 0,1,2. /c = 0; xi = cos 0° + * sin 0° = 1. ft. h = \; X2 = cos 120° + i sin 120° -i+iv3. k =2.; X3 = cos 240° + i sin 240° To find the cube roots of 8, we have -^/S = 2 'x/l = 2; - 1 + i V3; - 1 - i Vs. (We here use v'S to denote any cube root of 8, not merely the principal root.) Exercises. 1. Solve the equations x^ - 1 =0 and a;^ - 8 = algebraically and com- pare with above results. Solve the following equations by the trigonometric method and give a figure for each case: 2. x4 = 1 ; 4. x5 = 1 ; 6. x^ = 1 ; 3. x4 = 81; 5. x5 = 32; 7. x^ =27. 260. To express sin n6 and cos nQ in terms of powers of sin 6 and cos 6, w being a positive integer. We have (cos 6 -\- i sin 6Y = cos nd -f i sin nd. 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 COS nd + i sin nd -\- iln cos"~^ |2 n {n — l)(n 2) ^ 2G1] EXPONENTIAL VALUES OF SIN X AND COS X 239 This equation has the form P + iQ = P' + iQ'. Hence by the corollary in (258) we have cos nQ = cos" d - ^ |~ cos"-- 6 sin^ d -\- - - - . sin nd = n cos"- ^ g sin - "^ ^"^ ~ j^^'^ ~ ^^ cos"-^ 6 sin^ g + • • -. sin 4 9 = 4 cos3 sin i9 - 4 cos sin^ 0. cos 5 19 = cos5 0-10 cos3 d sin2 e + 5 cos ff sin^ e. Exercises. Expand in powers of sin 9 and cos 9: 1. sin 39; 3. cos 40; 6. sin 60; 2. cos30; 4. sin 5 0; 6. cos70. 261. Exponential Values of sin £c and cos a?. — We have the expansions, (219), e^ = l + a: + | + |+ • • • ; sinx = a:-^ + ^- • • • , ■ cosx=l-|2 + |4- • • • • In the first series replace x by ix and define the result to be e*^: noting that i2 = _ 1 ^-3 = - i, i^ = 1, • - - , we obtain eix = 1 + a:2 4^r^- - • • 4- 12+ L* - • ■ >,•(.- S* X5 |5 Hence > e'- = coscc + i sin X. Replacing x by -x; = cos a? — i sin x. 240 HYPERBOLIC FUNCTIONS [262 From these equations we find 3 i These formulas are useful in many appHcations of the trigono- metric functions. Exercises. Using the exponential values of sin x and cos x, show that: 1. sin2 X + cos2 X = I. 3. cos 2 a; = cos^ x — sin2 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 of (261) replace x by ix and define the results to be sin ix and cos ix respectively. We obtain smix x^ cos ix = 1 + ^ + T^ +. • . • . These equations we consider as defining the sine and cosine of the imaginary quantity ix. Multiply the first equation by i and subtract the result from the second. We obtain cos ix — i sin ix = e*. Change a; to —a:; cos ix -\- 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 cos IX = ^ ; sm IX = I 2 ' 2 We now define Hyperbolic cosine of x (= cosh x) = cos ix; Hyperbolic sine of x ( = sinh x) = - sin ix. ^ Then cosh a? = , sinh a? = ^ — 262] HYPERBOLIC FUNCTIONS 241 These functions are related to the hyperbola somewhat as the circular functions to the circle. The remaining hyperbolic functions are defined by the equa- tions sinh a? . ^, 1 . coth a? = :: — ; — ; tanh coshic sech 1 cosh a; Exercises. Show that: 1. sinhO = 0; cosh = 1. 2. sinhiri = 0; cosh irl = — ] 3. sinh- = i; cosh|'=0. csch £c = sinh 6. cosh ( — x) = cosh x. ^- 6. cosh2 X - sinh2 x = 1. 7. sech2 X = 1 — tanh2 x. 4. sinh ( — x) = — sinh x. 8. — csch2 x = 1 — coth2 i. Draw the graphs of the equations (see tables) : 9. y = e^. 11. y = cosh x. 10. y = e-^. 12. 2/= sinh X. 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 „P,. Theorem 1. The number of permutations of n things r ata time is «Pr ^ n{n-l) . . . (»i - i> 4- 1). Proof. Evidently „Pi = n. Now with each of the n objects we may pair any one of the remain- ing n — 1 objects. Hence „P2 = 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 nPz = n{n — I) {n — 2). Proceeding in this way we obtain the formula stated. When r = n we have nPn = I n. 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? y; 3. How many numbers greater than 1000 can be formed from the digits i, 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 /) things r at a time, and is denoted by „Cr. Theorem 2. The number of combinations of ?i things r at a time is _ „F,. _ n(n -1) ■ • • {n - r + 1) 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, |_r permutations. Hence there are \r times as many permutations as combinations, or \r^nCr = nPr = ^ (n - 1) . . . (^i - r + 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 committees of 3 persons each can be formed from a club of 10 persons? 4. Show that nCr = rvCn-r- (This is a convenient formula when r is nearly as large as n. It is then shorter to calculate nCn-r-) 6. Show that „Co + nCl + «C2 + • • • + nCn = 2". (Expand (1 + x)" 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. 265. Theorem 3. — The number of permutations of n things n at a time, when p things are alike, is \n \P Proof. Let P be the number 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 and must equal n^n 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 \n \p\q 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 from 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 cent to a dollar? 8. 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 proba- bility. Definition. The measure of the probability of the occurrence of an event is taken to be the quotient, number of favorable ways total number of possible ways 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 = • ^ 3 Similarly the chance for a black ball is =• 2G7] . 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-^, and that it a + 6 will fail is a + 6 Since the event must either happen or fail, the probability for which is — -^ -j :— 7- = 1 , we have 1 as the mathematical a-\-b a + 6 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: 7(^2 = 21. 6 2 Hence the required chance is: p = ^ = ^• (6) What is the chance that at least one be white? Favorable cases: both white, 4(^2 = 6; one white, other black, 3 X 4 = 12. .'. Total number of favorable cases is 18. Number of possible cases, as before, 21. 18 6 Hence ^ "" 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 < ' Exam-pie 2. From 12 tickets, numbered 1, 2, . . . 12, four are drawn at random. (o) What is the probability that they bear even numbers? Since 6 tickets bear even numbers, the number of favorable cases is 6C4. The total number of ways of drawing 4 tickets from 12 is 12C4. Hence = ^= 6-5-4.3 ^2 ^ 12C4 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 gG ways; also two bear- ing odd numbers in 6^2 ways. Combining any one of the first with any one of the second gives 6C2 X &C2 favorable ways. Hence 6C2 X 6C2 5 V 12C4 11 246 CHANCE [268,269 268. Exercises. 1. If 5 coins are tossed, 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 ball? 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 independent events shall happen is the product of their separate probabilities. Proof. Suppose the first event happens in a ways and fails in h ways, out of a a + 6 possible ways, and that the second happens in a' ways and fails in b' ways, out of a total of a' + 6' 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 + b) (a' + 6')- Hence _ ^^' _ ^ V ^' ^ ~ (a + 6) (a' + b') ~ M^ a^T&^ which is the product of the separate probabilities of the two events. As an immediate extension, we have the Theorem 5. // the probabilities of several independent events be Pi, P2, ' ' • Vnj ihe probability that all will happen is P =Pl Xp-iX ' ' ' 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 probabihty 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 will happen he p\, the chance that the second then follows he p2, that the third then follows he 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 Proof. The chance that r trials out of n shall succeed is p*", and that the other n — r trials shall fail is (1 — pY'"". Hence the probability of success in r particular trials and of failure in the n — r other trials is p" {\ — pY~\ But of the n trials, any r may be the successful ones, which gives „Cr possibilities, each having a probability p'' (1 — ?))""'". 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 _ A^MZ 500 500' 2. Two bags each contain 5 black balls, and a 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 5 . The chance 1 "^ that a white ball be now drawn from this bag is ^ . Hence the probabihty 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 nCrV^ (1 -p)"-'- = loCsQJ^l -0 = 270. Exercises. 128' 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? 5. How many different sets of throws can be made with a coin, each set consisting of 5 successive throws? 6. The chance that a person aged 25 years will live to be 75 is 2;| . 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 balls 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 of Equations 271. We shall refer to the equation (1) i^o-^" +i>i^"-'+i>2x"-^+ • • • +i>„-ia7+i>„ =0 as the standard form of the equation of « th degree ; pox" is called the leading term and p„ the constant (or absolute) term. The coefficient of the leading term may be made equal to unity by dividing the whole equation by this coefficient. 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 coefficients. We shall represent the polynomial forming the left member of equation (1) hyf{x); f (a) shall denote the value of this poly- nomial when X = a,f (b) the value when x = h, 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, thenfix) is divisible by (x - 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 R = 0, since / (a) = by hypothesis. Hence / (x) is divisible by {x - a) without a remainder. Conversely, assume f(x) = (x-a)Q. Fxit x = a and we have / («) = 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 f(x) = {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 called the depressed equation. Exercises. Show that each of the following equations has the root indi- cated, and find the other roots: 1. x3 - 9x2 + 26a; - 24 = 0; x = 2. 2. x4 + 3a:2_8x -24 = 0; x = - 3. 3. 3x3-14x2 + 17x-6 = 0; x = f . 274. Number of Roots. — We assume that every equation of the form (1), (271), has at least one root, that is, 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, he fiix)= 0. By the above assumption, this equation has a root, say 02. Removing from /i(x) the factor x — a-y, 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 Oi, a-z, as, . . . an- Obviously the required equation is Ai,x- ai)(x - a2){x - a^) . . . {x - aj = 0, A being an arbitrary constant. Exercises. Form the equations whose roots are: 1. 1, 2, 3. 3. 2, 2, -2, 0. 5. ±1, h i- 2. l!-l, 2. 4. -1,-2,-3,-4. 6. -h h 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, (.r — ai){x — ao) = x^ — (ai + 02) -k + ai«2. {x - ax){x- a2){x - 03)= x^ -(oi + a-z -{- a^) x- -{-{aiao -\- aoaz -\- a\a:i)x — aiaoaz- {x - ai){x — a2)(,x - as) (a; - a^) = x"^ - (01+0-2 + 03 + 04)^3 + (• • •) x^ —{■•■) x -\- 01020304. We here observe three facts, namely : 1. The 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 n — 1 roots; then if the equation whose roots are Oi, 02, . . . o,j_i, be a;'^-i+5ix"-2+ . . . +g„_i =0, we have by hypothesis, 9i = -(ai+«2+ • • • +an-i); gn-i = (-l)""^aia2 • . . a„. Multiplying the above equation by {x — a„), which introduces the new root o„, we find on collecting in powers of x: x" -{-(qi - a„) a;"-i + • • • - o„9„_, = 0, or, X" -(oi 4- 02 + • • • + a„_i + a„) x"- 1 + . . . + (- l)"oia2 . . . o„_iO„ = 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 5, 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 ivhich 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 7 ' where a and h are integers, relatively prime. Putting x = rwe have, Multiplying through by 6""^ and transposing, Here we have a fraction in its lowest terms equal to an integer, which is impossible. Hence y cannot be a root. Corollary. Any rational root of an equation whose coefficients are integers and whose leading coefficient is unity must he an integer. 278. Imaginary Roots. — // the general equation of nth degree, with real coefficients, has an imaginary root a + ib, then also the conjugate imaginary, a — ih, is a root. Proof. Assume that a + ih is a root of the equation f{x)=X''-\-piX''-'^-{-p2X''-^-\- ■ • • + p,^.iX + Pn= 0. Then (a + ib)^ + Pi{a + ih)''-'' + P2 (a + ih^-^ + . . • +p„_i(a + i6)+p, = 0. Expanding the binomials, reducing all powers of i to ± 1 or ± i, and collecting terms, we have a result of the form f{a + ih) =P-\-iQ = 0. Hence P = and Q = 0. (258.) Now substitute a — ib for x and proceed as before. The result will be / (a - ih) =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 — ib is a root. 279-281] THEORY OF EQUATIONS 253 ■ We may state our result as follows: Imaginary roots, if present at all, always occur in conjugate pairs. 279. Multiple 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 f{x)=0 has m roots, each equal to a. Then f{x) = {x-arQ, where Q is a new polynomial. Letf'ix) denote the first derivative of f (x) with respect to x; then f(x) = (x- a)-^ + m (x - ar-^Q. This shows that/'(x) contains the factor {x - a)'^''^, and hence that, if f{x)=0 contains a root "a" repeated m times, f'{x) = will contain this root repeated m— \ times; f {x) andf'{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) of f {x) andf'{x); to a factor {x - a^-"^ of the H. C. F. corresponds a factor {x — a)*" of f{x). 280. Exercises. Test for multiple roots and find all the roots of the equation.s: 1. 2-3 _ 3 j2 -(- 4 = 0. 6. x4 - 3 x3 - 7 x2 + 1.5 X + 18 = 0. 2. j3 - 3 X - 2 = 0. 6. x" + 4 x3 - 16 x - 16 = 0. 3. x4-2x3-llx2+12x + 36=0. 7. x" - x3 - 3 x2 + 5 x - 2 = 0. 4. 2.4-2 x3-39 x2 + 40x + 400 = 0. 8. 4 x" + 6 x3 + 5 x2 - 6 x + 4 = 0. 9. 9 x4 - 54 x3 + 80 x2 + 6 X - 9 = 0. LO. 72 x5 - 276 x-i + 278 x3 + 4.5 x2 - 108 x - 27 = 0. 281. Transformation of Equations. — In the following discus- sion we assume that any missing powers of x are inserted, supplied with zero coefficients, so as to make the equation formally com- plete. We consider the equation / (x) = PqX^ + Pl-T""^ + p^X""-- + • • • + Pn-lX + p„ = 0. 254 THEORY OF EQUATIONS [281 I. To change the signs of the roots. Put X = — y. We obtain, vq{- yY+ vxi- yY-'+V2{- yT-'-+ • • • +P«-i(-2/)+Pn=o, or, on multiplying through by (— 1)", Por-Piy"-' + ?>2r-'- • • • +(-i)"-i?>n-iy+(-irPn=o. Hence, to change the signs of the roots, change the signs of alternate coefficients, beginning with the second term. II. To multiply the roots by a constant factor, m. Replace xby— (so that i/ = mx). Then -(iT+-(r'+-(»r+ •••+--©+-=«• Multiplying through by m", we have. Hence, to multiphj the roots by a constant factor m, multiply the coefficients in order, beginning with the second, by w, m^, m^, . . . m". When w = — 1 we obtain the preceding rule for changing the signs of the roots. III. To increase the roots by a constant quantity, fi. Replace xhy y — h(so that y = x -]- h). Then Po (y - hr + Pi (y - hr-' + P2 (y-hr-^+ ■ • . + p„_i(t/-/o + p. = 0. Expanding the binomials and collecting in powers of y, we obtain a result of the form, We shall now show how to obtain the coefficients Pi, Po, • • • Pn- Replacing y in the last equation by x -\- h, the result must be the original equation, / (x) = 0. Hence fix) = po(x-{-hr + Pi{x + hr-'-\-P2(x + hr-^--\- ■ • . + P„_i(.r+/0+P.. This shows that if / (x) be divided hy x -\- h, the remainder is P„. If the quotient be divided hy x -\- h, the remainder is P„- 1 ; divid- ing the second quotient by {x + /i), the remainder is Pn-2, and so on. 282' THEORY OF EQUATIONS 255 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 + h, and so on. The successive remainders are, in order, Pji, Pn-l, Pn P\. 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 8x-15 = by 2. Performing the first division at length, we have: x^+Ox^ -8x- 15 x3 + 2a;2 X +2 x'^ — 2x — 4 quotient. -2x2 -8a: -2^2 -4x -4a;- 15 -4a;- 8 — 7 remainder. We first shorten this operation by omitting to write the powers of X, using only the detached coefficients, thus: 1 + 1 + 2 -2 -2 15 15 This may be written more compactly as follows: 1+0-8-15 I +2 +2-4- 8 1st quotient 1-2-4 7 remainder. 256 THEORY OF EQUATIONS [282 Dividing the quotient by a; + 2 we have, 1-2-4 I +2 + 2-8 2nd quotient 1-41 + 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 1 +2 + 2 1-2 + 2 7 1st remainder 1-4 + 2 + 4 2nd remainder ij — 6 3rd remainder. Then the transformed equation is: a:3 - 6 a;2 + 4 X - 7 = 0. To diminish the roots of an equation by h, proceed as above with X — h in place of x -\- h. As an example, we diminish by 4 the roots of the equation a;4 - 5 a;3 + 7 a;2 - 17 a; + 11 = 0. 1- 5+ 7-17 + 11 [-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 - 4 + 43 3rd remainder 1|+11 Hence the transforr a;4 + 4th remainder. ned equation is: Ilx3 + 43.c2 + 55a; -9 = 0. 283, 284 ] 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 = /(x), when f{oc) = p^Jc"-}-2>ix"-^-\-2>'i^"~^-^ ' ' ' +i>«-i-^+/>«. To construct the graph which shall represent the fluctuating values of y as X varies, we assume a series of numerical values for x, 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 = x^ — 2x — 1. y = 3? - Here a set of corresponding values of x and y are : x= 0, 1, 2, . . . , - 1, - 2, . ^ o l_X m 2x y 2,3, 0, Since the curve crosses the rc-axis when y = 0, we see that the abscissas of the points where the graph of the equation y = f (x) 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;^ — 2a:— 1=0 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 — I = 0. Hence the exact values of the other two roots are i (l ± Vs), or approximateh', +1.62 and —0.62. 284. Effect of Changing the Constant Term. — Suppose that we add a quantity k to the constant term of / {x)^ so that the equation y = f (x) becomes y = f (^) + ^>-'- 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 [284 k he added to the constant term of the equation y =f{x), the graph is displaced vertically through the distance k, upward if k is plus, downward if k is minus. As an example, consider the equations (1) (2) (3) 1 ^3 _ -2a; + 2, -2a; + 4, -2x + 6. The graphs are shown in figure (a). The curves are of precisely (o) the same form, but (2) lies two units higher than (1), and (3) two units higher than (2). (6) VY f ^ j \ r // ^ \ / // M \ / r '/ / ^ \ V ^\ // \^ \^ f" // A \ Ji X / \ \ / \ J ^ / O/ \ I / " \ o^ \ J X, o. A. Instead of displacing the curve vertically, say upward, the same effect is produced in the graph by moving the a:-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 OiZi, O2X2, O3X3 respectively. 285,286] THEORY OF EQUATIONS 259 285. 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 . ^x-^ -x^ -2x-^2 = has three real unequal roots; replacing 2 by 4, the two positive roots become equal; that is, the equation ia;3_a^2_2a; + 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 Ix^ -x^ -2x + Q = has one real root and two imaginary roots. In general, by changing the constant term in f{x), the graph of y = f{x) 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. x3 + .t2 -X - 1 =0 by 2; 2. a;3 _2x + l =0 by -2; 3. x3 -48x - 112 = by i; 4. x44-6x3 + 3x2 -26x -24 = by -J. Multiply the roots of the following equations by the smallest factor which will make all coefficients integers 6. 2;2 _|_ x + i = 0. 8. X3 - .1 X2 + .01 X = 0. 6. \ x3 - x2 + 3V =0. 9. x3 + » x2 - ,\ = 0. 7. a;2 _ i X - i = 0. 10. x4 4- 1.2 x2 - .225 x + .015 = 0. Increase the roots of the equation 11. x3 - 3 x2 + 4 = by 2. 12. 4 x3 - 3 X - 1 = by 3. 13. x4-2x3-llx2 + 12x + 36 =0 by -2. 14. x* - 2 x3 - 39 x2 + 40 X + 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. 17. x3 - 3 a:2 - 6 X + 1 = 0. 18. x" - 4 x3 - 8 X + 32 = 0. 15. x3 - 3 x2 + 2 = 0. 16. x3 - 2 x2 + 1 = 0. In the following equations change the constant so that two roots shall become imaginary. 19. x3 - x2 - 2 X = 0. 21. x3 - 3 X - 2 = 0. 3 x2 + 3 = 0. X + 1 = 0. Solve the following equations, given one root. 23. x3 -2x2+x -2 = 0; x=V^- 24. 2 x4 - 3 x3 + 5 x2 - 6 X + 2 = 0; x = - 2 V^. 25. x5 - 8 x3 - 8 x2 + 64 = 0; x = - 1 - V^. y / \ / \ / \ o \ \ \ \ / \ / V / ^ 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) /(a;)=x3- 40^2 + 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, +2, +3, 20, 1, +4, +1, +4. +4. f(x) = When / (x) changes sign, the graph crosses the a:-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 -fl and +2, and a third between +3 and +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. 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 inte- gral part of the root required (281). 1 -4+0+4 [-1 -1+3+3 -3-3 -1+2 + 1 1 -2 - 1 ^ The transformed equation is (2) a;3 - a;2 - 5 a: + 1 = 0. Since (1) has 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) x^ - 10 a:2 - 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 x^ and x- in (3) we get an approximate value, x = 2. Putting x = 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) x3 - 10 x2 - 500 X + 1000 = 0. (a) Locate the root between and 10. Neglect terms in x^ 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 (6) Diminish roots by figure found in (a) . 1 - 10 - 500 + 1000 |j - 1+ 9+509 1 - 9 - 509 + 491 -1+8 1- 8 - 1 -517 Transformed equation: x^ — 7 x"^ — 517a; + 491 = 0. (c) Multiply roots hy 10. x3 - 70 a; - 51,700 x + 491,000 = 0. Repeat these operations on the last equation. (a) re = 491,000 4- 51,700 = 9+ . By trial the sign of the left member is + when a: is 9 and 8, but changes when x is 10. Hence the root is between 9 and 10. (6) 1 - 70 - 51,700 + 491,000 |^ - 9+ 549 + 470,241 1 - 61 - 52,249 - 9+ 468 + 30,759 1-52 53,717 (c) lj-43 x^ - 43a;2 - 52,717 a; + 20,759 = 0. - 430 a; - 5,271,700 a; + 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, the root is between 3 and 4. (6) 1 - 430 - 5,271,700 + 20,759,000 |j-_3 - 3 + 1,281 + 15,818,943 hence 1 - 427 - 5,272,981 - 3+ 1,272 + 4,940,057 424 3 5,374,353 1|-431 a;3 -~421 a;2 - 5,274,253 x + 4,940,057 = 0. We thus have the required root of (1) as a? = 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 r'^ and x- in the last equation. This gives x = .9+, and our root is, finally, X = 1.1939 + . A convenient arrangement of the whole operation of finding this root is as follows: 1-4 + + 4 |- 1 -1+3+3 3-3 1 + 2 + 1 1-2 - 1 1 - 10 - 500 + 1000 |- 1 - 1 + 9+509 1 - 9 - 509 + 491 -1+8 1- 8 - 1 - 517 u- 1 - 70 - 51,700 + 491,000 - 9+ 549 + 470,241 1 - 61 - 52,249 - 9+ 468 + 30,759 1 - 52 - 53,717 - 9 If 1-43 1 - 430 - 5,271,700 + 20,759,000 | - 3 - 3 + 1,281 + 15,818,943 ' 1 - 427 - 5,272,981 - 3 + 1,272 + 4,940,057 1-424 - 3 - 5,274,253 1 - 421 Root, 1.193 9+. 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 = -\-'X and x = — oo.) (2) When an even number of roots lie between x = a and x = b, f (a) and / (b) 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. x3 - 24 X - 48 = 0. 12. 4 x3 - 3 X - 1 = 0. 2. x3 - 7 x2 + 4 X + 24 = 0. 13. x" + x3 - 2 x2 - 3 x - 3 = 0. 3. x3 - 2 X + 1 = 0. 14. x4 - 2 x3 - 8x2 + 24x - 48 = o. 4. x3 - x2 + X - 1 = 0. 15. x4 - 4 x3 - 8 X + 32 = 0. 5. x3 + x2 + X + 1 = 0. 16. x4 + 2 x3 + X + 2 = 0. 6. x4- 6x2 + 5=0. 17. 3x4-2x3 -16x2-56x + 96 = 0. 7. x3 - 7 X - 5 = 0. 18. x3 - 7 X - 7 = . 8. x3 - 31 X - 19 = 0. 19. 8x4 + 16 x3 + 18 x2 + X + 7 = 0. 9. x3 - 48 X - 112 = 0. 20. 7 x3 + 8 x2 - 14 X - 16 = 0. 10. 2 x3 - 18 x2 + 46 X - 30 = 0. 21. 2 x* - 5 x3 - 32 x + 80 = 0. 11. 7 x3 - 9 X + 5 = 0. 22. 2 x5 - 4 x3 + 3 x2 - 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 cannot be done for equations of degree higher than the fourth. We give here a solution of the cubic equation (1) aox3 + 3 aia;2 + 3 a2X + ag = 0. We first obtain a new equation containing no term of second degree. To do this, put X = ij -{- h. Expanding and collecting in powers of y, aoy^ + 3 (ao/i + aO if + 3 (00/1^+ 2 a^h + 02) 2/ + aoh^ + 3 aih- + 3 02/1 + 03 = 0. The term in y- drops out if aoh + ai=0, or h= ^• With this value of h the equation becomes o , 3 (0002 — ar) , ao^as — 3 aoaia2 + 2 ai^ "''^ + ^ ^ + a? =^- Putting y = —' we have 2^+3 (aotta - ai2) z + (ao^ag - 3 aoaia2 + 2 ai^) = 0. Let H = 0002 - fli^; G^ = floras - 3 ao«ia2 + 2 Oi^. Then the equation becomes (2) z^-\-3Hz-\-G = 0. To solve this equation let 2 = Vr + Vs. Then 23 = r + s + 3 \/rs (Vr + -s/s), or, 2^ - 3 's/rs • 2 - (r + .s) = 0. If this is to be identical with (2), we must have yfrs = — H, and r + s = — G; or, rs = — H^, and r -{- s ^ — G. 266 . CUBIC EQUATIONS [290 Solving for r and s, -G+ \/W+nP -G- VG2 + 4 ^3 ' = 2- ' '= 2 Then z = ^r-^ 0, one real root, two imaginary roots. By direct calculation, for which we shall not take space, we find {z, - z-i) {Z2 - 23) (23 - 21) = V-27((?2 + 4/^3), or, {zi - Z2Y {Z2 - zzY {zs - z,Y = - 27 (G2 + 4 H^). , When the roots are all real, their differences are real, hence the left member of the last equation is positive; therefore G^ + 4 H^ must be negative. When two roots are eqYial, their difference is zero; hence G^ + 4 H^ = 0. When two roots are imaginary, they must be conjugate imaginaries; suppose them to be Z\ = a + ih and Z2 = a — ib. Let the third root be 23 = c, where c is real [(1), (288)]. Then we show directly that (z^ —22)^ is negative, and that (22—23)^(23—21)^ is positive, hence the left member of the above equation is nega- tive; therefore G^ -{- 4: H^ must be positive. The quantity G^ -\- 4 H^ is called the discriminant of the cubic z^ -j- Z Hz -\- G = 0, 291] QUARTIC EQUATIONS 267 When all the roots are real, i.e., G--\- 4 H^ <0,r and s are con- jugate complex quantities; let them be r = A -\- iB; s = A — iB. In this case Vr and '^ s cannot be evaluated algebraically. The roots may then be obtained in trigonometric form. Let A = u CO?, v; B = u sin v. Then r = u (cos V -\-i&m.v); s = u (cos v — i sin v). Hence Vr = Vm(cos ^ h *sm ~ 1, / v-\-2kT!r . . v-\-2kTr\ , „ , „ Vs = Vw ( cos ^ I sm ^ J ; A; = 0, 1, 2. Here ^u denotes the real cube root of u. We now find z= sm ft- 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) tan r then / sin (s — a) sin (s — b) sin {s — c) , Sins ,_, , 1 ^ tanr (9) tan-^1 = -:— 7 r ^ ^ 2 sin (s — a) /,.^^ . 1 _ tanr (10) tan-l?=-^^ -Ti ^ ^ 2 sin (s — b) /,,N .^ 1 ^ tanr (11) tan- C = 2 sin {s — 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) 2S = A + B-\-C and ' (13) tan Il = \/- ''''^''^ cos {S — A) cos {S — B) cos {S — €) (14) (15) — c os S cos {S — A) sin B sin € 1 , /cos is - B) cos (,S' - C) sin B sin C ,_, ^1 ./ - cos -S cos (-S' - ^) (17) tan|« = tani?cos(« - ^), (18) tan|6 = tan JJ cos (*S' - jB), (19) tan I c = tan 72 cos {S - C). 301. Napier's Analogies. — Dividing tan|A by tan | 5 and reducing, we have tan I ^ _ sin (s — h) tan I B sin (s — a) By composition and division, tan I A + tan | B _ sin (.s — 6) + sin (s — a) _ tan I A — tan | B "" 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 t; on the right, we have sin Ty{A-\- B) tan | c (20) or, sini (^ — B) tan| {a — b) 1 , , sin 5 (^ - -B) ^ 1 (200 tanj (« - 6) = ^-^^-^i^-c. Multiplying tan h A by tan h B and reducing, tan ^ A tan ^ B _ sm(s — c) 1 sin c By composition and division, and reduction as above, cos I (^ + ^) _ tan I c ^^^^ cos 1{A- B) ~ tan|(« + 6)' or, ^ 1 ^ , . cos i (A - B) 1 (210 tan-(. + 6)= ^^^^^^^^^ tan-c. These formulas determine the other two sides when two angles and their included side are given. Proceeding as above with tan ^ a and tan ^ b, or by the principle of duality applied to formulas (20) to (21'), we obtain sin I {a -\-b) _ cot| C ^^^^ sin lia-b) ~ tan r,{A-B)' or, 1 , sin :; (a — b) i (220 tan-(^-^) = ^.^;^^^_^^^ cot-C, (23) or, cos k (« + b) cot I (7 cos I (a - fe) tan i (^ + B) 1 cos i (a — b) 1 (230 t^°i(-^ + ^) %osi(« + >) '°'^^- 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. K = ^ ^'!rg^^'^ X 4 TT 72^ or, K = E (radians) X R". To obtain E, we may first calculate the angles. E may also be obtained by one of the following formulas, which we add without proofs. 1 tan 5 a tan r, b sin C tan-E 3 1 + tan r, a tan r, h cos C ' tan-^ = y 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, b, A. Calculate B by (1), then C by (22) or (23), and c by (20) or (21). „, , sin b sin B Check: -. — = -. — rz> sm c sm C which involves the computed 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 + = — • II. Given two angles and an opposite side. Denote the given parts by ^, B, a. Calculate b by (1), then proceed as in I. Ambiguous Case. Formula (1) gives two values of b. Retain the value or values ivhich make A — B and a — b of like sigii. III. Given the three sides. Calculate the angles by (9), (10), (11). ^, , sin A sin B sin C Check: = -. — j- = sm a sm o sm c 278 SPHERICAL OBLIQUE TRIANGLES [304 IV. Given the three angles. Calculate the sides by (17), (18), (19). Check: As in III. V. Given two sides and their included angle. Denote the given parts by a, h, C. Calculate ^ {A -\- B) by (23'), ^ {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 ^, B, c. Calculate | (a + 6) from (21'), I (a - 6) 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', h = 62° 25', A = 120° 48'. Formulas. . „ sin 6 . . sm B = - — sm A , sm a ^ . ^ sin I (a + fo) ^ , , . D\ cot I C = -^-7 — —~ tan i(A - B), ^ sm I (a — 6) . sin I (A + B) , . , , . ^, , sin 6 sin B ^^''^' sinc'sinC Computations. log sin b = 9.9476 a = 100° 37' A =120° 48' log sinyl= 9.9340 b= 62° 25' B= 50° 46' colog sin a = 0.0075 a + 6 = 162° 62' A-\-B = 170° 94' log sin j5 = 9.8891 a-b= 38° 12' A - 5 = 70° 2' B= 50°46'.5 Ha + &)= 81° 31' ^{A-\-B)= 85° 47' or 129° 13^5 |(a-&)= 19° 6' HA -5)= 35° 1' Reject the larger value of B by the rule in I. log tan i (A - 5) = 9.8455 log tan ^ (a - b) = 9.5395 log sin i (a + 6) = 9.9952 log sin ^ (A ■{- B) = 9.9989 colog sin 1 (a - 6) = 0.4852 colog sin ^ {A - B) = 0.2412 log cot ^ C = 0.3259 log tan | c = 9.7796 iC= 64°43'.5 ic = 31°3' C = 129° 27' c = 62° 6' Check: log sin b = 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. a = 53° 18'.3, b = 36° 5'.6, c = 50° 24'.9. 2. a = 42° 15'.3, h = 33° 18'.8, c= 60°32'.l. 3. a = 84° 14' 30", b = 44° 13' 46", c = 51° 6' 20". 4. .1 = 116° 8'.5, B = 35° 46'.6, C = 46° 33'.7. 5. A= 97° 53', 5= 67°59'.7, C= 84°46'.7. 6. A = 53° 42' 34", B= 62° 24' 26", C = 155° 43' 12" 7. a = 89° 0', , & = 47° 30', . C = 36° 0'. 8. a = 70° 20', b = 38° 28', C = 52° 30'. 9. b = 19° 24', c = 41° 36', A = 84° 10'. 10. a= 88° 24' 3", c = 120° 10' 55", B = 49° 27' 50". 11. a = 102° 22', B = 84° 30', , C = 125° 28'. 12. h = 76° 40' 48' A= 84° 30' 20' C = 130° 51' 33' 13. c = 104° 13'.4, A = 63° 48'.6, B= 51°46'.2. 14. c = 108° 39' 10", A= 64° 48' 52", B = 40° 23' 17". 15. , 6 = 54° 18' 16", A = 127° 22' 7" C = 72° 26' 40" 16. a = 88° 27' 50' , b = 107° 19' 52' . C = 116° 15' 0' 17. b = 83° 5' 36", c = 64° 3' 20", A = 57° 50' 0". 18. b = 68° 45', B = 58° 5', C = 50° 51'. 19. a = 56° 37', A = 123° 54', B = 57° 47'. 20. a = 48°, b = 67°, A = 42°. 21. 6= 81°, A= 72°, B = 119°. 22. a = 69° 34'. 9, c = 70° 20'.3, C = 50° 30'. 1. 23. a = 69°ll'.8, b = 56° 3S'.5, A = 68° 40'. 24. a = 151° 01' 5" b = 134° 10' 52" A = 144° 20' 45" 25. a= 40° 8' 28", b = 118° 20' 8", A= 29° 45' 32", 26. , a = 88° 12'.3, , A= 63°15'.2, , B = 132° 18'. 27. c = 100° 49' 30", B = 95° 38' 11", C= 97° 26' 28". 28. A = 45°, a = 10°, b = 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 0. 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 Ao the two stations (figure). Let the positions of the two stations be Xi, ! and X2, 02 respectively. 280 CELESTIAL SPHERE [307 Then in AA1PA2, PAi P 90 ' -cj^i, PA2 = 90° -9^2, and Z A1PA2 = X2 — Xi- Hence in AA1PA2 two sides and their included angle are known, and A1A2 (in degrees) maybe calculated as in V of (303). Problem 2. A ship is to sail from A I to A 2 by the shortest path (great circle). On what course (at what angle with the meridian) will she depart from ^1; on what course will she Assuming the positions of A I and A2 given, we have two sides and the included angle of the triangle A1PA2. We must calculate angles Ai and 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: New York -I = 4 55 54, <;& = 40° 45' N. San Francisco 8 9 43, 37° 47' N. Mexico City 6 36 27, 19° 26' N. 2. A vessel sails on a great circle from San Francisco, ^ = S*" 9° 43', ^= 37° 47' N. to Sydney, /I = 13" 55" 10', (J) = 33° 52' S. Find the courses of departure and arrival and the distance sailed. 3. If the vessel in exercise 2 makes 12 knots an hour, what is her position (>l 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 equator or simply equator (QwQ'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 sivne, 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', 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 azimuth, denoted by A; ZS, or 90° - h, is the zenith distance of .S and denoted by z. Thus h and ^, or z and A, completely define the position of S with reference to horizon and zenith. With reference to the equator and pole, ^*S 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° — is the polar distance of S, and denoted by p. Thus the position of S is defined by 3 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° PS ^ p = 90° - 8; Z ZPS = t; c/); ( -, u' , :;■ «o. X — ■ ■. ; y — ; • At, 2 bg — af bg — af mqr + ps ps + mqr _r>fq^^ _nmq_^ 29. No solution. 31. No solution. 33.20,17,5. 35. mq — np mq — np 3,2,1. 37. 3,4,5. 39. i, |, oo. 41. 1,2,3,4. 43. 1, .8, .2, .6. 45. 16t\ hrs-, 7Hhrs. 47. S4000; 4i%. 49.36,9. 51.89,35. 53.13,17,20. 55. 1, If , li. 57. 2, 3, 6 hrs. 59. $9150, $8600, $7550. ANSWERS 287 Article 75 1. 2, - 6. 3. 2, - 8. 6. - 2, 7. 7. 3, - 4f. 9. 5, Ij 13. -2^,5. 15. 3, -4^. 17. 2 6, -6. 19. 2, c. 11. t, -^. Article 86 1.6. 3. 0or3. 5.1. 7.13. 9.4. 11. _ Vl441-29 ^^^ ^ 15. 3or-? 25. 3. 27. ±\/-mp. 29, 17. li 19. 15. 21. ± i VI- 23. 4 or - 1 ±6V«2^^fc2. 31. ±ia. 33. ±mV^. 35. ^^'^^^, , " • 37. b±\fW^^ 39. a y/^r 41. ± 8 or ± a /^ 2-V^ a + 1 43. ^^±^v^r:r4. 46 4 ^4 or - 8. 47. 4 or - 9. 49. 27 or 64. 51. or 9. 53. 14, 16, 18; or, b - a±Va2+62-6a6 ■14, -16, -18. 55. 30X60. 57. ^ \/ab - A 10 61. I 63. . 571 -9; a;< -1 { , n - 1 and < - 9; - 1 and > -9. 71. i(Vl7-l); ^(Vl7 + l). Article 93 1. ± W2; ± I ^/2. 3. I V2; - ^ V2. 5. -6± VTi.3±2 vn or If; 3 or U- 9. 6 ± V6. - 2 ± 3 V6 20 20 11. m =±2. 7. Article 95 1. ±4=; t4=. 3. -^-:-±.. 5. 0, 2; 1, 0. 7. l^^ j: 3^^1559 Vl3 Vl3 Vl3 Vl3 145 -162±2 V-1559 g 4^ 145 ■ 23 13 9± V-23 13 11. ± I V5- Article 97 1. 0, 1;0, 1. « 65±Vl29 l±Vl29 _ 7±4V^^ 2T4 o. t:?^ ; 77; — • 5. 7^ : 32 16 1±V5;^^^- 9. -4±2V3; -7±4 V3. 11. - 1. 288 ANSWERS Article 99 1. 1,-1; 0,-1. 3. W5; h 5. 35 ' 35 54±V66 -.12T3V66 25 25 ). ± V v"^ ± f V- 7. 11. db Vs. Article 105 1. X = ± I \^, ±1 VS, y = ±h V2, + W5. 3. X = 0, 3, ± r% Vl3; 7/ = 2, 0, T A Vi3. 5. x = 0, 9, V; 2/=0, -6, V- Article 106 1. ±1 V29±V41, ± W7TV41. 3. ±x«5 V-5; ±V-¥. 5. ±V3, ± j\ V57, 0, T t\ V57. Article 107 1. ±V3;±1. 3.±f;±|. 5.±^;±i^'. Article 111 l.x=±25;y=±6. 3.±5;±4. 5. i^; ^-^- 7. f, i; i, |. 9. 7^ 499. g^ 923|_ 11. 13- 7_ 13. ^ 13; ^ 7; two solutions. 15. 7, - 1; - 3, 17i 17. 37rt, 4; 43t\, 7. 19. 4, 5; 4, 3; two answers. 21. 14fi, 5; 15H. 2. 23. 8, 9; 9, 8. 25. ±2, l. 3. 1 + x2 + x3 + 2 x4 + • • • . 5. i X - 2 X 4 ^ + 2 ^ -t- • '• 3 X 9 ^ 27 81 ^ Article 239 . 3 _ 1 3 2 2_ _1_ g 2 V3+3 ^■8(3x + l) 8(x + 3)' 'x x+2'^x-2' •6(x-2-V3) _ 2 V3 - 3 - _JL 1 1__ 9 3 5-3x 6 (X _ 2 + V3) 4 (X - 1) 4 (X + 1) 2 (x2 + 1) ■ ^- X "^ x2 + 4* n. 1-1 +.,AF^.. 13. -2+2 _ 3, ^g_ ^ 3 X ^ 3 (x2 + 3) X ' X - 2 (x - 2)2 *"' 3 (x + 1) 2 17... 1 ,.+.^^ + ..^^T^- 19. 4 1 3(x2-x + l) 2(x-l) ' 5(x-2) ' 10(x + 3) x-2 x-1 21 _1 2_ ^_, X + 1 • X + 2 ^ X - 2 294 ANSWERS Article 241 3. X = 11, y = I, 2 = t\. 5. X = H, 2/ = 1. 2 = ft- 7. Not independent. Article 249 1. 0. 3. 0. 5. 8. 7. 398. 9. 832. 11. aihcsdi. 33. u = j%%, v = - i, 10 = If. 35. Inconsistent. Article 257 1. V2, - 45°; 5, 36° 52'; ^146, 114° 27'; 2, 90°; 2, 0°; 2, 0°; 6, 30°; 36, -60°; 4,90°. Article 259 3. ± 3; ± 3 i. 5. xi = 2; X2 = 2 (cos 72° + i sin 72°)| X3 = 2 (cos 144° + isin 144°); etc. 7. xi = V3; X2 = -^^-^= — ; X3 = — ^^"2 ; etc. Article 260 1. 3 cos2 dBmO - sin3 0. 3. cos* ^ — 6 cos2 sin2 d + sin* 0. 5. 6 cos^ sin 6 - 20 cos3 sin3 6 + Qcose sin^ 0. Article 263 1. 24. 3. 240. Article 264 1. 20. 3. 120. 7. 190. Article 266 1. 1260. 3. 360. 5. eC* X 14C7 + eCs X nCe + eCs X 14C5 = 71500. 7. 73. 9. 4; there will be three different throws. 11. 36; there will be 21 different throws. Article 268 1. j%. . 3. j\. 6. i§. 7. ^h- 9- .'/s- 11. i^lhz- Article 270 1. A. 3. /j; i 5. 6. 7. x*5. 9- i- Article 275 1. a;3 - 6 x2 + 11 .T - 6 = 0. 3. x4 - 2 x3 - 4 x2 + 8 X = 0. 6. 6 x* - 5x3-5x2+5x -1=0. Article 280 1. - 1, 2, 2. 3. 3, 3,-2, - 2. 5. 3, 3, - 1, - 2. 7. 1, 1, 1, - 2. 9. 3, 3, ± i ANSWERS 29^ Article 286 1. x3 + 2 a;2 - 4 X - 8 = 0. 3. x3 - 12 x - 14 = 0. 5. x2 + 2 x + 1 = 0. 7. j;> _ 2 X - 2 = 0. 9. x3 + x2 - 9 = 0. 11. x3 - 9 x2 + 24 x - 16 = 0. 13. x* + 6 x3 + x2 - 24 X + 16 = 0. 15. /i = 1; x3 - 3 x = 0. 17. /i = 1 ; a;3 _ 9x - 7 = 0. 23. ± V^, 2. 25. 2, ± 2 V2, - 1 ± V^- Article 291 1.2,2,-1. 3. 1, -§, -|. 5. 3,2 ±2 Vs. 7.-1,-2,3. 9.1,-1, - 1 ± V^^. llj (- 1 ± VB), H5 ± V37). 13. 2, - 2, - 2, - 2. 15. 4, 2, -1 ± V-3. Article 305 1. A = 79° aO'.S, B = 46° 15'.3, C = 70° 55'.6. 3. A = 130° 5'.4, 5 = 32° 26'.1, C = 36° 45'.8. 5. 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'. 13. a = 76° 25', 6 = 58° 19', C = 116° 31'. 15. 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. 23. c = 84° 30', B = 56° 20', C = 97° 19'. 25. 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' 20", b = 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. 3. X = g'' 34'« 15«, ^= 22° 6' N; course, S 44° 28' VV. Article 307 1. A = ± 92° 50'; < = ± 5^ 4^^ 12«. 3. /i = 43° 27'; A = 70° 3'. 5. N 1° 33'.6 E or W; < = ± S'' 55'" 54*. 7. 9° 46'.4. INDEX PAOB 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 196-7 297 PAGB Coordinates 42 polar 231 Cosecant 95, 100 Cosine 95, 100 Cotangent 95, 100 Co versed sine Ill Cubic equation 264 Declination 281 Degree of a term 64 of a polynomial 64 De Moivrc'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 sodving equations 226 Differences 203 Difference quotient 180 Discriminant of quadratic oquii- 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 18 298 INDEX PAGE Exponent, irrational 20 laws 17-21 negative 17 positive integral 7 rational 19 zero 18 Exponential equations 86 values of sin a; and cos x . . . . 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 PAGE 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 comple:: 21 imaginary 21 irrational 19 principal root of 22 rational 5 INDEX 299 PAGE 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 RooLs of unity 237 Secant 95, 100 Series, alternating 173 binomial 196 geometric 173 infinite 171 PAGE 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 Sjmthetic division 255 Tangent, trigonometric . . . 95, 100 to a curve 181 Terminal line 99 Terrestrial sphere 279 Triangles, plane right 98 plane obUquc 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 IS APPENDIX A The Greek Alphabet Letters. Name. Letters. Name. Letters. Name. A, a, Alpha 1,1, Iota P,P, Rho B,/3, Beta K, ., Kappa S,a, Sigma r,7, Gamma A, X, Lambda T, T, Tau A, 5, Delta M, M, Mu Y,v, Upsilon E,., Epsilon N, ., Nu *I^ , Phi z,r, Zeta H, ^, Xi X, X, Chi H, r/, Eta 0,0, Omicron ^.'A, Psi e, d, ■&, Theta n,7r, Pi n, CO, Omega List of Formulas Factors of a" ± b", n being a positive integer (9). a" — 6" is divisible by (a — 6) and by (a -\- b) when n is even, a" — 6" is divisible by (a — 6), not by (a -\-b), when w is odd. o" + &" is divisible by (a + 6), not by (a — b), when w is odd. a" + 6" is not divisible by (a + b) or by (a — b) when n is even. Special Cases. a2 _ 52 = (a + &) (a 6) . a^ + 6^ has no real factors. a3-b^=(a-b)(a^-\-ab-i-b^). a3 + 63 = (a + b) (a^ - a6 + 62). 64=(a2 + 62) (a2-62). a. + 6"* has no real factors. i5 _ 55 = (a _ I) (^4 _|_ ^36 + a-b~ + a63 + 6-*). o5 + 65 = (a + 6) (a4 + o262-a63+64). Factor Theorem. — If / (x) reduces to zero when x = a, f (x) contains the factor (x — a). (11), (272). 301 302 FORMULAS Exponents. (20) to (25). 1 i xn (a-)y = a'y. {ahY = a'h\ if) =f,' Imaginary or Complex Numbers. (26.) t^V^; i2=-l; 1-3 =-i; 2;4=+l, etc. V^ = i yfa. a^ + h^ ={a + ib) (a - ib). X + iy = r (cos d -\- i sin 6) = re^^. Surds. — If a -\- ^ = c-{- Vd, where V6 and V5 are surds, then a = c and b = d. (29.) Logarithms. (37), (39), (226). If a^ = m, then x = loga rn. 171 loga mn = logo m + loga n. loga ^ = loga w - logo n. loga mP = p loga m. loga V W = - loga W, loga a = 1. loga 1=0. k)ga = - », if a > 1. Change of Base, loga w = logb m X loga &• If a = 10 and b = e, then loga b = login e = M. (Table V.) Hence logio m = M loge m. Binomial Theorem. (42), (220-1). / , 7x„ „ , « 11. , w(w— 1) „_o72 I n(n— l)(n— 2) „_„ „ (a+6)" = a"+na"-i6H — ,^ ' a'^' -b^ -{- -^ r~ -'a" ^b^+- n(n-l)(n-2) . . ■ (n-r+D ^,,_,^, j \r FORMULAS 303 Quadratic Equation, a.c- -\-bx-\-c = 0. (74), (76), (78). Roots real and unequal if 6^ — 4 ac > 0. Ellipse: 5; + ^ = 1. Hyperbola: ^ - ^ = ± 1. := -h±^b^-4ac^ Roots real and equal if b^ - 4 ac = 0. ^ Roots imaginary if 6^ — 4 ac < 0. b c Sum of roots = . Product of roots = - • a a Graph of ?/ = ax^ + 6a: + c is a parabola. Standard Equations of Conic Sections. Circle: x'^ i- y^ = r^- Parabola: y^ = 4ax; rc^ = 4 ay. ^-r^ = L Hyperbola: ^; - |. 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) a-{-b: a — b = c -]- d: c — d; (4) a" : 6" = c'^ : d\ If ai : bi = a2 : bo = as : bs = • • ■ , ,, . ^, , . pai + ga2 + ras + ■ then anv of these ratios = -r — ; — r — , — r — , — pbi + 962 + ?-63 + ■ where p, q, r are any multipliers; 1 r iu ^- ,7^1" + ap" + «3" + ■ • • also any of these ratios = V/ , „ , , „ , , ^ 1 T Oi -f- 02 -r 03 -+-••• If y l; P + 2P 3^ ' ' ' (div. whenp=l. Derivatives. (210.) _dij _ y Ay ^ = slope of tangent to curve y =f(x). ^^ ^ dx^ Ax™o A^ ( = rate of change of y relative to x. FORMULAS 305 Formulas for Differentiation. (211-2.) du _ du dy ^_n ^ ^^y^ — ^^ dx dy dx dx ' dx dx d (u -\- V -\- w -\- ■ • ■) _ du dv dw dx dx dx dx , /u\ du _ dv^ d (uv) _ dv du \v/ _ dx dx dx dx dx dx v- dy _dy du dx du dx when y is a function of u, and u a function of x. dx dx 1 da' , , - • -T- = a log a. X dx ^ dsin X — y — = cos X. dx d COS X — 3 = — sin X. dx dtanx „ — J = sec^ X. dx d cot X „ dsec X , , — -. — = sec X tan x dx. dx d cscx — J — = — esc a; cot x. dsm-^x 1 d cos" ^ X — 1 dx Vl - x^ dx Vl - x^ d tan" ^ X 1 dcot-^x -1 dx l+a;2 dx l-\-x^ dspc-irc 1 dcsG-'^x -1 dx a: Va:2 - 1 dx x Vx2 - 1 Maclaurin's Series. (218.) / Cr) = / (0) + xf (0) + -j^ /"(O) + ^^/'" (0) + . . . . Some Standard Series. e' = 1 -j- X -\- -.^ -{- j^z -{- ■ • • . Always convergent. sina: = x — jir + -jT— ••' . Always convergent. \6 [5 306 FORMULAS loge (l + x) = x-'j-\-j- Always convergent. Convergent only if - l cos A = -} tan A = r' ceo A C ^ C . A ^ esc A = - > sec A = r > cot A = - • aba vers A. = 1 — cos A. covers A = 1 — sin A. ]\Iore generally, if x be an angle of any magnitude, as XOP in the figure of (132), ordinate abscissa ^ ordinate sm X = -7T-, 1 cos X = -p— > tan x = —, — ; — ■ > distance distance abscissa distance distance ^ abscissa cscx = — p — — ) seca; = -i — ; > cota; = — p — -— • ordinate abscissa ordinate Relations between the Functions of an Angle. Formulas, Group A. (137.) 1 • 1 o 4- 1 e * COS a: 1. sinx = 3. tanx = — ^- 5. cot a: = ^ CSC X cot X Sin X 1 sin.r 6. sin- .r + cos-a; = 1. 2. cosa;=- 4. tana: = '- ■ « . , ^ 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 . . , ,...,,, (even ,,. < . ,. of a: increased 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. 310 FORMULAS Functions of + aj and - x. (140.) /(+ x) =/(— x), when/ = cosine or secant." /(+x) = — S {— x), when/ = sine, cosecant, tangent, cotangent. Angles Corresponding to a Given Function. (146.) Let 6 denote the smallest positive angle having a given func- tion equal to a given number a. Then all angles such that ^ ( sin x = a , , I. ] are X = 2 WTT + ^ and (2 n + 1) tt - ^; ■■■) esc a: = a cos x = a sec X = a are x = 2mT ± 6] ^_._. tan X = a 111, < are x = nir -{- 6. (cot X = a Formulas, Group B. (155.) 9. sin {x -\- y) = sin x cosy + cos x sin y. 10. cos (x + y) = cos X cosy — sin x sin y. 11. sin {x — y) = sin a: cos ?/ — cos a: sin y. 12. cos (x — y) = cos a; cos ?/ + sin a; sin ?/. , tanx + tany 13. t'"'(^ + !/) = i-tanxtani/ - 14. cot(x + y)^ "°V"°*^7' - ^ ^ cot a; + cot ?/ tan a; - tan ?/ 15. t^^(^-^) = l+tanxtant/- cot X cot y + 1 16. cot(x-^)= ,3ty-cotx Formulas, Group C. (157.) Double Angle. Half-Angle. 14. sin 2 X = 2 sin x cos x. 17. sin .=±v/^- 15. cos 2x = cos^x - sin2 x, 18. cos ^ x = ± v/ ^ +cosx ^ = l-2sin2x, j9_ tan|x=±v/iZ^, T 1 + cos X = 2cos2x-l. _^ l-cosx ^ sin X 16. tan2x = ^-^*J54_. = ^^" ^ ! 1 — tan^ X 1 — cos X FORMULAS 311 Fonnulas, Group D. (158.) 20. sin w + sin y = 2 sin — - — cos — ^ — 21. smw — sm y = 2cos — ^ — sin — ^r — 22. cos w + cosy = 2 cos — ^ — cos — - — oo ^ . U-\- V . U — V 23. cosw — cosy = — 2 sin — - — sin — ^ — Solution of Plane Triangles Right Triangles. — By means of the definitions of the trigo- nometric functions write an equation involving the two given parts and a required part; solve this for the required part. Oblique Plane Triangles. (169-172.) Law of Sines: 1. a : 6 : c = sin ^ : sin B : sin C (169) Law of Cosines: 2. a^ = 62 .+ c^ -2 be cos A. (170) Law of Tangents: 3. ^ = ^"f^^Tm - (171) "^ a-\-b tan ^ {A -\- B) Half-Angles. (172.) Let s = |(a + 6+c) and , = y/ (^ - «) (^ - ^) (^ " ^) . 4. sin 5 A ' oc ▼ s (s — a) ^^5^. 7. tan*4=-^- ▼ 6c " s — a Solution of Oblique Plane Triangles. (173-8.) Case L Given two angles and a side. (174) Use law of sines. Case II. Given two sides and the included angle. (175) Use law of tangents, then law of sines. 312 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 = | a6 sin C = -^sis-a) (s-h){s-c). (178) Spherical Trigonometry Spherical Right Triangle. (313-6.) — Let A, B,C be the angles, and a, h, c the sides. Arrange the five parts a, b, co-B, co-c, co-A in circular order. These parts are then connected by Napier's Rules : , . , ,, ( product of cosines of opposite parts ; sme of middle part = j ^^^^^^^ ^^ ^^^^^^^^^ ^^ ^^.^^^^^ p^^^^ 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 & : sin c = sin A : sin B : sin C. Law of Cosines: cos a = cos 6 cos c + sin 6 sin c cos A. Half-Angles. -(a + 6 + c); tanr . / sin (s — a) sin (s — b) sin (s — c) ^ V sin ,s 4. , 1 . . /sm (s — b) sm (s — c) 2 T sm 6 sm c 1 . . /sm s sm (s — a) 5. cos 7; A. = V -. — j-\ ' 2 V sm b sm c 6. tan p: A 1 / sin (s — b) sin (s — c) ^ 2 ~ V sin s sin (s — a) 1 . tan r tan^A = 2 sin (s — a) I FORMULAS 313 Half-Sides. S = l(A+B + C); tan 72= y ^^^ ^ 2' ' ' ^' V cos (S - A) cos (S-B) cos (S-C) 13. . 1 J-cosScos(S- A) sin ;^ a = y - 2 V sin B sin C 14. cos o a = V • D • ^ 2 > sin 5 sin C 1 . / cos (.S - B) cos (>S - C) sin 5 sinC /— cos S cos (*S — ^ ) cos {S - B) cos {S-C) , _ ,1 4 / — COS »S COS (*S — ^ ) 15. tan^ra = V ~ 16. tan -rt = tan R cos {S — A). Napier's Analogies. in + 1 ^ j,A sin H^ - -6) , 1 19 tan - (a - 6) = . {) . , 'z tan - c. 2 sin I (vl + i?) 2 or> J. 1 / , r\ cos I (^—5), 1 20. ta„2(«+6) = ,„4;^^g; tan^o. 21. tani(A-B)=$4f^cotic. 2 sin I (a + 6) 2 22. tan -{A + B)= f^. — —~ cot ^ C. 2 ' cos ^ (a + &) 2 Spherical Excess. E =(A-j-B-^C)- 180°. 1 „ tan I a tan § 6 sin C 23. tan^S = 2 1 + tan I a tan ^ & cos C 24. tan 2 -2^ = Vtan | s tan !(« — «) tan |(s - 6) tan^ (s - c). Area = ^ ^^^Iq^^^'' X 4 7ri?^ = 7^; (radians) X R'. 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. 314 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 not 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 (35). 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 — 7i, 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 302 = 2.4800. This is of course not the exact logarithm of 302, but only its value to four decimal places. Writing the kst equation in exponential form, we have 302 = los-isoo^ 315 316 EXPLANATION OF TABLES Multiplying both sides by 10, 3020 = 10 X 102-4800 = 103.4800, Hence, log 3020= 3.4800. Multiplying again by 10, 30200 = 10 X 103-4800 = i04-48oo_ 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 = 10^-4800 4- 10 = 100-4800, .302 = 100-4800 ^ 10 = 100-4800-1^ .0302 = 100-1800-1 -^ 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 - 10, 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 = l.-\- • • (3) 3 digits to the left of the point, log w = 2.+ • • (4) 4 digits to the left of the point, log m = 3.+ • • and so on.. Hence rule (a). EXPLANATION OF TABLES 317 In the case of decimal numbers, when m Hos between log m lies between (1) 1 and 0.1, and - 1, (2) O.I and 0.01, - I 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 ?/i = 8.+ ■ • • — 10 (3) 2 ciphers follow the point, log /« = ?.+ • • • — 10 (4) 3 ciphers follow the point, log //i = 6.+ • • • — 10 and so on. Hence rule (b). Interpolation. — Exam-pie. Find log 3024. From the table, mantissa of log 302 =.4800; ^^^^^^^^^ ^ 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 differences 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 desired, 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 oflog 164 = .2148; d = 27, Correction for .3 = S log 164.3 = 2.2156 2. log (164.3) ' = ? log (164.3)^ = I log 164.3. = I (2.2156) = 1.4771. 318 EXPLANATION OF TABLES 3. log .01047 = ? Mantissa of log 104 = .0170; d = i2, Correction for .7 = 29 log .01047 = 8.0199 - 10 log V(.01047)4 = ?_ -V^.01Q47^ = (.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) = 1 (38.2448 - 40) = 9.5612- 10. Anti-logarithm. — The number whose logarithm is x is called the anti-log aritlnn of x. Thus, a X = log 7n, then m = anti-log x. Given a logarithm, to obtain the corresponding number {anti-loga- 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 I ^ m .4806 i [ 14 ' 303 .4814 ) Hence, without regard to the decimal point, m = 302 + fj = 3024 + . Pointing o£f properly, TO = anti-log 0.4806 = 3.024+. 2. log TO = 7.0959 - 10. TO = ? mantissa of log 124 = .0934 j ^ mantissa of log to = .0959 \ [35 mantissa of log 125 = .0969 > Hence m has the sequence of figures 124 + U = 1247 +. Pointing off properly, TO = anti-log (7.0959 - 10) =.0012474-. Note. The value of the quotient 1 1 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 319 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 (lifference is largo, it is possible to obtain correctly more than four significant figures of a number when its four-place logarithm is given. Cologarithm. — The cologaritkm of a number is the logarithm of the reciprocal of the number. Thus: colog m = log— = log 1 — log wi = — log m. In practice we usually write it in the form colog 771 = — log m = (10 — log 7n) — 10. Rule. To form the cologarithm of a number, kubtract its logarithm from 10 and write — 10 after the result. Exmnples. 1. colog 302 = (10 - log 302) - 10 = (10 - 2.4S00) - 10 = 7..5200 - 10. 2. colog .003024 = (10 - log .003024) - 10 = (10 - [7.4806 - 10]) - 10 = 2. 5194. Use of the Cologarithm. Exa7nple. Calculate the value of • 541 X • 0o2o 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 = 9.6 180 - 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 = 302x.415X5|jX^3- 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 = 1.082 log 771 = 20.4468 - 20 m = 2.7975. 320 EXPLANATION OF TABLES As a last example, we calculate the value of the quantity, =n/: (■00812)i X (-47L2)3 (-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 logm = I [log (.00812)3 -|_ log (471.2)3 + colog (522.3)3 + colog(.01242)U 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)5 = 8.6064 - -10 log (471.2)3 = 8.0196 colog (522.3)3 = 1.8463 - - 10 colog (.01242)^ = 1.4294 2 |19.9017 - -20 log m = 9.9508 - -10 m = .8929. Exercises. Verify the following equations: 1. log 7 = 0.8451. 2. log 253 = 2.4031. 3. log 253.5 = 2.4040. 4. log .0253 = 8.4031 - 10. 5. log .002533 = 7.4036 - 10 6. log 6544 = 3.8158. 7. log 4.007 = 0.6028. 8. log .9995 = 9.9998 - 10. 9. log V766 = 1.4421. 10. log yig = 7.1158 - 10. 11. log (.0022)3 = 2.0272 - 10. 12. log V Cv