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 i.JIRY
 
 PLANE 
 TRIGONOMETRY 
 
 \ 
 
 BY 
 
 ARNOLD DRESDEN, Ph.D. 
 
 Professor of Mathematics, Swarthmore College 
 
 SECOND PRINTING, CORRECTED 
 
 NEW YORK 
 
 JOHN WILEY & SONS, Inc. 
 
 London: CHAPMAN & HALL, Limited
 
 : 
 
 Copyright, 1921 
 
 By 
 
 ARNOLD DRESDEN 
 
 Printed in U. S. A. 
 
 tbchna ' i a )mpo8ition oo. 
 
 CAMBRIDGE, MASS., U. S. A. 4 1 1
 
 
 raO- 
 
 _ -1 v 
 
 PREFACE 
 
 While the importance of the function concept for elementary 
 mathematics has become recognized by many writers of college 
 algebra texts and of "unified freshman mathematics " books, it 
 has received little recognition from writers on elementary trigo- 
 nometry. To emphasize this importance has been the leading 
 motive in writing the present book. A somewhat detailed study of 
 the graphs of the trigonometric functions (Chapter V) and of 
 the inverse functions (Chapter VIII) has been introduced for 
 this purpose. Much more could and should be done in this di- 
 rection; perhaps the present effort may suffice as a first step. 
 
 The opportunity afforded by the writing of a new text has been 
 used to make some changes in the presentation of the traditional 
 material. Circular measurement of angles is introduced in the 
 first chapter so as to be available for use throughout the course. 
 The fundamental theorems on projections are presented early and 
 are used subsequently so that the student may be familiar with 
 them when they are applied in a general proof of the addition 
 theorems, based on a method quite generally followed by conti- 
 nental writers. Recognizing the value of the "solution of tri- 
 angle-," a good deal of space has been devoted to this subject, 
 and an attempt has been made to develop it in such a manner 
 that the student can appreciate the reasons for the different 
 methods that are discussed. 
 
 On the question of " applied problems," I have taken a definite 
 position. I do not think it feasible to introduce into an ele- 
 mentary text technical material from the applied sciences, impor- 
 tant though such material may be. Without such material, how- 
 ever, applications cannot well be anything but problems which 
 use the language of the applied sciences without really belonging 
 to them. An elementary text can render useful service, even to 
 applied science, by stressing the fundamental concepts of trigonom- 
 etry and by setting problems which connect with the student's 
 actual experience and which suggest ways in which these concepts
 
 IV PREFACE 
 
 may be applied, leaving actual applications to the fields to which 
 they belong. 
 
 It has not seemed desirable to add to the number of tables of 
 logarithms already available. The elementary treatment of log- 
 arithms in Chapter III and the problems scattered throughout 
 the book call for the use of a set of five-place tables, of which 
 there are many excellent ones in existence. 
 
 No attempt at logical completeness has been made, but rather 
 has it been my aim to adapt the treatment to the stage of logical 
 development which may be expected of students who begin the 
 study of trigonometry. I am aware of the fact that a fuller dis- 
 cussion might be made in several instances and I shall be happy 
 if the treatment as given should arouse the critical powers of some 
 students and develop in them a desire for more penetrating analysis. 
 
 The material as here presented was used originally in mimeo- 
 graphed form by a few classes in the University of Wisconsin. 
 I am under a debt of gratitude to the Department of Mathematics 
 for allowing the material to be thus tried out. And it is with 
 special pleasure that I recognize my indebtedness to colleagues in 
 that department and to Professor T. M. Simpson, now of the 
 University of Florida, to some for suggestions and criticisms, to 
 others for assistance in the reading of the proofs. If I add to this 
 my appreciation of the courtesy shown by the publishers of the 
 book, I am ready to rest my case with the jury consisting of the 
 teachers and students of trigonometry. 
 
 ARNOLD DRESDEN 
 
 University of Wisconsin 
 March, 1921
 
 CONTENTS 
 
 CHAPTER I 
 
 POSITIVE AND NEGATIVE LINES AND ANGLES. COORDINATES. 
 RADIAN MEASUREMENT 
 Art. Page 
 
 1. Directed magnitudes 1 
 
 2. Points on a line. Directed lines 1 
 
 3, 4. Points in a plane 2 
 
 5. A fundamental theorem 3 
 
 6, 7. Projections of line segments 4 
 
 8, 9. Directed angles 5 
 
 10, 11. Radian measure 6 
 
 CHAPTER' II 
 
 THE TRIGONOMETRIC RATIOS. SIMPLE IDENTITIES 
 
 12, 13. Standard position 9 
 
 14-17. Trigonometric ratios 9 
 
 18, 19. Reduction to the first quadrant 11 
 
 20. Ratios of acute angles 13 
 
 21, 22. Ratios of 30 , 60 , and 45 13 
 
 23. Ratios of 90, 180, 270, and 360 15 
 
 24, 25. Trigonometric functions 16 
 
 26. Periodicity 17 
 
 27, 28. Relations between the trigonometric functions 18 
 
 CHAPTER III 
 LOGARITHMS 
 
 29. Theory of exponents 20 
 
 30. The use of exponents in calculation 21 
 
 31, 32. Definition of logarithms 21 
 
 33, 34. Fundamental theorems on logarithms 22 
 
 35. Common logarithms 24 
 
 36, 37. Use of a table of logarithms 26 
 
 38, 39. Calculation by means of logarithms 2S 
 
 v
 
 VI CONTENTS 
 
 CHAPTER IV 
 
 SOLUTION OF RIGHT TRIANGLES. APPLICATIONS 
 
 Art. Page 
 
 40, 41. The right triangle 32 
 
 42. Accuracy of the calculation. Checking the results 33 
 
 43, 44. Isosceles triangles 34 
 
 45-47. Projection 36 
 
 48, 49. Applications 38 
 
 CHAPTER V 
 THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS 
 
 50. Graphs of sin and cos 6 41 
 
 51, 52. Examples of graphs 42 
 
 53, 54. Operations on graphs 45 
 
 55, 56. Applications 47 
 
 57, 58. Graphs of tan 6 and cot 6 48 
 
 59. Graphs of tan (a6 + b) and cot (ad + b) 50 
 
 60, 61. Applications 51 
 
 62, 63. Graphs of sec 6 and cosec 52 
 
 CHAPTER VI 
 THE ADDITION FORMULAE 
 
 64. A special case 55 
 
 65-67. Addition and subtraction formulae for the sine and the cosine . 55 
 
 68, 69. Addition formulae for the tangent and the cotangent 58 
 
 70, 71. Double angle and half-angle formulae 59 
 
 72, 73. Factorization formulae 61 
 
 74. Summary 63 
 
 75. Exercises on Chapter VI 63 
 
 CHAPTER VII 
 THE SOLUTION OF TRIANGLES 
 
 76. The Law of Sines. The area of a triangle 65 
 
 77, 78. Two angles and one side 66 
 
 79, 80. Two sides and an angle opposite one of them 68 
 
 81, 82. The Law of Cosines 73 
 
 83. Summary and critique 76 
 
 84. The Law of Tangents 76 
 
 85, 86. Mollweide's [equations 78 
 
 K7, 88. Two sides and the included angle 79 
 
 89. The half-angle formulae for the angles of a triangle 80
 
 CONTENTS Vll 
 
 Art. Paee 
 
 90, 91. Threesides 81 
 
 92. Inscribed and circumscribed circles. Area 82 
 
 93. Summary 84 
 
 94, 95. Exercises en Chapter VII 86 
 
 CHAPTER VIII 
 
 INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC 
 EQUATIONS 
 
 96. Inverse functions 90 
 
 97, 98. The graphs of a pair of inverse functions 92 
 
 99. The inverse sine function 94 
 
 100. The other inverse trigonometric functions 95 
 
 101. Exercises 97 
 
 102. Relations between multiple-valued and single-valued inverse 
 
 functions 98 
 
 103-106. Trigonometric equations 98 
 
 List of Answers to the Exercises 105 
 
 Index 109
 
 PLANE TRIGONOMETRY 
 
 CHAPTER I 
 
 POSITIVE AND NEGATIVE LINES AND ANGLES. CO- 
 ORDINATES. RADIAN MEASUREMENT 
 
 1. Directed magnitudes. The use of positive and negative 
 numbers is a familiar method of indicating temperatures either 
 above or below a certain fixed temperature, which is taken as the 
 zero of the scale. In the same way, it is convenient to use posi- 
 tive and negative numbers to designate the measures of other 
 magnitudes, the values of which may fall on either side of a 
 certain fixed value, taken as a point of departure. For example, 
 northern latitude may be designated as positive, southern lati- 
 tude as negative; eastern longitude as positive, western longi- 
 tude as negative; a credit balance as positive, a debit balance as 
 negative; altitude above sea level as positive, altitude below sea 
 level as negative, etc. 
 
 2. Points on a line. Directed lines. A simple graphical rep- 
 resentation of such magnitudes is obtained by means of a straight 
 line, called the axis, upon which are indicated a fixed point, 
 called the origin, a unit distance and a positive direction. 
 
 O (Origin) 1 Positive Direction 
 
 1 1 . >- 
 
 U Unit >i Axis 
 
 Fig. 1 
 
 For every number, positive or negative, there is a single cor- 
 responding point, P, on this line. P is the point whose distance 
 from the origin, 0, measured in terms of the unit distance, is, in 
 magnitude and direction, equal to the given number. Conversely, 
 to every point, P, on the line there corresponds a real number, 
 viz.. the measurement of the distance OP in terms of the unit 
 distance, prefixed by a plus or minus sign. With the conventions 
 
 1
 
 2 POINTS IN A PLANE 
 
 of Figure 1, positive numbers will correspond to points to the 
 right of 0, negative numbers to points to the left of 0. The 
 number which is in this way associated with a point P is called 
 the coordinate of P. 
 
 3. Points in a plane. A simple extension of this method 
 enables us to designate the position of a point in a plane by means 
 of a pair of numbers. We take two mutually perpendicular 
 lines, called the axes of coordinates, one horizontal and one verti- 
 cal. Their point of intersection, called the origin of coordinates, 
 is used as origin on each axis, as explained in the preceding section. 
 Moreover, a unit distance and a positive direction are specified 
 on each axis. The horizontal axis is called the axis of abscissae 
 or the X-axis ; the vertical line the axis of ordinates or the Y-axis. 
 The position of an arbitrary point P in the plane is now deter- 
 mined by the horizontal distance from the F-axis to P, meas- 
 ured according to the unit and positive direction on the X-axis, 
 and by the vertical distance from the X-axis to P, measured in 
 accordance with the specifications on the F-axis. In this way 
 two real numbers are obtained, called respectively, the abscissa 
 
 or ^-coordinate (or simply the 
 j "*") and the ordinate or 
 
 y-coordinate (or simply the 
 
 -i "jr") of the point P. The 
 
 position of the point P is in- 
 dicated by means of these 
 ~*x two numbers in parentheses, 
 
 I the ^-coordinate being placed 
 
 first, and the ^/-coordinate 
 IV second. Thus the points A, B 
 
 p 2 and C of Figure 2 are repre- 
 
 sented by the symbols (3,5) 
 (3,-2) and (2,-3) respectively. Conversely, for every pair 
 of real numbers, such as ( 2,4), a point is uniquely determined 
 of which these numbers are the x- and ^-coordinates. 
 
 The four parts into which the two axes of coordinates divide 
 the plane are called the four quadrants; they are numbered as 
 indicated in Figure 2. We say that A lies in the first quadrant 
 (in I), B in the third quadrant (in III) and C in the fourth quad- 
 rant (in IV). 
 
 // 
 
 111
 
 A FUNDAMENTAL THEOREM 3 
 
 4. Exercises. 
 
 1. Locate the points A (0,0), AT (3,5), Y (-4,1), Z (0,3), V (-2, -2), 
 W (5,-2), U (-2,0), T (4,-3). 
 
 2. Determine the coordinates of the third vertex of an equilateral triangle 
 of which the origin and the point (6,0) are the other two vertices. 
 
 3. Determine the coordinates of the center of the circle which passes 
 through the points (0,0), (4,0), and (0,-4). 
 
 4. Show, by construction, that the points (0,2), (1,5), (3,11) and ( 2,-4) 
 lie on a straight line. 
 
 5. Show, by construction, that the points (1,5), (6,4), (1,-3) and 
 (6,-2) lie on the circumference of a circle of which the point (2,1) is the center. 
 
 6. Determine the coordinates of the center of the circle which passes 
 through the points A (5,0), B (0,-5) and C (-5,0). 
 
 7. Determine the coordinates of the points midway between the points 
 A (2,7) and B (4,3); P (3,-4) and Q (7,4); X (-1,3) and Y (3,-1). 
 
 8. Find the distance from the origin of each of the following points: 
 A (3,-4), B (-5,4), C (-4,-6), D (5,5). 
 
 9. If r denote the distance of a point from the origin, determine both 
 coordinates for each of the following points: 
 
 A: r = 5, y = 3; B: r = 13, x = -12; C: r = 3, y = -4; D: r = 2, x = 1. 
 
 10. The point P lies on a circle about the origin as a center and with a 
 radius equal to 10; its ordinate is twice its abscissa, the two coordinates hav- 
 ing like signs. Determine the coordinates of P. 
 
 5. A fundamental theorem. 
 
 Theorem I. If three points A, B and C be taken arbitrarily on a 
 directed straight line, the sum of the three segments AB, BC and 
 CA is equal to zero. 
 
 Proof. Reading from left to right, the three points must lie 
 in one of the following six orders : A B C- A C B- 
 
 b a c } B,A,C; B,C,A; 
 
 Fig. 3 C, A, B; C, B, A. 
 
 In the third case, illustrated in Figure 3, the segments BA, BC 
 and AC are positive segments, satisfying the relation BA-\- AC 
 
 = BC, or -BA +BC -AC = 0. But, -BA = AB and -AC 
 
 = CA. We conclude therefore 
 
 ,4 B + BC + CA = 0. 
 
 In all the other cases the proof proceeds in entirely analogous 
 manner: the details are left to the reader.
 
 4 
 
 PROJECTIONS OF LINE SEGMENTS 
 
 Corollary. If any number of points be taken arbitrarily on a 
 directed straight line, the sum of the segments from the first point 
 to the second point, from the second point to the third point, etc., 
 and from the last point to the first point, is equal to zero. 
 
 Proof. 
 then 
 
 Let the points be A, B, C, D, E and F. We know 
 
 AB + BC + CA = 0, 
 EF + FA + AE = 0, 
 
 and 
 
 CD + DE + EC = 0, 
 
 CA + AE + EC = 0. 
 
 If we now add the first three of these equalities and subtract the 
 last one from their sum, we obtain 
 
 AB + BC + CD + BE + EF + FA = 0, 
 
 which proves the corollary. 
 
 6. Projections of line segments. In elementary geometry, 
 the projection of the segment AB of a line I upon a line m is de- 
 fined as the segment A-JBi of the line m determined by the feet 
 Ai and B x of the perpendiculars dropped from A and B respec- 
 tively upon m. We prove now the following important theorem. 
 
 Theorem II. The sum of the projections upon a directed line m of 
 the directed segments of a closed broken line is equal to zero. 
 
 Proof. Let the brok- 
 en line be ABCDEFA 
 (see Fig. 4). The pro- 
 jections of the directed 
 i segments AB, BC, CD, 
 DE, EF and FA up- 
 on the directed line 
 m are the directed 
 segments AiB\, Bid, 
 
 Fig. 4 
 
 CiDi, D 1 E U ExF lt and FA X 
 Theorem I, 
 
 ArBt + B,C X + CiD, + /),#! + tfiF, + FiAi 
 
 hence the theorem is proved. 
 
 In virtue of the Corollary to 
 
 0; 
 
 Corollary. The sum of the projections upon a directed line m 
 of the directed segments of an open broken line ABC . . . XP is equal 
 to the projection upon m of the directed line AP.
 
 DIRECTED ANGLES 5 
 
 Proof. If the projections of A, B, C, ... X, P upon m be denoted 
 by A h Bi> d, . . . Xi, Pi, we have to show that 
 
 A1B1 + B1C1 + + X1P1 = A1P1, 
 or that A1B1 + Bid + + X1P1 + PiA x = 0. 
 
 But this follows, in virtue of Theorem II, from the fact that 
 ABC . . . XPA is a closed broken line. 
 
 7. Exercises. 
 
 1. Plot the points A (3,4), B (5,7) and C (7,2). Project the segments AB, 
 BC and CA upon the X-axis and prove that the sum of the projections is 
 equal to zero. 
 
 2. Proceed in a similar manner with the points P (1, 3), Q (3,2), R (5,4) 
 and S (7,-4). 
 
 3. Determine the length of the projections upon the X- and F-axes of the 
 broken line O (0,0), A (4,0), P (4,4). 
 
 4. Prove that the sum of the projections upon the F-axis of the segments 
 AB, BC and CA of Exercise 1 is equal to zero. 
 
 5. Show that the projections upon each of the coordinate axes of the 
 
 straight line PS of Exercise 2 is equal to 
 the sum of the projections of the segments 
 of the broken line PQRS. 
 
 6. The points 0(0,0), R (1,2) and Q (-3,4) 
 (see Fig. 5) are the vertices of a right tri- 
 angle. Show that the projection of the 
 
 i . \^ > jy hypotenuse OQ upon each of the coordinate 
 
 axes is equal to the sum of the projections 
 of the legs OR and RQ. 
 
 7. Plot the points A (-4,3) and B (-2, 
 7). Show that the projections upon each 
 
 Y IG 5 of the axes of the line AB is equal to the 
 
 sum of the projections of AO and OB. 
 
 8. Plot the points Q ( 3, 7) and R(l, 3). Show that the projections 
 upon the coordinate axes of OQ is equal to the sum of the projections of OR 
 and RQ. 
 
 9. Proceed as in S for the points Q (4,-4) and R (-2,-3). 
 
 8. Directed angles. In Iho preceding paragraphs we have 
 enlarged the concept of line segment by attributing to it not only 
 magnitude, but also direction. We proceed now to extend the 
 concept of angle in a similar way, viz., by giving an angle sense 
 as well as magnitude. We distinguish between the two sides of 
 an angle by calling one the initial side, the other the terminal 
 
 Q
 
 6 RADIAN MEASURE 
 
 side. The angle is now to be thought of as having been generated 
 by rotating a line from the position of the initial side to the position 
 of the terminal side; and the angle is measured by the amount of 
 this rotation. If the rotation is clockwise, the resulting angle is 
 called negative; if the rotation is counterclockwise, the angle is 
 called positive. The initial and terminal sides of an angle, as well 
 as the sense of rotation, are indicated by means of an arrow, as 
 in Figure 6. 
 
 The most common unit of measurement for angles is the degree, 
 which is the 360th part of the angle obtained by one complete 
 revolution. An angle of 180 (180 degrees) is called a straight 
 angle, an angle of 90 (90 degrees) is called a right angle. Angles 
 smaller than a right angle are called acute ; angles greater than a 
 right angle are called obtuse. Two angles whose sum is equal to 
 a straight angle are called supplementary angles; two angles 
 whose sum is equal to a right angle are called complementary 
 angles. 
 
 9. Exercises. 
 
 1. Draw angles of -45, 457, -312, 583, 1080, -630. 
 
 2. Determine the complements of the following angles: 25, 78, 23 J , 
 154, 217, -112, 325, 427, -508. Construct these angles and their com- 
 plements. 
 
 3. Determine the supplements of the following angles: 89, 127, 212, 
 195, -315, 287, 513. -459. 
 
 4. Draw the following angles: 120 + 70, 65 + 145, 180 - 25, 
 270 - 105, 180 + 55, 270 + 132. 
 
 10. Radian measure. The degree, in which the angles pre- 
 sented so far have been measured, is the unit commonly used in 
 practical work.* For theoretical purposes, another unit, called 
 the radian is to be preferred. A radian is an angle, which sub- 
 
 * In France, the "grade," one four-hundredth part of a complete revolu- 
 tion, is frequently used as a unit of measurement. It has the advantage of 
 fitting into the system of decimal notation.
 
 RADIAN MEASURE 7 
 
 tends on the circumference of any circle described with the vertex 
 of the angle as a center, an arc equal to the radius of that circle. 
 
 Fig. 7 
 
 Since the circumference of a circle of radius r is equal to 2 irr, 
 the radius of any circle will be contained 2 x times in its cir- 
 cumference. Hence the following relations hold : 
 
 360 = 2 7T radians; 
 
 1 = ~ radians = 0.01745 . . . radians ; 
 lot) 
 
 180 
 1 radian = = 57.29578 = 57 17' 44".81 .... 
 
 T 
 
 By means of these relations the measurement of an angle can 
 be converted from radian measurement into degree measure- 
 ment and vice versa. The radian measurement of an angle is 
 frequently called the circular measure of the angle. When no 
 unit is indicated, it is understood that circular measure is meant. 
 Whenever possible, the circular measure of an angle is expressed 
 by means of multiples and submultiples of t* 
 
 The radian measurement of an angle at the center of a circle 
 bears a simple relation to the arc which this angle intercepts on 
 the circumference of the circle. For every radian in the angle, 
 we will have on the circumference an arc equal to the radius. 
 Hence, if the radian measurement of the angle be denoted by d 
 and the radius of the circle by r, we shall have 
 
 length of arc = r. 
 
 * A submultiple of a number is the quotient of that number by an integer; 
 e.g., a/5 is a submultiple of a; 7r/3 is a submultiple of w.
 
 EXERCISES 
 
 11. Exercises. 
 
 1. Determine the circular measure of the following angles: 45, 30, 
 -225, 330, 540. -150, 60, -270, -810, 210, 480, -650. 
 
 2. Determine the number of degrees in each of the following angles: 
 
 3 7T 4 IT 2 7T 5 TV ; 
 
 2 3 ' 5 4 2 
 
 , 1.5, 2.3. 
 
 3. Determine the circular measures of the complements and of the sup- 
 plements of the following angles: 2 tt/3, -3x/4, 75, 7 tt/6, -112, -5 tt/3, 
 135, tt/8, 2 tt/9, -325, 7 tt/6, -3 /2. 
 
 4. Construct the following angles: w/2, 5 tt/6, 7 tt/3, llx/4, ir/6, 
 3 x/4, -2 tt/3, 2 tt, - 11 tt/6, -3 tt, 2, -5.2. 
 
 6. A wheel makes 10 revolutions per second. Determine the circular 
 measure of the angle described by one of the spokes in 15 seconds. 
 
 6. How large an arc will a central angle of 2.5 radians subtend on a 
 circle whose radius is 4 feet? 
 
 7. How large an angle, at the center of a circle whose diameter is 10 
 feet, will subtend an arc whose length is equal to 4 feet? 
 
 8. An angle of 2 radians placed at the center of a circle intercepts an arc 
 of 4 feet. What is the radius of the circle ? 
 
 9. If a wheel makes 50 revolutions per second, how large an angle does 
 its radius describe in 1 minute? 
 
 10. A carriage wheel covers a distance of 1 mile in 1000 revolutions. How 
 large is its radius?
 
 CHAPTER II 
 THE TRIGONOMETRIC RATIOS. SIMPLE IDENTITIES 
 
 12. Standard position. For the purpose of comparing differ- 
 ent angles we place them in such a way that their vertices coin- 
 cide with the origin of a system of coordinate axes, and that their 
 initial sides fall along the positive half of the X-axis of this sys- 
 tem. Angles so placed are said to be in standard position. Ac- 
 cording as the terminal side of the angle in standard position 
 falls in I, II, III or IV, the angle is said to lie in the first, second, 
 third or fourth quadrant respectively. 
 
 13. Exercises. 
 
 1. Draw the following angles in standard position: 7. r i" > , tt/3, 120, v/2, 
 -165, 5 tt/6, -21.3, -9 v/5, 325, 195, -3 t, 155. 
 
 2. Draw, in standard position, the complements of the angles given in 
 Ex. 1. 
 
 3. Draw, in standard position, the supplements of the angles given in 
 Ex. 1. 
 
 *-X 
 
 14. Trigonometric ratios. On tho terminal side of the angle 
 6, placed in standard position, an arbitrary point P is taken. 
 The coordinates of P and its distance 
 OP from 0, called the radius vector of 
 P, are denoted by x, y and r respec- 
 tively; it is agreed that the direction 
 from to P shall always be the posi- 
 tive direction along the terminal side, 
 so that r is always a positive number. 
 While now the numbers x, ij and r will 
 be different for different positions of 
 the point P on the same terminal side, 
 
 it follows from the properties of similar triangles that any one 
 of the six ratios which subsist among these three numbers will 
 be the same for one position of the point P as for any other posi- 
 tion and will depend upon the position of the terminal side only. 
 
 9 
 
 Fig. 8
 
 10 SIGNS OF THE RATIOS 
 
 These six ratios are called the trigonometric ratios of the angle 0, 
 and are fundamental in the whole field of mathematics. They 
 are known by the following names: 
 
 - Is called the sine of the angle 6, written sin o, 
 
 - is called the cosine of the angle 9, written cos 9, 
 r 
 
 - is called the tangent of the angle 0, written tan o, 
 
 x 
 
 - is called the cotangent of the angle 0, written cot 0, 
 y 
 
 - is called the secant of the angle o, written sec 0, 
 
 X 
 
 - is called the cosecant of the angle 0, written cosec 0. 
 y 
 
 Notice that the two ratios occurring in each of the three groups 
 are obtained, one from the other, by replacing x by y and y by x, 
 and that interchanging x and y replaces each ratio by its co-ratio.* 
 
 15. Exercises. 
 
 1. Draw each of the following angles in standard position, measure the 
 ooordinates and the radius vector of some point on the terminal side of each 
 and determine the trigonometric ratios: 25, ir/6, 2 n-/3, 145, 7 tt/5, 220, 
 7 jr/4, 315\ 11 ir/5. 425*, 21 tt/8, 700. 
 
 2. Proceed as in Exercise 1 with the following angles: 30, x/4, 130, 
 -5tt/6, -210, -4r/3, -320, -5x/4, -112, -19 x/6, -670, -17x/4. 
 
 16. Signs of the ratios. Because the radius vector is always 
 positive, and because the signs of the coordinates of a point 
 depend only upon the quadrant in which the point lies, the alge- 
 braic signs of the trigonometric ratios of an angle are determined 
 by the quadrant in which the angle lies. The agreements as to the 
 signs of the coordinates of a point (see 3) lead to the following 
 table of signs for the trigonometric ratios at the top of p. 11. 
 
 We observe, furthermore, that the numerical values of the ratios 
 of an angle in any quadrant are equal to the ratios of some angle 
 in the first quadrant. A careful inspection of the diagram will 
 enable the student to determine, for any given angle, an angle 
 in the first quadrant whose ratios are numerically equal to those 
 
 * For the origin of the names of the trigonometric functions see, for in- 
 stance, Ctijori, "A History of Mathematics," p. 109.
 
 REDUCTION TO FIRST QUADRANT 
 
 n 
 
 
 I 
 
 n 
 
 III 
 
 IV 
 
 
 + 
 
 + 
 
 - 
 
 
 
 
 
 + 
 
 - 
 
 - 
 
 + 
 
 
 
 tangent 
 
 + 
 
 - 
 
 + 
 
 - 
 
 cotangent 
 
 + 
 
 - 
 
 + 
 
 - 
 
 secant 
 
 + 
 
 - 
 
 + 
 
 cosecant 
 
 + 
 
 + 
 
 
 - 
 
 of the given angle. By the use of a table of the trigonometric 
 ratios of acute angles the ratios of an arbitrary angle can there- 
 fore be found. 
 
 17. Exercises. 
 
 Draw each of the following angles in standard position; determine the 
 trigonometric ratios by measurement, as in 14. Determine an angle in the 
 first quadrant whose trigonometric ratios are numerically equal to those of 
 the given angle, and verify the results of the graphical determination by 
 means of a table of the trigonometric ratios of acute angles: 
 
 1. 130. 2. 215. 3. IItt/6. 4. -493. 5. 1300. 6. -9 x/5. 7. 4 tt/5. 
 8. -1105. 9. 600. 10. llx/8. 11. 310. 12. 17 x/6. 
 
 18. Reduction to first quadrant. We turn now to the problem 
 of determining for every arbitrary angle 9 an angle 9' in I, such 
 that the ratios of 9' shall be numerically equal to the ratios of 9. 
 
 Consider, as an example, the angle 9 in III, in Figure 9a. If 
 
 
 i 
 
 Y 
 
 
 
 
 p" 
 
 q _e 
 
 
 \y" 
 
 i , 
 
 
 x y 
 
 O X" 
 
 Q" ' 
 
 y 
 
 /r 
 
 
 
 i 
 
 3 
 
 
 
 Fig. 9a 
 
 Fig. 9b 
 
 we construct in I an angle 6' equal in magnitude 1 to the acute 
 angle between the terminal side of 9 and the X-axis, and take a
 
 12 EXERCISES 
 
 point P' on the terminal side of 0', such that OP' = OP, we find 
 by considering the two triangles OP'Q' and OPQ, that 
 
 r' = r, y' = y, and x' = x. Why? 
 
 It follows from this that the ratios of 6 are numerically equal 
 to those of 6', the acute angle made by the terminal side of 6 
 with the X-axis. 
 
 If we construct in I an angle 6" equal to the acute angle be- 
 tween the terminal side of and the F-axis, and take on the 
 terminal side of 6" a point P" such that OP" = OP (see Fig. 9b), 
 we find upon considering the triangles OP"Q" and OPQ, that 
 
 r = r", x = y", and y = x". Why? 
 
 It follows from this that the ratios of 8 are numerically equal 
 to the co-ratios of the angle 6", the acute angle made by the 
 terminal side of 6 with the F-axis. 
 
 These facts are summarized in the following theorem: 
 
 Theorem I. The numerical values of the trigonometric ratios of 
 any angle 6 are equal to the ratios of the positive acute angle 6' which 
 the terminal side of d makes with the X-axis; they are also equal to 
 the co-ratios of the positive acute angle 0" which the terminal side of 
 6 make> with the I -axis. 
 
 The algebraic signs of the ratios are determined by the quad- 
 rant in which the angle 6 lies (see 16). 
 
 19. Exercises. 
 
 1. Apply the theorem of 18 to angles in I and show that the trigonometric 
 ratios of a positive acute angle are equal to the co-ratios of its complement. 
 
 2. Apply the theorem of 18 to angles in II and show that the trigonometric 
 ratios of an angle that is less than 180, are numerically equal to those of its 
 supplement. 
 
 3. Apply the theorem of 18 to angles in IV and show that the trigonometric 
 ratios of a negative acute angle are numerically equal to those of the corre- 
 sponding positive angle. 
 
 4. Determine all the trigonometric ratios for the following angles: 
 
 (a) 237, (b) 3tt/5, 
 
 
 (C) 11 TT 8, 
 
 (d) 338, (r) 1(33, 
 
 
 (/) 17 x/9. 
 
 Also for the following angles: 
 
 
 
 i) -248, (b) -512, (c) ^, (<D 
 
 -11 TT 
 
 ~~6~ ' 
 
 (*>x (/) nr
 
 RATIOS OF 30, 60, 45= 
 
 13 
 
 >A' 
 
 20. Ratios of acute angles. For acute angles, the definitions 
 of 14 may be put in a special form equivalent to that of 14, but 
 frequently better adapted to the use that is to be made of them. 
 The abscissa, ordinate and radius vector of any point P on the 
 terminal side of the angle 9 are, in this 
 case, the sides and hypotenuse of a 
 right triangle OPQ, in which the given 
 angle 6 actually occurs as an acute angle. 
 The scaffolding, constituted by the 
 coordinate axes, may now be removed, 
 and the ratios of the angle 6 may be 
 defined with reference to this triangle 
 OPQ instead of with reference to the 
 coordinate axes, by putting 
 
 "side opposite the angle" in place of y, 
 "side adjacent to the angle" in place of x, 
 "hypotenuse" in place of r. 
 
 In this way we obtain the following form for the definition of 
 the trigonometric ratios of acute angles : 
 
 sine 9 = 
 
 tangent 9 = 
 
 secant e 
 
 21. 
 
 side opposite 
 hypotenuse 
 
 side opposite e 
 side adjacent to 
 
 hypotenuse 
 side adjacent to 9 
 
 cosine 9 = 
 
 > cotangent 9 = 
 
 cosecant 9 
 
 side adjacent to 
 hypotenuse 
 
 side adjacent to 
 side opposite 9 
 
 hypotenuse 
 side opposite 9 
 
 Ratios of 30, 60, 45. This special form of the definitions 
 Y enables us to find simple numerical 
 
 values for the ratios of certain angles: 
 (a) To determine the ratios of an 
 angle of 30, or of an angle of (30, a 
 ~^* x diagram like the one in Figure 11 is 
 constructed, in which QP' = PQ, and 
 ZOQP = Z.OQP' = 90. It follows 
 that ZPOP'= Z()!>P'= Z OPT = 60, 
 so that A OPP' is equilateral. Choos- 
 ing PQ as the unit of measurement, we find OP = 2. PQ = 1, OQ = 
 V3. The definitions of 20 lead then to the following results:
 
 14 
 
 EXERCISES 
 
 Theorem II. The trigonometric ratios of angles of 30 and 60 are 
 given by the following formulae: 
 
 sin 30= 1/2; cos 30 = V3/2; tan 30 = 1/Vjj = V3/3 cot 30 = V3; 
 
 sec 30 = 2/ V3 =[2 V3/3; cosec 30 = 2; 
 Sin 60 = V3/2; cos 60 = 1/2; tan 60 = V3 ; cot 60 = I/V3 = V3/3; 
 
 sec 60 = 2; cosec 60 = 2/ V3 = 2 V3/3. 
 p 
 
 >x 
 
 Fig. 12 
 
 (6) For the ratios of 45, a diagram like the one in Fig. 12 is 
 used, in which OQ is taken as the unit of measurement. The 
 results are summarized in 
 
 Theorem III. The trigonometric ratios of an angle of 45 are given 
 by the following formulae: 
 
 sin 45= 1/V2 = V2/2; cos 45= 1/V2 = V2 2; tan 45= cot 45= 1; 
 
 sec 45 = cosec 45 = V2. 
 
 22. Exercises. 
 
 1. Determine the trigonometric ratios of the following angles, without the 
 use of the tables: 
 
 (a) 120, 135 and 150. (b) 210, 225 and 240, (c) 300, 315 and 330 
 
 2. Determine the value of each of the following expressions, without the 
 use of tables: 
 
 (a) sin 30 cos 60 + cos 30 sin 60; 
 
 (b) cos 120 cos 30 - sin 120 sin 30; 
 C ) ta n 120 + tan 00 . 
 
 {C) 1 - tan 120 tarTGO ' 
 
 (d) cos 150 sin 120 - cos 210 sin 150; 
 
 3. Similarly for: 
 
 (a) sin 315 cos 2 45 + tan 2 30 sec 135; 
 (6) sin 2 240 cot 225 - cos 2 330 tan 315;* 
 
 (c) cosec 2 300 sin 60 tan 150 + sec 2 210 cot 240 cos 2 30; 
 
 (d) cof 150 cosec 240 - tan 330 sec 3 150. 
 
 * The notation sin 2 means (sin 0) 2 . 
 powers of the trigonometric ratios. 
 
 Similar notations are used for other
 
 RATIOS OF 90, 180, 270, 360 15 
 
 23. Ratios of 90, 180, 270, 360. For the ratios of an angle 
 whose terminal side coincides with one of the coordinate axes, a 
 special consideration is necessary, because application of the 
 definitions of 14 would require division by 0, an operation which 
 is impossible. To divide a number a by would require the 
 determination of another number b such that b multiplied by 
 would yield a, which is obviously impossible, unless a were 0. 
 
 When 8 = 0, we have x = r, y = 0. Hence: 
 
 sin = 0/r = 0, cos = r/r = 1, tan = 0/r = 0, 
 cot = r/0 = ?, sec = r/r = 1, cosec = r/0 = ?. 
 
 For an angle of 90, x = 0, and y = r, so that we now find: 
 
 sin 90 = 1, cos 90 = 0, tan 90 - r/0 = ?, 
 cot 90 = 0, sec 90 - r/0 - ?, cosec 90 = 1. 
 
 Similar results are obtained for angles of 180 and 270; but 
 it should be noticed, that for an angle of 180, the x and the r 
 are equal numerically but opposite in sign, in accordance with 
 the agreements as to the signs of these quantities, so that we 
 have in this case x = r. For a similar reason, we find that 
 for an angle of 270, y = r. Consequent^ the ratios of these 
 angles have the following values: 
 
 sin 180 = 0, cos 180 = -1, tan 180 = 0, 
 
 cot 180 = - r/0 = ?, sec 180 = -1, cosec 180 = r/0 = ?. 
 
 sin 270 = -1, cos 270 = 0, tan 270 = -r/0 = ?, 
 
 cot 270 = 0, sec 270 = r/0 = ?, cosec 270 = -1. 
 
 The question marks put after cot 0, cosec 0; tan 90, sec 90, 
 cot 180, cosec 180; and fan 270, sec 270 are intended to indi- 
 cate that these ratios cannot be determined because a division 
 by would be involved therein. 
 
 These facts are expressed in the following theorem: 
 
 Theorem IV. The cotangent and the cosecant of angles of and 
 180 do not exist; the secant and the tangent of angles of 90 and 
 270 do not exist. The remaining ratios of these angles are given 
 by the following formulae: 
 
 sin = tan = 0, 
 
 
 cos = sec = 1 : 
 
 sin 90 = cosee 90 = 1, 
 
 
 cos 9(T = cot 90 = 0; 
 
 sin 180 = tan 180 = 0, 
 
 
 cos 180 = sec 180 = -1; 
 
 sin '170 = cosec 270 = 
 
 -1, 
 
 cos 170 = cot 270 = 0.
 
 16 
 
 TRIGONOMETRIC FUNCTIONS 
 
 24. Trigonometric functions. When an angle is changing, as, 
 e.g., the angle described by the spoke of a revolving wheel or the 
 angle described by a swinging pendulum, there is in general a 
 value of each of the trigonometric ratios corresponding to every 
 value of the variable angle. Whenever such a correspondence 
 exists between two varying magnitudes, whereby to each value 
 of one there corresponds a value of the other, it is said that one of 
 the variables is a function of the other. In the present case, the 
 trigonometric ratios are functions of the angle; they are called 
 trigonometric functions. 
 
 If we consider a variable angle 9, which while steadily remain- 
 ing in I, increases towards r/2, the variation of the tangent of 9 
 may best be studied by keeping either x or y fixed as the angle 
 varies, as in Figures 13a and 13b respectively. In either case, the 
 
 *~x 
 
 Fig. 13a 
 
 Fig. 13b 
 
 ratio y/x remains positive; its numerical value increases indefi 
 nitely. These facts are expressed by the following statement: 
 " As 9 tend* towards 90, being always less than 90, tan 9 
 
 increases indefinitely, and by the formula: 
 
 lim tan 9 = + oo ." 
 e 9o - 
 
 In a similar manner it is seen that if an angle tends towards 
 90, while steadily remaining in II, the ratio y/x is negative 
 throughout, while the numerical value increases indefinitely; i.e., 
 
 " As 9 tends towards 90, being always greater than 90, tan 9 
 decreases indefinitely, expressed by the formula: 
 
 lim tan 9 = oo." 
 
 e > no + 
 
 The fact that lim tan 9 = +oo, and that lim tan 9 = *= oc 
 flvon" - e oo 4- 
 
 corroborates our former conclusion that for an angle of 90 the 
 
 tangent does not exist .
 
 PERIODICITY 
 
 17 
 
 25. Exercises. 
 
 1. Interpret and prove the following statements: 
 (a) lim sec = -fao. 
 
 >90- 
 
 (c) lim cot = -f-ao. 
 
 >180 + 
 
 (e) lim cosec = +oo. 
 
 e > 180 - 
 
 2. Also: 
 (a) lim tan = +oo. 
 
 > 270 - 
 
 (c) lim sec = -f- go. 
 > 270 + 
 
 (e) lim cot = +ao. 
 > 360 + 
 
 (</) lim cosec 
 _^ 360 + 
 
 + *. 
 
 (6) 
 
 0- 
 
 lim sec = ao. 
 ->90 + 
 
 (d) 
 0- 
 
 lim cot = oo. 
 -> 180 - 
 
 
 
 lim cosec = or 
 
 -> 180 -f 
 
 (6) 
 
 
 lim tan = oc. 
 
 270 + 
 
 (d) 
 
 
 lim sec = 00. 
 
 -> 270 - 
 
 
 
 lim cot = x. 
 -> 360 - 
 
 (h) 
 
 e 
 
 lim cosec = a 
 > 360 - 
 
 3. Determine the value of the following expressions: 
 (a) sin 2 jt/2 cos 2 ir/6 tan 2 3 7r/4 sec 2 11 x/3. 
 (6) cos 2 7r/2 sin 2 3 w/2 + cot 2 11 tt/4 cosec 2 11 tt/6. 
 
 (c) sin 3 w cot 3 tt/2 tan 5 tt/6 cos 3 ir. 
 
 (d) sec 3 ir sec 2 9 7r/4 + cosec 3 w/2 cosec 2 17 ir/6. 
 
 26. Periodicity. In the preceding paragraph we have dis- 
 cussed the properties of the trigonometric functions of the vari- 
 able angle 6 for values of d which increase or decrease towards 
 0, 90, 180 and 270. We consider next some other properties 
 of the trigonometric functions, the first to be considered being 
 the property of periodicity. A function is said to be periodic if 
 there exists a constant number a, such that the function assumes 
 the same value for 6 + a as for 0, no matter what value d may have. 
 The number a is called the period of the function. Examples of 
 periodic functions are furnished by many natural phenomena, 
 such as the swinging pendulum, the motion of the tides, the 
 movement of a vibrating string, the wave motion of sound, light 
 and electricity, etc. The fact that the trigonometric functions 
 possess this property of periodicity is one of the reasons for their 
 great importance in the study of natural phenomena. 
 
 If two angles which differ by an integral multiple of 360 (2 w) 
 are placed in standard position, they have the same terminal 
 side; hence their trigonometric ratios are equal; e.g., 
 
 sin (0 + m 360) = sin (6 + m 2tt ) = sjn 6,
 
 18 RELATIONS BETWEEN TRIGONOMETRIC FUNCTIONS 
 
 for every value of 6, and for every positive or negative integral 
 value of m. We have therefore the following theorem: 
 
 Theorem V. The trigonometric functions are periodic; they have 
 a period of 2 w. 
 
 In Chapter V we shall see how the trigonometric functions, in 
 particular the sine and the cosine, may be used for the repre- 
 sentation of more general periodic functions. 
 
 27. Relations between the trigonometric functions. From the 
 definitions of the trigonometric ratios in 14, there follow imme- 
 diate^ the following theorems: 
 
 Theorem VI. The sine and the cosecant, the cosine and the secant, 
 the tangent and the cotangent of any angle are reciprocals of each 
 other; i.e.: 
 
 sin e cosec = 1; cos sec = 1; tan cot = 1. 
 
 Theorem VII. The tangent of any angle is equal to the quotient 
 of its sine by its cosine; the cotangent of any angJa is equal to the 
 quotient of its cosine by its sine; i.e.: 
 
 , sin , cos e 
 
 tan = -; cot = . - - 
 
 cos0 sin0 
 
 Since (see Fig. 8), the abscissa, ordinate and radius vector of 
 any point P are respectively the legs and the hypotenuse of a 
 right triangle, we have for any angle 6: 
 
 x 1 + y 2 r 2 . 
 
 Dividing both sides of this equality in succession by r 2 , by x 2 
 and by ?y 2 , and making use of the definitions given in 14, we 
 obtain : 
 
 Theorem VIII. The sum of the squares of the sine and the cosine of 
 any angle is equal to unity; the square of the secant of any angle 
 diminished by the square of the tangent, and the square of the cose- 
 cant of any angle diminished by the square of the cotangent are 
 each equal to unity; i.e.: 
 
 sin 2 + cos 2 = 1; sec 2 tan 2 = 1; cosec 2 - cot 2 6 = 1. 
 
 Theorems VI, VII and VIII may be used to determine all the 
 trigonometric ratios of an angle as soon as one of them is known. 
 By means of these theorems a great many other relations may be 
 proved to hold between the trigonometric functions.
 
 EXERCISES 19 
 
 28. Exercises. 
 
 1. Given sin = . Determine the remaining ratios of 0. 
 
 Since sin is negative, may lie in III or in IV. In either case, we can 
 draw the terminal side of 0, by determining a point for which y = 2 and 
 r = 3. We find then x VB or x = Vo according as is in III or IV. 
 Knowing x, y, and r, we can obtain the trigonometric ratios of 0. 
 
 Proceeding without direct application of the definitions, we can use the 
 formula sin 2 + cos 2 9 = 1 to find cos 9; after that the other ratios can all 
 be determined by means of Theorems VI and VII of 27. 
 
 2. Determine all the ratios of 9, when it is known that tan = 4. 
 
 3. Similarly, when cos 9 = | and 9 lies in IV. 
 
 4. Also when cot 9 = 7 and 9 lies in II. 
 
 Prove that the following identities result from the formulae of 27: 
 
 5. sin 9 = tan 9 cos 0. 6. cot 9 = cos 9 cosec 0. 
 
 7. tan 2 9 + sin 2 9 + cos 2 = sec 2 9. 8. cosec 9 - sin 9 = cot 9 cos 0. 
 9. sin 2 - sin 4 = sin 2 cos 2 0. 10. sec 2 + cot 2 = cosec' + tan 2 9. 
 11. sin 2 (9 cos 2 = sin 4 cos 4 9. 
 tan 2 (9 sec 2 9-1 
 
 12. sin 2 = 
 
 13. cos 2 9 = 
 
 1 + tan 2 sec 2 
 
 cot 2 cosec 2 1 
 
 1 + cot 2 cosec 2 
 
 14. tan + cot = sec cosec 0. 
 
 15. sec 2 + cosec 2 9 = sec 2 cosec 2 0. 
 
 16. cosec 2 sec 2 = cot 2 sec 2 + cosec 2 tan 2 9. 
 
 17. 
 
 1 + sin cos0 4/1+ sin 
 
 v/f 
 
 cos 1 sin 1 sin 
 
 cos 
 
 18. sec - tan = 
 
 1 + sin 
 sin 
 
 19. cosec cot 9 = , 
 
 1 + cos 
 
 20. 2 (cos 6 + sin 6 0) - 3 (cos 4 9 + sin 4 0) = - 1. 
 
 tan a + tan 8 , . a 
 
 21. - = tan a. tan 0. 
 
 cot a + cot 8 
 
 = \/| 
 
 99 i/ * v,)S0_ sin _ 1 cos 
 
 ' + cos ~ 1 + cos ~ sin0 
 
 _ cot , . 
 
 23. + sin tan = sec 0. 
 
 cosec 
 
 24. sin + cot cos = cosec 0. 
 
 25. sin (cot + tan 0) = sec 0. 
 
 26. cos (cot + tan 0) = cosec 0. 
 
 27. sec - cos = tan sin 0.
 
 CHAPTER III 
 
 LOGARITHMS 
 
 29. Theory of Exponents. In order to facilitate the calcula- 
 tions involved in the application of the trigonometric ratios to 
 the solution of problems, we take up the study of logarithms. We 
 recall first the following parts of the theory of exponents. 
 
 (a) A continued product, all of whose factors are equal to the 
 same number, a, is called a power of that number. The number 
 a is called the base of the power; the number of factors in the 
 product is called the exponent of the power. Powers are classified 
 according to their exponents; they are represented in abbreviated 
 form by the base and the exponent. 
 
 Thus a a a a a, the 5th power of a, is represented by a 5 ; 
 a being the base of the power and 5 the exponent. 
 
 (b) Besides powers whose exponents are positive integral num- 
 bers, defined in (a), we consider powers whose exponents are zero, 
 negative or fractional. Such powers are defined as follows: 
 
 The zero-th power of any number is equal to 1; e.g., a = 1. 
 A power of a whose exponent is negative is the reciprocal of the 
 power of a, whose exponent is the corresponding positive number: 
 
 - = \- 
 
 a 
 
 A power of a positive number, a, whose exponent is a unit frac- 
 tion is the real -positive root of a whose index is equal to the de- 
 nominator of the fraction: 1/a = Vo. 
 
 Thus, e.g., 2~-> = 1/32; 36* = 6; 7 = 1. 
 
 (c) From these definitions are derived the following funda- 
 mental theorems, usually referred to as the "Laws of Exponents": 
 
 I. The product of two powers of a number a is equal to that 
 power of a, whose exponent is equal to the sum of the exponents 
 of the factors: a p X a" = a p+Q . 
 
 II. The quotient of two powers of a number a is equal to that 
 power of a, whose exponent is equal to the exponent of the dividend 
 diminished by the exponent of the divisor: a? -f- a Q a p ~ a . 
 
 20
 
 DEFINITION OF THE LOGARITHM 21 
 
 III. A power of a power of a number a is equal to that power 
 of a whose exponent is equal to the product of the exponents of 
 the given powers: (a p ) Q a va . 
 
 Thus, e.g.: V2 X \ = 2* X 2" 2 = 2r = V^; 
 {/& = (3 4 )s = 3*; 
 25 -T- j$ z = 5 2 -z- o- 3 = 5 5 = 3,125. 
 
 30. The use of exponents in calculation. The fundamental 
 idea underlying the use of logarithms in calculations is that 
 the laws of exponents may be utilized for the purpose of mul- 
 tiplication and division of numbers, for raising numbers to a 
 power, or for extracting roots. If, e.g., we had a list of the suc- 
 cessive positive integral powers of 2 from 2 1 to 2 150 , the product 
 and quotient of two of them, their powers and roots, could, 
 within certain limits, be found by addition, subtraction, multi- 
 plication and division. For example, we would have: 
 
 2 49 1 
 
 OS v 9 17 98+17 _ 025. _ _ 049-63 _ 0-14 _ . 
 
 Z. A 6 - . - . , ^63 * ~ & 214' 
 
 (2 17 ) 6 = 2 17x6 = 2 102 ; v 7 ^ 44 = (2 114 ) A = 2 12 . 
 
 If, therefore, we could write every number as a power of some 
 fixed number, all multiplications and divisions could be reduced 
 essentially to additions and subtractions, while powers and roots 
 of numbers could be found by simple multiplications and divisions. 
 
 31. Definition of the logarithm. Accordingly, we lay down 
 the following definition : 
 
 Definition I. The logarithm of any number p with respect to 
 the base is the exponent of that power of which is equal to p. 
 
 This logarithm is designated by the symbol log a p. We shall con- 
 sider only logarithms of positive numbers with respect to bases 
 which are positive.* 
 
 As a consequence of this definition, we can say: 
 
 logj 16 = 4, because 2 4 = 16; log :f \ = 2, because 3~ 2 = J; 
 
 log, Vo = ',, because 5- = V5; log 7 1 = 0, because 7 C = 1. 
 
 * It does not fall within the scope of this book to consider whether for any 
 number p and any base a, there always exists an unique logarithm of p with 
 respect to the base a. We take for granted that if a and p are positive, there 
 always exists uniquely a real number loga p. The further study of this ques- 
 tion belongs to the Theory of Functions.
 
 22 FUNDAMENTAL THEOREMS ON LOGARITHMS 
 
 32. Exercises. 
 
 1. Determine log3 27, log 3 2 V, log 3 a/3, logs 1, log 3 y/ll, logs 3. 
 
 2. Determine log 10 10, log 10 1000, log 10 .01, logio .0001, logio 1. 
 
 3. Determine log 2 8, log 2 \/2, log 2 ye, log 2 1/V2, log 2 4/\/8. 
 
 4. Determine logs 4, log 4 8, log 9 27, log 27 J, logie h 
 
 5. It is known that logio 17 = 1.23045, logio 29.5 = 1.46982, logio 83 = 1.91908. 
 Determine 10 1 - 46 * 2 , 10 1 - 23045 , 10 1 - 91908 , 10 3 - 23045 , 10- 46982 
 
 6. Determine 10 logl 7 , 5 Iogi 7 , 2 log ^ 7 , a log a 7 . 
 
 33. The fundamental theorems on logarithms. From the 
 "Laws of Exponents," quoted in 29 (c), we derive now, by the aid 
 of the definition of 31, the following fundamental theorems on 
 logarithms, in which we prove what was merely suggested in the 
 last paragraph of 30. 
 
 Theorem I. The logarithm of the product of two numbers with 
 respect to the base a is equal to the sum of the logarithms of the 
 factors: 
 
 lOga (pq) - lOga p + lOga q. 
 
 Proof. The theorem evidently says nothing else than that the 
 exponent of that power of a which equals pq is equal to the sum 
 of the exponents of the powers of a which are equal to p and q. 
 For, let 
 
 loga p = x, and log a q = y; 
 
 i.e., let a x = p, and a v = q. 
 
 Then a xJrV = pq; i.e., logs pq = x + y = log a p + log a q. 
 
 Theorem II. The logarithm of the quotient of two numbers with 
 respect to the base a is equal to the logarithm of the dividend dimin- 
 ished by the logarithm of the divisor: 
 
 lOga - = lOgapJ- lOga q. 
 
 Theorem III. The logarithm of a power of a number with respect 
 to the base a is equal to the exponent of the power multiplied by the 
 logarithm of the number: 
 
 loga p n = n log,, p. 
 
 The proofs of Theorems II and III are left to the student. 
 We proceed to illustrate the use of these theorems: 
 (a) To determine logio 14 when it is known that logio 2 = .30103 
 and logio 7 = .84510.
 
 EXERCISES 23 
 
 From Theorem I, it follows that logio 14 = logio 2 + logio 7 
 = 1.14613. 
 
 14 X 25 
 (6) To determine logio ^ , when log l0 2 = .30103, logio 3 
 
 = .47712, logio 5 = .69897 and logio 7 = .84510. 
 
 By the use of Theorems I and II, we find: 
 
 logio 27 = lo & 10 14 "+* lo & 10 25 ~~ logw Q7. 
 
 But, from Theorem III it follows that logio 25 = logio 5 2 = 2 
 logio 5 = 1.39794, and that logio 27 = logio 3 3 = 3 logio 3 = 1.43136. 
 
 .-. logio H * 2 = logio 14 + 2 logio 5-3 logio 3 = 1.11271. 
 
 3/25 x 32 
 (c) To determine log y ^ -, when logio 2 = .30103, logio 3 
 
 = .47712 and logio 5 - .69897. 
 
 By the use of Theorems I, II and III, we find: 
 
 logio y ^ = 3 [2 log 10 5 + 5 logio 2-4 logio 3] 
 
 = I [1.39794 + 1.50515 - 1.90848] = | .99461 
 = .33154. 
 
 34. Exercises. 
 
 Given, that log 10 2 = .30103, log l0 3 = .47712, and logio 7 = .84510, determine 
 
 t 5 /27 r >< 
 
 1. 
 
 logio 5. 
 
 2. 
 
 logio 21. 
 
 3. 
 
 logm 35. 
 
 4. 
 
 log!-, 70. 
 
 5. 
 
 1 40 
 
 logio gj 
 
 6. 
 
 logio \/32. 
 
 7. 
 
 , 125 X 32 
 logio 27 
 
 8. 
 
 , t /28 X 25 
 logio V 81 
 
 10. logio 
 
 - X49 
 9. lognj V ^ 
 
 81 X 64 
 
 25 X </2 
 
 ... , \/8 x vT) 
 
 11. logio rji 
 
 V < 
 
 10 1 V 32 X 75 
 12. logio , 
 
 V245 X 54
 
 24 
 
 COMMON LOGARITHMS 
 
 35. Common logarithms. It will now be recognized that the 
 plan suggested in 30 can actually be carried out, provided we can 
 "write every number as a power of some fixed number," i.e., 
 provided we can find the logarithm of every number with respect 
 to some fixed base. The Differential Calculus furnishes methods 
 for solving this problem and for constructing a "table of 
 logarithms." Por the purpose of computation, logarithms with 
 respect to the base 10, called common logarithms are most useful. 
 Without considering here the problem of constructing a table of 
 common logarithms, we shall see how a comparatively small table 
 of logarithms can be made to serve our purpose.* 
 
 From now on, the symbol "log" will be used to designate log w , 
 i.e., common logarithm. 
 
 We know then that log 1000 = 3, log 100 = 2, log 10 = 1, 
 log 1 = 0, log .1 = - 1, log .01 = - 2, log .001 = - 3, etc.; i.e., 
 if we arrange numbers in the geometrical progression of scale A, 
 their common logarithms will form the arithmetical progression of 
 scale B. 
 
 Scale A 
 
 JV 
 
 .001 
 
 .01 
 
 .1 
 
 1 
 
 10 
 
 100 
 
 1000 
 
 Scale B 
 
 log N 
 
 -3 
 
 -2 
 
 -1 
 
 
 
 1 
 
 2 
 
 3 
 
 We assume now, without proof, that as the number N increases, 
 its logarithm will also increase. It will then follow that if a num- 
 ber N lies between two successive terms of scale A, log N will lie 
 between the two corresponding integers of scale B. Hence if we 
 have determined the place of a number N in scale A, we shall know 
 the integral part of log N, and conversely, if we know the integral 
 part of log N, we shall know the position of N in scale A. 
 
 The integral part of the logarithm of a number is called the 
 characteristic; the fractional part, taken positively, is called the 
 mantissa. Here it is to be understood that if the logarithm of a 
 number is negative, it will be written as the sum of a negative 
 integer (the characteristic) and a positive fraction (the mantissa). 
 1 
 
 For example, log 
 
 V10 
 
 log 10 2 = 2 will be written in the 
 
 * Any set of five-place tables of logarithms of numbers and of trigonometric 
 functions may be used in connection with this text.
 
 COMMON LOGARITHMS 25 
 
 form 1 + .50000, or 9.50000 10; the characteristic is 1 (or 
 9-10), the mantissa is .50000.* 
 
 We conclude that the position of a number in scale A and the 
 position of its logarithm in scale B are related to each other in 
 the way expressed by the following theorem: 
 
 Theorem IV. The characteristic of the common logarithm of a 
 number .V is the lesser of the two integers in scale B, corresponding 
 to the two numbers in scale A between which V lies; conversely, a 
 number V will lie between those two numbers in scale A, which cor- 
 respond to the characteristic of log N and the next greater integer 
 in scale B. 
 
 This theorem may be reduced at once to the following working 
 rule : 
 
 The characteristic of log A is equal to the integer in scale B, corre- 
 sponding to that number in scale A which is pointed off like A"; con- 
 versely, JV is to be pointed off like that number in scale A, which 
 corresponds to the characteristic of log A r in scale B. 
 
 It remains to determine the mantissa of a logarithm; it is found 
 from a table. It is important to observe however that the 
 mantissa of the logarithm is the same for all numbers which differ 
 only in the position of the decimal point. Let us compare, for 
 instance, log 475.3 and log .04753. Since 475.3 = 10 4 X .04753, 
 it follows that log 475.3 = 4 + log .04753. But the addition of 
 4 to log .04753 will simply increase its characteristic by 4, with- 
 out affecting the mantissa. In general, if A and B differ only in 
 the position of the decimal point, the greater one, say A, can be 
 obtained by multiplying the smaller one, B, by a power of 10, i.e. 
 .1 = 10* X B, where k is a positive integer. Therefore log 
 A = k + log B, so that log A is obtained from log B by adding 
 to it the integer k. This however will merely increase the charac- 
 teristic by k and will not alter the mantissa. This fact finds expres- 
 sion in the following theorem: 
 
 Theorem V. The mantissa of the common logarithm of a num- 
 ber is determined by its sequence of digits only; conversely, the 
 sequence of digits of a number is determined by the mantissa of its 
 logarithm. 
 
 ' : When a five-place table is used, all mantissas are written out to five 
 places of decimals.
 
 26 USE OF A TABLE OF LOGARITHMS 
 
 36. Use of a table of logarithms. Theorems IV and V enable 
 us to find from a five-place table of logarithms the logarithm to 
 five places of any number of five significant figures; * and to 
 determine five significant figures of a number whose mantissa is 
 given to five places. It only remains to become familiar with the 
 arrangement of the tables. This is best explained by means of 
 examples. 
 
 Example 1. To find log 47.316. 
 
 Since 47.316 lies between lOjand 100, (or, since 47.316 is pointed 
 off like 10) the characteristic of log 47.316 is 1. To determine the 
 mantissa we look up in the tables the sequence of digits 47316. 
 The first three digits are found in the left-hand column of the 
 table of logarithms of numbers, usually headed N, the fourth digit 
 is found in the line at the top or bottom of the page. In this way 
 we find that 
 
 for 47320 the mantissa is 67504; while 
 for 47310 the mantissa is 67495. 
 
 For an increase of 10 in the number, there is therefore an in- 
 crease of 9 in the mantissa. Now we approximate to the value 
 of the mantissa for 47316 by assuming that as the number in- 
 creases by ten equal steps from 47310 to 47320, the mantissa 
 will also increase by ten equal steps from 67495 to 67504. We 
 must then add to 67495 six tenths of 9. The table of proportional 
 parts (usually headed P.P.) shows that T %- of 9 is equal to 5.4, 
 which we round off to 5. We conclude that the mantissa for 
 47316 is 67500 and that log 47.316 = 1.67500. 
 
 Example 2. To find log .089327. 
 
 Since .089327 lies between .01 and .1, (i.e. since .089327 is pointed 
 off like .01) the characteristic of the logarithm is 2 or 8 10. 
 
 The mantissa for 89330 is 95100; the mantissa for 89320 is 
 95095. A difference of 10 in the number makes a difference of 
 5 in the mantissa. 
 
 From the table of proportional parts, we find that -^ of 5 is 3.5, 
 which we round off to 4, so that the mantissa for 89327 is 95099 
 and therefore log .089327 - 8.95099 -10. 
 
 * The figures of a number significant for its logarithm are the digits left 
 after the ciphers at the beginning and end of the number have been removed.
 
 USE OF A TABLE OF LOGARITHMS 27 
 
 Example 3. Given log A T = 2.23130. To determine N. 
 
 We now reverse the process followed in the preceding examples 
 and begin by searching among the mantissas given in the table 
 for 23130. We find that 
 
 23147 is the mantissa for 17040, and that 
 
 23121 is the mantissa for 17030, so that a difference of 
 
 26 in the mantissa makes a difference of 10 in the number. 
 
 The given mantissa 23130 exceeds 23121 by 9; from the table 
 of proportional parts, we find that -^ of 26 equals 7.8, while 
 yV of 26 equals 10.4. Of these, the former is closer to 9 than 
 the latter; hence 23130 is the mantissa for 17033, which must there- 
 fore be the sequence of digits for the required number N. Since 
 the characteristic is 2, N must be pointed off like 100 so that the 
 decimal point must be placed between the and the 3. We con- 
 clude that the required number N is 170.33. 
 
 Example 4. Given log A" = 9.07025 10. To determine N. 
 
 We find that 07041 is the mantissa for 11760, while 07004 is 
 the mantissa for 11750, so that a difference of 37 in the mantissa 
 makes a difference of 10 in the number. Hence a difference of 21 
 in the mantissa corresponds to a difference of 6 in the number; 
 the sequence of digits of N must therefore be 11756. Since the 
 characteristic is 1, the number must be pointed off like .1, so 
 that we conclude that N = .11756. 
 
 The table of the logarithms of the trigonometric functions gives 
 directly the logarithms of the trigonometric ratios, both charac- 
 teristic and mantissa, except that from the logarithms of sines 
 and cosines, and from the logarithms of tangents of angles less 
 than 45 and of cotangents of angles greater than 45, the term 
 10 is usually omitted. The arrangement of the table will be 
 readily understood from the following examples. 
 
 Example 5. To determine log tan 37 23' 45". 
 
 We find log tan 37 23' - 9.88315 - 10 and log tan 37 24' 
 = 9.88341 10, so that a difference of 1 minute in the angle 
 causes a difference of 26 in the logarithm of its tangent. Accord- 
 ingly a difference of 45" in the angle will cause a difference of 
 20 in the logarithm of the tangent. Hence log tan 37 23' 45" = 
 9.88335 - 10.
 
 28 CALCULATION BY MEANS OF LOGARITHMS 
 
 Example 6. To determine log cos 54 29' 13". 
 
 We find log cos 54 29' = 9.76413 - 10, while log cos 54 30' 
 = 9.76395 10, so that as the angle increases by 1 minute, the 
 logarithm of its cosine decreases by 18. Consequently for an 
 increase of 13" in the angle there will be a decrease of {$ of 18, 
 i.e., of 4 in the logarithm of the cosine; hence log cos 54 29' 13" 
 = 9.76409 - 10. 
 
 Example 7. Given log sin 6 = 9.47468 10. To determine 
 
 We find that log sin 17 21' = 9.47452 - 10, while log sin 17 22' 
 = 9.47492 10, so that a difference of 40 in the logarithm of the 
 sine corresponds to a difference of 1 minute in the angle. Conse- 
 quently, a difference of 16 in the logarithm of the sine corresponds 
 to a difference of \% of a minute, i.e., of 24" in the angle. We 
 conclude that d = 17 21' 24". 
 
 37. Exercises. 
 
 Determine: 
 
 1. log 56.387. 5. log 978.94. 9. log cos 47 58' 15". 
 
 2. log .084923. 6. log .00073299. 10. log cot 15 47' 50". 
 
 3. log 1.0576. 7. log sin 27 15' 20". 11. log sin 78 29' 40". 
 
 4. log .20458. 8. log tan 68 37' 35". 12. log cos 36 35' 45". 
 
 Determine the number N, when : 
 
 13. log JV = 1.65783. 15. log N = .27586. 17. log AT= 7.80880 -10. 
 
 14. log N = 9.04987 - 10. 16. log N = .09675. 18. log AT =3.97538. 
 
 Determine the angle 0, when: 
 
 19. log tan 9 = .27725. 22. log cot 9 = 8.83225 - 10. 
 
 20. log cos 9 = 9.88247 - 10. 23. log cos = 9.68552-10. 
 
 21. log sin 9 = 9.48030 - 10. 24. log tan = 9.96795 -10. 
 
 38. Calculation by means of logarithms. We are now prepared 
 to apply the results of the preceding articles. In all calcula- 
 tions, it is important to arrange the work in such a way as to 
 secure the greatest possible accuracy with least effort. This is 
 accomplished best by making a plan of the entire calculation 
 before looking up the logarithms, as illustrated in the following 
 examples:
 
 CALCULATION BY MEANS OF LOGARITHMS 29 
 
 Example 1. To determine A r = \7 - : ^=-^ 
 
 First we apply Theorems I, II and III; in this way we find 
 that log N = f [log 47.321 + log .015732 - log .9763]. 
 Accordingly we make the following plan for the calculation: 
 
 log 47.321 = 1 
 
 log .015732 = 8 -10 
 
 A 
 
 log .9763 = 9 -10 
 
 S 
 
 3 
 
 \ogN = 
 
 N = 
 
 After having completed the plan, we turn to the tables to deter- 
 mine the mantissas and to complete the calculation; this gives 
 the following results: 
 
 log 47.321 = 1.67505 
 
 log .015732 = 8.19678 - 10 
 
 A 
 
 19.87183 - 20 
 log .9763 = 9.98958 - 10 
 
 S 
 
 29.88225 - 30 
 
 3 
 
 log A r = 9.96075 - 10 
 
 A r = .91358. 
 
 In order to make the subtraction possible without introducing 
 a negative mantissa, we wrote the characteristic of the minuend 
 in the form 19 20 instead of 9 10: to make the division possible 
 without introducing a negative fraction, we wrote the mantissa 
 of the dividend in the form 29 - 30, instead of 9 - 10. 
 
 v 7 o t i + ^ f356.12x C56836)-7' 
 
 Example 2. lo determine A = 
 
 51.834 X \ .0843
 
 30 EXERCISES 
 
 We find that log N = 5 [log 356.12 + 2 log .56836 -(log 51.834 
 + ^ log .0843)], so that the calculation is carried out in the fol- 
 lowing form: 
 
 log 356.12 = 2.55159 
 log .56836 = 9.75462 - 10; 2 log .56836 = 19.50924-20 
 
 2.06083 
 log 51.834 = 1.71461 
 
 log'\0843 = 28.92583 -30; | log .0843 = 9.64194 - 10 
 
 1.35655 1.35655 
 
 .70428 
 
 log A r = 3.52140 
 N = 3322.0 
 
 7 _ _ , A . .075869 X sin 47 15' 36" 
 
 Example S. I o determine J\ = ,. 00 nn , 
 
 ^ tan 68 23 
 
 log N = log .075869 + log sin 47 15' 36" - log tan 68 23'. 
 log .075869 = 8.88006 - 10 
 
 log sin 47 15' 36" = 9.86596 - 10 
 
 A 
 
 8.74602 - 10 
 log tan 68 23' = .40201 
 
 S 
 
 log N = 8.34401 - 10 
 N = .022081 
 
 39. Exercises. 
 
 Calculate: 
 
 14.03 X 1-028 
 
 (.005734)* 
 
 3 A 5 ^15 7 4 - 3857 X .1296 
 
 2 ' V~507~' .00097135 
 
 4.357 X (.08356) 3 * 7 6.7248 X .098376 
 
 \/XJ57 V 57 - S4y X -0001574 ' 
 
 . (.4316)* X 52.07 _ 51.86 X sin 35 IS' 
 
 C083) 2 sin 49 23' 
 
 4 6 1 54.7 X -0287 10. tan 28 37' 15" X cos 56 13' X 
 
 V 15.8 
 
 cosec 75 20' 20".
 
 EXERCISES 31 
 
 11. 9.63 X sin 42 17' X sin 18 i /57.8 X 67.84.5 X 23.593 
 
 29' 30". ' V 98.627 
 
 1875 X tan 65 34' 10" * /.0024856 X .57321 X .009847 
 
 sin 17 54' 20" 17 - V~ sin 9 15' 15" 
 
 i / 4.268 X sin 81 20' t 7 .08734 X .09586 X .06792 
 
 ' V tan 14 38' ' V .15647 
 
 14 p8.643 X cos 21 17' ~| 3 .5487 X sin 38 27' 15" 
 
 " 1.9.5064 X cot 68 23'J " .98346 
 
 , c 1.0567 X cos 28 43' 50" nn 3.75 X sin 53 27' 30" 
 
 !*> ^ ^, 20. 
 
 12 
 
 13 
 
 2.3981 sin 145 35' 40"
 
 CHAPTER IV 
 THE SOLUTION OF RIGHT TRIANGLES. APPLICATIONS 
 
 40. The right triangle. If enough elements of a right triangle 
 are given to determine it completely, every other magnitude con- 
 nected with the triangle can be determined by means of the trigo- 
 nometric ratios of the angles. A right triangle is completely 
 determined by any one of the following sets of data: 
 
 (a) two legs, (6) one leg and the hypotenuse, (c) one acute 
 angle and a leg, (d) one acute angle and the hypotenuse. 
 
 In the case of the right triangle the remaining elements can be 
 determined in each case by the use of the special form of the 
 definitions for acute angles, given in 20. It is customary to 
 demote the lengths of the sides of a triangle by small Roman 
 letters, the vertices of the opposite angles by the correspond- 
 ing Roman capitals, the vertex of the right angle and the hy- 
 potenuse in a right triangle usually being denoted by C and c, 
 respectively. 
 
 In cases (a) and (6) the knowledge of the sides enables us to 
 find one of the ratios of either of the acute angles, and hence 
 these angles themselves. In cases (c) and (d) we proceed in the 
 opposite manner: from the given side and appropriate trigono- 
 metric ratios of the given angle the other sides are found. The 
 second acute angle is the complement of the given angle. 
 
 From the definitions given in 20 we obtain immediately the 
 following theorem: 
 
 Theorem I. In a right triangle, the following relations hold be- 
 tween the sides and angles: 
 
 the side opposite an acute angle = hypotenuse X the sine of the 
 angle; 
 
 the side adjacent to an acute angle = hypotenuse X the cosine of 
 the angle; 
 
 the side opposite an acute angle = the adjacent side X the tangent 
 of the angle. 
 
 32
 
 ACCURACY OF THE CALCULATION 33 
 
 41. Example (see Fig 14). 
 
 Given. Z.C = 90, ZA = 47 13', b = 14.35. 
 
 Required. B, a, c. 
 
 Solution. We find ZB = 90 - 47 13' = 42 47'. 
 
 Moreover, we have by Theorem I, 
 a = b tan A ; 
 also, b c cos A , 
 
 whence c = b/cos A. 
 
 The unknown elements of the triangle have 
 now been expressed in terms of the known 
 elements. It remains to calculate a and c. A *~ b~u.i r o 
 This may be done by the use of logarithms Fig. 14 
 
 (see 38) as follows: 
 
 log& = log 14.35 = 1.15685; log 6 = log 14.35= 1.15685 
 log tan A = log tan 47 13'= .03364; 
 
 log cos A = log cos 47 13' = 9.83202 - 10 
 ___ __^4 S 
 
 loga= 1.19049 logc= 1.32483 
 
 a =15.506 c = 21.127 
 
 Using tables giving the values of the ratios themselves (the 
 so-called natural values) instead of those of their logarithms, we 
 obtain : 
 
 a = 14.35 X tan 47 13' = 14.35 X 1.0805 = 15.505. 
 b = 14.35 -f- cos 47 13' = 14.35 -r- .6792 = 21.128. 
 
 42. Accuracy of the calculation. Checking the results. The 
 
 values of the logarithms and of the trigonometric ratios found in 
 a table are correct only to within the limit of accuracy of the table, 
 i.e., to within .000005 if 5-place tables are used, or .00005 in the 
 case of 4-place tables. 
 
 Therefore, since the sum and difference of two approximate 
 numbers are more readily determined and are frequently more 
 nearly accurate than their product or quotient, it follows that in 
 most cases the calculation by means of logarithms is to be pre- 
 ferred, particularly if the data of the problem are themselves 
 approximations.* 
 
 * A comparison of the advantages, of calculations with and without the use 
 of logarithms requires a much more detailed discussion than can here he de- 
 voted to it. Most of the calculations in this hook have heen made hy means 
 of logarithms, on account of the greater convenience of this method.
 
 34 
 
 ISOSCELES TRIANGLES 
 
 To secure a higher degree of certainty as to the correctness of 
 the final numerical results of a calculation, these results should 
 be checked. A rough check can be obtained by drawing the 
 figure to scale by use of ruler and compass and then measuring 
 the required elements. A more accurate, numerical check in- 
 volves the testing of the results, together with the data, in other 
 relations between the sides and angles of the triangle than those 
 used in the solution of the problem. 
 
 For the right triangle, the Pythagorean relation c 2 = a 2 -\- b 2 
 serves the purpose especially well, since it involves all three sides 
 of the triangle. If written in the form 
 
 a 2 = c 2 b 2 = (c 6) (c + 6) or b 2 = c 2 a 2 = (c a) (c + a), 
 
 it is well adapted to logarithmic calculation. For the example of 
 the preceding section, it furnishes the following check: 
 
 c = 21.127 
 a = 15.506 
 
 A 
 
 c + a = 36.633 log (c + a) = 1.56388 
 
 c-a= 5.621 log (c - a) = .74981 
 
 log (c + a) (c- a) = 2.31369 
 
 2 
 
 log b = 1.15685 
 
 This value should agree to within four units of the last decimal 
 place with the value of log b derived from the value of b as given. 
 
 43. Isosceles triangles. An isosceles triangle is divided into 
 two congruent right triangles by a perpendicular from the vertex 
 to the base. The methods explained in 40, 41 and 42 suffice 
 therefore for the treatment of such triangles. 
 
 a 
 
 
 \h 
 
 ^v" - -251B2 
 
 y^fd 
 
 
 
 ^Nq 
 
 V 
 
 
 1 
 
 Fig. 15 
 
 
 Example (see Fig. 1 5) 
 
 
 
 
 Given. ZP = ZQ = 
 
 37 
 
 28', a = 
 
 .23152. 
 
 Required. V, h, b. 
 
 

 
 EXERCISES 35 
 
 Solution. We find Z V/2 = 90 - 37 28' = 52 32'; .'. Z V 
 = 105 4'. 
 Furthermore 
 
 sin P = h/a, i.e., /i = a sin P, 
 
 and cos P = \ b/a, i.e., ^ 6 = a cos P, 
 
 whence 6 = 2 a cos P. 
 
 log a = 9.36459 - 10 log a = 9.36459 - 10 
 
 log sin P = 9.78412 - 10 log cos P = 9.89966 - 10 
 
 A 
 
 log h = 9.14871 - 10 log 2 = .30103 
 
 h = .14087 log 6 = 9.56528 - 10 
 
 b = .36752 
 
 Check, /i 2 = a 2 - (6/2) 2 = (a + 6/2) (a - 6/2) 
 a = .23152 
 6/2 = .18376 
 
 a + 6/2 = .41528; log (a + 6/2) = 9.61834 - 10 
 a - 6/2 = .04776; log (a - 6/2) - 8.67906 - 10 
 
 A 
 
 log [a 2 - (6/2) 2 ] = 8.29740 - 10 
 
 2 
 
 log/i = 9.14870- 10 
 
 44. Exercises. 
 
 Calculate the unknown elements of the triangles ABC in which Z C = 90 
 and in which the following elements are given; check the results: 
 
 1. a = 373, 6 = 52G. 7. A = 84 35', c = 378. 
 
 2. a = .1432, 6 = .0756. 8. A = 44 35', c = 378. 
 
 3. a =2.146, c = 4.292. 9. .1 = 45 3'. a =.08512. 
 
 4. b = 13.071, c = 19.603. 10. a = .06891, c = .09004. 
 
 5. A = 68 25', b = 8732. 11. b = 13.683, a = 3.9857. 
 
 6. B = 27 13', 6 = .06315. 12. B = 5 2', c = 1.0059. 
 
 Determine the remaining elements of the following isosceles triangles (the 
 notation being in accordance with Fig. 15) and check the results: 
 
 13. b = 26.804, P = 57 13'. 17. b = 24.192, h = 16.387. 
 
 14. b = 35.96, V = 128 46'. 18. h = .05831, P = 10 19'. 
 
 15. a = 12.05, h = 8.041. 19. b = 9.0834, a = 9.9457. 
 
 16. h = 1.0203, V = 44 52'. 20. a = 6.8032, P = 25 27'. 
 
 21. a = 16584, V = 90.
 
 36 
 
 PROJECTION 
 
 45. Projection. The projection of a directed segment AB of 
 a directed line I upon another directed line m, which makes with 
 I an angle d can now be readily determined. Let the length of 
 AB (see Fig, 16) be equal to a; then, since the direction of AB 
 
 Fig. 16 
 
 > m 
 
 is opposite to the positive direction upon I, AB = a. The 
 projection of A B upon m is AiBi. Through A draw a line m' 
 parallel to m, cutting BB X in C. Then AC = A,B, (Why?), 
 and Z CAD = d (Why?). Moreover, cos 6 = -cos 6' (see 18). 
 Since now A\B X is in the positive direction upon m, we have 
 
 AiBi = AC = a cos 6' = a cos d = AB cos 6. 
 
 This result finds expression in the following theorem: 
 
 Theorem II. The projection of a directed segment AB of a directed 
 line /, upon a directed line m, Is equal, in magnitude and direction, 
 to AB cos d, where 9 is the angle which I makes with m. 
 
 Fin. 17 
 
 This theorem is illustrated by the following examples: 
 
 (a) In Figure 17, / makes with m an angle of 240; since AB 
 
 is a negative segment of the directed line I, we have AB = 4. 
 
 Hence, AJi, = Vvo] m AB = - 4 cos 240 = + 2.
 
 APPLICATION OF THEOREM II 
 
 37 
 
 (6) In Figure 18, I makes with m an angle of 135 and AB = 3. 
 
 3\/2 
 
 Hence A 1 B l = ~Proj m AB = 3 cos 135 = n = - 2.1. 
 
 Fig. 18 
 
 46. Application of Theorem II. We return now to Theorem II 
 of 6 and calculate the projections of the segments b} T means of 
 Theorem II of the present chapter. Consider, for example, the 
 equilateral triangle ABC of Figure 19, of which the side AC 
 
 Fig. 19 
 
 makes an angle of 15 with the positive X-axis. 
 
 The side AB then makes with OX an angle of 75, 
 the side BC makes with OX an angle of 45, 
 the side CA makes with OX an angle of 1!)."). 
 
 If the length of the side of the triangle be denoted by a and the 
 projections of the vertices upon the A'-axis by A\, B h and C\, then 
 we find: 
 
 ArB, + B 1 C i + CiA 1 = Proj^l B + Proj^C + Proj^C^ 
 
 = a (cos 75 + cos 45 - cos 15) 
 = a (.2088 + .7071 - .<H>r>9) = 0.
 
 38 
 
 APPLICATIONS 
 
 47. Exercises. 
 
 1. Determine the projection upon the X-axis of a segment, 5 feet long, of 
 a line which makes with the X-axis an angle of 30. 
 
 2. Determine the projection of the same line upon the F-axis. 
 
 3. Determine the projections upon the X- and F-axes of a segment, 7 feet 
 long, of a line parallel to the F-axis. 
 
 4. Show that the sum of the projections upon the X-axis of the sides of 
 the square of Figure 20 is equal to zero. Show that the same result holds 
 for the projections upon the F-axis. 
 
 5. (See Fig. 21.) Show that the projection of AC upon the X-axis (F-axis) 
 is equal to the sum of the projections of AB and BC upon the X-axis (F-axis). 
 
 -s-X 
 
 Fig. 20 
 
 Fig. 21 
 
 48. Applications. Numerous problems in different fields of 
 science can be solved by means of the methods developed in the 
 preceding paragraphs. To solve such problems there is required, 
 however, besides a knowledge of trigonometry, an understanding 
 of the technical terms used in those different fields. We explain 
 below a few technical terms which will be used in the exercises of 
 the following paragraph and in Chapter VII. For further appli- 
 cations the student is referred to books on surveying, navigation, 
 astronomy, artillery, etc. 
 
 The angle made by the line along which an object is sighted 
 with a horizontal line through the point of observation and in 
 the same vertical plane as the line of sight, is called the angle of 
 elevation or the angle of depression of the object, according as the 
 object is higher or lower than the point of observation. 
 
 The surveyor frequently designates the direction of a line by 
 means of its bearing, i.e., the acute angle which the line makes 
 with a N. S. line through the point of observation, indicating at 
 the same time whether the line runs to the east or the west of the
 
 EXERCISES 
 
 39 
 
 N. S. line. The bearings of the lines OA, OB and OC in Figure 
 22 are denoted as N 23 E (read 23 degrees East of North), 
 
 Fig. 22 
 
 Fig. 23 
 
 N 47 W (read 47 degrees West of North) and S 18 W (read 
 18 degrees West of South) respectively. 
 
 The points of the compass, as used by the navigator, are indi- 
 cated on Figure 23. 
 
 49. Exercises. 
 
 1. The angle of elevation of the top of a mountain from a point A, situ- 
 ated in a plane 1500 feet below the top, is 19 27'. Determine the distance 
 from A to the top in an air line; also the horizontal distance from A to the 
 foot of the mountain. 
 
 2. From a lightship L, at a distance of 500 feet from a point ^4 on shore, 
 the angle of elevation of a water tower vertically above A is 28 33'. Deter- 
 mine the height of the tower above the level of the ship. 
 
 3. The angle of depression of a point P as seen from an airship 1S00 feet 
 above the ground is (33 27'. What is the straight line distance of the air- 
 ship from P? 
 
 4. An observation tower T is 40 feet high and stands 20 feet from the 
 edge of a vertical cliff, whose top is 400 feet above sea level. A ship S is in 
 the vertical plane through T and on a line at right angles to the shore line; 
 its angle of depression from T is 9 2)5' 10". How far is <S from the shore? 
 
 5. From a point .4 on the edge of a stream and 5 feet above the ground, 
 the angle of elevation of the top of a tree straight across the stream is 20 13'. 
 The height of the tree above the ground is estimated to be 15 feet. How 
 wide is the stream at that point?
 
 40 EXERCISES 
 
 6. A vessel is observed directly south from a lighthouse L and S 43 27' 
 W from a lighthouse M known to be 50 miles due East from L. What are 
 the distances of the vessel from each of the lighthouses? 
 
 7. From points P and Q, 150 feet apart and both lying in the same ver- 
 tical plane through a spire S, the angles of elevation of the spire are observed 
 to be 12 13' and 28 36' respectively. How high is the spire if P and Q are 
 on the same side of the spire; if P and Q are on opposite sides of the spire? 
 
 8. Two points, P and Q, known to be a mile apart on a level road which 
 lies in a vertical plane through the top of a mountain A are observed from A. 
 The angles of depression of P and Q are 23 17' and 32 27' respectively. 
 How high is the mountain top above the road? 
 
 9. The bearing of the line CA is N 12 E; the bearing of the line CB is 
 S 78 E, while A bears N 15 W from B. The distance CB is known to be 
 257 feet. Determine the distances from A to B and to C. 
 
 10. From a point B on one side of a stream and 5 feet above the ground, 
 the angle of elevation of a point P directly across on the opposite shore is 
 found to be 34 13'. A level line BC, 100 feet long is laid off at right angles 
 to the stream, and from C the angle of elevation of P is found to be 20 43'. 
 Determine the height of P and the width of the stream. 
 
 11. A railroad track consists of a horizontal piece followed by a down- 
 ward grade 4 miles long, making an angle of 12 with the horizontal. It is 
 proposed to replace the 12 grade by a 4 grade. How much of the hori- 
 zontal track must be removed to accomplish this change? 
 
 12. How much track mileage would be saved by the change proposed in 
 Problem 11? 
 
 13. The Washington monument is 555 feet high. From a point P on a 
 level with the base, the angle of elevation of the top is 60. How far is P 
 from the bottom of the monument? How far is P from the top? 
 
 14. Raising the ridgepole of a roof 3 feet, changes the angle under which 
 the rafters slope from 32 c to 40. How high was the roof originally, and how 
 long are the tie beams? 
 
 15. A lightship is observed due West at 10 a.m. and X40W at 2 p.m. by 
 a vessel that is traveling due South at 20 miles per hour. How far is the vessel 
 from the lightship at 10 a.m., at noon, at 2 p.m.?
 
 CHAPTER V 
 THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS 
 
 50. Graphs of sin 8 and cos 8. Many properties of the trigo- 
 nometric functions appear in a new light if we consider their 
 graphical representation. In this chapter we shall be concerned 
 principally with the graphical representations (usually called the 
 graphs) of the functions sin 8 and cos 8 and of the sine and cosine 
 functions of expressions of the form ad + b, i.e., of sin (ad + b) 
 and cos (ad + b): For their construction we make use of the 
 graphical representation of pairs of numbers used in Chapter I. 
 
 Angles 8 measured by an arbitrary unit will be represented by 
 distances along the X-axis (abscissae) ; with each abscissa will 
 be associated as ordinate a line representing the value of that 
 function of the angle 8 whose graph we wish to obtain. The 
 determination of these ordinates can be carried out graphically 
 by the following simple device. 
 
 After having established the unit of measurement on the X- 
 and F-axis, we describe a circle whose radius is equal to this 
 
 A I 
 
 O U 1 
 
 -e- H 
 
 Fig. 24 
 
 ->>-X 
 
 unit and whose center is on the X-axis (see Fig. 24). At the 
 center of this circle we construct an angle 8 whose initial side lies 
 in the positive direction along the X-axis. From the point P, 
 where the terminal side of 8 meets the circle, we drop a perpen- 
 
 41
 
 42 
 
 EXAMPLES OF GRAPHS 
 
 dicular PQ upon the X-axis. Since for P the radius vector is 
 equal to unity, the sine of angle d is measured in magnitude and 
 direction by the ordinate of P, i.e., by the line QP. Through P we 
 draw a line PP', parallel to the X-axis. We then lay off on the 
 X-axis a distance OA, representing the angle 6 in magnitude and 
 in sense, through A we draw a line A A' parallel to the F-axis, 
 meeting the line PP' in B; this point B is then a point on the 
 graph of sin 0. 
 
 If the angle be laid off with its initial side in the positive 
 direction along the F-axis and on the vertical diameter of the 
 
 Fig. 25 
 
 circle (see Fig. 25), we obtain an ordinate QP which measures in 
 magnitude and direction the cosine of 0. 
 
 51. Examples of graphs. 
 
 (a) Construct the graph of sin for 6 varying from to 360 (see 
 Fig. 26). 
 
 >X a 
 
 
 , 
 
 r 
 
 >"i 
 
 >2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 "-> 
 
 \ 
 
 v~ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 /-V-W 
 
 ///\ P 2 
 
 
 
 
 
 
 
 
 
 N 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 rww 
 
 ////>, 
 
 
 ~7Bi 
 
 
 
 
 
 
 1 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^^ Si 
 
 /f l 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 [^-^^M 
 
 ok, 
 
 A{ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 WW/ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \\\y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ? 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Fig. 26 
 
 At the center of the auxiliary circle we lay off angles of 0, 15, 
 30, . . . 330. 345, 360, with initial sides in the positive direc- 
 tion along the X-axis and with terminal sides cutting the circle
 
 EXAMPLES OF GRAPHS 
 
 43 
 
 in points P , Pi, P2, etc. On the X-axis we lay off distances 0A a 
 = 0, OAi = ir/ 12, OA 2 = x/6, etc., representing the radian meas- 
 urements of these angles. Through P , Pi, P2, etc. we draw lines 
 PoX , P1X1, P 2 X 2 , etc. parallel to the X-axis; through A , A h A 2 , 
 etc. we draw lines A Y , A\Y\, A 2 Y 2 , etc. parallel to the F-axis. 
 The intersections B , B 1} B 2 , etc. of the lines A Y and PoX , 
 A\Yi and P1X1, A 2 Y 2 and P2X2, etc. respectively, are points whose 
 :r-coordinates measure angles and whose ^-coordinates measure 
 the sines of these angles. A smooth curve drawn through the 
 points Bo, Bi, B 2 , etc. is the graph of the function sin 6 for 6 vary- 
 ing from to 360. 
 
 (6) Construct the graph of cos 6 for 6 varying from to 360 (see 
 Fig. 27). 
 
 
 r 
 * 
 
 \ 
 
 *2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 X 0l 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / /-s > 
 
 
 ui?>. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 a2S\- 
 
 i//sS\ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 p^ 
 
 jj^~~\ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i^^? 
 
 %^r~~^ I 
 
 oki-i 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 y-v/y 
 
 \\V\/ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \\y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i ^ 
 
 
 
 , 
 
 
 
 
 
 
 
 
 
 
 
 ~i 
 
 
 
 h- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Fig. 27 
 
 We construct at the center of the auxiliary circle angles of 0, 
 15, 30, . . . 345, 360 with initial sides in the positive direc- 
 tion along the Y-axis and with terminal sides cutting the circle 
 in points Q , Q u Q 2 , etc. On the X-axis we determine the points 
 A A h A 2 , etc. as in example (a). Lines Q0X0, Q1X1, Q 2 X 2 , etc. 
 parallel to the X-axis will meet lines A Y , A x Y h A 2 Y 2 , etc. 
 parallel to the F-axis in points B 0i B u B 2 , etc. whose ^-coordi- 
 nates measure angles and whose ^-coordinates measure the cosines 
 of these angles. A smooth curve drawn through these points 
 Bo, 7>i, B 2 , etc. is the graph of the function cos 6 for 6 varying 
 from to 360. 
 
 (c) Construct the graph of sin (2 9 30) for 6 varying from 
 to 360. We construct as in example (a) the lines P0X0, P1X1, 
 P 2 X 2 , etc. and A Y A^Y h A 2 Y 2 , etc. By means of these lines 
 we determine a point B , whose ^'-coordinate measures an angle 
 of and whose ^-coordinate measures the sine of 2 30,
 
 44 
 
 EXERCISES 
 
 i.e., the sine of 30; a point B h whose ^-coordinate measures 
 an angle of 15, and whose y-coordinate measures the sine of 
 2 15 30, i.e., the sine of 0; a point B 2 , whose z-coordinate 
 measures an angle of 30 and whose ^/-coordinate measures the 
 sine of 2 30 30, i.e., the sine of 30, etc. A smooth curve 
 through the points B , B 1} B 2 , etc. is the required graph. 
 
 (d) Construct the graph of cos (0/3 + 45) for 6 varying from 
 to 360 (see Fig. 28). 
 
 Q, Q J 
 
 3 
 i 
 
 Qo 
 
 r 
 ] 
 
 * 
 
 y> 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 5To v. 
 
 2^" 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 TF 1 
 
 /\\ 
 
 / /\ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ' x ?, 
 
 /\\\ 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 WW 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 l^^a 
 
 UtZ^~~\ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 +-JC 
 
 
 1^~ --J 
 
 A X A 2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \///l \\ 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \// 1 \ \\ y 
 
 
 
 
 
 
 
 
 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 H 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Fig. 28 
 
 We construct, as in example (b) the lines QoXq, QiX\, Q 2 X 2 , 
 etc. and A Y , AiY h A 2 Y 2 , etc. By means of these lines we 
 determine a point B , whose ^-coordinate measures an angle of 
 and whose ^/-coordinate measures the cosine of 0/3 + 45, 
 i.e., cos 45; a point B x whose aj-coordinate measures an angle 
 of 15 and whose ^-coordinate measures the cosine of 15/3 
 + 45, i.e., cos 50; a point B 2 , whose x-coordinate measures an 
 angle of 30, and whose 1 y-coordinate measures the cosine of 
 30/3 + 45, i.e., cos 55, etc. A smooth curve 1 drawn through 
 the points Bo, B h B 2 , etc. is the required graph. 
 
 52. Exercises. 
 
 Construct the graphs of the following functions, for 9 varying from to 
 
 360: 
 
 1. cos 2 9. 
 
 2. cos (9/2. 
 
 3. sin 2 9. 
 
 4. sin 9/2. 
 
 5. 2 cos. 
 
 6. 2 sin 9. 
 
 7. \ sin 9. 
 
 8. | cos 9. 
 
 9. sin 3 9. 
 10. 3 sin 9. 
 3 cos 9 3. 
 cos 3 9. 
 \ cos 3 9. 
 3 sin 9 3. 
 I sin 3 9. 
 I cos 3 a. 
 
 11. 
 12. 
 13. 
 14. 
 15. 
 16. 
 
 17. 
 
 cos ( e). 
 
 18. 
 
 sin (0 + 90). 
 
 19. 
 
 cos 1.9 - ISO 3 ) 
 
 20. 
 
 cos (9 + 180) 
 
 21. 
 
 cos (6 - 90). 
 
 22. 
 
 sin (0 - 00). 
 
 23. 
 
 cos (0 + 270) 
 
 24. 
 
 sin '9 - 270).
 
 OPERATIONS ON GRAPHS 45 
 
 53. Operations on graphs. The graph of the function sin 3 
 associates with any particular value of the same ordinate that 
 the graph of sin associates with 3 0, a point three times as far 
 from the origin as 0. Hence to every point A on the graph of 
 the function sin there corresponds a point B on the graph of the 
 function sin 3 9, three times as near to the F-axis as the point A. 
 It follows from this that the graph of the function sin 3 9 ma\' 
 be obtained by contracting the graph of the function sin in 
 the direction of the X-axis in the ratio 3:1. If we apply these 
 same considerations to the general case, we obtain the following 
 theorem : 
 
 Theorem I. The graphs of the functions sin ad, cos ad may be 
 obtained by contracting the graphs of the functions sin 9 and cos 9 
 respectively towards the Y-axis in the ratio a : 1, if a is a positive 
 rational number. 
 
 Here it is to be understood that if a < 1, the contraction 
 becomes an enlargement in the ratio 1 : 1/a. 
 
 The graph of sin ( 9) associates with each value of the 
 same ordinate that the graph of sin associates with 0. Hence 
 to every point A on the graph of sin there corresponds a point B 
 on the graph of sin (8) obtained by reflecting A in the F-axis 
 as a mirror.* This leads to the following theorem: 
 
 Theorem II. The graphs of the functions sin ( 9), cos ( 0) may 
 be obtained by reflecting the g aphs of the functions sin 9 and cos 9 
 respectively in the Y-axis as a mirror. 
 
 The graph of the function sin (2 9 + 180) associates with any 
 particular value of 9 the same ordinate that the graph of sin 2 9 
 associates with 9 -f- 90, a point on the X-axis 90 farther to the 
 right than 9. Hence, to every point A on the graph of sin 2 9 there 
 will correspond a point B on the graph of sin (2 9 -\- 180), a dis- 
 tance of 90 to the left of A . It follows from this that the graph 
 of sin (2 9 + 180) can be obtained by translating the graph of 
 sin 2 9 in the direction of the negative X-axis a distance of 90. 
 In the same way, we see that the graph of sin (2 9 180) may 
 be obtained by translating the graph of sin 2 9 in the direction of 
 the positive X-axis a distance of 90. 
 
 * 11 is said to be the reflection of .1 in the F-axis as a mirror if the V-axis 
 is the perpendicular bisector of the line AB.
 
 46 
 
 OPERATIONS ON GRAPHS 
 
 If we carry through the same considerations for the general 
 case, we obtain the following theorem: 
 
 Theorem III. The graphs of the functions sin (ad + b), cos (ad + b) 
 may be obtained by translating the graphs of sin ad and cos ad respec- 
 tively a distance of | b/a \,* in the direction of the negative X-axis, if b/a 
 is positive; in the direction of the positive X-axis, if b/a is negative. 
 
 Theorems I, II and III may be used to obtain the graphs of the 
 functions sin {ad + b) and cos (ad + b), where a and b are arbi- 
 trary rational numbers by translating, reflecting and contracting 
 (enlarging) the graphs of the functions sin 6 and cos respectively. 
 This will now be illustrated by examples. 
 
 (1) To obtain the graph of the function sin (3 + 30), we first 
 contract the graph of the function sin horizontally in the ratio 
 3:1, obtaining in this way the graph of the function sin 3 
 (see Fig. 29), in virtue of Theorem I. 
 
 *x 
 
 Fig. 29 
 
 We now translate the graph of sin 3 a distance of 10 to the 
 left and obtain in this way, in virtue of Theorem III, the graph 
 
 of sin (3 [6 + 10]) = sin (3 + 30). 
 
 +~X 
 
 Fig. 30 
 
 (2) To construct the graph of the function sin (90 6), we re- 
 flect the graph of sin in the F-axis as a mirror, which gives us 
 the graph of sin { d), in virtue of Theorem II (see Fig. 30). 
 
 * The symbol | b/a | designates the numerical value of b/a. Thus, if a 1, 
 b = 90, | b/a | = | -90 | = 90; similarly, | -4 | = 4, | -f | =|.
 
 APPLICATIONS OF GRAPHS 47 
 
 Translating this latter graph a distance of 90 to the right, we 
 shall obtain, in virtue of Theorem III, the graph of sin ( [6 90]) 
 = sin (90 - 6). 
 
 54. Exercises. 
 
 Construct the graphs of the following functions: 
 
 1. cos (90 + 0). 7. cos (0 + 180). 13. cos (3 - 60). 
 
 2. sin (90 - 0). 8. cos (180 - 0). 14. sin (0 - 180). 
 
 3. cos (90 - 6). 9. sin (270 + 0). 15. sin (3 + 45). 
 
 4. sin (0 - 90). 10. sin (270 - 0). 16. cos (90 - 2 d). 
 
 5. sin (90 + 0). 11. sin (2 +60). 17. sin (60 + 3 0). 
 
 6. cos (0 - 90). 12. cos (2 - 90). 18. cos (180 - 4 0). 
 
 55. Applications of graphs. The graphs of the trigonometric 
 functions sin (ad + b) and cos (ad + b), which we have learned 
 to construct in the foregoing paragraphs, will now be used to 
 illustrate and verify some of the important properties of the func- 
 tions sin d and cos d. 
 
 (1) The functions sin ax and cos ax are periodic functions 
 whose period is equal to 2 t/ | a ( , where a represents any rational 
 number. (See footnote on p. 46 and 26.) 
 
 (2) The graphs of the functions sin (d) and sin 6 are symmet- 
 ric with respect to the X-axis, i.e., for any value of d, the corre- 
 sponding values of these functions are equal but opposite in sign. 
 We conclude therefore: 
 
 (1) sin (-0) = -sine, for every value of 6. 
 
 The graphs of the functions cos ( 6) and cos coincide, so 
 that we have: 
 
 (2) cos (-0) = cos0, for every value of 0. 
 
 Formula (1) shows an analogy between the sine function and an 
 odd power of a variable; for we know that, e.g., ( d) lh = 15 . 
 On the other hand formula (2) shows an analogy between the 
 cosine function and an even power, for, e.g., ( 0) lfi = 6 ir '. On 
 account of this analogy, formulae (1) and (2) are frequently 
 expressed in the form " the sine is an odd function," and " the 
 cosine is an even function." 
 
 (3) The graph of the function sin (90 6) coincides with the 
 graph of cos 6; from this we conclude: 
 
 (3) sin (90 - 6) - cos 6, for every value of 6. (See 19, Ex. 1.)
 
 48 
 
 GRAPHS OF TAN AND COT 8 
 
 Formula (3) as well as most of those which appear in the exer- 
 cises below can also be obtained by the use of Theorem I in Chap- 
 ter II. The advantage gained by deriving them by the present 
 method lies in the fact that it emphasizes more sharply the fact 
 that these relations hold for every value of the variable, i.e., that 
 they are properties of the trigonometric functions. 
 
 56. Exercises. 
 
 Prove graphically: 
 
 1. sin (180 -0) = sin(9. 
 
 2. sin (0 - 180) = -sin0. 
 
 3. sin ((9 + 180) = -sin 0. 
 
 4. cos (90 + 6) = -sin0. 
 
 5. cos (0 - 90) = sin 0. 
 
 6. cos (180 + 6) = -cos 0. 
 
 7. sin (90 +6) = cos 0. 
 
 8. sin(0 - 90) = -cos0. 
 
 9. cos (90 - 0) = sin 0. 
 10. cos (180 - 0) = -cos0. 
 
 11. cos (0 - 180) = -cos0. 
 
 12. sin (0 + 270) = -cos0. 
 
 13. cos (0 + 270) = sin 0. 
 
 14. sin (270 - 0) = -cos0. 
 
 15. cos (270 - 0) = -sin 0. 
 
 16. cos (0 - 270) = -sin 0. 
 
 17. sin (0 - 270) = cos 0. 
 
 18. sin (0 - 45) + sin (0 + 45) = V2 sin 9. 
 
 19. sin (0 + 45) - sin (0 - 45) = V2 cos 0. 
 
 20. sin + cos = V2 sin (0 + 45). 
 
 57. Graphs of tan 6 and cot 6. We proceed now to a brief 
 consideration of the graphs of he functions tan 9 and cot 6. For 
 this purpose we begin by developing a graphical method for 
 determining an ordinate which measures the tangent of a given 
 angle 6, analogous to the method developed for the sine in 50. 
 
 *~x 
 
 Fig. 31 
 
 We return to the auxiliary circle used in 50 and 51 and draw a 
 line TT' tangent to this circle at the right hand extremity of 
 its horizontal diameter (see Fig. 31). At the center of the circle
 
 GRAPHS OF TAN d AND COT 6 
 
 49 
 
 we draw an angle 9 whose initial side lies in the positive direction 
 along the X-axis. From the point S, where the terminal side 
 of 9 or, for angles in II and III, the terminal side of 9 produced 
 through the origin, cuts the tangent line TT', we draw a perpen- 
 dicular SR upon the X-axis. Since for S the abscissa is equal 
 to unity the tangent of 9 is measured by the ordinate of S, i.e., by 
 the line RS. Through S we draw a line SS' parallel to the X-axis. 
 Furthermore, we lay off on the X-axis, as in 50, a distance OA, 
 representing the angle 9; through A we draw a line A A' parallel 
 to the F-axis, meeting the line SS' in B; this point B is then a 
 point on the graph of tan 9. 
 
 Now, following the general method explained in 51, we readily 
 construct the graph of the function tan (ad + 6). 
 
 To obtain an ordinate measuring the cotangent of 9, we draw a 
 line UU' tangent to the circle at the left hand extremity of its 
 horizontal diameter. We lay off the angle 9 at the center of the 
 circle, with its initial side in the positive direction along the 
 F-axis, and determine the point S where the terminal side of 9, or 
 the terminal side produced through the origin, meets this tangent 
 line; from S we drop a perpendicular SR upon the X-axis. It is 
 clear (see Fig. 32) that if we consider this position of 9 as the 
 
 Fig. 32 
 
 standard position, the ordinate for the point S is equal to unity, 
 so that the cotangent of 6 is measured by the abscissa, i.e., by 
 the line RS, no matter how large the angle 9 may be. Through 
 S we draw a line SS' parallel to the X-axis; we lay off, as before, 
 a distance OA on the X-axis, measuring the angle 9, draw the 
 line A A' parallel to the F-axis and determine the point B, in
 
 50 GRAPHS OF TAN AND COT (ad+b) 
 
 which the lines SS' and A A' meet. This point B is a point on 
 the graph of cot 8. The graph of the general function cot (ad -f- 6) 
 is now constructed by the methods used in 51. 
 
 58. Exercises. 
 
 Construct the graphs of the following functions, for varying from to 
 360: 
 
 1. tan 2 9. 7. tan 0/3. 13. 2 tan 9. 
 
 2. cote/2. 8. cot 3 9. 14. 3 tan 9/3. 
 
 3. tan (9 + 90). 9. cot (9 + 270). 15. tan (2 6 + 90). 
 
 4. cot (6 - 90). 10. tan (0 - 270). 16. cot {2 9 - 180). 
 6. tan (9 + 180). 11. tan (180 - 6). 17. cot (180 + 2 9). 
 6. cot (9 + 180). 12. cot (180 - 9). 18. tan (180 - 2 9). 
 
 59. Graphs of tan (ad + b) and cot (ad + b). Mere repetition 
 of the arguments of 53 will enable us at once to establish the 
 following theorems, analogous to Theorems I, II, III of 53. 
 
 Theorem la. The graphs of the functions tan ad, cot a9 may be 
 obtained by contracting the graphs of the functions tan 9 and cot 9 
 respectively towards the Y-axis in the ratio a : 1, if a is a positive 
 rational number; if a < 1, the contraction becomes an enlargement 
 in the ratio 1 : 1/a. 
 
 Theorem Ha. The graphs of the functions tan (-0), cot (9) may 
 be obtained by reflecting the graphs of the functions tan e and cot e 
 respectively in the Y-axis as a mirror. 
 
 Theorem Ilia. The graphs of the functions tan (a9 + b), cot 
 (ad + b) may be obtained by translating the graphs of the functions 
 tan a9 and cot a9 respectively a distance of | b/a \ , in the direction of 
 the negative X-axis, if b/a is positive; in the direction of the positive 
 X-axis, if b/a is negative. 
 
 These theorems enable us to obtain the graphs of functions of 
 the form tan (ad + /;) and cot (ad + b), where a and b are arbitrary- 
 rational numbers, by translating, reflecting and contracting (en- 
 larging) the graphs of tan 6 and cot 6 respectively. 
 
 (a) To construct the graph of tan (2 d + 45), we first contract 
 the graph of tan d horizontally in the ratio 2 : 1, so as to obtain
 
 APPLICATIONS 
 
 51 
 
 the graph of tan 2 d. Then we translate the latter graph a dis- 
 tance of 22| in the direction of the negative X-axis. (See Fig. 33.) 
 
 Fig. 33 
 
 (b) To construct the graph of cot (180 26), we reflect the 
 graph of cot d in the F-axis as a mirror, and contract the latter in 
 the ratio 2 : 1 in the direction of the X-axis, obtaining in this 
 way the graph of cot (2 6). If we now translate this graph a 
 distance of 90 in the direction of the positive X-axis, we obtain 
 the graph of the function cot (-2 [6 - 90]) = cot (180 - 2 6), 
 which proves to be identical with the graph of cot ( 2 6). (See 
 Fig. 34.) 
 
 Fig. 34 
 
 60. Applications. The graphs of the functions tan (ad + b) 
 and cot (ad -f- b) may now be used to derive some of the prop- 
 erties of the functions tan 6 and cot 6, by the method explained 
 in 55. We obtain in this way the following results: 
 
 (1) the functions tan ad and cot ad are periodic functions whose 
 period is w/\ a |.
 
 52 
 
 GRAPHS OF SEC AND COSEC 
 
 It is to be observed that the period of the tangent and cotangent 
 functions is half as large as that of the corresponding sine and 
 cosine functions. 
 
 (2) the tangent of ir/2 and 3 w/2, and the cotangent of and w 
 do not exist (see Theorem IV, Chapter II). Moreover 
 
 Lim tan = + c 
 
 >90- 
 
 Lim cot = oo . 
 
 0^>18O- 
 
 Lim tan = oo . 
 
 > 90+ 
 
 Lim cot = + oo . (See 24, 25.) 
 
 fl >.180+ 
 
 (3) the tangent and cotangent are odd functions, i.e., 
 
 tan ( 0) = tan 0, cot ( 0) = cote. 
 
 (4) tan (180 - 0) = - tan 0, etc. 
 
 tan (90 0) = cot 0, etc. (See exercises below.) 
 
 61. Exercises. 
 
 Construct the graphs of the following functions: 
 
 1. tan (90 + 0). 5. tan 2 0. 
 
 2. eot(-0). 6. tan (180 + 0). 
 
 3. tan (90 - 0). 7. cot (0 - 180). 
 
 4. cot (90 + 0). 8. cot 0/2. 
 
 Prove graphically: 
 
 13. tan (9 + 180) = tan 0. 
 
 14. cot (0 + 180) = cot 0. 
 
 15. tan (90 + 0) = -cot0. 
 
 16. cot (90 - 0) = tan 0. 
 
 17. tan (0 + 270) = -cot 6 
 
 18. cot (90 +0) = -tan0. 
 
 9. tan (3 - 90). 
 
 10. tan (0/2 + 90). 
 
 11. cot (2 - 90). 
 
 12. tan (90 - 2 0). 
 
 19. tan (0 - 180) = tan 0. 
 
 20. cot (270 + 0) = -tan 6 
 
 21. tan (270 - 0) = cot 0. 
 
 22. cot (0 - 270) = -tan 6 
 
 23. tan (45 + 0) = cot f4o c 
 
 24. tan (45 - 0) = cot (45 
 
 62. Graphs of sec0 and cosec 0. For the sake of complete- 
 ness, we add a method for the construction of the graphs of the 
 functions sec and cosec 0. For the former, we use the diagram 
 of Figure 31. Since for the point S, the abscissa is equal to unity, 
 the secant of is measured by the radius vector MS. If we 
 describe an arc with M as center and MS as radius we may deter- 
 mine on the vertical diameter of the circle a point Si, whose
 
 GRAPHS OF SEC0 AND COSEC 6 
 
 53 
 
 perpendicular distance from the X-axis is equal in magnitude 
 and direction to sec 6* Through $1 we draw a line SiSi' parallel 
 to the X-axis meeting the line A A' in a point B, which is a point 
 on the graph of sec 6. The graph is completed by means of the 
 method of 51 (see Fig. 35). 
 
 Fig. 36 
 
 For the graph of the cosecant function we use a diagram similar 
 to the one used for the cotangent (see Fig. 32). The ordinate of 
 
 * Here it is to be understood that Si is to be determined on the positive 
 half of the diameter, if S lies on the terminal side of 0; on the negative half of 
 the diameter if S lies on the terminal side of produeed through the origin.
 
 54 
 
 EXERCISES 
 
 being equal to unity, the cosecant of 6 is measured by the radius 
 vector MS. Describing an arc with center at M and radius 
 equal to MS, we find on the vertical diameter of the circle a point 
 Si whose distance from the X-axis is equal in magnitude and 
 direction to the cosecant of 6 (see footnote on p. 53). Drawing 
 a line SiSi parallel to the X-axis, we determine upon the line 
 AA' , drawn parallel to the F-axis, a point B which is a point of 
 the graph of cosec 8. Repeating this construction for various 
 values of 0, as explained in 51, we obtain the graph of the function 
 cosec 9 (see Fig. 36). 
 
 For the functions sec 8 and cosec 8 we can now develop theorems 
 analogous to theorems I, II, III of 53, by means of which the 
 graphs of the general functions sec (ad + b) and cosec (ad + b) 
 can be obtained by translating, reflecting and contracting (enlarg- 
 ing) the graphs of sec 8 and cosec 8. This in turn enables us to 
 prove graphically various properties of these functions. 
 
 63. Exercises. 
 
 Construct the graphs of the following functions: 
 
 1. sec 2 0. 5. cosec 0/2. 
 
 2. sec (90 + 0). 6. cosec (180 + 0). 
 
 3. cosec (0). 7. cosec (0 90). 
 
 4. sec (-0). 8. sec (3 - 270). 
 
 9. cosec 3 9. 
 
 10. sec (0 + 270). 
 
 11. cosec (0/2 + 90). 
 
 12. sec (180 - 0). 
 
 Prove graphically: 
 
 13. The secant function is an even function. 
 
 14. The cosecant function is an odd function. 
 
 15. sec (90 - 0) = cosec 0. 
 
 16. cosec (90 - 0) = sec 0. 
 
 17. sec (180 + 0) = -sec0. 
 
 18. cosec (180 + 0) = - cosec 0. 
 
 19. sec (90 + 0) = - cosec 0. 
 
 20. sec (0 - 90) = cosec 0. 
 
 21. cosec (90 + 0) = sec 0. 
 
 22. cosec (0 - 90) = - sec 0. 
 
 23. sec (180 -0) = - sec 0. 
 
 24. sec (9 - 180) = - sec 0. 
 
 25. sec (270 + 0) = cosec 0. 
 
 26. cosec (270 +0) = - sec 0. 
 
 27. cosec (180 - 0) = cosec 0. 
 
 28. cosec (0 - 180) = - cosec 9. 
 
 29. sec (0 270) = - cosec 0. 
 
 30. cosec (0 - 270) = sec 0.
 
 CHAPTER VI 
 THE ADDITION FORMULAE 
 
 64. A special case. In the preceding chapter we have proved 
 formulae like sin (0 + 90) = cos 6; cos (90 - 6) = sin 0, etc. 
 We next inquire how the trigonometric functions of the sum and 
 difference of any two angles may be expressed in terms of the 
 functions of these angles; i.e., we ask in the first place for formulae 
 for sin (a + 0), sin (a 0), cos (a + 0) and cos (a 0) in terms 
 of sin a, cos a, sin and cos 0. 
 
 We begin by deriving in a slightly different manner some of 
 the formulae mentioned above. From the graphs of the functions 
 sin (a + 90) and cos a; cos (a -f 90) and sin a, we have already 
 concluded, that 
 
 (1) sin (a + 90) = cos a; and that (2) cos (o + 90) = -sin a. 
 
 If in these formulae we replace a by a 90, we obtain 
 (3) sin a = cos ( - 90), and (4) cos a = -sin (a- 90), 
 
 which may also be derived directly from the graphs. 
 
 65. Addition formulae for the sine and the cosine. We pro- 
 ceed now to the general case. We place the angle a in standard 
 position and we bring the initial side of into coincidence with the 
 terminal side of a; let OA then be the terminal side of 0. 
 
 The angle a + /3 is then in standard position with respect to 
 the axes OX and OY (see Fig. 37), but not the angle /3. In order 
 to make possible the discussion of the trigonometric ratios of 0, 
 we introduce as auxiliary axes the lines OP and OQ. The posi- 
 tive direction on OP is that of the initial side of 0; the positive 
 direction on OQ is so determined that when XOY rotates about 
 until OX coincides with OP, then OY will coincide with OQ. 
 From this it follows that 
 
 Z XOP = Z YOQ = a. 
 
 We take now a point on the terminal side of a -f- /3 (which is at 
 the same time terminal side of /3) and construct its coordinates 
 
 55
 
 56 ADDITION FORMULAE FOR THE SINE AND THE COSINE 
 
 x and y with respect to the axes OX and OY, and also its coordi- 
 nates x' and y' with respect to the auxiliary axes OP and OQ. 
 Moreover, we select the line OA as our unit of measurement. 
 In this way we find: 
 (1) sin /3 = OE/OA = OE, and cos = OD/OA = OD. 
 
 Furthermore, we see that 
 
 sin (a + 0) = OC/OA = OC = Projy(M. 
 In order to express sin (a + /3) in terms of the trigonometric 
 ratio of a and /3 separately, we proceed as in 46 and 47, making use 
 
 of the Corollary of 6, with triangle 
 OB A. Thus we find: 
 (2) sin(a + /3) = ProjyOA 
 
 = FrojyOD + FrojyDA 
 = Proj Y 0D + Proj y 0E. 
 But, OD is a segment of the direct- 
 -*~ x ed line OP, which makes with the Y 
 axis the angle a 90; hence, by 
 Theorem II of 45, 
 Proj yOD = OD cos (a - 90) 
 
 = cos cos (a - 90). Why? 
 And, OE is a segment of the 
 directed line OQ, which makes with 
 OY the angle a; hence Proj yOE = OE cos a = sin /3 cos a. 
 Why? 
 
 Substituting the last two results in (2), we find that 
 
 sin (a + /3) = cos /3 cos (a 90) + sin j8 cos a. 
 Finally, we make use of 64 to obtain the result 
 
 sin (a + /3) = sin a cos /3 + sin /3 cos a. 
 To obtain a formula for cos (a + /3) we proceed in the same way, 
 projecting on the X-axis instead of on the F-axis. This leads to 
 the following development: 
 
 cos (a + j8) = OP/0.4 = OB = Proj^Oyl 
 
 = Proj^OP + Proj z rM - Proj ^07) + Proj*0P 
 = OD cos a + OP cos (a + 90) 
 = cos /S cos a + sin /3 cos (a + 90) = cos a cos /3 
 sin a sin jS 
 Thus we have obtained the following important result: 
 cos (a + /3) = cos a cos /3 sin a sin 13, 
 
 (3) 
 
 sin (a + $) = sin a cos (3 + cos a sin /3;
 
 EXERCISES 57 
 
 i.e., the cosine of the sum of two angles is equal to the product of the 
 cosines of these angles diminished by the product of their sines; 
 the sine of the sum of two angles is equal to the product of the sine 
 of one by the cosine of the other plus the product of the sine of the 
 other by the cosine of the first. 
 
 Formulae (3) are known as the addition formulae for the sine 
 and cosine functions. The proof given here is entirely inde- 
 pendent of the quadrant in which the angles lie. The student 
 should however, carry the proof through for various positions of 
 the terminal sides of the angles. 
 
 These addition formulae may now be used in the first place to 
 derive some of the other results of Chapter V. We have, for 
 instance: 
 
 sin (a + 180) = sin a cos 180 + cos a sin 180 = -sin a, 
 cos (a + 180) = cos a cos 180 - sin a sin 180 = -cos a. 
 
 66. Subtraction formulae. In the second place we use the 
 addition formulae to express the sine and cosine of a /3 in terms 
 of the sine and cosine of a and /3. We know from 55, (2) that the 
 sine is an odd function and that the cosine is an even function, i.e., 
 that 
 
 sin ( /3) = sin /3, cos ( (3) = cos j3. 
 
 We replace /3 by /3 in the addition formulae, and make use of 
 the above formulae. In this way we find: 
 
 COS (a 8) = cos [a + ( &)] = cos a cos ( 8) sin a sin ( /3) 
 
 = cos a cos 8 + sin a sin /S, 
 
 sin (a 8) = ' s i [ + ( 8)] = s 'n a ( 'OS ( 8) + cos a sin ( 8) 
 = sin a cos 8 eos a sin 8- 
 
 67. Exercises. 
 
 1. Express sin 7.5 and cos 75 in terms of the ratios of 4.") and 30, and 
 use the results for calculating the ratios of 75. 
 
 2. Calculate the ratios for 105. 
 
 3. Calculate the ratios for 15. Compare the results with those obtained 
 from the tables. 
 
 4. Determine the angles in the third quadrant whose ratios may be 
 calculated without the use of tables, by-means of the addition and subtraction 
 
 formulae.
 
 58 ADDITION AND SUBTRACTION FORMULAE 
 
 Verify the following formulae by means of the addition and subtraction 
 formulae: 
 
 5. cos (180 - 0) = -cos 0. 7. cos (270 0) = sin 0. 
 
 6. sin (180 - 0) = sin 0. 8. sin (270 - 6) = - cos 0. 
 9. sin (0 + 45) + sin (0 - 45) = V2 sin 0. 
 
 10. cos (9 + 45) + cos (0 - 45) = V2 cos 0. 
 
 11. sin (0 + 30) + cos (9 + 00) = cos 9. 
 
 12. sin (0 - 60) = -cos {9 + 30). 
 
 13. cos (45 +9) = sin (45 - 9) = V2 (cos 9 - sin 9). 
 
 14. sin (45 + 9) = cos (45 - (9) = V2 (cos 9 + sin 0). 
 16. COS (a + 0) cos (a Id) = COS 2 a -f- cos- 1. 
 
 16. sin (a -4- 0) sin (a 0) = sin 2 a sin 2 /3 = cos 2 cos 2 a 
 
 17. cos (nTr +0) = ( -1)" cos 0. 19. cos (nir - 0) = (-1)" cos0. 
 
 18. sin (nw + 9) = (-1)" sin 0. 20. sin (titt - 0) = (-1)"+' sin d. 
 
 21. cos [(2 n 4- 1) x/2 - 9] = ( - 1) sin 0. 
 
 22. sin [(2 n + 1) tt/2 - 0] = ( - 1) cos 0. 
 
 23. cos [(2ra + 1) tt/2 +0] = (-1)" +1 sin 0. 
 
 24. sin [(2 n + 1) tt/2 4- 0] = ( - 1) cos 0. 
 
 68. Addition and subtraction formulae for the tangent and 
 cotangent. Since tan 6 = sin 0/cos 9 and cot 6 = cos 6 /sin 9 
 (Why?), the addition and subtraction formulae for the sine and 
 cosine enable us to obtain corresponding formulae for the tangent 
 and cotangent, viz., 
 
 tan (a + |8) 
 
 sin (a + 0) 
 cos (a 4- /3) 
 
 sin a cos /3 + cos a sin /3 
 
 cos a cos j8 sin a sin 13 
 
 and 
 
 
 
 cot (a 4- j3) 
 
 cos (a + /3) 
 sin (a + 0) 
 
 cos a cos /3 sin a sin /3 
 sin a cos /3 4- cos a sin /3 
 
 The results may be expressed in terms of tangents alone by 
 dividing the numerators and the denominators of each of these 
 fractions by cos a cos /3, or in terms of cotangents alone by 
 dividing them by sin a sin (3. In this manner we find: 
 
 sin a cos cos a. sin 
 , , . cos a cos cos a cos tan a 4- tan 
 
 tan (a + 0) = 
 
 cos a cos sin a sin 1 tan a tan 
 
 COS a cos cos a COS
 
 DOUBLE AND HALF-ANGLE FORMULAE 59 
 
 and 
 
 cos a cos i3 sin a tan 
 . _ sin a sin ft sin a sin fl _ COt a cot 1 
 
 a si n a cos . cos sin cot a + cot 
 
 sin a sin sin a sin 
 
 Starting with the formulae 
 
 ON sin (a /3) , , , rtS cos (a /S) 
 
 tan (a - 0) = t ^ and cot (a - 3) = ^ r (' 
 
 cos (a |3) sin (a 3) 
 
 and proceeding in exactly the same way as above, we find: 
 
 -v tan - tan /J . . a . cot a cot + 1 
 
 tan (a - /s) = r-r - : j ^ and cot (-) = j ^-r 1 
 
 1 -(- tan a tan 3 cot - cot a 
 
 69. Exercises. 
 
 1. Determine can 75 and cot 75 by means of the addition formulae for 
 the tangent and cotangent. 
 
 2. Determine tan 15 ' and cot 15 by means of the subtraction formulae. 
 
 3. Determine tan 105 and cot 105. (It would not be advisable to 
 write 90 + 15 in place of 105 in this problem. Why not?) 
 
 , 1 + tan 9 cot 9 + 1 
 
 4. Prove that tan (45 + 9) = cot (45 - 9) 
 6. Also that tan (45 - 9) = cot (45 + 9) = 
 
 1 - tan 9 cot 9-1 
 1 - tan 9 cot 9 - 1 
 
 1 + tan 9 cot 9 + 1 
 
 6. Express tan (a + 0) in terms of the cotangents of a and 0. 
 
 7. Express cot (a + 0) in terms of tan a and tan 0. 
 
 8. Prove that tan (30 + 6) = cot (60 - 9) = L_ 3 tan 9 . 
 
 v 3 - tan 6> 
 
 9. Prove that tan (60 + 0) = cot (30 - 6) = 1 + 3cotfl 
 
 cot - V3 
 
 70. Double angle formulae and half-angle formulae. From 
 the addition formulae we derive, by putting a = @ = 9, the fol- 
 lowing formulae, by means of which our knowledge of the ratios 
 of any angle enables us to find the ratios of an angle twice as 
 large; in other words, formulae which express the ratios of any 
 angle in terms of the ratios of an angle half as large; we find:
 
 60 DOUBLE AND HALF-ANGLE FORMULAE 
 
 cos 2 = cos (0 + 0) = cos cos e - sin sin e = cos 2 - sin 2 e. 
 
 Sin 2 = sin (0 + 0) = sin cos + cos sin = 2 sin COS 0, 
 tan + tan 2 tan 
 
 tan * ~ 1 - tan tan ~ 1 - tan 2 
 If we put 8 = #/2, and therefore 2 = <, we obtain an equiva- 
 lent form of these formulae, bringing out more vividly the fact 
 that they express the ratios of an arbitrary .angle in terms of the 
 ratios of one half of that angle, viz., 
 
 sin 8 = 2 sin - cos ~ , cos 8 = cos 2 ~ sin 2 = , tan 8 = > 
 
 22 2 2 i_tan 2 ^ 
 
 where we have again written 8 in place of <f>. 
 
 The second of these formulae leads to another important set of 
 results, in the following manner: 
 
 To the two members of the identity 
 
 cos 8 = cos 2 - sin 2 -) 
 we add the corresponding members of the identity 
 
 1 o I 9 ^ 
 
 1 = cos 2 ^ + sin--) 
 
 a 
 
 yielding the result 1 + cos 8 = 2 cos 2 - (1) 
 
 If we subtract the first of these identities from the second we 
 
 obtain: 1 . J , , 
 
 1 cos 8 = 2 sin 2 j: (2) 
 
 Upon solving the resulting identities (1) and (2) for cos 8/2 
 and sin 8/2 respectively, we obtain: 
 
 e i /l + COS 6 , 9 , 4 /l - COS 
 
 formulae which express the ratios of one half of an angle in terms 
 of the ratios of that angle. The plus or minus sign is to be used 
 according to the quadrant in which 8/2 falls. 
 
 Dividing the second of the latter formulae by the first, we find: 
 
 VF 
 
 ,0 i /l cos0 sin 1 - cos 
 tan - = * 
 
 2 ~ V l + cos 1 + cos sin 
 the last two forms being derived from the first by multiplying 
 the numerator and the denominator of the fraction under the 
 radical sign by 1 + cos 8 and 1 cos 8 respectively. In the last 
 two expressions the double sign is not necessary, because 1 +- cos 6
 
 FACTORIZATION FORMULAE 61 
 
 and 1 cos are always positive, while tan 0/2 and sin always 
 have the same sign. 
 
 71. Exercises. 
 
 1. Determine the functions of 30 by means of those of 60. 
 
 2. Determine the ratios of 30 by means of those of 15, found in 67, 3. 
 
 3. From the ratios of 15, derive those of 7 30'. 
 
 4. Obtain a formula which expresses cot 0/2 rationally in terms of sin 9 
 and cos 0. 
 
 5. Prove that cosec 19 = \ sec 9 cosec 9. 
 
 6. Prove that sec x = \/ r r 
 
 2 1 + sec d 
 
 7. Derive the double angle formula for the tangent from the double 
 angle formulae for sine and cosine. 
 
 8. Derive the half angle formula for the tangent from the double angle 
 formula for the tangent. 
 
 72. Factorization formulae. We return once more to the addi- 
 tion formulae for the sine and cosine in order to derive from them 
 a set of formulae which will enable us to convert the sum (or the 
 difference) of the sines or cosines of two angles into a product. 
 Such a conversion is of great importance, as is evident from 
 Theorem I of Chapter III, whenever we wish to carry out loga- 
 rithmic calculations with expressions which involve sums or differ- 
 ences of sines or cosines, and also for many theoretical purposes. 
 
 We know: 
 
 sin (a + (3) = sin a cos /3 + cos a sin 0, 
 and sin ( /?) = sin a cos /3 cos a sin /3. 
 
 Adding the corresponding members of these two identities, we 
 find: 
 
 (1) sin (a + J3) + sin (a (3) = 2 sin a cos /3. 
 Subtracting them, we find: 
 
 (2) sin ( a + /3) sin (a 0) = 2 cos a sin (3. 
 
 These formulae can be put in a slightly different form, more 
 useful for the purpose for which we are deriving them, by putting
 
 62 EXERCISES 
 
 a + ft = 6 and a ft = <, whence we obtain by addition and 
 subtraction 
 
 e + <f> 0-0 
 
 a = ^-^ and /3 = s 
 
 Substitution of these values for a + /3, a ft, and /3 in formulae 
 (1) and (2) gives us the following results: 
 
 sin 6 + sin <t> = 2 sin "t cos 
 
 and sin sin <f> = 2 cos 5-^ sin 5-^ 
 
 We now proceed in exactly the same manner with the formulae 
 
 cos (a + ft) = cos a cos ft sin a sin /3, 
 and cos (a ft) = cos a cos ft + sin a sin ft, 
 
 and find the formulae: 
 
 cos + cos 0=2 cos ~- cos 
 
 2 2 
 
 and cos cos 4, = 2 sin "I sin 
 
 73. Exercises. 
 
 Convert the following sums and differences into products: 
 
 1. sin 55 + sin 65. 4. cos 87 + cos 42. 
 
 2. cos 75 - cos 15. 5. cos 312 - cos 252. 
 
 3. sin 27 - sin 18. 6. sin 213 + sin 237. 
 Calculate by means of logarithms: 
 
 7. (sin49 + sin35)(cos37-cos51 ). 9. (cos 137 + cos 84) 3 . 
 
 8 "in 57 + sin 24 . Q Vcos 306 - cos 246. 
 
 sin 57 - sin 24 
 
 Prove the following identities: 
 
 sin + sin <j> , 6 + <j> <t> 
 11. r^ ; ~ : = tan -= cot 5 
 
 + <P . <t> 
 12. - = tan - tan ^ 
 
 sin 
 
 9 - 
 
 sin 
 
 <t> 
 
 cos 
 
 9 - 
 
 cos 
 
 4> 
 
 cos 
 
 9 + 
 
 cos 
 
 <i> 
 
 sin 
 
 + 
 
 sin 
 
 <t> 
 
 cos 
 
 9 + 
 
 cos 
 
 <f> 
 
 cos 
 
 - 
 
 COS 
 
 <t> 
 
 13. = tan 
 
 . COS0 COSc/> 6 + d> 
 
 14. - . = tan t-' 
 
 sin d sin (f> 2
 
 74. 
 
 1. 
 2. 
 3. 
 
 4. 
 
 MISCELLANEOUS EXERCISES 
 Summary of formulae proved in Chapter VI. 
 
 cos (a + jS) = cos a cos /3 sin a sin j3, 
 sin (a + /3) = sin a cos /3 + sin /3 cos a, 
 cos (a j8) = cos a cos /3 + sin a sin /3, 
 sin (a j8) = sin a cos /3 sin /S cos a, 
 tan a + tan /? 
 
 tan ( a + /3) = 
 tan (a /3) = 
 
 1 tan a tan /3 ' 
 tan a tan j8 
 
 1 + tan a tan /3 ' 
 
 7. cos 2 = cos 2 5 - sin 2 0, 
 
 8. sin 2 9 = 2 sin cos 0, 
 
 2 tan 
 
 9. tan 2 9 
 
 10. 
 11. 
 
 12. 
 13. 
 14. 
 15. 
 16. 
 
 1 - tan 2 0' 
 cos 0/2 = j=V (l + cos0)/2 , 
 sin 0/2 = V(1 - cos 0)/2, 
 
 + 9 l 
 tan - = - 
 
 2 sm 
 
 cos 
 
 sin 
 
 sin + sin </> = 2 sin - 
 
 sin sin = 2 cos 
 
 cos + cos = 2 cos 
 
 cos0 cos^ = 2 sin 
 
 1 + cos 
 
 + 
 
 2 
 
 + 
 
 cos 
 
 2 Sm_ 2~ 
 
 + 
 
 cos 
 
 + 
 
 63 
 
 see 65. 
 see 65. 
 see 66. 
 see 66. 
 
 see 68. 
 
 see 68. 
 
 see 70. 
 see 70. 
 
 see 70. 
 
 see 70. 
 see 70. 
 
 see 70. 
 see 72. 
 see 72. 
 see 72. 
 see 72. 
 
 75. Miscellaneous Exercises on Chapter VI. (The more diffi- 
 cult examples are marked with a *.) 
 
 1. Determine the ratios of 165 (a) by means of the addition formulae: 
 (b) by means of the half-angle formulae. 
 
 2. Prove that tan tt/8 = V2 - 1. 
 
 3. Prove the identity: sin (60 + 6) - cos (30 + 0) = sin 0. 
 
 a /-i i i . (- sm 67 sin 34) sin 23 
 
 4. Calculate: z-^z 
 
 cos 17 
 
 Prove the following identities: 
 
 5. cos 3 = 4 cos 3 d 3 cos 6 
 
 6. sin 3 0=3 sin 0-4 sin 3 6. 
 sin 2 a sin 2 fi 
 
 *7. tan ( a + p) = . 
 
 sin a cos a sin (i cos tf 
 
 8. sin 2 3 sin 2 2 6 = sin 5 5 sin 6.
 
 Mn 1 + sin0 
 
 10. - = = tan 
 
 1 sin 
 
 64 MISCELLANEOUS EXERCISES 
 
 ft _ 3 tan - tan 3 
 
 9. tan 3 = i - 
 
 13 tan 2 
 
 (5+0- 
 
 11. sin (a + + 7) = sin a cos /3 cos 7 + sin /3 cos a cos 7 + sin 7 cos a 
 
 cos j3 sin a sin /3 sin 7. 
 
 12. cos (a + /3 + 7) = cos a cos /3 cos 7 sin a sin /3 cos 7 sin /3 sin 7 
 
 cos a sin a sin 7 cos /3. 
 
 *13. tan a tan /3 + tan /3 tan 7 + tan 7 tan a = 1, provided a + + 7 = 90. 
 
 * . sin 2 a cos a a 
 
 14. tr-: 7r - =. = tan 5 
 
 1 + cos 2 a 1 + cos a 2 
 
 16. cos (0 + 30) + sin (6 + 240) = - sin 6. 
 
 16. sin (60 +6) + cos (6 + 30) = V3 cos 6. 
 
 2 
 
 17. -r = sin a. 
 
 cot a/2 -j- tan a/ 2 
 
 18. sin + sin (0 + 2 jr/3) + sin (0 + 4 tt/3) = 0. 
 
 19. cos + cos (0 + 2 tt/3) + cos (0 + 4 tt/3) = 0. 
 
 20. sin 4 = 4 sin cos 3 4 sin 3 cos 0. 
 
 21. cos (a+/?)+ cos (a /3) + cos ( a + /3)+ cos ( a 0) = 4 COS a cos 0. 
 
 1 - tan 2 a/2 
 
 22. - - -- = cos a. 
 1 + tan 2 a/2 
 
 23. tan (0 + 45) + tan (0 - 45) = 2 tan 2 0. 
 
 *24. tan 2 a tan a = ; s 
 
 cos a + cos 6 a 
 
 25. cos 4 = 8 cos 4 0-8 cos 2 + 1. 
 
 *26. sin 2 a + sin 2 j3 + sin 2 7 = 4 sin a sin sin 7, if a + + 7 = 180. 
 
 27. sin 3 = 4 sin sin (tt/3 + 0) sin (tt/3 - 0). 
 
 28. Calculate the functions of 3 w, 8. 
 
 *29. tan (tt/4 + 0/2) + cot (V/4 + 0/2) = 2 sec 0. 
 
 *30. cos 2 a + cos 2 (3 + cos 2 7 = 1 4 cos a cos /3 cos 7, if a + /3 + 
 7 = 180.
 
 CHAPTER VII 
 
 THE SOLUTION OF TRIANGLES 
 
 76. The Law of Sines; the area of a triangle. We consider 
 in this chapter the problem of " solving an arbitrary triangle 
 ABC," i.e., the determination of the unknown elements of a tri- 
 angle ABC of which sufficient elements are given. The problem 
 is solved by the use of various relations subsisting between the 
 sides and angles of a triangle. 
 
 *~x 
 
 Fig. 38a 
 
 Denoting the length of the perpendicular dropped from the 
 ^vertex A upon the opposite side BC by h a , and using analogous 
 notations for the other perpendiculars, we have, 
 (1) h a = c sin B = b sin C, hb = a sin C = c sin A, 
 h c = b sin A = a sin B. 
 For h a in Fig. 38a, the above result is obtained most readily, if 
 we remember that Z B is placed in standard position with refer- 
 ence to the axes indicated in the diagram. 
 From these formulae we derive: 
 
 (a) Theorem I. The sines of the angles of a triangle are propor- 
 tional to the sides opposite the angles.* (Law of Sinks.) 
 
 * It is a familiar theorem of plane geometry that of two unequal sides of 
 a triangle the greater side lies opposite the greater angle. The law of sines 
 may be looked upon as completing this theorem by stating how the unequal 
 sides are related to the angles opposite them.
 
 66 TWO ANGLES AND ONE SIDE 
 
 To prove this law, we divide the two expressions for h a in equa- 
 tions (1) by be, those for h b by ac. In this way we obtain: 
 
 sin B/b = sin C/c = sin A /a, 
 
 which was to be proved. 
 
 (b) Theorem II. The area of a triangle is equal to one half of the 
 product of any two sides multiplied by the sine of the angle included 
 by them. 
 
 Proof. Denoting the area of triangle ABC by A, we have: 
 
 A = a/2 h a = 6/2 h b = c/2 h c . 
 
 If in these expressions we substitute for the altitudes h a , h b and 
 h c the values which they have in equations (1), we find 
 
 A = \ ab sin C = \ be sin A = \ ca sin B, 
 
 which was to be proved. 
 
 77. Two angles and one side. A triangle is determined when 
 two angles and a side are given. The law of sines suffices to deter- 
 mine the remaining elements in this case. For, suppose that A y 
 B, and a are given; we have then C = 180 (.4 + B). 
 
 Moreover sin B/b = sin A/a and sin C/c = sin A /a; hence 
 
 , a sin B , a sin C 
 
 b = : -7- , and c . 7- 
 sin A smi 
 
 which completes the solution. 
 
 If the given elements are measured by short numbers the cal- 
 culation may be carried out directly by means of a table of natural 
 values; the expressions for b and c, however, are well adapted to 
 calculation by means of logarithms. 
 
 Example. 
 
 Given, a = 43.257, A - 57 23', C = 49 47'. 
 
 Required. B, b and c. 
 
 Solution. B = 180 - (A + C) = 180 - 107 10' = 72 50'. 
 
 b a . _ a sinB _ 43.25 7 sin 7 2 50' 
 
 sin B ~ sin A ' 1,e '' ~ sin A ' sin 57 23'
 
 a 
 sin A ' 
 
 EXERCISES 67 
 
 asinC 43.257 sin 49 47' 
 
 i.e., c = 
 
 sin C 
 
 log 43.257= 1.63606 
 log sin 72 50'= 9.98021-10 
 
 sin A sin 57 23' 
 
 log 43.257= 1.63606 
 log sin 49 47'= 9.88287-10 
 
 11.61627-10 
 log sin 57 23' = 9.92546-10 
 
 S 
 
 11.51893-10 
 log sin 57 23'= 9.92546-10 
 
 78. 
 
 1. 
 
 2. 
 3. 
 4. 
 5. 
 
 log 6= 1.69081 
 6 = 49.069 
 
 Exercises. 
 
 A = 39 27', B 
 
 logc = 1.59347 
 c =39.216 
 
 108 51', b = .43215. Determine C, a, c. 
 C = 30, A = 45, c = 123. Determine B, a, b. 
 B = 27 45' 15", C = 89 19' 20", c = 14.302. Determine A, a, b. 
 A = 37 12' 30", C = 58 26' 40", a = 103.47. Determine B, b, c. 
 We wish to determine the distance from a point A to a point B, situ- 
 ated in a marsh, visible but not accessible from A (see Fig. 39). For this 
 
 purpose, we measure the distance from A to a point C, from which both A 
 and B are visible and we measure the angles BAC and ACB. Calculate AB, 
 if AC = 750 feet, ABAC = 32 27' and ZACB = 47 12'. 
 
 6. Devise a method for finding the distance from a point A on one bank 
 of a river to a point B on the opposite bank. 
 
 T 
 
 Fig. 40 
 
 7. To determine the height of a tower 7'>S^ (see Fig. 40) we measure the 
 angle of elevation of its top T from two points A and B lying on the same 
 side of T, in the same vertical plane with T and a known distance apart.
 
 68 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 
 
 Calculate the height of the tower, if we find that AB = 257 feet, Z TAB = 
 19 43' and Z TBS = 47 29'. (Compare the present method of solving 
 this problem with the method used for solving problem 10 in 49.) 
 
 8. Solve problem 10 in 49 by the method used in the preceding problem. 
 
 79. Two sides and an angle opposite one of them. If two 
 
 sides of a triangle and an angle opposite one of them are given, 
 the triangle is not completely determined. Let there be given 
 the sides a and b and the acute angle A (see Fig. 41). To con- 
 
 struct the triangle ABC we lay off a line equal to b on one of the 
 legs of the angle A. The vertex C is then located. We then 
 strike an arc with C as a center and a as a radius. The intersec- 
 tion of this arc m with the second leg AX of angle A determines 
 the third vertex of the triangle, B. 
 
 It is now clear that if a is shorter than the perpendicular dis- 
 tance p from C to AX, as in Figure 42a, then the arc m will not 
 
 Pig. 42a 
 
 Fig. 42b 
 
 intersect AX at all, so that no triangle can be constructed. If a 
 is equal to the perpendicular p, as in Figure 42b, we obtain a single 
 right-angled triangle. If a is greater than p, but less than b (see 
 Fig. 42c), then the arc m will cut AX in two points, B and B' ,
 
 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 69 
 
 and we will obtain two triangles, ACB and ACB' ', both of which 
 will satisfy the requirements of the problem. Since CB = CB', 
 Z CB'B = Z CBB', and therefore the angles B and B' occurring 
 in triangles ABC and AB'C respectively are supplementary angles. 
 If a > b (see Fig. 42d) the arc m will meet AX again at two points 
 
 Fig. 42d 
 
 Fig. 42e 
 
 B and B' ', but one of these points will fall on XA produced, so 
 that the triangle AB'C, to which it gives rise, does not contain 
 the given angle A. Hence in this case there is only one triangle 
 Avhich satisfies the requirements of the problem, viz., triangle ABC. 
 If the given angle A is obtuse the construction of triangle ABC 
 proceeds in the same way as before (see Fig. 42e). It is evident 
 that in this case there is no solution possible, unless a> b. We 
 observe moreover that in all the cases which we have considered 
 the perpendicular p is equal to b sin A. Hence we can put the 
 results of this discussion in the following form: 
 
 Theorem III. If two lines and an angle, such as a, b, A are given, 
 then there may be no triangle, one triangle or two triangles of which 
 the given lines are sides and of which the given angle is an angle 
 opposite one of these sides, viz., 
 
 If a < b sin A, there is no triangle; 
 
 if a = b sin A and A is acute, there is one triangle; 
 
 if a > b sin A, a < b and A is acute, there are two triangles; 
 
 if a = b and A Is acute, there is one triangle; 
 
 if > b, there is one triangle. 
 
 The different cases mentioned in this theorem are illustrated 
 in the examples which follow. The solution proceeds in each case 
 according to the following plan:
 
 70 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 
 
 From the law of sines, we find: sin. B = , which enables 
 
 a 
 
 us to determine B. Next we can calculate C, since C = 180 
 
 (A + B) . Finally we use the law of sines to find c, viz., c = j- 
 
 sin A 
 
 If a < b sin A, we find that sin B > 1; hence no angle B can 
 be found, and therefore no triangle exists. If a = b sin A, then 
 sin B 1, B = 90 and we have a single right-angled triangle. 
 
 If a > 6 sin A, then sin B < 1 and we can determine angle B 
 from the tables. But, besides the acute angle B found from the 
 tables, we must also consider the supplement of B, whose sine is 
 equal to the sine of B; we have therefore B' 180 B. More- 
 over if a < b, then we must have B > A (see footnote on p. 65); 
 hence both angles, B and B', can be used, if A is acute, while 
 neither can be used if A is right or obtuse. If, on the other hand, 
 a > b, then we must have B < A, so that only the acute angle B, 
 found from the tables, can be used. 
 
 Example 1. 
 
 Given, a - 4.73, b = 18.65, A = 43 27'. 
 
 Required, c, B, C. 
 
 Solution . 
 
 sin B sin A xU , . D b sin A 18.65 sin 43 27' 
 
 j = , therefore sin B = = -7-== > 
 
 6 a a 4.73 
 
 log 18.65 = 1.27068. 
 
 log sin 43 27' = 9.83741 - 10. 
 
 A 
 
 log b sin A = 11.10809 - 10. 
 
 log a = log 4.73 = .67486. 
 
 We notice that log b sin A > log a; therefore n < b sin A. 
 Hence no triangle can be found with the given elements. 
 
 Example 2. 
 
 Given, a = 14.73, b = 18.65, A = 43 27'. 
 
 Required, c, B, C.
 
 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 71 
 
 Solution. We have now 
 
 . D 6 sin A 18.65 sin 43 27' 
 
 sin B = = rj-=^ 
 
 a 14.73 
 
 log 18.65 = 1.27068, 
 
 log sin 43 27' = 9.83741 - 10, 
 
 A 
 
 log b sin A = 11.10809 - 10, 
 
 log 14.73 - 1.16820, 
 
 S 
 
 log sin B = 9.93989 - 10. 
 
 Here a > b sin A, a < b and A is acute. Hence there are two 
 triangles in this case. We find: 
 
 First solution Second solution 
 
 B = 60 32' 43". B' = 119 27' 17". 
 
 We know A = 43 27'. We know A = 43 27'. 
 
 Hence C = 180 -(A + B) Hence C" = 180 - (A + B) 
 = 76 0' 17". - 17 5' 43". 
 
 Furthermore : Furthermore : 
 
 sin C sin A . A , . sin C" sin A . ,. , 
 
 - = } and thererore , = > and therefore 
 
 c a c a 
 
 _ osin C _ 1 4.73 X si n 76 17 / r , _ a sin C _ 14.73 X sin 17 5' 43" 
 
 sin A sin 43 27' sin A sin 43 27' 
 
 log 14.73 = 1.16820 log 14.73 = 1.16820 
 
 log sin 76 17" = 9.98691 - 10 log sin 17 5' 43"= 9.46829-10 
 
 A A 
 
 11.15511 - 10 10.63049 -10 
 
 log sin 43 27' = 9.8374 1 - 10 log sin 43 27' = 9.8374 1-10 
 
 S s* 
 
 logc= 1.31770 logc'= .79908 
 
 c = 20.783 c'= 6.2963
 
 72 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 
 Figure 43 is a drawing to scale of the triangles ABC and AB'C. 
 
 Example 3. 
 
 Given, a = 24.73, b = 18.65, A = 143 27'. 
 
 Required, c, B, C. 
 
 , . . b sin A 18.65 sin 143 27' 
 
 solution, sin i) = = nA no 
 
 a 24.73 
 
 log 18.65 = 1.27068 
 
 log sin 143 27' = log sin 36 33' = 9.77490 - 10 
 
 A 
 
 log b sin A = 11.04558 - 10 
 log 24.73 = 1.39322 
 
 S 
 
 log sin B = 9.65236 - 10 
 B = 26 41' 14" 
 
 Since a > b and A is obtuse, there is only one solution and B 
 must be acute. Hence the value found from the table is the 
 only one that can be used in this case. 
 
 C = 180 - (143 27' + 26 41' 14") = 9 51' 46" 
 asinC 24.73 sin 9 51' 46" 
 
 c = 
 
 sin A sin 143 27' 
 
 log 24.73 = 1.39322 
 log sin 9 51' 46" = 9.23373 - 10 
 
 A 
 
 log a sin C = 10.62695 - 10 
 log sin 143 27' = 9.77490 - 10 
 
 S 
 
 log c = .85205 
 c= 7.1130
 
 THE LAW OF COSINES 
 
 73 
 
 80. Exercises. 
 
 1. a = 42.3, b = 57.03, A = 35 35'. Determine c, B, C. 
 
 2. c = 507.8, b = 751.3, B = 23 47'. Determine a, A, C. 
 
 3. b = 5.5, c = 4.3, C = 75 29'. Determine a, A, B. 
 
 4. a = 3.207, c = 7.831, C = 137 18'. Determine b, A, B. 
 
 5. a = 37.052, 6 = 49.312, .4 = 19 25'. Determine c, B, C. 
 
 6. c = .047031, b = .047031, C = 28 31'. Determine a, A, B. 
 
 7. An inland, known to be 75 miles wide, subtends an angle of 40 17' 
 from a point P, 40 miles distant from one extremity of the island. How far 
 is P from the other extremity of the island? 
 
 8. A flagpole, 10 feet high, subtends an angle of 2 37' from a point A. 
 If A is 200 feet from the foot of the pole, how far is it from the top? 
 
 9. Two lighthouses, M and L are 40 miles apart. At 8.30 a.m. a ship S 
 leaves M and travels at the rate of 12 miles per hour. At 11 a.m. the distance 
 between M and L subtends an angle of 33 from the ship. How far is S 
 from L at that moment? 
 
 81. The Law of Cosines. Two sides and the included angle. 
 Three sides. If three sides of a triangle are given, and also if 
 two sides and the included angle are given, the triangle is deter- 
 mined. The law of sines does not suffice, however, to calculate 
 the unknown elements in these cases. We therefore proceed to 
 derive a new relation between the sides and angles of the triangle. 
 
 Fig. 44a 
 
 In Fig. 44a, the obtuse angle B is in standard position with 
 reference to the axes indicated in the diagram. We have, there- 
 fore, in both figures: 
 
 BD = c cos B, h a = c sin B, BC = a. 
 
 Using now Theorem I of 5. we find that BD + DC + CB = 0. 
 Therefore, DC = - BD - CB = BC - BD = a - c cos B. (1) 
 Moreover b 2 = h a 2 + W" 1 = c 2 sin 2 B + L>C 2 . (2)
 
 74 THE LAW OF COSINES 
 
 Substituting (1) in (2), we find: 
 
 b 2 = c 2 sin 2 B + (c cos B a) 2 
 
 = c 2 sin 2 B -\- c 2 cos 2 5 + a 2 2 ac cos 5; 
 i.e., b 2 = e 2 + a 2 - 2 oc cos B, 
 
 We have therefore proved the following theorem: 
 
 Theorem IV. The square of one side of a triangle is equal to the 
 sum of the squares of the other two sides diminished by twice the 
 continued product of these two sides and the cosine of the angle 
 included by them. (Law op Cosines.) 
 
 By dropping perpendiculars from each of the other two vertices 
 of triangle ABC and proceeding in an exactly similar manner, we 
 obtain two formulae analogous to the one derived above, viz., 
 
 c 2 = a 2 -f b 2 2 ab cos C and a 2 b 2 + c 2 2 be cos A. 
 
 In the form in which they are here given, these formulae enable 
 us to calculate the third side of a triangle of which two sides and 
 the included angle are known. If we solve them for the cosines 
 of the angles, we obtain: 
 
 . tf+c 2 a 2 c 2 +o 2 /> 2 a 2 -\-b 2 c 2 ._. 
 
 cos A = ; , cos B = - , cos C= ^. (3) 
 
 2 be 2 ca 2 ab 
 
 in which form they are well adapted to determining the angles of a 
 triangle whose sides are given. 
 
 We observe that none of the formulae derived in the present 
 section are suited to logarithmic calculation. For this reason 
 they are useful only for computations which involve short numbers. 
 
 Example 1. 
 
 Given, a = 14, b = 27, C = 35. 
 
 Required, c, A, B. 
 
 Solution. By means of Theorem IV we find: 
 
 C 2 = a 2 + ?,2 _ 2 ab cos C = 14- + 27 2 - 2 14 27 cos 35; 
 
 C 2 = 196 + 729 - 756 X .81915 = 925 - 619.27740 = 305.72260. 
 
 c = V 305.72260 = 17.484.
 
 THE LAW OF COSINES 75 
 
 To complete the calculation, we use the law of sines, and find: 
 
 . a sin C , . b sin C 
 
 sin A = j and sin B = 
 
 c c 
 
 log 14 = 1.14613 log 27 = 1.43136 
 
 log sin 35 = 9.75859 - 10 log sin 35 = 9.75859 - 10 
 
 -A -A 
 
 S 
 
 10.90472 - 10 11.18995 - 10 
 
 log 17.484 = 1.24264 log 17.484 = 1.24264 
 
 log sin A = 9.66208 - 10 log sin B = 9.94731 - 10 
 
 A = 27 20' 28" or 152 39' 32" and B = 62 20' 34" or 1 17 39' 26". 
 
 Since a < c, we must have A < C; hence A = 27 20' 28". 
 
 Since b > a and b > c, we must have B > A and B > C; both 
 values of B satisfy this condition; it is clear, however, that with 
 the smaller value of B, the condition that A + B + C = 180 
 would not be satisfied; therefore B = 117 39' 26". 
 
 Example 2. 
 
 Given, a = 10, b = 15, c = 19. 
 
 Required. A, B, C. 
 
 Solution. Using formula (3), we find: 
 
 r a 2 + 6 2 - & 1 00 + 225 - 361 -36 tonnn 
 
 cosC = -2flr ^ = m=- J20oa 
 
 Since cos C is negative, Z C is obtuse and cos (180 C) = 
 -cos C = .12000. 
 We find from the tables, that 180 - C = 83 6' 29"; and hence 
 C = 96 53' 31". 
 Furthermore, 
 
 c 2 + a 2 - b 2 361 + 100 - 225 236 __ 
 
 2 ca 
 
 380 
 
 
 380 
 
 .UiilUfJ, 
 
 therefore 
 
 B = 51 36' 27". 
 
 
 
 
 And finally, 
 
 
 
 
 
 h 1 + ( .2_ 
 
 a 2 225 + 361 - 
 
 100 
 
 486 
 ~ 570 " 
 
 .85263; 
 
 ( "- 1 2 be 
 
 570 
 
 
 whence we obtain A - 31 30' 4". 
 
 Check. A + B + C = 180 0' 2".
 
 76 EXERCISES 
 
 82. Exercises. 
 
 1. a = 120, b = 150, C = 60. Determine c, A, B. 
 
 2. p = 1.3, q = 1.4, r = 1.5. Determine P, Q, R. 
 
 3. o = 17, 6 = 15, c = 29. Determine A, B, C. 
 
 4. r = .45, s = .78, T = 45. Determine t, R, S. 
 
 5. To determine the width of a lake, the distances of its extreme points 
 A and B from a point P and the angle subtended by AB at P are measured. 
 It is found that AP = 750 feet, BP = 600 feet, and LP = 32. 
 
 6. It is desired to make a triangle out of sticks that are 5, 8, and 9 inches 
 long. What angle should the first two of these sticks make, in order that 
 the third one may be just long enough to join their free ends? 
 
 7. Prove that a- = b 2 + c 2 2 be cos A. 
 
 8. Prove that a 2 + b 2 + c 2 = 2 ab cos C + 2 be cos A + 2 ca cos B. 
 
 a 2 + b 2 - c 2 
 
 9. Prove that cos C = 
 
 2 ab 
 
 83. Summary and critique. In the preceding sections we have 
 learned to calculate the unknown parts of triangles of which are 
 given (a) one side and two angles (77); (6) two sides and an 
 angle opposite one of the sides (79) ; (c) two sides and the included 
 angle (81) ; or (d) three sides (81) . We know, moreover, from the 
 study of plane geometry, that if a triangle is to be determined 
 by sides and angles, then the given elements must form one of 
 the four sets (a), (6), (c) or (d) enumerated above. Hence the 
 general problem of "solving a triangle" has been solved in sec- 
 tions 77-81, in so far as it relates to triangles determined by 
 means of sides and angles. There arc 1 however two criticisms to 
 be made of the theory developed so far, viz. : 
 
 (1) The methods developed in 81 for cases (c) and (d) are not 
 suited to the use of logarithms and are not very useful, therefore, 
 in problems involving long numbers. 
 
 (2) There are no convenient methods for accurately checking 
 the calculations in cases (a), (b) and (c). 
 
 In order to meet these criticisms some further relations between 
 the sides and angles of a triangle will now be derived. 
 
 84. The law of tangents. From Theorem I, we conclude that 
 
 sin A _ a 
 
 siiT# ~ b ' ( j
 
 THE LAW OF TANGENTS 77 
 
 Adding 1 to both sides of this equation, we obtain: 
 
 sin A -f sin B _ a -f- b 
 sin B ~b~' 
 
 subtracting 1 from both sides of equation (1) gives us: 
 
 sin A sin B _ a b 
 sin B b 
 
 (2) 
 
 (3) 
 
 If we divide the sides of (2) by the corresponding sides of (3), we 
 find: 
 
 sin A + sin B a + b 
 
 sin A sin B a b 
 
 (4) 
 
 We now make use of formulae (13) and (14) of 74 to factor 
 respectively the numerator and denominator on the left hand 
 side of (4); we also divide the numerator and denominator by 2 
 and obtain: 
 
 a + b _ sin \ (A + B) cos \ (A - B ) 
 
 a - b ~ sin \ (A - B) cos \ (A + B) 
 
 tan %(A+B) cot \ C 
 
 tan \{A -B) tan |(A - B) 
 To justify the last step we observe that A -J- B + C = 180. 
 
 Hence 
 
 \{A + B) = 90 - \ C. (5) 
 
 .-. tan | (A + B) = tan (90 - \ C) = cot \ C (see 60, (4)). 
 The result is most conveniently written in the following form, 
 
 tan | {A - R) = ^- cot * C ' 
 
 a -\- b 
 
 which formula is expressed in the following theorem: 
 
 Theokkm V. The tangent of one half the difference of two angles 
 of a triangle is equal to the quotient of the difference of the sides 
 opposite these angles by their sum, multiplied by the cotangent of 
 one half the angle included by them. (Law of Tangents.)
 
 78 MOLLWEIDE'S EQUATIONS 
 
 85. Mollweide's Equations. If we multiply both sides of equa- 
 tions (2) and (3) of 84 by the corresponding sides of the equation 
 
 sin B b c , , 
 
 . 7= = - , we nnd that 
 sin C c 
 
 sin A + sin B _ a -f- b 
 
 sin C c 
 
 and 
 
 sin A sin B a b 
 
 sin C 
 
 (1) 
 
 (2) 
 
 We factor the numerators on the left hand sides of these equa- 
 tions by means of Formulae (13) and (14) of 74. The denomina- 
 tors we change by writing sin C = 2 sin \ C cos \ C, which is a 
 consequence of formula (8) of 74. In this way equations (1) and 
 (2) will assume the following form: 
 
 (3) 
 (4) 
 
 sin j (A + B) cos \ (A - B) _ a + b 
 
 sin \ C cos \ C c 
 
 and 
 
 sin % (A - B) cos % (A + B) a -b 
 
 sin \ C cos | C c 
 
 Moreover, from equation (5) of 84, it follows that 
 sin i (A + B) = cos \ C 
 and cos \ (A + B) = sin \ C, (see 19 and 55). 
 
 Consequently, equations (3) and (4) may be simplified to the 
 final form: 
 
 a + b _ cos \ (A - B) a - b _ sin \ ( A - li) 
 
 -. z 7; and ^ 77 
 
 c sin C c cos C 
 
 These equations are known as Mollweide's equations. 
 
 86. Exercises. 
 
 1. Prove: tan \ (C-A) = C -^-- cot \ B. 
 
 c + a 
 
 cos -J yl a 
 
 3. Derive a proof of the law of tangents from Mollweide's equations.
 
 TWO SIDES AND THE INCLUDED ANGLE 79 
 
 87. Two sides and the included angle. The law of tangents 
 enables us to meet the criticism brought forward in 83 with refer- 
 ence to case (c). Moreover, Mollweide's equations are well 
 adapted to serve for checking the calculations in cases (a), (6) 
 and (c), because they involve all the sides and all the angles of 
 the triangle. 
 
 Example. 
 
 Given, a = .4503, b = .7831, C = 43 48'. 
 Required, c, A, B. 
 
 Solution. We use the law of tangents to determine angles A 
 and B: 
 
 tan i( p-A)-*= cot | C = jfH COt 21 34 '- 
 log .3328 - 9.52218-10 
 log 1.2334 = .09110 
 
 S 
 
 9.43108-10 
 log cot 21 54' = .39578 
 
 A 
 
 log tan \ (B - A) = 9.82686-10 
 \ (B - A) =33 52' 11", 
 but, 
 
 \ (+A) = 90-i C = 90-21 54' = 68 6' 
 
 A and S. 
 
 Therefore, 
 
 Z B = h(B -i- A) + | ( - A) = 101 58' 11" 
 and 
 
 Zi = f(5 + i)-|(B-4)= 34 13' 49". 
 
 Furthermore: 
 
 _ a sin C _ .4503 X sin 43 48 r 
 
 sin A ~ sin 34 13' 49" 
 
 log .4503 = 9.65350-10 
 
 log sin 43 48' = 9.84020-10 
 A 
 
 19.49370-20 
 log sin 34 13' 49" = 9.75014-10 
 
 logc = 9.74356-10 
 c = .5.5406
 
 80 THE HALF-ANGLE FORMULAE 
 
 Ckeck. aM(B -C) = L=i. 
 
 cos f A a 
 
 \{B-C) = 29 5' 6", I A = 17 6' 55", 6 - c = .22904 
 
 log sin H#-C) = 9.68673 - 10 log (6 - c) = 9.35992 - 10 
 
 log cos A = 9.98032 - 10 log a = 9.65350 - 10 
 
 S S 
 
 9.70641 - 10 9.70642 - 10 
 
 88. Exercises. 
 
 1. c = 27.04, b = 84.31, A = 112 44'. Determine a, B, C. 
 
 2. a = 3152, c = 4281, B = 88 27'. Determine 6, A, C. 
 
 3. p = .0432, <? = .0586, R = 47 36'. Determine r, P, Q. 
 
 4. x = .8132, z = .5817, F = 120. Determine y, X, Z. 
 
 6. A hill slopes at an angle of 17. A point P is 39.3 feet up the hillside; 
 a point Q is in the plane, 173.5 feet from the foot of the hill and in the same 
 vertical plane as P. How far is P from Q? 
 
 6. The distance from the earth to the sun is approximately 92.9 million 
 miles, the distance from the earth to the moon approximately 239,000 miles. 
 What is the distance from the sun to the moon at a moment when the line 
 sun-moon subtends at the earth an angle of 24 36'? 
 
 89. The half-angle formulae for the angles of a triangle. We 
 
 start with the first formula for tan \ A given in 70 and multiply 
 numerator and denominator of the fraction under the radical 
 sign by 2 be. In this way we find : 
 
 , , . /l cos A . /2 be 2 be cos A /1N 
 
 tan \ A = \/ rVc ^ A = \/ 2bc + 2bc cosA ' (D 
 
 If we multiply both sides of formula (3) of 81 by 2 be, we obtain: 
 
 2 be cos A = b 2 + c 2 - a 1 . (2) 
 
 We substitute (2) in (1); this gives us: 
 
 i /2 bc + a 2 - b 2 - c 2 = /a 2 - (ft 2 -~2~bc + c 2 ) 
 
 tan 2 A ---- y 2 bc + f)2 + c , _ -., y 2 5c + _ fl2 
 
 v/ 
 
 (q + 6 - c) (a - b + c) 
 
 (6 + c + a) (6 + c -a)' W
 
 THREE SIDES 81 
 
 Now, we introduce the abbreviation 2 s for the perimeter of 
 the triangle; i.e., a + b + c = 2 s. Subtracting in turn 2 a, 2 b 
 and 2 c from both sides of this equality, we obtain : 
 a-\-b-{-c = 2 (s a), 
 a - 6 + c = 2 (s - 6), 
 and a 4- 6 c = 2 (s c). 
 
 These expressions are now substituted in equation (3), which 
 then becomes: 
 
 s (s a) s a s * a 
 
 where P = \ / (s ~ q) (s ~ b) {s ~ c) 
 
 In a similar way, we obtain the analogous formulae: 
 
 tan|B = -^-r and tan | C = 2- . 
 s b s c 
 
 These formulae will be referred to as the half-angle formulae 
 for the triangle. 
 
 90. Three sides. The results of the preceding section make it 
 possible to remove the one remaining point contained in the 
 criticism of 83, viz., to devise a treatment for case (d) adapted to 
 calculation by logarithms. 
 
 Example. 
 
 Given, a 
 
 = 14.931, b 
 
 = 16.902, < 
 
 : = 24.315. 
 
 
 Required. 
 Solution. 
 
 A, B, C. 
 
 a = 
 b = 
 
 14.931 
 16.902 
 
 
 
 
 c = 
 
 2 s = 
 
 24.315 
 
 A 
 
 56.148 and therefore s = 
 
 28.074; 
 
 s a = 
 
 13.143 
 
 and 
 
 log a) = 
 
 1.11870 
 
 s -b = 
 
 11.172 
 
 and 
 
 log (.s b) = 
 
 1.04813 
 
 s c = 
 
 3.759 
 
 A* 
 28.074 
 
 and 
 
 log (s C) 
 log s = 
 
 .57507 
 A 
 2.74190 
 1.44830 
 
 1.29360 
 
 
 
 
 
 * It is well to use this chock upon the calculation of the quantities s, 
 a, s b, and s c before proceeding with the rest of the work.
 
 82 INSCRIBED AND CIRCUMSCRIBED CIRCLES 
 
 Since p = y/^ZSZS5Z5 1 we haye lQg v = M68Q 
 
 Subtracting log (s a), log (s b) and log (s c) in turn from 
 log p, we find by means of the half -angle formulae: 
 
 log tan A/2 = 9.52810 - 10, log tan B/2 = 9.59867 - 10, 
 
 log tan C/2 = .07173; 
 .-. A/2 = 18 38' 33", B/2 = 21 38' 52", C/2 = 49 42' 37", 
 
 and A = 37 17' 6", = 43 17' 44", C = 99 25' 14". 
 Check. A + B + C = 180 0' 4". 
 
 91. Exercises. 
 
 1. a = .96834, 6 = .94572, c = .95902. Determine A, B, C. 
 
 2. a = 453.67, b = 112.34, c = 369.85. Determine A, B, C. 
 
 3. x = 1.004, ?/ = 1.705, z = 1.526. Determine X, Y, Z. 
 
 4. a = 4500, b = 5400, c = 6300. Determine A, B, C. 
 
 5. A triangular piece of land is to be staked off, so that its sides measure, 
 respectively, 73.84 rods, 68.701 rods, and 32.503 rods. How may this be 
 done? 
 
 6. Two vessels leave the same harbor at the same moment, both going 
 at the rate of 12 miles per hour. After 2| hours, the vessels are 24 miles 
 apart. If one of the vessels was sailing in a due easterly course, in what 
 direction was the other vessel going? 
 
 92. Inscribed and circumscribed circles. Area. The results 
 of the preceding articles enable us to derive other important 
 relations, viz.: 
 
 (1) A formula for the radius of the inscribed circle (see Fig. 45). 
 
 The points of tangency of the inscribed circle of radius r divide 
 the total perimeter of the triangle into six parts, which are equal 
 in pairs. Hence the sum of the segments of any set of three, of
 
 INSCRIBED AND CIRCUMSCRIBED CIRCLES 83 
 
 which no two are equal, must be equal to s, one half of the peri- 
 meter; e.g., 
 
 AF + BD + DC = s or AF + a = s; 
 
 hence AF = EA = s a. 
 
 In similar manner we show that FB = BD = s b and DC 
 = CE = s - c. 
 Moreover A AOF is a right-angled triangle (Why?) and LOAF 
 
 = A/2. (Why?) 
 
 Therefore tan 
 
 2 AF s - a 
 
 On the other hand, we have from the half-angle formulae for 
 
 the angles of a triangle : tan = - . From this we conclude 
 
 2 s a 
 
 that r = p, i.e.: 
 
 Theorem VI. The radius of the Inscribed circle of a triangle ABC 
 Is given by the formula: 
 
 -V 
 
 (s - a) (s - b) (s - c) 
 
 where s designates one half of the perimeter of the triangle. 
 
 (2) A second formula for the area of a triangle. 
 We see from Fig. 45, that 
 
 A ABC = A AOB + A BOC + A COA 
 
 /o i /o , . /o a + b + c 
 = r c/2 -f- ?' a/2 + r 0/2 = r = rs. 
 
 Using the formula for the radius of the inscribed circle found 
 above we get: 
 
 Theorem VII. The area of a triangle ABC Is given by the formula: 
 
 A = V s (s - a) (s - b) (s - c). 
 
 This formula, known as Hero's formula, expresses the area of 
 the triangle in terms of the sides only; it is well known from the 
 study of plane geometry. Together with the formula derived
 
 84 SUMMARY OF RESULTS OF CHAPTER VII 
 
 in 76, it suffices to determine the area of the triangle in all cases 
 which we have considered. 
 
 (3) A formula for the radius R of the circumscribed circle. 
 
 We draw (see Fig. 46) the diameter AOD and connect C with D. 
 
 Fig. 46 
 
 Then Z ACD is a right angle (Why?) and Z ADC = Z ABC 
 
 AC b 
 = B (Why?). Hence sin B = sin ADC = .-r. = ?m From this 
 J ' AD 2 li 
 
 it follows hat R = ^. ^. By means of the law of sines, we 
 2 sin B 
 
 derive from this wo other expressions for R, so that we can now 
 
 state the following theorem: 
 
 Theorem VIII. The radius of the circumscribed circle of a triangle 
 ABC is given by the formulae: 
 
 R = 
 
 2 sin A 1 sin li 2 sin C 
 
 93. Summary of the results of Chapter VII. 
 
 i. Law of sines: . = - 75 = . ^ see 76. 
 
 sm A sin B sin C 
 
 2. Law of cosines: a 2 = b 2 + c 2 2 6c cos A, see 81. 
 
 52 = c 2 + ft2 _ 2 m cos . 
 c2 = a 2 + 52 _ 2 a 6 cos C.
 
 SUMMARY OF RESULTS OF CHAPTER VII 85 
 
 3. Law of tangents: tan h (A B) = 7- cot h C, see 84. 
 
 a + b 
 
 tan \ (B - C) = ^~ cot \ A. 
 
 tan|(C - A) = ^cot^B. 
 c + a 
 
 4. Half -angle formulae for the triangle: 
 
 see 89. 
 
 tan \ A 
 
 V 
 sa 
 
 tan| 
 
 V 
 s-b 
 
 tan \ C 
 
 = p 
 s c 
 
 where p 
 
 {s a) (s b) (s c) 
 
 5. Mollweide's equations: 
 
 a + b cos h (A - B) a-b sin \ (A - B) 
 
 c 
 
 siniC 
 
 b + c 
 
 cos h{B-C) 
 
 a 
 
 sin ^ A 
 
 c + a 
 
 cosHC-^) 
 
 c 
 
 cos C 
 
 b -c 
 
 sin J (5 - C) 
 
 a 
 
 cos h A 
 
 c a 
 
 sin | (C - A) 
 
 > see 85. 
 
 b sin 7* # b cos 1 5 
 
 6. The area of triangle: 
 
 a. A = \ ab sin C = | 6c sin yl = | ca sin 5, see 76. 
 
 b. A = Vs (s - a) (s - 6) (s c), see 92. 
 
 7. The radius of the inscribed circle: 
 
 ,. = J (s-a)(8-b)(8-c) 
 V s 
 
 8. The radius of the circumscribed circle 
 
 see 92. 
 
 lii n . . r> ' n n si ' See i/Z. 
 
 2 sin yl 2 sin # 2 sin C
 
 86 MISCELLANEOUS EXERCISES AND APPLICATIONS 
 9. Methods for solving triangles: 
 
 
 Solved by use of: 
 
 
 Case 
 
 
 Checked by use of: 
 
 
 
 Without logs 
 
 With logs 
 
 
 Two angles and one 
 
 Formulae 1, 6a 
 
 Formulae 1, 6a 
 
 Formulae 5 
 
 side. 
 
 
 
 
 Two sides and an 
 
 Formulae 1, 6a 
 
 Formulae 1, 6a 
 
 Formulae 5 
 
 angle opposite one 
 
 
 
 
 of them. 
 
 
 
 
 Two sides and the 
 
 Formulae 2, 6a 
 
 Formulae 3, 6a 
 
 Formulae 5 
 
 included angle. 
 
 
 
 
 Three sides. 
 
 Formulae 2, 6b 
 
 Formulae 4, 6b 
 
 LA + LB + LC 
 = 180 
 
 94. Exercises. 
 
 1. Prove the law of sines by means of tne method indicated in 92 (3). 
 
 2. Show, in Figure 45, that BD + CE + EA = s. 
 
 3. Prove, in Figure 45, that CE = DC = s - c. 
 
 abc 
 
 4. Use 92, (3) to show that the area of a triangle is equal to 
 
 4/r 
 
 Determine the area, and the radius of the circumscribed circle for each of 
 the following triangles: 
 
 5. a = .473, 6 = .586, C = 23 47' 12". 
 
 6. x = 3.045, Y = 47 28', Z = 65 34'. 
 
 7. a = 41.35, b = 36.78, A = 35 27'. 
 
 8. c = 632, a = 571, B = 30. 
 
 Determine the area, and the radius of the inscribed circle for each of the 
 following triangles: 
 
 9. a = 40.37, b = 31.56, c = 27.08. 
 
 10. a = .971, b = .506, c = .683. 
 
 11. a = 437, c = 856, C = 38 41'. 
 
 12. a = .0456, b = .0731, C = 74 26'. 
 
 95. Miscellaneous exercises on Chapter VII and applications. 
 
 1. The area of a triangular piece of land is 43 acres. One side measures 
 440 yards, and the angle at one extremity of this side is 43. What must the 
 remaining sides and angles be? (1 acre = 4840 sq. yds.)
 
 MISCELLANEOUS EXERCISES AND APPLICATIONS 87 
 
 2. A lighthouse is observed N 15 W from a vessel which is sailing 15 
 miles an hour in a due northerly course. Half an hour later the bearing of 
 the same lighthouse is N 37 W. How far is the lighthouse from the second 
 position of the vessel and how long will it be before the lighthouse is sighted 
 due West? 
 
 3. From the top of a mountain the angles of depression of two consecu- 
 tive milestones in the same vertical plane with the top of the mountain are 
 10 and 15. How high is the mountain ? 
 
 4. A forester observes that the angle of elevation of an observation tower 
 from a point P is 5; after walking towards the tower along a horizontal road 
 for a distance of 500 feet, he observes that the angle of elevation has changed 
 to 35. How much farther will he have to go to reach the foot of the tower? 
 
 Calculate the unknown parts and the area of each of the triangles indicated 
 in Exs. 5-7: 
 
 5. a = 47.032, b = 35.614, A = 27 45' 16". 
 
 6. a = 3, b = 5, c = 7. 
 
 7. a = 23. c = 17, C = 42 23'. 
 
 Determine the area and the radii of the inscribed and circumscribed 
 circles in each of the triangles indicated in Exs. 8-10: 
 
 8. a = 256, C = 17 13', B = 45 16'. 
 
 9. b = 2.25, c = 1.75, A = 54. 
 
 10. a = 15, b = 17, c = 25. 
 
 11. The ratio of the lengths of two sides of a triangle is 5 : 8, the included 
 angle is 35. Determine the other angles of the triangle. 
 
 12. A wireless tower is built on the edge of a cliff. From a boat at sea, 
 the angle of elevation of the top of the tower is 30. After rowing towards 
 the shore for a distance of 400 feet it is found that the angles of elevation 
 of the bottom and top of the tower are 45 and 57 respectively. How high 
 is the cliff and how high is the tower? 
 
 200 ft 
 
 13. To determine the distance between two points A and B, situated on 
 the surface of a lake, two points C and I) an; selected in such a manner that 
 both .4 and B are visible from C and 1) (see Fig. 47). It is found that VI)
 
 88 MISCELLANEOUS EXERCISES AND APPLICATIONS 
 
 = 200 feet and that /ACB = 33, / BCD = 54, ^ADB = 40 and 
 
 snr^A -o ,, . An 200 sin 65 . .. . __ 200 sin 75 
 
 /CD A = 6o . Show that AC = : ^^ , and that BC = : ^ 5 . 
 
 sin 28 sin 21 
 
 Then calculate the distance AB. 
 
 14. A vessel is sailing due east at the rate of 20 miles per hour. At 10 
 a.m. a lighthouse L is bearing N 10 W, while a second lighthouse M is bear- 
 ing N 40 E; at 2 p.m. the bearings of L and M are N 50 W and N 5 E 
 respectively. How far are L and M apart? 
 
 15. Determine also from the observations recorded in Problem 14 the 
 direction of the line from L to M. 
 
 16. To measure the height of a mountain AF, above a horizontal plane P, 
 we measure the distance between two stations B and C (see Fig. 48), so selected 
 
 Fig. 48 
 
 that at least one of them, say B, lies in the plane P, that each is visible from 
 the other and that A is visible from both. The angles ACB, ABC and the 
 angle of elevation a of .4 as seen from B are measured. It is found that 
 BC = 500 feet, /.ACB = 85 25', I ABC = 84 33' and a = 40 17'. Deter- 
 mine the height of the mountain above the plane P. 
 
 17. The angle of elevation of a church steeple T from a point R is 17 25'. 
 At a point S, 250 feet from R, the line TR subtends an angle of 73 47', while 
 from R the line TS subtends an angle of 65 3 8'. Determine the height of T 
 above the horizontal plane through R. 
 
 18. A flag staff on top of a monument subtends an angle of 3 at a point 
 P, 400 feet above the ground and at a horizontal distance of 300 feet from 
 the foot of the monument. From the same point P the monument itself sub- 
 tends an angle of 43. Determine the height of the flag staff and the height 
 of the monument. 
 
 19. One side of a triangle is 75 feet and the angle opposite this side is 34. 
 The sum of the other sides is 125 feet. Determine all the sides and angles of 
 this triangle. 
 
 20. A building 20 feet high is surmounted by a steeple 30 feet high. How 
 far from the foot of the building must an observer stand in order that the 
 building and steeple may subtend the same angle at his eye, which is 5 feet 
 above the grtuind? 
 
 21. A pole 10 feet tall is divided into two parts, a lower part of 6 feet and 
 an upper part of 4 feet. Frtnn a point P. 7 feet above the bottom of the pole,
 
 MISCELLANEOUS EXERCISES AND APPLICATIONS 
 
 89 
 
 these two parts subtend the same angle, the pole being held vertically. How 
 far is the pole from P? 
 
 22. The angle of elevation of a church steeple from a point A, due south 
 of it, is 27; and from a point B, due west of the steeple, and in the same hori- 
 zontal plane as A, the angle of elevation is 35. Moreover the distance AB 
 is 150 yards. Determine the height of the steeple above the plane AB. 
 
 23. From a point A the angle of elevation of the top T of a flagpole which 
 stands on top of a building is 37 47'. From a point B, 100 feet nearer to the 
 building and lying in the vertical plane through T and A, the angle of eleva- 
 tion of T is 47 32'. How high is T above the ground? 
 
 24. The angle of elevation of the bottom of the flagpole described in 
 Problem 23, as seen from B, is 45 25'. Determine the height of the pole. 
 
 25. To determine the height XY of a wireless-tower X above a horizontal 
 plane P two points A and B are selected in this plane P (see Fig. 49). The 
 
 Fig. 49 
 
 angles XAY, XBY, YAB and ABY and the distance AB are measured. It 
 is found that Z XAY = 67, Z XBY = 58 3'. 1, Z YAB = 27, Z YBA = 1S C 
 and AB = 81.6 feet. Determine the height XY and check the calculations.
 
 CHAPTER VIII 
 
 INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONO- 
 METRIC EQUATIONS 
 
 96. Inverse functions. The concept function which we have 
 used in preceding chapters will now be studied in somewhat 
 greater detail. The concept may be denned for our purpose in 
 the following way: 
 
 Definition I. If a relation between two variables, x and y, is given 
 in such a way that to every value of either there correspond one or 
 more values of the other, then .t is a function of v, and y is a function 
 
 of X. 
 
 If these two functional dependences of y upon x and of x upon 
 y are written down explicitly, we obtain two functions, one in 
 terms of x, yielding y, the other in terms of y, yielding x. In 
 the first of these functions x is the independent variable and y the 
 dependent variable; in the second y is the independent variable 
 and x the dependent variable. Two such functions which result 
 from the same relation between the two variables are called 
 inverse with respect to each other. This may be expressed as 
 follows : 
 
 Definition II. If a relation between two variables, x and y, be 
 solved in turn for r in terms of y and for y in terms of .r, we obtain 
 two functions which form a pair of inverse functions. 
 
 The important connections between them are recognized more 
 readily if one letter be used for the independent variable in both 
 functions and another letter for the dependent variable in both 
 functions. 
 
 Example 1. The relation between the variables .r and y de- 
 termined by the equation 
 
 3x + 2?y-7 = (1) 
 gives rise to the functions 
 
 7 3x /n . 7 2 y ,_. 
 
 y = - (2) and x = ^ "- (3). 
 
 90
 
 INVERSE FUNCTIONS 
 
 91 
 
 The graphical representation of the relation between x and y 
 can readily be obtained from equations (2) or (3); in either case 
 we will obtain the straight line of Figure 50. 
 
 )\ 
 
 J 
 
 
 3 
 
 \ 
 
 
 2 
 
 - \ 
 
 
 1 
 
 
 
 O 
 
 1 2 2H 3 
 
 
 Fig. 50 
 
 Denoting the independent variable by t and using u to denote 
 the dependent variable in both cases, equations (2) and (3) will 
 become 
 
 u 
 
 Zt , 7-2* 
 
 and u ~ 
 
 3? 
 
 <u 
 
 
 
 
 21 s 
 
 \ u 
 
 7-3 t 
 
 
 
 2 
 
 
 
 
 
 1 
 
 1 
 
 
 <T 
 
 > T 
 
 O 
 
 1 
 
 
 2\ 3 3fS. 4 
 
 
 Fig. 51 
 
 Here we have a pair of inverse functions, whose graphs are repre- 
 sented in Figure 51.
 
 92 
 
 GRAPHS OF INVERSE FUNCTIONS 
 
 Example 2. The relation y 2 x = gives rise to the pair of 
 inverse functions 
 
 u = t 2 and u = vt, 
 whose graphs are given in Figure 52. 
 
 u 
 
 \ 
 
 Fig. 52 
 
 In the first function of this pair, one 'positive value of u corre- 
 sponds to every value of t; but in the other function two values of u 
 correspond to every positive value of t, and no value of u to negative 
 values of t. The function u t 2 is a single-valued function of t, 
 defined for all values of t; the inverse function u = x^t is a two- 
 valued function of t defined for positive values of t only. It may 
 happen that the inverse function of a single-valued function is 
 3-valued, or, in general, multiple-valued. 
 
 97. Graphs of inverse functions. Let the functions u = f (t) 
 and u = g (t) be a pair of inverse functions whose graphs are the 
 curves drawn in Figure 53. If the point P (a, b) belongs to the 
 graph of u = f (t), then the point Q (b, a) must be on the graph of 
 u = g {t), because the two functions may be thought of as having 
 been obtained from one and the same equation between x and y, 
 x and y being replaced by t and u in one case, and by u and t in 
 the other. Now it is readily seen that two such points P (a, b) 
 and Q (6, a) are symmetrically situated with respect to the 45 
 line MN (see Fig. 54), i.e., that MN is the perpendicular bisector 
 of the line PQ. For As ROP and QOS are congruent, (Why?)
 
 GRAPHS OF INVERSE FUNCTIONS 
 
 93 
 
 Hence PO = QO and Z ROP = Z SOQ. Consequently Z POT 
 = Z QOT (Why?) and A POT and A QOT are congruent. 
 (Why?) Therefore PT = QT and Z PTO = Z QTO = 90, which 
 
 *-r 
 
 Fig. 53 
 
 I 
 s 
 a 
 
 /! X 
 
 > r 
 
 / 
 
 O a ft 
 
 
 Fig. 54 
 
 was to be proved. Consequently from points A, B on the graph 
 of any function, points A', B' , ... on the graph of the inverse 
 function may be obtained by determining A', B', ... as points 
 which are symmetrically situated with .1, B, . . . with respect 
 to the 45 line MN, i.e., by reflecting the points A, B, . . . in
 
 94 
 
 THE INVERSE SINE FUNCTION 
 
 MN as a mirror*. These results may be summarized in the fol- 
 lowing theorem: 
 
 Theorem I. If u =/ (t) and u = g (t) are a pair of Inverse func- 
 tions the graph of either function may be obtained by reflecting the 
 graph of the other function in the 45 line as a mirror. 
 
 98. Exercises. 
 
 Determine the inverse function associated with each of the following func 
 tions: 
 
 1. u = 4 t - 3. 
 
 < 2 /3. 
 
 3. u = logio t. 
 
 4. u = 
 
 Construct the graphs of the following pairs of inverse functions: 
 5. u = Vt -4, u = t 1 + 4. 6. u = t/3, u = 3 t. 
 
 7 At -7 
 
 7. u = V9 - t 2 , u = Vo - t 2 . 
 
 8. u = V 
 
 bt -4' 
 
 5< -3 
 
 99. The inverse sine function. The equation y = sin x estab- 
 lishes a relation between the variables x and y, represented 
 graphically by the sine curve (see dashed curve in Fig. 55). For 
 
 Fig. 55 
 
 every value of x there is one corresponding value of y. If the 
 ('filiation be solved for x in terms of y, and x and y be replaced by 
 u and / respectively, we obtain the inverse sine function; it is 
 expressed in the form u = arc sin t, which is read "u is the angle 
 * See footnote on page 45.
 
 THE OTHER INVERSE TRIGONOMETRIC FUNCTIONS 95 
 
 whose sine is t." The graph of the inverse sine function of t, i.e., 
 of arc sin t, may be obtained, in virtue of Theorem I, by reflecting 
 the sine curve in the 45 line as a mirror. In this way we obtain 
 the full-drawn curve of Figure 55. 
 
 The inverse sine function of t cannot be expressed in a simple 
 manner in terms of algebraic or trigonometric functions of t. It 
 must be regarded as an entirely new function, defined as the in- 
 verse of the sine function. From this definition its properties 
 will be derived by the aid of its graph. 
 
 We observe that the function u = arc sin t gives an indefinitely 
 large number of values of u for every value of t, which lies between 
 1 and 1, and that it gives no value of u at all for values of t 
 which lie outside the range ( 1, 1). The inverse sine function 
 is an infinitely multiple-valued function, defined only for values of 
 t between 1 and 1. 
 
 So, e.g., arc sin \ = 30, 150, 510, -210, . . . just as sin 30 
 = sin 150 = sin 510 - sin (-210) = = \. 
 
 Definition III. The symbol Arc sin t, called the principal value 
 of arc sin t, is used to designate the numerically smallest angle 
 whose sine is equal to t. 
 
 For example Arc sin \ = 30 = ir/6. 
 
 It is clear that Arc sin t is a single-valued function of t, defined 
 for values of t between 1 and +1. Its graph is given by the 
 heavily drawn portion of the graph of arc sin t in Figure 55. 
 
 When t is between and 1, Arc sin t will be between and -; 
 
 when t is between 1 and 0, Arc sin t will be between - and 0. 
 
 100. The other inverse trigonometric functions. In a similar 
 manner the equations u = arc cos t, u = arc tan t, u = arc cot t, 
 u = arc sec t and u = arc cosec t represent the inverse functions 
 associated with the remaining trigonometric functions cos t, tan t, 
 cot t, sec t, and cosec / respectively. These equations are read 
 "u is the angle whose cosine is equal to t," etc. The graphs of 
 the inverse trigonometric functions are obtained from the graphs 
 of the associated trigonometric functions by the use of Theorem I, 
 i.e., by reflecting the latter in the 45 line as a mirror. In 
 Figures 56 and 57, the graphs of the functions arc cos / and arc 
 tan / arc indicated by the full-drawn curves. We observe thai 
 both of these functions are infinitely multiple-valued for those
 
 96 THE OTHER INVERSE TRIGONOMETRIC FUNCTIONS 
 
 
 / 
 
 u 
 
 
 M=arc cos 
 
 \ 
 
 s2 
 
 
 s 
 
 / 
 
 
 /v 
 
 \ 
 \ 
 
 -7T AVi-1 
 \ ' 2 
 
 
 A 
 
 
 -i 
 
 
 
 
 -IT 
 
 2 
 
 
 
 V 
 
 -7T 
 
 
 Fig. 56 
 
 u arc tan t 
 
 Fig. 57
 
 EXERCISES 97 
 
 values of the independent variable for which they exist at all; i.e., 
 for every value of t between 1 and 1 there is an infinitely large 
 number of values of arc cos t; while for every value of t there is 
 an infinitely large number of values of arc tan t. 
 
 If cot u = t, then tan u = 1/7; hence arc cot t = arc tan 1/t, 
 i.e., an angle whose cotangent is equal to t is equal to an angle 
 whose tangent is equal to 1/t. For an analogous reason we have 
 arc sec t arc cos 1/t and arc cosec t = arc sin 1/t. Since the 
 functions arc cot t, arc sec t, and arc cosec t are so readily expres- 
 sible by means of the functions arc tan t, arc cos t, and arc sin t 
 respect ively, we shall limit our study to the latter three functions. 
 
 As in the case of arc sin t, we define single-valued functions 
 corresponding to the multiple-valued functions arc cos t and arc 
 tan t. 
 
 Definition IV. The symbol Arc cos t, called the principal value 
 of arc cos t, designates the least positive angle whose cosine is equal 
 to t. 
 
 Definition V. The symbol Arc tan t, called the principal value 
 of arc tan t, designates the numerically smallest angle whose tangent 
 is equal to t. 
 
 The graphs of the single-valued functions Arc cos t and Arc 
 tan t are given by the heavily drawn portions of Figures 56 and 
 57 respectively. 
 
 When t lies between and 1, Arc cos t is between and 7r/2; 
 when t lies between 1 and 0, Arc eos / is between t/2 and t; 
 when t is positive, Arc tan t is between and 7r/2; 
 when / is negative, Arc tan t is between tt/2 and 0. 
 
 101. Exercises. 
 
 1. Determine arc cos l\ Arc tan 1; Arc cot ^3: Arc cos 1; arc sin 0; Arc 
 cos 1 V2. 
 
 2. Evaluate sin (arc sin \); cos (arc co.s ); tan (arc tan V7). 
 
 3. Determine: cos ('arc tan 1/V3); tan (arc sin v); sec (arc cos J); 
 sin (arc cos .4321); cos (Arc sin 0); cot (Arc tan Vs). 
 
 4. Evaluate :_ cos (Arc cos ] + Arc sin 1); tan (Arc sin 1/V2 Arc tan 1); 
 sin (Arc tan \ 7 3 + Arc cos 1 /\ A 2\ 
 
 5. Construct the graphs of the functions u = arc sin 2 /; u = arc sin / .'!; 
 u = arc cos t/'i.
 
 98 MULTIPLE- AND SINGLE-VALUED INVERSE FUNCTIONS 
 
 102. Relations between multiple-valued and single-valued in- 
 verse functions. There is a simple algebraic relation between 
 the multiple-valued function arc sin t and the single-valued func- 
 tion Arc sin t. If u = Arc sin t, we know that 
 
 t = sin u = sin(w 180 + u), when n is an even integer, and that 
 t = s'mu = sin(n 180 u), when nis an odd integer (see Chap- 
 ters V and VI). 
 
 Hence arc sin t = n 180 + Arc sin /, when n is even, 
 and arc sin t = n 180 Arc sin t, when n is odd. 
 
 These two equalities, which may be derived directly from the 
 graph of arc sin t (see Fig. 55) , are combined in the single formula : 
 
 arc sin t = n 180 + (-1)" Arc sin t. 
 
 In entirely analogous manner we obtain the further formulae: 
 
 arc cos t = n 360 Arc cos t, 
 arc tan t = n 180 + Arc tan t. 
 
 103. Trigonometric equations. Let us consider the following 
 problem: To construct a rectangle whose perimeter is 10" and 
 whose diagonal is 4". 
 
 Fig. 58 
 
 If A BCD be the required rectangle and 6 the angle between the 
 diagonal AC and the side AB (see Fig. 58). we have: 
 
 AB = AC cos = 4 cos 
 
 and 
 
 BC = AC sin 6 = 4 sin 0. 
 
 The conditions of the problem are therefore expressed by the 
 equation 
 
 8 cos 9 + 8 sin 6 = If). 
 
 This equation involves trigonometric functions of the unknown 
 6; it is called a trigonometric equation. Equations of this sort
 
 SOLUTION OF TRIGONOMETRIC EQUATIONS 99 
 
 occur in various branches of mathematics and in its applications. 
 We consider their solution. 
 
 104. Solution of trigonometric equations. " To solve a trigo- 
 nometric equation" means to determine all possible angles which 
 satisfy the equation. It is important to notice the difference 
 between a trigonometric identity, which is true for all angles for 
 which it has a meaning, and a trigonometric equation, which is 
 satisfied by certain angles only. The determination of these 
 special angles constitutes the "solution of the trigonometric 
 equation." 
 
 The work of solving a trigonometric equation may be divided 
 into three parts, viz. : 
 
 (1) To express all the trigonometric ratios occurring in the 
 equation in terms of a single ratio of a single angle, 
 
 (2) To solve the resulting equation obtained in (1), as an 
 algebraic equation for that ratio, and 
 
 (3) To determine all the angles corresponding to the values 
 for that ratio found in (2) . 
 
 Returning to the example of 103, we proceed as follows: 
 
 (1) We express cos in terms of sin by means of the formula: 
 cos = Vl sin 2 0. Substituting this in the equation derived 
 in 103, we obtain the equation 
 
 sin + Vl - sin 2 = f , 
 or 
 
 Vl - sin 2 = i - sin 0. 
 
 (2) We square both sides of this equation, collect terms and 
 clear of fractions. In this way we obtain the following quadratic 
 equation in terms of sin 0: 
 
 32 sin 2 - 40 sin + 9 = 0. 
 
 This equation we solve by means of the quadratic formula, 
 which gives us: 
 
 40 VlGOO - 1152 40 V448 40 21.1660 
 
 sm 
 
 64 64 64 
 
 sin = 61.1660/64 or 18.8340/64. 
 
 (3) From this we conclude: 
 
 = arc sin 61.1660/64 or arc sin 18.8340/64.
 
 100 FURTHER EXAMPLES 
 
 By means of the tables, we find 
 
 Arc sin 61.1660/64 = 72 53' and Arc sin 18.8340/64 = 17 7'. 
 .-. 6 = n 180 + (-1)" 72 53' or n 180 + (-1)" 17 7'. 
 
 The nature of the problem is such that only acute angles 9 
 are applicable, so that we need only consider the principal values 
 of 9. Using these, we find : 
 
 AB = 4 cos 9 = 1.177 or 3.823, 
 
 and BD = 4 sin 9 = 3.823 or 1.177; 
 
 we verify that 2 AB + 2 BD = 10. 
 
 105. Further examples. 
 
 1. To solve the equation 
 
 tan 9 tan 2 9 + cot 9 + 2 = 0, 
 we use the formulae: 
 
 (1) tan 2 6 = t 2 ^ n i a and cot 9 = . l (see 74, 9). 
 
 1 tan- 6 tan 9 
 
 Substituting these expressions in the original equation, it 
 becomes : 
 
 2 tan 2 9 1_ 9 = 
 
 1 - tan 2 9 tan 9 
 
 (2) This is an algebraic equation in terms of tan 9; we clear 
 the equation of fractions and collect terms. This leads us to the 
 following quadratic equation: 
 
 tan 2 9-2 tan 0-1=0. 
 
 If we solve this equation for tan 9 by means of the quadratic 
 formula, we find: 
 
 tan 9 = o = 1 - = 2 " 4142 or -- 4142 - 
 
 (3) It remains to determine arc tan 2.4142 and are tan .4142. 
 From the tables, we find: Are tan .4142 = 22 30', and therefore 
 Are tan (-.4142) = -22 30', whence follows are tan (-.4142) 
 = n- 180 -22 30'.
 
 FURTHER EXAMPLES 101 
 
 Furthermore Arc tan 2.4142 = 67 30', and therefore arc tan 
 2.4142 = n 180 + 67 30'. 
 
 2. To solve the equation 
 
 3 sin 2 6 - 4 sin - 4 = 0. 
 
 In this problem the object of step (1) has already been accom- 
 plished, since the equation in its original form is an algebraic 
 equation in terms of sin 0. 
 
 Solving it by means of the quadratic formula, we find: 
 sin0 = 2 or - = -.6667. 
 
 There are no angles whose sine equals 2 ; we have only to deter- 
 mine therefore arc sin ( .6667). From the tables, we find that 
 Arc sin .6667 = 41 49'; hence Arc sin (-.6667) = -41 49', 
 from which we obtain the final result: 
 
 = arc sin (-.6667) = n 180 + (-1)' (-41 49'). 
 For: n = 1, we find 6 = 180 + 41 49' = 221 49'; 
 n = 2, 6 = 360 - 41 49' = 318 11' ; 
 
 n = -1, = -180 + 41 49' = -138 11'; 
 
 n = -2, = -360' - 41 49' = -401 49'; 
 
 n = 0, = -41 49'; etc. 
 
 The method outlined above for the solution of trigonometric 
 equations is not always the shortest or the most convenient one. 
 Frequently other special methods enable us to obtain a solution 
 more quickly or in better form. The student will do well to bear 
 this in mind, particularly in cases in which the algebraic equa- 
 tion, to which the method described above leads, is such that he 
 cannot solve it. One such special method is illustrated by the 
 following example. 
 
 3. To solve the equation 
 
 cos 2 x sin 2 x = 1 . 
 
 Since tan 45 = 1, we may write this equation in the form 
 tan 45 cos 2 x sin 2 . r = 1 , 
 
 from which we derive, by multiplying both sides of the equation 
 by the corresponding sides of the equality cos 45 = 1 / \ / 2, 
 sin 45 cos 2x cos 45 sin 2 x = 1 'v A 2, 
 
 or sin (45 - 2 x) = 1/^2. 
 

 
 102 FURTHER EXAMPLES 
 
 From this we conclude that we must have 
 
 45 - 2 x = arc sin 1/V2 = n 180 + (-1)" 45, 
 i.e., 2 x = 45 - n 180 - (-1)" 45, 
 
 or finally: x = 45 ~ (~l) n - 4g; _ n g()0> 
 
 For n = 0, 1, 1, 2, 2, we obtain the special values 
 3 = 0, x = -45, x = 135, x = -180, a; = 180 respec- 
 tively. 
 
 The same method may be applied in solving any equation of 
 the form 
 
 a cos m 6 + 6 sin m 6 = c, 
 provided 
 
 \c\ < Va'+b\ 
 
 We divide both sides of the equation by b and determine the 
 angle a by the formula a = Arc tan a/6. Then tan a = a/b, and 
 
 cos a = , the square root being taken with the sign of 6; 
 
 v^ + 6 2 
 then the equation will take the following form: 
 
 tan a cos m 8 + sin m 6 = c/b. 
 
 We now multiply both sides of the equation by cos a and make 
 use of the addition formula for the sine; in this manner we find 
 
 . . nS C COSa C 
 
 sin (a + m 6) = r - , ;/ 
 6 v a ' + 6- 
 
 Since now 'c 1 S Va 2 + 6 2 , the right hand side of this equation is 
 always numerically less than 1. It may therefore be solved for 
 a + m 6, so that we obtain the final result: 
 
 = - \ -a + arc sin ,= = 
 m L \'o- + 6-J 
 
 where a = Arc tan a/b, and where the square root is to be taken 
 with the sign of 6.
 
 EXERCISES 103 
 
 106. Exercises. 
 
 1. Solve the equation: tan x + 3 cot x = 4. 
 
 2. Solve the equation: sin -f cos = V2/2. 
 
 3. Solve the equation: tan 3 cos + sec = 0. 
 
 4. Solve the equation: cos 2 a; = 3 sin x + 2. 
 
 5. Solve the equation: sin = tan 0/2. 
 
 6. Solve the equation : cos 2 A = 2 sin 2 ^1 . 
 
 7. Solve the equation: 4 sec 2 = 3 (tan + 2). 
 
 8. Determine all the angles whose tangent is the reciprocal of their sine. 
 
 9. Determine all the angles whose secant is the reciprocal of their tangent. 
 
 10. The graphs of the functions sin and tan are drawn with respect to 
 one set of axes and units. Determine all those values of for which the 
 ordinate on the tangent curve is twice as long as the corresponding ordinate 
 on the sine curve. 
 
 11. Construct an isosceles triangle whose perimeter is 20 inches and 
 whose altitude is 6 inches. (Hint. Take the base angle of the triangle as 
 the unknown.) 
 
 12. Construct a right triangle of which the perimeter is 36 inches and one 
 leg is 12 inches. 
 
 13. Determine a point P on the circumference of a circle of radius 5 inches, 
 the sum of whose distances from the extremities of a fixed diameter A B is 
 equal to 14 inches. 
 
 14. Construct a triangle ABC such that the perimeter is equal to 30 
 inches, the side c is equal to 10 inches and the angle B is equal to 60. 
 
 15. Resolve a force of 100 pounds into two mutually perpendicular com- 
 ponents, whose sum is equal to 120 pounds. 
 
 16. Construct a triangle ABC of which the perimeter is 60 feet the side 
 a is equal to 20 feet and the angle B is equal to 45.
 
 ANSWERS 
 
 Art. 4, Page 3 
 2. (3,3 V3). 3. (2,-2). 6. (0,0). 7. (3,5); (5,0); (1,1). 8. 5, Vfl, 
 
 V52, 5 y/2. 9. 4; 5; impossible; Vs. 
 
 Art. 9, Page 6 
 
 2. 65, 12, 113, -64, -127, 202, -235, -337, 598. 3. 91, 53, 
 392, -15, 495, -107, -333, 639. 
 
 Art. 11, Page 8 
 
 1. x/4, x/6, -5 x/4, 11 x/6, 3x, -5 x/6, x/3, -3 tt/2, -9x/2, 7 x/6, 
 8 tt/3, -65 tt/18. 2. 270, -240, 72, 225, -90, 270/x, 414 /x. 6. 
 300 x. 6. 10 feet. 7. 144 /x. 8. 2 feet. 9. 6000 x. 10. .84 feet. 
 
 Art. 22, Page 14 
 2a. 1. 26. -V/2. 2c. 0. 2d. (-3 + V3)/4. 3a. -7^2/12. 36. 
 3,2. 3c. (-2 + V3)/3, 3d. 46/9. 
 
 Art. 25, Page 17 
 3a. -13/4. 36. 4. 3c. -I/V3. 3d. -6. 
 
 Art. 28, Page 19 
 
 2. sine = 4/Vl7, cose =1/Vl7, cote = 1/4, sec = Vl7, cosec 9 
 
 = v / l7/4. 
 
 Art. 32, Page 22 
 
 1. 3, -3, 1/3, 0, 3/5, 1. 4. 2/3, 3/2, 3/2, -2/3, -1/4. 
 
 Art. 34, Page 23 
 8. .46831. 9. .28452. 10. 2.21638. 11. .60061. 12. -.93404. 
 
 Art. 39, Page 30 
 
 1. 3.1335. 2. 1.9851. 3. .0052026. 4. 5712. 5. .63014. 6. 4.2266. 
 7. 583.35. 8. 8.524. 9. 39.478. 10. .31365. 11. 2.0550. 12. 1.3425. 
 13. 4.0198. 14. 7360.5. 15. .38638. 16. 30.628. 17. .0093403. 18. 
 .0602s.-,. 19. .31(197. 20. .1.3321. 
 
 105
 
 106 ANSWERS 
 
 Art. 44, Page 35 
 
 1. A = 35 20' 28", B = 54 39' 32", c = 644.83. 2. A = 62 10' 8", 
 = 27 49' 52", c = .16193. 3. A = 30, B = 60, 6 = 3.717. 4. A = 
 48 10' 51", B = 41 49' 9", a = 14.609. 5. B = 21 35', a = 22,073, c = 
 23,737. 6. A = 62 47', a = .12279, c = .13808. 7. B = 5 25', a = 
 376.32, 6 = 35.682. 8. B = 45 25', a = 265.33, b = 269.22. 9. = 44 
 57', b - .08497, c = .12027. 10. A = 49 56' 5", B = 40 3' 55", b = 
 .057956. 11. A = 16 14' 26", B = 73 45' 34", c = 14.252. 12. A = 
 84 58', a = 1.0020, b = .088254. 13. F = 65 34', a = 24.751, h = 20.809. 
 14. P = 25 37', a = 19.940, h = 8.621. 15. P = 41 51' 32", V = 96 
 16' 56", fe = 17.950. 16. b = .84246, a = 1.1038, P = 67 34'. 17. P = 
 53 34', V = 72 52', a = 20.368. 18. b = .64065, a = .32559, F = 159 
 22'. 19. P = 62 49' 45", V = 54 20' 30", /i = 8.8480. 20. b = 12.286, 
 h = 2.9235, 7 = 129 6'. 21. b = 23,454, h = 11,727, P = 45. 
 
 Art. 47, Page 38 
 
 1. 4.3301 feet. 2. 2.5 feet. 3. feet; 7 feet, 
 
 Art. 49, Page 39 
 
 1. 4,504.8 feet; 4,247.7 feet, 2. 272.04 feet. 3. 2,012.2 feet. 4. 
 2,641.4 feet. 5. 27.155 feet. 6. 52.781 miles; 72.705 miles. 7. 53.868 feet; 
 23.246 feet. 8. 1.3313 miles. 9. 566.09 feet; 504.38 feet. 10. 90.21 feet; 
 125.30 feet. 11. 7.980 miles. 12. .059 miles. 13. 320.43 feet; 640.86 feet. 
 14. 8f feet; 28 feet. 15. 67.127 miles; 78.142 miles; 104.43 miles. 
 
 Art. 67, Page 67 
 1. (V6 + V2)/4, (Vg - V2)/4. 
 
 Art. 69, Page 69 
 1. 2 + V3, 2 - V3. 
 
 Art. 71, Page 61 
 
 3. V8-2 V6-2 V2/4, Vs + 2 V6 + 2 V2/4, (V2 - 1) (Vs - V2). 
 
 Art. 73, Page 62 
 
 1. 2 sin 60 cos 5. 2. -2 sin 45 sin 30. 3. 2 sin 4 30' cos 22 30'. 
 4. 2 cos 64 30' cos 22 30'. 5. 2 cos 12 sin 30. 6. -2 sin 45 cos 12. 
 7. .22490. 8. 2.8834. 9. - .24629. 10. .99726. 
 
 Art. 76, Page 63 
 
 la. (Vo-v / 2)/4,-(V6+V2)/4,V3-2. lb. V2-V3/ 2,-^2+^3/2, 
 V3-2. 4. .14762.
 
 ANSWERS 107 
 
 Art. 78, Page 67 
 
 1. a = .29015, c = .23995, C = 31 42'. 2. a = 173.95, b = 237.62, fi = 
 105. 3. a = 12.736, 6 = 6.6607, A = 62 55' 25". 4. b = 170.27, 
 c = 145.80, fi = 84 20' 50". 6. 559.40 feet. 7. 137.18 feet. 
 
 Art. 80, Page 73 
 
 1. c = 72.612, B = 51 40' 30", C = 92 44' 30"; c' = 20.149, 5' = 128 
 19' 30", C = 16 5' 30". 2. a = 1187.5, .1 = 140 23' 56", C = 15 49' 
 4". 3. No solution. 4. 6 = 5.1662, A = 16 7' 27", 5 = 26 34' 33". 
 5. c = 79.735, B = 26 15' 36", C = 134 19' 24"; c' = 13.280, #' = 153 
 44' 24", C" = 6 50' 30". 6. a = .082050, A = 122 58', B = 28 31'- 
 7. 100.91 miles. 8. 203.87 feet or 195.72 feet. 9. 61.671 miles. 
 
 Art. 82, Page 76 
 
 1. c = 137.48, A = 49 6' 22", B = 70 53' 36". 2. P = 53 7' 50", 
 Q = 59 29' 25", R = 67 22' 50". 3. A = 26 44', B = 23 23' 12", C = 
 129 52' 48". 4. t = .56081, R = 34 34' 6", S = 100 26'. 5. 399 feet. 
 6. 84 15' 37". 
 
 Art. 88, Page 80 
 
 1. a = 97.988, B = 52 31' 18", C = 14 44' 42". 2. b = 5247.1, A = 
 36 54' 20", C = 54 38' 40". 3. r = .04343, P = 47 16' 5", <2 = 85 7' 
 55". 4. y = 1.2136, A' = 35 28' 24", Z = 24 31' 36". 6. 211.40 feet. 
 6. 92.685 million miles. 
 
 Art. 91, Page 82 
 
 1. A = 61 6' 29", B = 58 46' 4", C = 60 7' 24". 2. A = 132 45' 4", 
 B = 10 28' 36", C = 36 46' 26". 3. X = 35 39' 50", Y = 81 56' 20", 
 Z = 62 23' 50". 4. A = 44 24' 50", B = 57 7' 20", C = 78 27' 50". 
 5. Angles should be 85 50' 24", 68 7' 4", 26 2' 28". 6. N 42 J 50' 40" E 
 or S 42 50' 40" E. 
 
 Art. 94, Page 86 
 
 5. A = .055898, ff = .3033. 6. A = 3.3797, R = 1.6544. 7. A = 697.4, 
 R = 35.648. 8. A = 90,218, R = 316.89. 9. A = 426.6, - = S.6172. 
 10. A = .16379, r = .15166. 11. A = 157,375, r = 128.71. 12. A = 
 
 .00 16056, r = .016573.
 
 108 ANSWERS 
 
 Art. 96, Pages 86-89 
 
 1. 1,387.1 yards, 1,106.7 yards, 121 16' 7", 15 43' 53". 2. 5.1818 
 miles; 16.6 min. 3. 2,722.7 feet. 4. 71.393 feet. 5. c = 75.527, B = 
 20 38' 55", C = 131 35' 49", A = 626.3. 6. A = 21 47' 12", B = 38 
 12' 46", C = 120, A = 6.4951. 7. b = 23.962, A = 65 47' 0", = 71 
 50' 0", A = 185.75; 6' = 10.016, A' = 114 13' 0", B' = 23 24' 0", A' = 
 77.642. 8. A = 7,768.8, r = 28.433, R = 144.33. 9. A = 1.5928, r = 
 .54269, R = 1.1556. 10. A = 124.44, r = 4.3665, R = 12.807. 11. 36 
 17' 58", 108 42' 2". 12. 239.93 feet, 129.53 feet. 13. 300.92 feet. 14. 
 106.84 miles. 15. N 75 E. 16. 1,849.6 feet. 17. 109.34 feet. 18. 16.07 
 feet, 346.40 feet. 19. 96.742 feet, 28.258 feet, 133 50' 14", 12 9' 46". 
 20. 15 Vf feet. 21. V23 feet. 22. 61.699 feet. 23. 266.88 feet. 24, 
 19.04 feet. 25. 84.01 feet. 
 
 Art. 98, Page 94 
 1. u (t + 3)/4. 2. u = V3l. 3. u = 10. 4. u = </i 
 
 Art. 101, Page 97 
 1. 60, 420, etc.; 45, 225, etc.; 30; 0; 0; 180; etc.; 45. 2. 1/3; 
 2/5; Vf. 3. V3/2; 1; 4; .9018; 1; I/V3. 4. 0; 0; (V3 + i)/2 VJF. 
 
 Art. 106, Page 103 
 
 1. 45 + n 180, 71 33' 54" -f n 180. 2. nir + (-1) n n/6 -tt/4. 
 3. n- 180 + (-1)" 41 48' 39". 4. nw +(-l) n + 1 7r/6, rnr + (-1) n + 1 tt/2. 
 5. 2n;r, (2 n + l)7r/2. 6. nx (-l) M x/6. 7. n- 180* + 49 
 
 36'34",n- 180 -23 2' 39". 8. n 360 51 49' 30". 9. n- 180 + (-l) r ' 
 38 10' 30". 10. nx, 2 nx x/3. 2 tt 2 tt/3. 11. Base angle = 61 
 55' 40"; base = 6.4. 12. Sides are 15 and 9; angles are 36 52' 12" and 
 53 7' 48". 13. Chords are 8 and 6. 14. A = 60, a = b = 10. 15. 97.42 
 and 22.58 pounds. 16. b = 16.7964, c = 23.204, C = 77 39'.
 
 INDEX 
 
 (The Numbers refer to Articles) 
 
 Abscissa, 3 
 Acute angle, 8 
 Angle, 8, 48 
 
 of depression, 48 
 
 of elevation, 48 
 Area of a triangle, 76, 92 
 Axis, 2 
 
 of abscissae, 3 
 
 of coordinates, 3 
 
 of ordinates, 3 
 
 Base of a power, 29 
 
 of a system of logarithms, 31 
 Bearing, 48 
 
 Characteristic of logarithms, 35 
 Checking calculations, 42, 83 
 Circle, 92 
 
 Circumscribed circle, 92 
 Circular measure, 10 
 Clockwise rotation. 8 
 Common logarithms, 35 
 Compass, 48 
 
 Complimentary angles, 8 
 Co-ratio, 14 
 Cosecant, 14, 62 
 Cosine, 14, 51, 81 
 Cotangent, 14, 57 
 Counterclockwise rotation, 8 
 
 Degree, 8 
 
 Dependent variable, 96 
 Depression, angle of, 48 
 Directed angle, 8 
 
 Elevation, angle of, 48 
 Equations, trigonometric, 103, 104 
 Exponent, 29 
 
 Fractional exponent, 29 
 Function, even, 55 
 
 inverse, 96, 97 
 
 multiple-valued, 96 
 
 odd, 55 
 
 single-valued, 96 
 
 trigonometric, 24 
 
 Grade, 10 
 
 Graphing, 50 
 
 Graphs, 51, 57, 62, 97, 100 
 
 Independent variable, 96 
 Initial side, 8 
 Inscribed circle, 92 
 Inverse cosecant, 100 
 
 cosine, 100 
 
 cotangent, 100 
 
 function, 96 
 
 secant, 100 
 
 sine, 99 
 
 tangent, 100 
 
 Law of cosines, 81 
 Laws of exponents, 29 
 
 logarithms, 33 
 Law of sines, 76 
 Law of tangents, 84 
 Logarithm, 30 
 
 Mantissa of a logarithm, 35 
 Mollweide's equation, 85 
 
 Negative exponents, 29 
 
 Obtuse angle, 8 
 Ordinate, 3 
 Origin, 2 
 
 of coordinates, 3 
 
 109
 
 110 
 
 INDEX 
 
 Period, 26 
 Periodicity, 26 
 Power, 29 
 
 Principal value, 100 
 Projection, 6, 45 
 
 Quadrant, 3 
 
 Radian, 10 
 Radius vector, 14 
 Right angle, 8 
 Rotation, clockwise, 8 
 counterclockwise, 8 
 
 Secant, 14, 62 
 Sine, 14, 51, 76 
 Standard position, 12 
 Straight angle, 8 
 
 Submultiple, 10 
 Supplementary angles, 8 
 
 Tangent, 14, 57, 84 
 Terminal side, 8 
 
 Trigonometric equations, 103, 104 
 Trigonometric functions, 24 
 ratios, 14 
 
 Unit distance, 2 
 
 Variable, 96 
 
 x-coordinate, 3 
 X-axis. 3 
 
 y-coordinate, 3 
 Y-axis, 3 
 
 Zero-th power, 20
 
 
 r
 
 3 1205 00260 0144 
 
 UC SOUTHERN REGIONAL LIBRARY FACILITY 
 
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