UC-NRLF 
 
 SB 27fl 
 
 1. 7L 
 
IN MEMORIAM 
 FLOR1AN CAJORI 
 
TEIGONOMETEY 
 
 FOR 
 
 SCHOOLS AND COLLEGES 
 
 FREDERICK AKDEREGG, A.M. 
 
 PROFESSOR OP MATHEMATICS 
 
 AND 
 
 EDWARD DRAKE ROE, JR., A.M. 
 
 ASSOCIATE PROFESSOR OF MATHEMATICS 
 
 IN OBERLIN COLLEGE 
 
 BOSTON, U.S.A., AND LONDON 
 
 GINN & COMPANY, PUBLISHERS 
 
 1896 
 
COPYRIGHT, 1896, BY 
 F. ANDEREGG AND E. D. HOE, JR. 
 
 ALL BIGHTS RESERVED 
 
 , CAJORI 
 
PKEFACE. 
 
 THIS book is a revision of lectures on trigonometry which, 
 we have given to our students for several years, and have 
 furnished them during the last two years in manifolded form. 
 The results have been so satisfactory, that we have been led 
 to hope that the work would prove useful in a larger field. 
 Our aim has been to adapt the work to the needs of students 
 and teachers, with reference to other mathematical subjects 
 both elementary and advanced. 
 
 Since in our own and a growing number of other institu- 
 tions, an explicit course on a book of tables precedes the 
 course in trigonometry, we have omitted that work, and do not 
 publish tables. We have used some fundamental principles 
 of algebra and geometry, with which students of trigonometry 
 can be assumed to be familiar, without giving references. 
 
 We have not given numerical solutions of examples under 
 the different cases of triangles, because the competent teacher 
 does not need them, and we wish to leave him free to use 
 such models as he prefers. 
 
 We have endeavored to lay down general conventions and 
 definitions, to emphasize consistency in the use of the con- 
 ventions, to give general demonstrations in which we have 
 carefully aimed at logical soundness, directness, and sim- 
 plicity, and to exhibit the unity of the subject as made up of 
 
 911316 
 
iv PREFACE. 
 
 its related parts. We believe that in this way the work is 
 simplified, the student gets the ground plan of the higher 
 analysis, and is saved from much subsequent intellectual 
 lameness, which results from an excessive absorption of 
 the attention on special cases. At the same time, when it 
 seemed pedagogically helpful, we have led up to- the general 
 demonstration by the consideration of special cases. 
 
 We have given numerous examples at appropriate points, 
 in order to admit of such selection as the teacher may find 
 necessary for different classes, and to secure a thorough 
 application of the theory. 
 
 The reader will find a number of original features in the 
 book. 
 
 Our views on trigonometry and related subjects have been 
 influenced by many American, English, French, and German 
 treatises, and especially by our respected teachers, Professors 
 J. M. Peirce, W. E. Byerly, and B. 0. Peirce, of Harvard 
 University. 
 
 The figures which appear in this book were drawn by 
 Mr. D. C. Churchill of this college. We have taken some 
 examples from other text-books. 
 
 F. A. AND E. D. E., JR. 
 
CONTENTS. 
 
 PLANE TRIGONOMETRY. 
 CHAPTER I. 
 
 PAGE 
 
 A. LINEAR AND ANGULAR MAGNITUDE ...... 1 
 
 DEFINITION OF TRIGONOMETRY ...... 2 
 
 B. MEASURE-UNITS .3 
 
 a. Linear Units and their Transformations ... 3 
 
 6. Angular Units and their Transformations ... 3 
 
 1. Sexagesimal ; 2. Centesimal ; 3. Circular . . 3 
 
 C. EXAMPLES. 1 5 
 
 CHAPTER II. 
 
 A. CONVENTIONS .......... 6 
 
 a. Conventions about Lines 6 
 
 6. Conventions about Angles 7 
 
 B. TRIANGLE OF REFERENCE ........ 7 
 
 C. TRIGONOMETRIC FUNCTIONS ....... 8 
 
 _D. MUTUAL RELATIONS OF FUNCTIONS . . . . . . 11 
 
 E. EXAMPLES. II 13 
 
 CHAPTER III. 
 
 A. VARIATION OF TRIGONOMETRIC FUNCTIONS . . . .16 
 
 B. LINE REPRESENTATIONS OF THE TRIGONOMETRIC FUNCTIONS 20 
 
 C. INVERSE TRIGONOMETRIC FUNCTIONS 22 
 
 D. EXAMPLES. III. 
 
CONTENTS. 
 
 CHAPTER IV. 
 
 A. TRIGONOMETRIC FUNCTIONS OF INTEGRAL MULTIPLES OP 
 
 -, 0, i.e., of m - <f> . . . -. . . .24 
 
 a. When m is even, and equal to 2 n . . . . . - 32 
 
 6. When m is odd, and equal to (2 n + 1) . . . .33 
 
 B. PERIODICITY OP THE TRIGONOMETRIC FUNCTIONS . . .36 
 
 C. THE FUNCTIONS OP SOME SPECIAL ANGLES . . . .37 
 
 D. EXAMPLES. IV. 38 
 
 CHAPTER V. 
 
 A. PROJECTION . .... t .... 42 
 
 a. When the Line and Axis are Co-planar ... . 42 
 6. When the Line and Axis are not Co-planar . . .43 
 
 B. FUNCTIONS OP SUMS AND DIFFERENCES OF ANGLES . . 44 
 
 C. FUNCTIONS OF MULTIPLE ANGLES ...... 46 
 
 D. FUNCTIONS OF SUB-MULTIPLE ANGLES . ... . . 46 
 
 E. FORMULAS FOR THE SUMS AND DIFFERENCES OP THE SINES OP 
 
 TWO ANGLES, ALSO OF THE COSINES OF TWO ANGLES . 47 
 
 F. EXAMPLES. V . . . . . . ' , . . . 48 
 
 G. EXAMPLES. VI. 49 
 
 CHAPTER VI. 
 
 A. RIGHT TRIANGLES 52 
 
 a. Cases .......... 52 
 
 b. Solutions 52 
 
 B. EXAMPLES. VII. , 53 
 
 C. PRACTICAL APPLICATIONS TO PROBLEMS ON HEIGHTS AND DIS- 
 
 TANCES .......... 54 
 
 D. EXAMPLES. VIII 64 
 
CONTENTS. v jj 
 
 CHAPTER VII. 
 
 PAGE 
 
 A. SOME GENERAL FORMULAS FOR OBLIQUE TRIANGLES . . 56 
 
 a. Theorem of Sines 56 
 
 b. Theorem of Tangents ....... 57 
 
 c. Theorem of Cosines ........ 57 
 
 d. Formulas for one Side of a Triangle in Terms of the Adja- 
 
 cent Angles and the other two Sides ... 58 
 A R c* 
 
 e. Formulas for the Functions of - , - , - . . . .58 
 
 /. Formulas for the Area of a Triangle, and the Radii of the 
 
 Inscribed and Circumscribed Circles .... 60 
 
 B. OBLIQUE TRIANGLES 62 
 
 a. Cases 62 
 
 6. Solutions . . . 63 
 
 C. EXAMPLES. IX. 64 
 
 SPHERICAL TRIGONOMETRY. 
 
 CHAPTER I. 
 
 A. INTRODUCTION 67 
 
 a. Geometrical Principles ....... 67 
 
 6. Sufficiency of the Convex Spherical Triangle . . .68 
 
 B. GENERAL THEORY OF SPHERICAL TRIGONOMETRICAL RELATIONS 69 
 
 C. SYSTEMS OF EQUATIONS DEMANDED BY THE PRECEDING THEORY 70 
 
 a. Equations in which the Parts enter Four at a Time . 70 
 6. Equations in which the Parts enter Five at a Time . . 75 
 c. Equations in which the Parts enter Six at a Time . 76 
 
 D. ON THE USE OF THE POLAR TRIANGLE IN ESTABLISHING RELA- 
 
 TIONS . . . .76 
 
Viii CONTENTS. 
 
 CHAPTER II. 
 
 PAGE 
 
 A. FORMULAS . ... . . . . . . 78 
 
 a. Spherical Right Triangles . ... . . .78 
 
 6. Quadrantal Triangles . 78 
 
 c. Deductions . . '. . . . . . .79 
 
 B. SOLUTIONS .......... 79 
 
 a. Right Triangle . . -..'-. . . . . .79 
 
 6. Quadrantal Triangle . . ^. ... . 79 
 
 C. SOME ASTRONOMICAL CONCEPTIONS ... . . ... .80 
 
 a. Celestial Sphere . . ... . . . 80 
 
 b. Spherical Coordinates ... . . . . . .82 
 
 D. EXAMPLES. X. 84 
 
 CHAPTER III. 
 
 A. FORMULAS FOR SPHERICAL OBLIQUE TRIANGLES . . .87 
 
 a. Formulas for Functions of | A, $ B, C . -. . 87 
 
 6. Gauss's Equations . 88 
 
 c. Napier's Analogies . . .- . ... . . 90 
 
 d. 1'Huilier's Formula . . . . . .91 
 
 e. Formula for the Polar Radius of the Circumscribed Circle 
 
 of a Spherical Triangle 92 
 
 /. Formula for the Polar Radius of the Inscribed Circle of a 
 Spherical Triangle . . 92 
 
 B. PLANE TRIGONOMETRY A SPECIAL CASE OF SPHERICAL TRIGO- 
 
 NOMETRY ....,.. . 93 
 
 C. SPHERICAL TRIANGLES . . . . . . . .94 
 
 a. Cases .,...' 94 
 
 6. Solutions . . . . . . . . . .95 
 
 D. EXAMPLES. XI. 106 
 
PLANE TEIGONOMETRY. 
 
 CHAPTER I. 
 
 A. Linear and Angular Magnitude. Definition of 
 Trigonometry. 
 
 1. Magnitude is duration in time, or extent of space, or 
 duration in time and extent of space of that which manifests 
 itself in time and space. 
 
 2. A line is the path of a moving point. It has one 
 dimension of space, length. The initial position of the gen- 
 erating point is the origin of the line. The final position is 
 its term. The line is taken from the origin to the term, and 
 is always determinate, unless the origin and term coincide, 
 when it may be indeterminate. Hence every line joining 
 two points represents at least two lines, opposites in direc- 
 tion, and negatives of each other. That is : 
 
 A B 
 
 It follows from this that if AI, A 2 , A n , be n points arranged 
 
 in any order on a line, 
 
 AiA 2 +A 2 A 3 +.A 8 A 4 + A^An+AA^I, 
 
 where k is either zero, or any positive or negative integer, 
 and I is the simplest length of the line. Linear magnitude 
 is the extent of a line, and may be unlimited. 
 
'2' 
 
 PLANE TRIGONOMETRY. 
 
 aiigle is the amount of turning in a plane of a 
 straight line about a fixed point through which it passes. 
 The initial position of the generating line is called an axis 
 of reference, and that part of it which forms the initial side 
 of the angle is called the initial line of the angle. The 
 whole of the generating line is called the rotating line, and 
 that part of it which forms the terminal side of the angle is 
 called the terminal line of the angle. The angle is taken 
 from the initial to the terminal 
 line. Hence every geometric 
 angle must represent at least two 
 angles, opposites in turning, and 
 negatives of each other. That is : 
 AOB+BOA^O, or 
 
 BOA = -AOB 
 AOB = BOA. 
 It follows from this that if a x , a 2 , a n , be any n lines in a 
 
 plane, and ' represent the angle from a k to a\ . then 
 a k 
 
 a ~^~ a ~^~ a ~^~ a + ^ = "360, 
 
 where n may equal zero or any integer, positive or negative. 
 Angular magnitude is the amount of turning of the generating 
 line, and may be unlimited. 
 
 4. It is to be noticed that the preceding definitions of a 
 line, and an angle, are extensions of the elementary concep- 
 tions ; for they require as essential to the line, and the angle, 
 two elements, magnitude and direction. The elementary 
 conceptions require only magnitude. From these definitions 
 corresponding extensions of all the elementary conceptions 
 into which lines, and angles enter, necessarily follow. 
 
 5. The subject matter of Trigonometry is lines, and angles. 
 Trigonometry is the investigation of the relations of the sides 
 and angles of a triangle. 
 
LINEAR AND ANGULAR MAGNITUDE. 
 
 B. Measure-Units. 
 
 6. The measure of a magnitude is expressed by the 
 number of measure-units the magnitude contains. Thus, the 
 measure of a line is 3 feet, 4 rods, 2 meters, etc. Both lines 
 and angles are measured by different measure-units. 
 
 (a) Linear Units and their Transformation. 
 
 7. The principal measure-units for lines are the English 
 and the French. The equation for transformation is : 
 1 meter ^39.37 inches, approximately. 
 
 (6) Angular Units and their Transformations. 
 
 8. The principal measure-units for angles are the sexa- 
 gesimal, the centesimal, and the circular. 
 
 1. In the sexagesimal system : 
 
 1 right angle = 90 degrees, written 90. 
 1 = 60 minutes, " 60'. 
 
 1' =60 seconds, " 60". 
 
 2. In the centesimal system : 
 
 1 right angle = 100 grades, written 100 g . 
 1 =100 minutes, 100 V . 
 
 T =100 seconds, 100". 
 
 3. The radian is that central 
 angle in any circle whose in- 
 tercepted arc is. equal in length 
 to the radius. Since in the same 
 circle, or in equal circles, central 
 angles are proportional to their 
 intercepted arcs, and if 8 repre- 
 sents the number of degrees in 
 the angle, r the radius, and s the 
 
 arc, then : FIG. i. 
 
PLANE TRIGONOMETRY. 
 
 _. _!_ a} 
 
 360 2 Trr ' 
 
 = ilo- 8 - V 
 
 \ s 
 
 This shows that f - J varies as 8, that is, varies as the angle. 
 Let <f> represent the angle in any system of measurement, 
 
 then d> = n ( - ) , where n is constant. When s = r, d> is con- 
 
 W 
 stant ; also when s = r, < is a radian, which we will take as 
 
 the angular measure-unit in the circular system, and represent 
 by r written like an exponent ; and we have, to determine n, 
 
 1 = n ( - ), or n = 1, which gives generally, < -, when the 
 
 angle <f> is expressed in radians. 
 
 The preceding demonstration shows that the radian is 
 
 180 
 invariable, and is equivalent to - , or 57.2957795+ . The 
 
 equation, <j> = - , r = - , or s r<^>, is of practical importance 
 r <f> 
 
 according as <, r, or s, respectively, is the unknown. 
 
 If 8 represents the number of degrees, y the number of 
 grades, and p the number of radians in an angle, we have, as 
 in the preceding demonstration : 
 
 _8_ y s 
 
 360 ~~ 400 ~~ 2 ?rr ' C 
 
 -L === JL. S= . 
 
 180 200 TT 
 
 From which follow these six equations of transformation : 
 
 8 = JL = ??5 - * - 
 
 10 TT 180 
 
LINEAR AND ANGULAR MAGNITUDE. 
 
 Making the symbols 8, y, p in the right members of these 
 equations successively equal to unity, we obtain the value of 
 each unit in terms of the others. This gives the following 
 table : 
 
 
 In terms of 
 
 Units 
 
 
 
 o 
 
 r 
 
 g 
 
 
 
 7t 
 
 10 
 
 1 
 
 1 
 
 180 
 
 9 
 
 
 180 
 
 
 200 
 
 l r 
 
 
 1 
 
 
 
 rt 
 
 
 Tt 
 
 
 9 
 
 it 
 
 
 l^ 
 
 10 
 
 200 
 
 1 
 
 Degree measure is used in practical, and circular measure 
 in theoretical work. 
 
 c. Examples. I. 
 
 1. Express 2523 17' 32" in grades, and radians. 
 
 2. Express 57s 17' 94" in degrees, and radians. 
 
 3. Express 2.3 r in degrees, and grades. 
 
 , tf r ?r r TC* . 
 
 4. Express TV, , , , in degrees. 
 
 5. Assuming the earth to be a sphere with a radius of 3963 miles, 
 compute the length of a geographical mile in statute miles. 
 
 6. Find the distance, measured on the arc of a great circle, in miles, 
 from Oberlin (lat. 41 17' N.) to the equator. 
 
 7. A man stands on the extremity of a radius of a circular platform 
 which is supported on a vertical axis through its center, and perpen- 
 dicular to its plane. The extremity of the given radius is opposite a 
 fixed arrowhead. The platform is made to rotate about its axis counter- 
 clockwise, and finally comes to rest. The man has moved through a 
 distance of mi. Find in degrees the actual angle through which the 
 platform has turned, and the apparent angle which the given radius 
 makes with the line of the fixed arrow, the radius being 15 ft. in length. 
 
CHAPTER II. 
 
 A. Conventions. 
 a. Conventions about Lines. 
 
 9. A plane is divided into four quadrants by two fixed 
 perpendicular straight lines. The plane may always be so 
 placed that one line is horizontal, and the other vertical. For 
 convenience we shall speak of the lines as the horizontal, and 
 the vertical line, or axis. 
 
 1O. The point of intersection of the two lines is called 
 the origin, and is denoted by the letter 0. The horizontal 
 axis is called the x-axis, the vertical, the y-axis. 
 
 11. The distance of any point measured from the y-axis 
 parallel to the x-axis, will be represented by x, and called its 
 abscissa ; and the distance of any point measured from the 
 x-axis parallel to the y-axis, will be represented by y, and 
 called its ordinate. The distance of any point from the 
 origin will be represented by r, and called its radius vector. 
 
 x = OM. 
 y = M P. X- 
 r = OP. 
 
 
 
 Y' 
 FIG. 2. 
 
TRIANGLE OF REFERENCE. 7 
 
 12. We will adopt the convention, that is, the AGREEMENT 
 that the direction FROM left TO right of FIXED horizontal lines 
 is POSITIVE, the OPPOSITE direction is NEGATIVE ; and that the 
 direction FROM below UPWARDS of FIXED vertical lines is 
 POSITIVE, the OPPOSITE direction is NEGATIVE. 
 
 13. By convention, the directions of the sides of an angle 
 are said to be positive if taken from the vertex outwards, the 
 opposite directions are negative. 
 
 b. Conventions about Angles. 
 
 14. The turning of a line which generates an angle may 
 be either counter-clockwise or clockwise. Counter-clockwise 
 rotation we will call, by convention, positive; the opposite, 
 negative. An angle generated by positive rotation we will call 
 positive ; an angle generated by negative rotation, negative. 
 
 15. The four quadrants are numbered the first, the 
 second, the third, and the fourth, respectively, beginning 
 with XOY, and taken in the order of positive rotation. 
 
 16. The initial line of any single angle will always be 
 taken as coinciding with OX, and the angle will be said to be 
 of that quadrant in which its terminal line lies. In any 
 composite angle the initial line of any angle coincides with 
 the terminal line of the angle preceding, and may, for the 
 moment, be considered as coinciding with OX. 
 
 B. Triangle of Reference. 
 
 17. If at any point of the x-axis, or of the y-axis, a per- 
 pendicular be erected and produced to intersect the rotating line 
 of a given angle at the point P, a right triangle will be formed, 
 which is called the TRIANGLE OF REFERENCE for the given 
 angle. The abscissa of P is always the base, its ordinate, the 
 perpendicular, and its radius vector, the hypotenuse of. the 
 triangle of reference. 
 
PLANE TRIGONOMETRY. 
 
 C. Trigonometric Functions. 
 
 18. The six ratios which can be formed by using the three 
 sides of the triangle of reference of a given angle, two at a time, 
 are called the six primary trigonometric functions of the angle. 
 
 The sine of the angle is the ratio of the perpendicular to the 
 hypotenuse. 
 
 The cosine of the angle is the ratio of the base to the hypote- 
 nuse. 
 
 The tangent of the angle is the ratio of the perpendicular to 
 the base. 
 
 The cotangent of the angle is the ratio of the base to the per- 
 pendicular. 
 
 The secant of the angle is the ratio of the hypotenuse to the 
 base. 
 
 The cosecant of the angle is the ratio of the hypotenuse to the 
 perpendicular. 
 
 19. The following figures show to the eye, how, by the 
 definition, triangles of reference are constructed for an angle 
 of each quadrant. 
 
 X- 
 
 Y' 
 FIG. 3. 
 
 x M 
 
TRIGONOMETRIC FUNCTIONS. 
 Y 
 
 Y' 
 
 FIG. 5. 
 
 2O. It is to be observed, and the student should carefully 
 prove, that the triangle of reference for any angle includes 
 four species, that all the triangles that can be constructed 
 are geometrically similar, and that in passing from a triangle 
 of one species to a triangle of any other species, either no 
 change, or two changes of sign occur in the terms of any 
 trigonometric function, and that, therefore, the function is 
 unaltered. This work should be done as is shown below for 
 the sine and cosine of an angle of the first quadrant. 
 
10 
 
 PLANE TRIGONOMETRY. 
 
 perpendicular 
 
 sine of <, written sin <, = *-r J -- = 
 
 hypotenuse 
 
 r 2 OP 3 
 cosine of <, written cos <, = 
 
 r 3 OP 4 
 
 base 
 hypotenuse 
 
 OM t 
 
 M 2 P 
 
 2 
 
 OP 2 
 
 OM 2 
 OP S 
 
 __ X2 _ M 3 P 8 _ Xg _ M 4 P 4 _ X4 ^ 
 
 ~ r 2 ~ OP 3 ~r 3 OP 4 ~ r 4 ' 
 
 We shall use as abbreviations for tangent of <, cotangent 
 of <, secant of ^, and cosecant of <j>, respectively, tan <, 
 ctn <^>, sec <, and esc <^>. 
 
 21. Since the trigonometric functions are abstract num- 
 bers, all the arithmetical operations can be performed upon 
 them. It is customary to indicate the nth power of sin <f> 
 by sin n <f> ; similarly for the other trigonometric functions. 
 
 22. In addition to the primary functions previously de- 
 fined, two secondary functions are usually defined as follows : 
 
 The versed-sine of <, written vrs <, =1 cos <j>, 
 The coversed-sine of <, written cvs </, = 1 sin <. 
 
MUTUAL RELATIONS OF FUNCTIONS. 
 
 11 
 
 Two others are more rarely defined as follows : 
 
 The suversed-sine of <j>, written svs <f>, = (2 vrs <), 
 
 1 + cos <. 
 The sucoversed-sine of <, written scs <, = (2 cvs <), 
 
 = 1 + sin <. 
 
 23. By referring to Figs. 3-6 we see that the sine, 
 cosine, secant, and cosecant of an angle of the third quadrant 
 are negative, because the terms of these ratios have opposite 
 signs, and that its tangent and cotangent are positive, because 
 the terms of these ratios have the same sign. Similarly, by 
 a careful examination of Figs. 3-6, the student should prove 
 the following table for the signs of the functions of angles 
 of the four quadrants. 
 
 Functions. 
 
 Quadrants. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 sine and cosecant .... 
 
 + 
 
 + 
 
 - 
 
 - 
 
 cosine and secant .... 
 
 + 
 
 - 
 
 - 
 
 + 
 
 tangent and cotangent . 
 
 + 
 
 - 
 
 + 
 
 - 
 
 Z>. Mutual Relations of Functions. 
 
 24. From the definitions of the trigonometric functions 
 it is evident that : 
 
 sn > = 
 
 CSC 
 
 cos < = 
 
 sec 
 
 ; tan < = 
 
 1 
 
 ctn 
 
 that is, the sine and the cosecant, the cosine and the secant, 
 the tangent and the cotangent of an angle <f>, are the recip- 
 rocals of each other, always. 
 
12 PLANE TRIGONOMETRY. 
 
 sin (f> cos < 
 
 0. tan <i = : ; ctn 6 = : > always. 
 
 cos <j!> ' sin < 
 
 0. Since, of whatever quadrant the angle <f> may be, that is, 
 whatever may be the intrinsic signs of x, y, and r, x 2 , y 2 , and r 2 
 are always positive, and y 2 + x 2 = r 2 , always, 
 
 y 2 x 2 
 .'.-] - = 1, that is, sin 2 </> + cos 2 </> = 1, always. 
 
 v 2 r 2 
 
 Similarly, Jg + 1 = -5 , that is, tan 2 < + 1 = sec' 2 <, always. 
 
 r 2 
 
 > 
 
 x r 
 
 And ^ + 1 = 1 > that is, ctn 2 </> -j- 1 = csc 2 ^>, always. 
 
 d. From the preceding relations, it is evident that some of 
 the functions may be expressed in terms of others. It can 
 be shown that all the functions may be expressed in terms of 
 any one, which may be chosen at pleasure as the independent 
 function. For example, all the functions may be expressed 
 in terms of the tangent as follows : 
 
 ctn cf> Since tan 2 </> + 1 = sec 2 <, 
 
 tan <p 
 
 .'. sec < = 
 
 Since cos <f> = , .*. cos <i = 
 
 sec </> 
 
 sin <f> . _ 
 
 cos </> ' 
 
 1 Vtan V + 1 
 
 esc <f> = , .'. csc <j> = 
 
 sin < tan < 
 
 It is left as an exercise for the student to complete the 
 following table. 
 
 1 
 
 
 
 Since 
 
 Vtan 2 </> -j- 1 
 
 
 tan ^> 
 
 
 Vtan 2 <^> -|~ 1 
 
 
EXAMPLES. II. 
 
 13 
 
 Func- 
 tions. 
 
 In terms of 
 
 sin 
 
 COS 
 
 tan 
 
 ctn 
 
 sec 
 
 CSC 
 
 sin 
 
 
 
 . tan0 
 
 
 
 
 "~ Vtan 2 + 1 
 
 COS 
 
 
 
 1 
 
 
 
 
 ^ Vtan 2 + 1 
 
 tan 
 
 
 
 tan0 
 
 
 
 
 ctn 
 
 
 
 1 
 
 tan0 
 
 
 
 
 sec 
 
 
 
 
 
 
 
 Vtan 2 + 1 
 
 CSC 
 
 
 
 Vtan 2 + 1 
 
 
 
 
 tan 
 
 E. Examples. II. 
 
 1. Of what quadrant is each of the following angles ? 38, 317, 2530, 
 - 1155, 275s, 31', - 13.4', -823, -423?, 625 43' 26", 761 e 87' 95", 
 -27.364'. 
 
 2. Determine the sign of each of the trigonometric functions of the 
 angles in example 1. 
 
 3. If is an angle of the first quadrant, find the values of the other 
 trigonometric functions, having given : sin = |, cos = if, tan = ^ 
 ctn = f , sec = 3, esc = |. 
 
 4. If is an angle of the second quadrant, find the other trigonometric 
 functions, having given : sin = |, cos = f , tan = |, ctn = 
 2, sec = 7, esc = 10. 
 
 5. If is an angle of the third quadrant, find the other trigonometric 
 functions, having given : sin = -, cos = V^, tan = VJJ, sec 
 = 5, ctn 0=1, esc = 2. 
 
 6. If is an angle of the fourth quadrant, find the other trigonometric 
 
 V3-1 
 
 functions, having given : sin = 
 
 ctn = 2 V6, sec = 
 
 2V2 
 V3 + 1 
 
 2V2 ' 
 
 = 1, tan0 = V8, 
 
 , CSC = 
 
14 PLANE TRIGONOMETRY. 
 
 7. Determine all the possible values of the other trigonometric func- 
 
 tions, having given : sin = V|, cos = Vfc, tan = - 
 ctn = 4 V3, sec = V2, esc = a. 
 
 8. If is an angle of any quadrant, draw triangles of reference for 
 
 the following angles : 4 n | + 0, (4 n + 1) | + 0, (4 n + 2) ^ -f 0, 
 
 Prove the following identities : 
 
 9. sin esc = cos sec = tan ctn = 1. 
 
 10. cos + vrs = sin + cvs = svs cos = scs sin = 1. 
 
 11. vrs svs 0=1 cos 2 = sin 2 0. 
 
 12. cvs scs = 1 sin 2 cos 2 0. 
 
 13. svs scs = 1 + sin + cos + sin cos 0. 
 
 14. cos tan = sin 0. 
 
 15. sin ctn = cos 0. 
 
 16. (sec + tan 0) (sec tan 0) = (esc + ctn 0) (esc ctn 0) = 1. 
 
 17. (sin + cos 0) 2 + (sin cos 0) 2 = 2. 
 
 18. tan + ctn = sec esc 0. 
 
 19. cos 2 sin 2 = 2 cos 2 1 = 1 2 sin 2 0. 
 
 20. (cos + sin 0) (esc sec 0) = ctn tan 0. 
 
 21. sin (1 + tan 0) + cos (1 + ctn 0) = sec + esc 0. 
 
 22. sin 4 + cos 4 = 12 sin 2 cos 2 0. 
 
 23. sin 4 cos 4 = sin 2 cos 2 0. 
 
 24. tan 2 ctn 2 = sec 2 csc 2 0. 
 
 25. (1 + tan 0) 2 + (1 - tan 0) 2 = 2 sec 2 0. 
 
 26. (1 + cos 0) 2 + (1 + sin 0) 2 = 3 + 2 (sin + cos 0). 
 
 27. (sin + cos 0) (tan + ctn 0) = sec + esc 0. 
 
 28. (sec + esc 0) 2 = (1 + tan 0) 2 + (1 + ctn 0) 2 . 
 
 29. tan (1 ctn 2 0) + ctn (1 tan 2 0) = 0. 
 
 30. sin 3 + cos 3 = (sin + cos 0) (1 sin cos 0). 
 
 31. sin 3 cos 3 = (sin cos 0) (1 + sin cos 0). 
 
 32. (.,, -*.,)= -=* 
 
 33. 
 
 34. esc + ctn = 
 
 esc ctn 
 35. 
 
EXAMPLES. II. 15 
 
 36. 
 37 
 
 ctn + ctn e 
 ctn4. + tanfl = ct 
 tan + ctn 
 
 tan + ctn 
 39. sin 2 tan 2 + cos 2 ctn" 2 = tan 2 + ctn 2 
 
 sin + sin _ cos + cos d 
 ' cos cos 6 ~ sin sin 
 
 , tan <j> + sec 1 
 41. - - - - = tan + sec 0. 
 1 + tan sec 
 
 1 sec + tan _ sec + tan 1 < 
 1 + sec tan ~~ sec + tan + 1 
 1 + esc + ctn _ esc + ctn 1 ^ 
 1 + esc ctn ctn esc + 1 
 44. (1 4- sec + tan 0) (1 + esc + ctn 0) 
 = 2 (1 + tan + ctn + sec + esc 0). 
 
CHAPTEK III. 
 
 A. Variation of Trigonometric Functions. 
 
 25. It is necessary to trace the changes in the values of 
 the trigonometric functions as the angle changes. The fol- 
 lowing two examples are sufficient to show how this is done. 
 
 Beginning with an indefinitely small angle, we will trace 
 the changes in the sine and tangent of the angle as the angle 
 increases. One of the triangles of reference which lie in the 
 same quadrant as the terminal line of the angle, will be used. 
 In the discussion of the sine, the hypotenuse of the triangle 
 of reference will be kept constant in length. The sign ~ 
 will be used for the words " approaches as its limit." is 
 the limit of an indefinitely small angle. It is evident that 
 the sine of an indefinitely small angle must be indefinitely 
 small, since the perpendicular of the triangle of reference 
 must be indefinitely small ; that is, sin a = e, where a = 0, 
 and e = 0. Since the variables sin a and e are always equal 
 however near they approach their limits, therefore their 
 limits, sin and 0, are equal ; that is, sin = 0. As the 
 angle increases between and 90, the perpendicular of the 
 triangle of reference increases, therefore the sine of the angle 
 increases, and vice versa. When an angle approaches 90 as 
 its limit, the length of the perpendicular approaches as its 
 limit the length of the hypotenuse ; therefore, as in the pre- 
 ceding case, sin 90 = 1. As the angle increases between 90 
 and 180, the perpendicular decreases, therefore the sine of 
 the angle decreases, and vice versa. 
 
 When an angle approaches 180 as its limit, its sine ap- 
 proaches as its limit, and sin 180 = 0. The sine of an 
 angle in the third and fourth quadrants is negative. As $he 
 
VARIATION OF TRIGONOMETRIC FUNCTIONS. 17 
 
 angle increases between 180 and 270, the perpendicular 
 increases in length, that is, decreases in algebraic value, 
 therefore the sine decreases, and vice versa. By the use of 
 limits it is evident that sin 270 = 1. As the angle in- 
 creases between 270 and 360, the perpendicular decreases 
 in length, that is, increases in algebraic value, therefore the 
 sine increases, and vice versa. And by the use of limits 
 sin 360 = sin = 0. If the angle should continue to in- 
 crease, the same set of values of the sine would recur period- 
 ically at intervals of 360. This recurrence of values is 
 called the periodicity of the sine, and will be noticed again. 
 
 26. In the discussion of the tangent, the base of the 
 triangle of reference will be kept constant in length. By 
 reasoning like that used for the sine it is evident that 
 tan = 0, and that as the angle increases between and 
 90 the tangent increases, and vice versa. As an angle ap- 
 proaches 90 as its limit, the perpendicular of the triangle of 
 reference increases indefinitely, therefore the tangent in- 
 creases in numerical value without limit, which is what is 
 meant when we say tan 90 oo . In the second quadrant 
 the base of the triangle of reference is negative, hence the 
 tangent is negative. As the angle increases between 90 and 
 180, the perpendicular decreases, therefore the tangent in- 
 creases in algebraic value, and vice versa. And by the doc- 
 trine of limits tan 180 = 0. In the third quadrant both 
 the base and the perpendicular of the triangle of reference 
 are negative, hence the tangent is positive. As the angle 
 increases between 180 and 270, the perpendicular increases, 
 therefore the tangent increases, and vice versa. As the angle 
 approaches 270 as its limit, the perpendicular increases 
 indefinitely, therefore the tangent increases in numerical 
 value without limit, and we say, as before, tan 270 = oo . 
 In the fourth quadrant, the perpendicular only of the triangle 
 of reference is negative, hence the tangent is negative. As 
 
18 
 
 PLANE TRIGONOMETRY. 
 
 the angle increases between 270 and 360, the perpendicular 
 decreases in length, therefore the tangent increases in alge- 
 braic value, and vice versa. And by the doctrine of limits 
 tan 360 = tan = 0. The tangent also has a recurrence 
 of values, at intervals of 180. 
 
 27. The following pictures, called graphs, of the pre- 
 
 f 
 
 7 
 
 Y' 
 FIG. 8. Graph of the Sine. 
 
 Y 
 
 FIG. 9. Graph of the Tangent. 
 
 ceding tracing of values are given to aid the imagination. 
 Abscissas represent values of the angle expressed in circular 
 measure ; ordinates represent the corresponding values of 
 the functions. 
 
VARIATION OF TRIGONOMETRIC FUNCTIONS. 
 
 19 
 
 The student should now trace the changes in the values of 
 the other trigonometric functions, and, independently, obtain 
 a table like the following. 
 
 Angle 
 
 sin 
 
 cos 
 
 tan 
 
 ctn 
 
 sec 
 
 CSC 
 
 
 
 
 TO; 
 
 i 
 
 TO 
 
 =j=co 
 
 1 
 
 = F co 
 
 -360 
 
 Qi 
 
 inc 
 dec 
 
 dec 
 inc 
 
 inc 
 dec 
 
 dec 
 inc 
 
 inc 
 dec 
 
 dec 
 inc 
 
 Qi 
 
 90 
 
 I 
 
 
 
 00 
 
 
 
 00 
 
 1 
 
 270 
 
 Q 2 
 
 dec 
 
 inc 
 
 dec 
 inc 
 
 inc 
 dec 
 
 dec 
 inc 
 
 inc 
 dec 
 
 inc 
 dec 
 
 Q 2 
 
 180 
 
 
 
 -1 
 
 TO 
 
 qpoo 
 
 -1 
 
 00 
 
 -180 
 
 Q 3 
 
 dec 
 inc 
 
 inc 
 dec 
 
 inc 
 dec 
 
 dec 
 inc 
 
 dec 
 inc 
 
 inc 
 dec 
 
 Q 3 
 
 270 
 
 -1 
 
 TO 
 
 00 
 
 
 
 T 
 
 -1 
 
 -90 
 
 Q 4 
 
 inc 
 
 dec 
 
 inc 
 dec 
 
 inc 
 dec 
 
 dec 
 inc 
 
 dec 
 
 inc 
 
 dec 
 inc 
 
 Q 4 
 
 360 
 
 =1=0. 
 
 1 
 
 TO 
 
 =FCO 
 
 1 
 
 =poo 
 
 
 
 
 sin 
 
 cos 
 
 tan 
 
 ctn 
 
 sec 
 
 CSC 
 
 Angle 
 
 In this table, for an angle generated by positive rotation, 
 i.e., an increasing angle, the upper word goes with the value 
 of the angle taken from the left-hand column ; and for an 
 angle generated by negative rotation, the lower word goes 
 with the value of the angle taken from the right-hand column. 
 The -f~ and. before and oo indicate how the function 
 changes sign as the angle passes through the quadrant limit. 
 
20 
 
 PLANE TRIGONOMETRY, 
 
 B. Line Representations of the Trigonometric 
 Functions. 
 
 28. If in each, of the primary trigonometric functions, 
 or ratios, a positive linear unit is taken for the denominator, 
 the number of linear units in the numerator will represent 
 the abstract number of units in the ratio, and the line which 
 is the numerator of the ratio will completely represent the 
 function. The student should remember that the line is not 
 the function, but represents it. This restriction gives the 
 following constructions. 
 
REPRESENTATIONS OF TRIGONOMETRIC FUNCTIONS. 21 
 
 MP 
 In these figures, for example, sin <f> = - -- , but OP is taken 
 
 equal to a linear unit. Hence, sin < is represented by M P. 
 cos (f> is represented by OM. Similarly the student should 
 prove for each quadrant, that 
 
 tan <f> is represented by M'P' ? 
 ctn <f> is represented by N Q, 
 sec (j> is represented by OP', 
 esc <f> is represented by OQ. 
 
 The secondary trigonometric functions are also represented 
 by lines as follows : 
 
 vrs <j> is represented by MM', 
 cvs <j> is represented by LN, 
 svs <f> is represented by R M, 
 scs <j> is represented by S L. 
 
 29. The student should ascertain the signs of the func- 
 tions in the four quadrants by means of the line representa- 
 tives. The accompanying figure 
 is intended forcibly to impress 
 on the student's mind that the 
 sine and cosecant of an angle 
 of a quadrant above the hori- 
 zontal line are positive, below 
 negative ; that the cosine and 
 secant of an angle to the right 
 of the vertical line are positive, 
 to the left negative ; and that 
 the signs of the tangent and 
 cotangent are given by the 
 combination of the signs of the 
 
 FIG. 14. 
 
 sine and cosine, quadrant by quadrant, 
 seen by line representation. 
 
 This is all readily 
 
22 PLANE TRIGONOMETRY. 
 
 C. Inverse Trigonometric Functions. 
 
 3O. If two variables are connected by any relation, either 
 one of them may be considered as the independent variable, 
 and the other as the dependent variable or direct function. 
 When the dependent variable is made independent, the in- 
 dependent variable becomes the dependent variable and 
 inverse function. 
 
 If x is any function of the angle <f>, </> is the angle whose 
 function is x. These two statements of the same relation, 
 viewed from the different standpoints alluded to, are symbol- 
 ized as follows : x = f (<), and < = f" 1 (x), and are read, x is 
 a function of <, and <j> is the anti-function of x. For example, 
 if x = sin <, <f> is the angle whose sine is x, which is written 
 <f> = sin^x, and is read, <j> is the anti-sine of x. Some writers 
 express the inverse functional relations as follows : <f> = arc 
 sin x, = arc cos x, and similarly for the other functions. 
 
 />. Examples. III. 
 
 1. By representing positive numbers by distances to the right of the 
 origin on the x-axis, and negative numbers by distances to the left of 
 the origin, and the values which functions can have by heavy lines, prove 
 that the following is a correct graphic representation of the possible 
 range of values of the primary trigonometric functions. 
 
 sin and cos X 
 sec and esc X 
 tan and ctn X' 1 
 
 oo 
 
 +00 
 
 -1 
 
 + 1 
 
 oo 
 
 +00 
 
 1 
 
 + 1 
 
 oo 
 
 +00 
 
 
 V 
 
EXAMPLES. III. 
 
 23 
 
 The results of this problem are very important. The student should 
 carefully remember both the range of possible and impossible values of 
 the functions. 
 
 Trace the changes in value of the following expressions : 
 
 6. vrs 0. 8. svs 0. 
 
 2. cos 0. 
 
 3. ctn 0. 
 
 4. sec 0. 
 
 5. esc 0. 
 
 7. cvs 0. 
 
 9. scs 0. 
 
 Construct an angle : 
 
 10. (a) Of the 1st, and of the 2d 
 " 4th 
 
 " 4th 
 
 3d 
 3d 
 
 " 4th 
 
 3d 
 4th 
 
 " 4th 
 
 3d 
 
 " 2d 
 
 " 4th 
 
 
 (b) 
 
 3d 
 
 11. 
 
 (a) " 
 
 1st 
 
 
 (b) 
 
 2d 
 
 12. 
 
 (a) 
 
 1st 
 
 
 (b) " 
 
 2d 
 
 13. 
 
 (a) " 
 
 1st 
 
 
 (b) " 
 
 2d 
 
 14. 
 
 (a) " 
 
 1st 
 
 
 (6) " 
 
 2d 
 
 15. 
 
 (a) " 
 
 1st 
 
 
 (6) " 
 
 3d 
 
 quadrant, whose sine is . 
 
 U t( It 1 
 
 " cosine is i. 
 
 -. 
 
 ". . " tangent is 2. 
 
 it i i . a, o 
 
 " ctn is*. 
 
 secant is V2. 
 
 -V2. 
 esc is V5. _ 
 
 " -Vs. 
 
 Construct all possible values of : 
 
 16. sin- 1 ( f ). 18. tan- 1 ( 3). 20. sec- 1 ( 5). 
 
 17. cos- 1 (S). 19. ctn-^if). 21. esc- 1 ( 2). 
 
 22. Can the following be constructed? (a) sin- 1 !, (&) cos- 1 ( |), 
 (c) tan- 1 m, (d) ctn- 1 ( 676), (e) sec- 1 (db i), (/) esc- 1 VS. 
 
 23. Are the following possible ? 
 
 (a) tan- 1 ( 36,947), (b) sec- 1 ( 0.7435), (c) sin- 1 ( 25). 
 
CHAPTER IV. 
 
 A. Trigonometric Functions of Integral Multiples 
 
 of |> <f>. 
 
 31. The complement of an angle is the angle obtained 
 by subtracting the given angle from 90. The supplement 
 of an angle is the angle obtained by subtracting the given 
 angle from 180. Thus, the complement of 50 is 40; the 
 complement of 125 is 35. The supplement of 110 .is 
 70; the supplement of 200 is 20. 
 
 Two cases of integral multiples of are to be distin- 
 
 a 
 
 guished ; first, when the multiplier is odd ; second, when it 
 
 is even. Under the first case the simplest multiples of 
 
 & 
 
 are, when the angle is expressed in degrees, 90 and 270, 
 the multipliers being 1 and 3, respectively. Under the 
 second case the simplest examples are, similarly, 0, 180, 
 and 360, the multipliers being 0, 2, and 4, respectively. 
 The examples enumerated lead us to seek the trigonometric 
 functions of the special angles <, (90 </>), (180db <), 
 (270 <), and (360 <j>) in terms of the functions of <j>. 
 
 In comparing the functions of (m 90 <) with the func- 
 tions of < the triangles of reference will, for convenience, 
 be made geometrically equal by keeping the hypotenuse con- 
 stant in length ; but it is to be observed that the demonstra- 
 tions in no wise require this restriction, but depend on the 
 underlying principle that the homologous sides of similar 
 triangles are proportional. Also in this discussion only the 
 primary functions will be considered. 
 
INTEGRAL MULTIPLES. 
 
 25 
 
 32. The functions of < have already been given. If <j> is 
 taken in any quadrant, the terminal lines of <j> and of ( <) 
 will lie on the same side of the y-axis and on opposite sides 
 of the x-axis. From the numerical equality of the angles, the 
 triangles of reference for </> and for ( <) can always be con- 
 structed geometrically equal, by drawing a single perpendicular 
 to the x-axis and producing it in opposite directions until it 
 cuts the terminal lines. Trigonometrically the bases and 
 the hypotenuses of the triangles of reference are equal, the 
 perpendiculars are negatives of each other. It follows that 
 all the functions of ( <) into which the perpendicular of the 
 triangle of reference enters, will be the negatives of the same 
 functions of </>, and that the others will be equal to the same 
 functions of <. 
 
 The following figure illustrates the preceding results. 
 
 Y' 
 
 FIG. 15. 
 
 We have here ^ = x, r l = r, and y r = y. Therefore, 
 
 sin ( 0) = ^=:^ = sin<, esc ( <) = = --= csc <. 
 i"i r YI Y 
 
 COS( <f>) = *l = '-'- = COS<, Sec (<) = -=: ;- = S6C <. 
 
 i"i r Xi 
 
26 
 
 PLANE TRIGONOMETRY. 
 
 33. In constructing the triangle of reference for the angle 
 (m 90 + <), the angle will be constructed as (<j> + m 90). 
 We will first consider the cases where m = 0, 2, and 4. 
 <f> + 0.90 = <f>, whose functions have been given by definition. 
 
 If 180 is added to < in any quadrant, the triangles of 
 reference for (< + 180) and < are geometrically equal. The 
 hypotenuses are equal, and the perpendiculars and bases are 
 respectively negatives of each other. It follows that those 
 functions of (< + 180) into which either the base, or the 
 perpendicular, enters singly, are the negatives of the same 
 functions of < ; and that those functions of (<j> -+- 180) into 
 which both the base and the perpendicular enter are equal to 
 the same functions of <f>. 
 
 The following figure illustrates the preceding results. 
 
 We have here r 1 = r, x l = x, and y x = y. Therefore, 
 
 sin (180 + <) = ^ = ^ = 
 TI 
 
 esc (180 - 
 
 sin 
 
 = esc d>, 
 
 = = = cos 
 
 sec (180 + <h) = = = sec <f>. 
 
 \ r/ 
 
INTEGRAL MULTIPLES. 27 
 
 tan (180 +$) = =-2 = tan <j>, 
 Xj x 
 
 ctn (180 + <#>) = - = ^ - = ctn <. 
 
 It is to be observed that the triangle of reference for 
 (< + 180) may be obtained by revolving the triangle of 
 reference for < about the origin in the plane of the figure 
 through a positive angle of 180. This reverses the direc- 
 tions, and, therefore, the signs of the base and the perpen- 
 dicular. 
 
 It is also to be observed that in the preceding discussion 
 < is an angle of any quadrant, therefore is unrestricted, and 
 may be positive or negative. 
 
 34. If < is replaced by ( <), the formulas "of 33 
 become, by the use of the formulas of 32, 
 
 sin (180 $) = sin ( <) = sin <j>, 
 .'.esc (180 <)=csc<. 
 
 cos (180 <) = cos (<)= cos <, 
 /.sec (180 <) = sec <. 
 
 tan (180 <) = tan ( <) = tan <, 
 .\ctn (180 </>)= ctn <j>. 
 
 Observing that (< 180) = (180 <), we have by the 
 formulas of 32, and the present section, 
 
 sin (< 180) = sin (180 <) = sin <, 
 .'. esc (< 180) = esc <. 
 
 Similarly for the other functions. Or, otherwise, the 
 triangle of reference for (< 180) is identical with the 
 triangle of reference for (</> -f- 180), and the functions of 
 (< 180) are the same as the functions of (<f> + 180). 
 
 35. If 360 is added to <, the triangles of reference for 
 <, and (< + 360) coincide, and the functions of <, and 
 (< + 360) are identical, and we have : 
 
sin (360 ^ 
 
 ) = sin ( 
 
 .'.esc (360 4 
 
 .) = esc < 
 
 cos (360 < 
 
 ,) = cos ( 
 
 .'.sec (360 4 
 
 .) = sec <. 
 
 tan (360 < 
 
 ) = tan ( 
 
 .\ctn (360 < 
 
 ) = ctn < 
 
 28 PLANE TRIGONOMETRY. 
 
 sin (360 + 4>) = sin <, .'. esc (360 + <) = esc <. 
 cos (360 + <) = cos <, .-. sec (360 + <) = sec <. 
 tan (360 -f 0) = tan<, .'. ctn (360 + <) = ctn <. 
 
 Keplacing < by ( <) these formulas give by 32, 
 
 = sin <, 
 
 = cos <, 
 = tan<, 
 
 Observing that (< 360) = (360 <), we have by the 
 formulas of 32, and the present section : 
 
 sin (< 360) = sin (360 <) = sin <j>. 
 
 Similarly for the other functions. 
 
 The functions of (<^> 360) can also be directly seen to 
 be identical with the functions of <, since the triangles of 
 reference of the two angles coincide. 
 
 36. We will now consider the cases of (m 90 =b <) where 
 m = 1, and 3. If 90 is added to the angle < of any quadrant, 
 the triangles of reference for (< + 90) and <, are geometri- 
 cally equal. The hypotenuses are equal, the perpendicular 
 and the base of the triangle of reference for (< + 90) are 
 equal respectively to the base and the negative of the perpen- 
 dicular of the triangle of reference for <. It follows that any 
 function of (< + 90) is the co-function of <, and that it has 
 the opposite or the same sign according as the base of the 
 triangle of reference for (< + 90) does, or does not, enter 
 into the ratio. The following figure illustrates the preceding 
 results. 
 
 We have here, r x = r, y^ = x, Xi = y. 
 
INTEGRAL MULTIPLES. 
 
 29 
 
 cos 
 
 tan 
 
 ctn(<+90) = - 
 Y 
 
 FIG. 17. 
 
 It is to be observed that the triangle of reference for 
 (< + 90) may be obtained by revolving the triangle of 
 reference for < about the origin in the plane of the figure, 
 through a positive angle of 90. This converts the base into 
 the perpendicular, and the perpendicular becomes the base 
 changed in sign. 
 
 37. If in the formulas of 36, < is replaced by ( <), 
 they become by the use of the formulas of 32 : 
 
 sin (90 <) = cos ( <) = cos <, 
 cos (90 <) = sin ( <) = sin <, 
 tan (90 <#>) = ctn ( ^>) = ctn <, 
 
 .'. esc (90 <f>) = sec <f>. 
 .'. sec (90 <f>) =csc <#>. 
 .-. ctn (90 <) = tan <#>. 
 
 These formulas make clear the meaning of the prefix " co." 
 They show, for example, that the cosine of an angle <f> is the 
 complement's sine, [sin (90 <)]. That is, the word " cosine " 
 is an abbreviation for " complement's sine." Similarly for 
 the other functions. 
 
30 
 
 PLANE TRIGONOMETRY. 
 
 Since (< 90) = (90 <), we have, by the formulas 
 of 32, sin (< 90) = sin (90 <)= cos <, 
 
 .'. esc (< 90) = sec <f>. 
 
 38. If 270 is added to the angle < of any quadrant, the 
 triangles of reference for (< -f 270) and < are geometrically 
 equal, the hypotenuses are equal, the base and the perpen- 
 dicular of the triangle of reference for (< + 270) are equal, 
 respectively, to the perpendicular and the negative of the 
 base of the triangle of reference for <. It follows that any 
 function of (< + 270) is the co-function of <j>, and that it has 
 the same, or opposite sign, according as the perpendicular of 
 the triangle of reference for (< -j- 270) does not, or does, 
 enter into the ratio. 
 
 The following figure illustrates the preceding results. 
 
 r l = r, x x = y, y t = 
 
 We have here, 
 
 sin (< + 270) = = 
 r i r 
 
 cos O + 270) = - = ^ = 
 r i r 
 
 x. 
 
 = cos 
 
INTEGRAL MULTIPLES. 31 
 
 tan(< + 270) = ^ = = - ctn <. 
 
 x i y 
 
 .'. esc (<f> + 270) = sec <, sec (< -f 270) = esc <, 
 ctn (< + 270) = tan <. 
 
 39. The angle (360 -f </>) may be obtained by adding 
 180 to the angle (<f> -f 180) of 33, whence the triangle of 
 reference for (< -f- 180) is turned as before, through a positive 
 angle of 180, and, therefore, its base and perpendicular suffer 
 a second reversal. And, in general, the addition of m 180 to 
 0, produces, or does not produce, a reversal of the signs of the 
 perpendicular and base of the triangle of reference, according 
 as m is an odd or even integer, positive or negative. 
 
 The angle (< + 270) may be obtained by adding 180 to 
 (< + 90) of 36. This turns the triangle of reference, as 
 before, through 180, and reverses the signs of both the base 
 and perpendicular of the triangle of reference for (< -}- 90). 
 The perpendicular and base of the triangle of reference for <j> 
 have been converted respectively into the base and perpen- 
 dicular of the triangle of reference for (< + 270), and have 
 suffered, respectively, two, and one, reversals of sign. And, in 
 general, the addition of (2 n + 1) 90 to <j>, converts the base 
 and perpendicular of the triangle of reference for <, into 
 the perpendicular and base, respectively, of the triangle of 
 reference f or < + (2 n + 1) 90. And the base of the result- 
 ing triangle of reference has, or has not, while the perpen- 
 dicular has not, or has, been reversed in sign, according as n 
 is even or odd. 
 
 4O. It should be carefully observed that in the preceding 
 results the functions of any angle combined with a multiple 
 of 90 are equal numerically to the same, or to the co-functions 
 of the angle, according as the multiplier is one of the even 
 numbers, 0, 2, 4 ; or one of the odd numbers, 1, 3. 
 
 It can be seen from the discussion of the next section that 
 the following general relations exist : 
 
32 PLANE TRIGONOMETRY. 
 
 ( TT \ 
 Function ( 2 n <#> ) = Function <, numerically, 
 
 Function ( (2 n -j- 1) < J = Co-function <, numerically, 
 
 where n is zero, or any integer, positive or negative. 
 
 41. We will now give a general demonstration of the 
 
 relations of the functions of ( m </> ) to the functions of <, 
 
 \ J / 
 
 where m is any integer, positive or negative. Let x^ yi, 
 r 1? and x, y, r, respectively, be the sides of the triangles 
 
 7T '" 
 
 of reference for < + m > and <#>. The problem divides itself 
 & 
 
 into two cases : 
 
 a. When m is Even, and Equal to 2 n. 
 We have by 39 : 
 
 r = tan <, 
 
 2V *i (-l) n x 
 
 n 77 + <A 1 = Ctn ^>- 
 
INTEGRAL MULTIPLES. 38 
 
 42. Replacing <f> by ( <) in the formulas of 41, we get 
 by the formulas of 32 : 
 
 sin 2 n * = ( 1) sm ( *) = ( 1) sm 
 
 . . esc 
 
 cos I 2 n - * 1 = ( 1) cos ( *) = ( 1) ' cos 
 \ / 
 
 .'. sec f2n-* ) = (-!)" sec*. 
 
 tan f 2 n - < ) = tan ( <f>) = tan 
 
 .'. ctn 2 n - e = ctn <. 
 A 
 
 Formulas for functions off* 2n-J are covered by the 
 formulas of 41, since n may be negative as well as positive. 
 
 ( ir\ 
 
 Or, otherwise, the triangle of reference for ( * 2 n ) is 
 
 identical with the triangle of reference f or ( * + 2 n ) 
 
 \ * / 
 
 b. When m is Odd, and Equal to (2 n + 1). 
 43. We have by 39 : 
 
 r = r. 
 
 ri 
 csc ( (2 n + 1) + *) = (- l) n sec 
 
34 PLANE TRIGONOMETRY. 
 
 .-. sec ((2 n 
 
 + 1) + = - (- 1)" esc </>. 
 
 .'. ctii ( (2 n + 1) | + < J = - tan <. 
 
 44. Replacing < by ( <) in the formulas of 43, we 
 have by the formulas of 32 : 
 
 sin (2 n + 1) - = (- l) n cos (- *) = (- l) n cos <#>, 
 
 esc ^(2 n + 1) | - ^ - (- l) n sec <j>. 
 
 cos (2 n + 1) - = - (- l) n sin (- <) - (- l) n sin <^> 
 
 /. sec n 
 
 tan f (2 n + 1) ^ <^> J = ctn (- </>) = ctn <fc 
 
 Formulas for functions of ( <#> (2 n -\- 1) 1 are covered 
 
 by the formulas of 43, since n may be negative as well as 
 positive. 
 
 45. The results of 31-44 inclusive, are contained in 
 the following formulas : 
 
INTEGRAL MULTIPLES. 35 
 
 Sin z n 
 
 cos 2 n < = ( l) n cos <, 
 
 tan ( 2 n ^ < ) = tan <, 
 \ J / 
 
 / TT \ 
 
 sin f (2 n + 1) - <H = (-l) n cos <#>, 
 
 cos f (2 n + 1) | </>") - =p (- l) n sin </>, 
 tan ^ (2 n + 1) ^ <#> j = + ctn <. 
 
 46. The formulas of 45 may be put into the following 
 form : 
 
 Since ( l) n sin <f> = sin [( l) n <] whether n is even or odd, 
 ( l) n cos < = cos [( l) n <] only when n is even, 
 ( l) n tan < = tan [( l) n </>] whether n is even or odd, 
 
 we have, dividing by the sign factor of the second member of 
 the first three formulas of 45, and writing the second 
 members first : 
 
 sn 
 
 = ( l) n sin f 2 n ^ 
 
 cos <j> = ( l) n cos 2 n 
 
 = cos ( 4 m <^> ) where 2 m = n, 
 \ z 
 
36 PLANE TRIGONOMETRY. 
 
 tan ( 2 n ^ 
 
 tan <f> = tan 2n< tan 2 n + <f> 
 
 Since n may be positive or negative at pleasure, the 
 preceding can be written more simply : 
 
 sin < = sin (n TT + ( l) n </>), 
 cos < = cos (2 m TT <), 
 tan < = tan (n TT -j- <). 
 
 47. If in the formulas of 46, (n7r+( 1), (2m7r<), 
 (n7r-]-<), be successively denoted by 3>, then $ represents all 
 possible angles whose sines, cosines, and tangents, are respec- 
 tively equal to the sine, cosine, and tangent of the particular 
 angle <. Thus, if sin < = -J-, the simplest angle may be 
 constructed whose sine is -J-, and the general solution is 
 $ = n TT + ( l) n <> where < is the angle constructed. 
 
 B. Periodicity of the Trigonometric Functions. 
 
 48. When f (x + na}= f (x), where n is zero, or any posi- 
 tive or negative integer, f (x) is said to be periodic, and " a " 
 is called the period of f (x). From the formulas of 41, it 
 appears that if, in them, n = 2 m, where m is zero, or any 
 positive or negative integer, 
 
 sin (< -f- 2 m TT) = sin <, esc (< + 2 m TT) = esc fa 
 cos (<j> + 2 m TT) = cos fa sec (< + 2 m TT) = sec fa 
 
 and that always, 
 
 tan (^ -f- n TT) = tan <, ctn (< -f- n TT) = ctn fa 
 
 and that, therefore, the period of the sine, cosine, secant, and 
 cosecant is 2 TT ; and of the tangent and cotangent is TT. 
 
 > 
 
THE FUNCTIONS OF SOME SPECIAL ANGLES. 
 
 37 
 
 r. The Functions of Some Special Angles. 
 
 49. To find the functions 
 of 30. Let XOP be equal to 
 30, and let OMP be its tri- 
 angle of reference. 
 
 Produce PM to P', making 
 MP' = PM, and draw OP'. 
 The triangle OPP' is geomet- 
 p/ rically equilateral, and MP 
 
 FIG. 19. =iOP. 
 
 OM = VOP 2 -MP 2 = 
 From this we get, 
 
 == ^L=I.^L = :, .-.esc 30 = 2. 
 
 ./q 
 
 Vs 
 
 sec 30 = f V3. 
 ctn30 = V5. 
 
 5O. To find the func- 
 tions of 45. Let XOP be 
 equal to 45, and let OMP 
 be its triangle of reference. 
 By geometry OM = MP, and 
 OM 2 + P 2 = 2 OM 2 = 2 M~P 2 
 
 = 
 
 V2 
 
 f 
 
 '/ 
 
 HP 
 
 Y 
 
 
 
 M 
 
 
 Y' 
 
 FIG. 20. 
 
38 PLANE TRIGONOMETRY. 
 
 From this we get, 
 
 -Up 
 
 . , eo MP V2 1 1 /- , eo /- 
 
 Sm45 ;= = = = V2 CSC45 = 
 
 -UP - 
 
 _ OM V2 11 /- ._ /- 
 
 cos 46"== = -- = = ;V2, ,. sec 45 = 
 
 tan 45 = = 1, /. ctn 45 = 1. 
 
 Or otherwise thus : Since 45 90 45, we have, 
 sin 45 = sin (90 45) = cos 45. 
 sin 2 45 + cos 2 45 = 2 sin 2 45 = 2 cos 2 45 = 1, 
 
 .'. sin 45 = cos 45 = 4- = ^ V2. 
 
 V2 2 
 
 The other functions of 45 may be obtained from the sine 
 and cosine of 45. Since the angles 60, 120, 150, 210, 
 240, 300, 330, are cases of (m 90 30); and the angles 
 135, 225, 315, are cases of (m 90 45), therefore their 
 functions can be at once obtained in terms of the known 
 functions of 30 and 45, respectively, by the use of the 
 formulas of 33-38, inclusive. 
 
 D. Examples. IV. 
 
 1. Find the complements and the supplements of the following angles: 
 57, - 210, - 65, - 116, 3 r , - 2M5. 
 
 Find the functions of the following angles : 
 
 2. 60; 3. 120; 4. 135; 5. 150; 6. 210; 7. 225; 8. 240; 
 9. 300; 10. 315; 11. 330. 
 
 For convenience of reference the following table is given : 
 
EXAMPLES. IV. 
 
 39 
 
 Angles. 
 
 Functions. 
 
 sin 
 
 cos 
 
 tan 
 
 ctn 
 
 sec 
 
 CSC 
 
 0, 360 
 
 180 
 
 
 
 1 
 
 
 
 QO 
 
 1 
 
 CO 
 
 30, 150 
 210, 330 
 
 (H 
 
 (+-HV3 
 
 (+-HV3 
 
 (+-)V3 
 
 (+-)*V3 
 
 ()2 
 
 45, 135 
 225, 315 
 
 ()iVi 
 
 (+-HV2 
 
 (+-)! 
 
 (-f-)l 
 
 (+_)V2 
 
 ()V2 
 
 60, 120 
 240, 300 
 
 ()iV3 
 
 ( + -H 
 
 (+-)V3 
 
 (+-)*V3 
 
 (+_)2 
 
 ()fV3 
 
 90 
 270 
 
 1 
 
 
 
 00 
 
 
 
 00 
 
 1 
 
 In this table () is an abbreviation for ( ), (+_) for ( =j=), and 
 
 12. Find the functions of the following angles : 30, 420, 765, 510, 
 - 585, 3360, 1260. 
 
 Prove that, 
 
 13. sin 295 = cos 155. 
 tan 217 = ctn ( 127). 
 sec (- 37) = - esc 233. 
 cos 11 45 = sin (-205). 
 ctn (- 127) = - tan 503. 
 esc ( 464) = sec 1606. 
 
 Given tan 200 = 0.364, find sec 70. 
 1 
 
 14. 
 
 15. 
 16. 
 17. 
 
 18. 
 19. 
 
 20. Given sin 165 = -=. ( VI 1), find tan 285. 
 
 2V2 
 
 If A, B, and C are the angles of a plane triangle, prove that : 
 
 21. sin A = sin (B+C), cos A = cos(B + C), tan A = tan (B+C), 
 sin A/2 = cos J (B + C), cos A/2 = sin \ (B + C), tan A/2 = ctn |(B + C). 
 
 22. sin A/2 sin B/2 sin C/2 = cos \ (B + C) cos $ (C + A) cos i (A+ B). 
 
 23. cos A/2 cos B/2 cos C/2 = sin -j- (B + C) sin i (C + A) sin i (A + B). 
 
 24. tan A/2 tan B/2 tan C/2 = ctn -J- (B + C) ctn -J- (C + A) ctn (A+ B). 
 
 25. sin (A + B + C) = 0, cos (A + B + C) = - 1, tan (A + B + C) = 0. 
 
 26. sini(A+B + C) = l, cos|(A+ B + C) = 0, tan(A+ B+ C) = oo . 
 
 27. sin A = - sin (2 A + B + C). 
 
40 PLANE TRIGONOMETRY. 
 
 28. sin | (A - B) = cos i (2 B + C), tan i (3 A + B) = ctn | (C - 2 A). 
 
 29. sin n (A + B) - ( l) n sin nC. 
 
 30. cosn(A'+ B)= (-l) n cosnC. 
 
 31. tan n (A + B) = tan nC. 
 
 Find an expression for all possible angles : 
 
 32. Whose secants are the same as sec 0. 
 
 33. Whose cosecants are the same as esc 0. 
 
 34. Whose cotangents are the same as ctn 0. 
 
 35. Obtain formulas for the functions of (0 2 n rt/2) in terms of 
 the functions of 0, where n is always a positive integer. 
 
 36. The same for [0 (2 n + 1) 7t/2~\. 
 
 37. Obtain the functions of 2 n jr/2, (2 n + 1) it/2, 4 n if/2, (4 n + 1) if/2, 
 (4 n -f 2) if/2, (4 n + 3) if/2. 
 
 NOTE ON THE GENERAL SOLUTION OF TRIGONOMETRIC EQUATIONS. 
 A general solution of a trigonometric equation is all possible angles which 
 satisfy the given equation. When the equation is algebraic in form, the 
 general solution may be conveniently expressed by a finite number of 
 general formulas, each of which gives an infinite number of values. 
 Thus if the given equation be sin = -, an expression containing all 
 possible angles, whose sines are i, is n?r + ( l) n if/Q (by formulas of 
 47), and t> = mt + ( l) n ?r/6 is the general solution. This is an 
 algebraic equation of the first degree in terms of sin 0, and there is one 
 general formula which contains an infinite number of values in itself. 
 
 If 2 sin 2 3 sin + 1 = is the given equation, we have : 
 
 (2 sin 1) (sin 1) = 0, sin = , sin = 1, 
 *i= nx+ (-l) n 7T/6, $2 = r\Tt + (- l) n if/2, 
 
 as the general solution. And it will be found that the number of general 
 formulas is not greater than the degree of the equation. 
 Obtain general solutions of the following equations : 
 
 38. sin = Vj. 
 
 39. cos = 3-. 
 
 40. tan = Vfr. 
 
 41. sec 2 tan 1 = 0. 
 
 42. 2 sin cos 2 sin + cos 1 = 0. 
 
 43. 2 sin = esc 0. 
 
 44. 3 tan ctn = 0. 
 
 45. 4 cos = sec 0. 
 
 46. 3 sin 2 + cos >2 sin = 0. 
 
 47. cos 3 6 cos 2 + 1 1 cos 6 = 0. 
 
EXAMPLES. IV. 41 
 
 Show that the following general formulas give the same sets of values, 
 though not corresponding value by value, for the same value of n. 
 
 48. $ = n?r + ( l) n 5 x/Q, and 2 ntf + 7t/2 it/?,. 
 
 49. $ = - 7T/4 - nx + (- l) n tf/4, 4> = 7T/4 + 2 ntf Tt/. 
 
 50. Show that one value of each of the expressions, sin- 1 x + cos- 1 x, 
 tan- 1 x + ctn- 1 x, sec- 1 x + esc- 1 x, is jf/2. 
 
 51. Find one value of each of the following expressions: sin 1 3/5 + 
 
 sin- 1 4/5, sec- 1 5 + sec- 1 -^-= , tan- 1 3 + tan- 1 1/3. 
 
 2V6 
 
 52. If is the simplest value of sin 1 x, sec~ l x, tan 1 x, successively, 
 prove that in general, 
 
 sin- 1 x + cos- 1 x = [n;r + ( l) n 0] + [2 tntf (a/2 0)] 
 = [2 (2 m + n) 1] | + [(- l) n 1] 0. 
 
 sec-i x + csc-ix = [ 2 mt 0] 4- [nit + (- l) m (?r/2 - 0)] 
 
 = [2 (2 n + m) + (- l) m j 7t/2 + [(- l) m+1 l] 0. 
 
 tan- 1 x + ctn- 1 x = [mt + 0] + [IUTT + 7t/2 0] 
 = [2 (m + n) + 1] 7T/2, 
 
 where m and n may be 0, or any positive or negative integers, and are 
 independent of each other. How may m and n and the double sign be 
 taken to give the particular solutions of problem 50 ? 
 
CHAPTER V. 
 
 A. Projection. 
 
 51. The orthogonal projection of a finite line upon an 
 axis of reference is the segment of the axis taken from the 
 foot of the perpendicular from the origin of the line upon the 
 axis, to that from the term. 
 
 It will be proved that such projection of a straight line is 
 equal to the length of the line multiplied by the cosine of the 
 angle between the positive directions of the axis and the line. 
 For convenience, projection will be considered under two 
 cases : 
 
 a. When the Line and Axis are Co-planar. 
 
 Thus, when the line OP and axes, OX and OY, lie in the 
 same plane, the projections of OP on the axes, or fixed 
 parallel lines, are the segments of the axes, or fixed parallel 
 lines, taken from the feet of the perpendiculars dropped upon 
 them from 0, to those from P. These segments are obviously 
 the base and the perpendicular, respectively, of the triangle 
 of reference for the angle XOP. 
 
 FIG. 22. 
 
PROJECTION. 
 
 43 
 
 \ 
 
 In Figs. 21-24 we have 
 
 YOP - YOX + XOP = - 90 + $ + n 360. 
 
 MP v 
 sin <> -^p- = L ; sin <j> = cos (<j> 90 -f n 360). 
 
 OM x 
 cos < = -^_- = - And these give 
 
 y = r cos (< 90 + n 360), and x = r cos <, 
 or MP = OP cos YOP, and OM = OP cos XOP. 
 
 Since n may be or any integer without affecting the 
 preceding result, it will be so chosen, for uniformity, that 
 the angle YOP shall be always the simplest positive angle. 
 
 b. When the Line and Axis are not Co-planar. 
 
 52. When the line and axis are not co-planar, the 
 angle between them is the angle taken from the positive 
 direction of the axis to the positive direction of a line 
 parallel to the given line, and co-planar with the axis. It 
 is evident that if two planes o and p be passed respectively 
 through the extremities and P of the line OP, and perpen- 
 dicular to the axis, the segment of the axis from plane o 
 to plane p, is the projection of OP on the axis. 
 
 In the figure, O'P' is parallel to OP, cuts the axis, and 
 pierces o and p in 0' and P', respectively. O'M and P'M" 
 lie in o and p, and cut the axis at M and M'. 
 
44 
 
 PLANE TRIGONOMETRY. 
 
 0' 
 
 FIG. 25. 
 
 By geometry, 0'P' = OP, 0'M"= MM'. O'M and P'M" are 
 perpendicular to MX. 
 ' By a. 0'M" = 0'P'cos<, or MM' = OPcos<. 
 
 53. It follows from the preceding work that the projec- 
 tion of the straight line P upon any axis is equal to the sum 
 of the projections of the parts of an uninterrupted broken 
 line going from to P. 
 
 The following notation will be used. The projection of a 
 line, I, on an axis, a, will be denoted by l a . 
 
 B. Functions of Sums and Differences of Angles. 
 
 54. In this figure, from the triangle of reference for /2, 
 
 FIG. 26. 
 
FUNCTIONS OF SUMS AND DIFFERENCES OF ANGLES. 45 
 
 cos (3 = - .-. OM = OP cos ft. 
 
 By 53, OP y = OM y +MP y . 
 
 These become : 
 
 OP sin (a + /5) = OM sin a -f M P cos a,* 
 or, by substituting the values of OM and MP, 
 
 OP sin (a + 0) = OP sin a cos (3 -f OP cos a sin /?. 
 
 .'. sin (a + /?) = sin a cos /8 + cos a sin /?. 
 OP cos (a + P) = OM cos a + M P cos (90 + a)* 
 
 which reduces to 
 
 cos (a + /?) = cos a cos |8 sin a sin /?. 
 
 55. If in the formulas of 54 ft is replaced by ( /?) 
 we obtain : 
 
 sin (a ft) = sin a cos /3 cos a sin /?. 
 cos (a /8) = cos a cos /3 -f- sin a sin /?. 
 
 56. From the formulas of 54 and 55 
 
 _^_ sin (a =b /3) _ sin a cos /3 cos a sin /8 
 
 ~~ ^ cos (a /3) cos a cos /S =p sin a sin y8 
 
 sin a cos /3 cos a sin /? 
 _ cos a cos /3 ~~ cos a cos /3 _ tan a tan ^8 
 
 cos a cos /3 sin a sin (3 1 =p tan a tan fi 
 
 cos a cos /8 ^~ cos a cos /? 
 
 * The values of these projections are in all cases justified by the 
 principle that if a line taken in the opposite direction, and for the sake 
 of the triangle of reference for /3, has its sign changed, then the projec- 
 tion becomes I cos (0 ip 180) = I cos 0, generally as before, with the I 
 essentially negative. 
 
46 PLANE TRIGONOMETRY. 
 
 / _4_ o\ 1 _ 1 =F tan a tan (3 
 tan (a /?) ~ tan a tan j3 
 _ ctn a ctn /? =F 1 
 ctn /3 ctn a 
 
 C. Functions of Multiple Angles. 
 
 57. If in the formulas of 54 and 56 ft is replaced by a, 
 
 sin (a -f- a) = sin 2 a = sin a cos a + cos a sin a = 2 sin a cos a. 
 cos (a -f" a) = cos 2 a = cos a cos a sin a sin a = cos 2 a sin 2 a 
 2 cos 2 a 1 = 1 2 sin 2 a. 
 
 tan a + tan a 2 tan a 
 
 tan ( a -+ a } = tan Z a = : = r- 
 
 1 tan a tan a 1 tan 2 a 
 
 ctn a ctn a 1 ctn 2 a 1 
 
 ctn (a + a) = ctn 2 a = 
 
 ctn a + ctn a 2 ctn # 
 
 />. Functions of Sub-Multiple Angles. 
 
 58. By the formulas of 57 
 
 cos 2 a' = 1 2 sin V = 2 cosV - 1. 
 If in these formulas 2 a' is replaced by a, 
 
 cos a = 1 2 sin 2 - > .'.2 sin 2 - = 1 cos a. 
 
 1 cos a 
 
 And cos a = 2 cos 2 - 1, .'. 2 cos 2 - = 1 -|-cosa. 
 
 w 
 
 a^Jl + cosa. 
 2 
 
 1 cos a 
 
 1 + cos a 
 
 cos a 
 
 cos a 
 
FORMULAS. 47 
 
 E. Formulas for the Sums and Differences of the 
 
 Sines of two Angles, also of the Cosines 
 
 of two Angles. 
 
 59. By 54 and 55 
 
 sin (a' + ft') + sin (a' ft') =2 sin a' cos '. 
 sin (a' + ft') sin (a 1 ft 1 ) = 2 cos a' sin /?'. 
 cos (a 1 -f )3') + cos (a' ft') = 2 cos a' cos 0'. 
 cos (a' ') cos (a 1 + ft') = 2 sin a' sin ft'. 
 
 In these formulas let a' + /?' = a, 
 
 and the formulas become : 
 
 sin a + sin /? = 2 sin -J (a -f- /3) cos % (a ft). 
 sin a sin /5 2 cos -J- (a + ft) sin -J- (a /8). 
 cos a -h cos ^8 = 2 cos |(a + ft) cos (a ft). 
 cos /3 cos a = 2 sin (a -f- )8) sin i (a ft). 
 
 sin a + sin ^ _ tan (a 
 
 sin a sin ^8 tan|-(a 
 
 sin a + sin ft 
 
 = 
 cos a + cos ^8 
 
 sin a + sin 
 
 cos ft cos a 
 sin a sin ft 
 cos a + cos ft 
 sin a sin ft 
 cos ft cos a 
 cos a -\- cos ft 
 cos 8 cos a 
 
 = ctn -J- (a j8). 
 
 N 
 = tan ^(a ft). 
 
 = ctn ^ (a + 0) ctn %(a 
 
48 PLANE TRIGONOMETRY. 
 
 F. Examples. V. 
 
 Find the functions of : 
 
 1. 15. 2. 22|. 3. 37. 4. 75. 
 
 Prove : 
 
 5. ctn 2 a = % (ctn a tan a). 
 
 6. esc 2 a = % (tan a+ ctn a). 
 
 7. tan a = esc 2 a ctn 2 or. 
 
 8. ctn a = esc 2 a + ctn 2 or. 
 
 9. tan (45 + $ a) = sec a + tan <* = ctn (45 -- a). 
 
 10. tan (45 - i a) = sec a - tan a = ctn (45 + i a). 
 
 11. tan (45 + i a) tan (45 - i a) = 1. 
 
 12. tan 50 + ctn 50 = 2 sec 10. 
 
 13. tan 50 ctn 50 = 2 tan 10. 
 
 14. cos (60 + a) + cos (60 a) = cos a. 
 
 15. sin (30 + or) + sin (30 a) = cos a. 
 
 16. sin (45 + a) = cos (45 - a). 
 Find: 
 
 17. sin (a + + 7). 
 
 18. cos (a + + 7). 
 
 19. tan(a + p + y). 
 
 20. sin 3 a in terms of sin a. 
 
 21. cos 3 a in terms of cos or. 
 
 22. tan 3 a in terms of tan a. 
 
 23. sin 4 a: in terms of sin a and cos a. 
 
 24. cos 4 a: in terms of cos a. 
 
 25. tan 4 <* in terms of tan a. 
 
 26. sin 5 a in terms of sin a. 
 
 27. cos 5 a in terms of cos a. 
 Prove : 
 
 28. sin a + sin /3 + sin 7 sin (a + + 7) 
 
 = 4sin(/3 + 7)sin|(7 + a)sini(a + /S). 
 
 29. cos a + cos /3 + cos 7 + cos (a + /3 + 7) 
 
 = 4 cos | (j8 -f 7) cos (7 + a) cos (a + /3). 
 
 30. tan a + tan p + tan 7 tan a tan /3 tan 7 = sm v* Zl 
 
 cos or cos /3 cos 7 
 
 31. ctn a + ctn fl + ctn 7 ctn a ctn fl ctn 7 = . -~ -' 
 
 sin a sin p sin 7 
 
 32. sin 2 x + sin 2 y + sin 2 z + sin 2 (x + y + z) 
 
 = 2 [1 cos (y + z) cos (z + x) cos (x + y)]. 
 
EXAMPLES. VI. 49 
 
 33. cos 2 x + cos 2 y + cos 2 z + cos 2 (x + y + z) 
 
 = 2 [1 + cos (y + z) cos (z + x) cos (x + y)]. 
 
 G. Examples. VI. 
 
 1. If 7 g , 5, and p r be the three measures of an angle, prove that 
 
 r -,=*>*. 
 
 7t 
 
 2. A bicycle is propelled on a circular arc of mile radius at the rate 
 of 30 miles per hour. Through what angle has it turned in two minutes ? 
 
 Prove the following identities : 
 
 3. 
 
 csc 2 sec 2 
 
 4. sin 2 (1 + n ctn 2 0) + cos 2 (1 + n tan 2 0) 
 
 = sin 2 (n + ctn 2 0) + cos 2 (n + tan 2 0). 
 
 5. sin 6 cos 6 = (2 sin 2 1) (1 sin cos 0) (1 + sin cos 0). 
 (tan 0-1) (esc - 1) ctn + 1 _ 
 
 ctn 1 (tan + 1) (esc 0+1) 
 
 7. Find sin 18 and cos 18. (Use the geometric properties of the 
 decagon.) 
 
 8. Find sin 54 and cos 54. 
 Prove the following identities : 
 
 9. sin 65 + cos 65 = V2 cos 20. 
 10. sin 78 sin 18 + cos 132= 0. 
 
 sin sin 2 + sin 2 sin 5 + sin 3 sin 10 _ 
 
 cos sin 20 + sin 2 cos 50 cos 3 sin 10 
 
 sin + sin 3 + sin 5 + sin 7 
 " cos + cos 3 + cos 5 + cos 7 ~ *" 
 
 3 sin sin 2 + sin sin 4 + sin 2 sin 7 _ 
 
 sin cos 2 + sin 2 cos 5 + sin cos 8 ~~ 
 
 eoe (a -8ffi--co.(8 a -fl _ 
 
 sm 2 + sin 2 (3 
 
 16. = 2co g _ 
 
 sm 2 a + sm 2 /3 
 
 a sin (a ft) + b sin or + a sin (or + ft) _ 
 a cos (a /3) + b cos a + a cos (a + j8) ~~ 
 
 17. sin (a + ft + 7) + sin ( a + /3 + 7) + sin (a /S + 7) 
 
 + sin (a + j8 7) = 4 sin a sin j3 sin 7. 
 
 18. cos (a + j8 + 7) + cos ( a + p + 7) + cos (a + 7) 
 
 + cos(a: + 7) = 4 cos a cos cos 7. 
 
50 
 
 PLANE TRIGONOMETRY. 
 
 19. 8 cos 
 
 a + 
 
 y 
 
 cos 
 
 7 
 - 
 
 2 2 2 
 
 = cos 2 a 4- cos 2 /3 + cos 2 7 + 4 cos a cos /3 cos 7 + 1. 
 
 ctn (n 2) ctn n + 1 
 
 20. s -A. li_ r = (ctn0 tan0). 
 
 ctn (n 2) ctn n 
 
 21. 
 
 sin n sin (n 2) _ 
 
 = ctn (n 1) 0. 
 
 cos (n 2)0 cos n 
 Justify the following statements : 
 
 22. sin- 1 a sin- 1 b = sin* 1 [a Vl b 2 b Vl - a 2 ]. 
 
 23. cos- 1 a cos- 1 b = cos- 1 [ab ^ V(l - a 2 ) (1 - b 2 ) ]. 
 
 ab 
 
 24. tan- 1 a tan- 1 b = tan- 1 
 
 ab 
 
 25. tan- 1 a + tan- 1 b + tan- 1 c = tan- 1 
 
 26. tan- 1 a + tan- 1 b + tan- 1 
 
 a 
 
 b + c-abc 
 
 1 ab be ca 
 1 a b ab_g 
 4* 
 
 1+a+b- 
 
 27. a VI - b 2 + b VI - a 2 = 1, given sin- 1 a + sin- 1 b = 
 
 28. sin- 1 a - sin- 1 b = cos- 1 [ab + Vl - a 2 - b 2 + a 2 b 2 ]. 
 
 29. a* = i, given tan- 1 |-| + tan- 1 1^| = * 
 
 30. a + b + c = abc, given tan- 1 a + tan- 1 b + tan- 1 c = it. 
 
 31. 2 sin ^ = + Vl + sin + Vl sin 0, given = 200. 
 
 32. 2 sin | = Vl + sin Vl sin 0, given 450 < < 630. 
 
 33. 2 cos | = Vl 4- sin + Vl sin between fixed limits. Deter- 
 mine the limits. 
 
 Solve for in the following equations : 
 
 34. a cos + b sin = 1. 
 
 Suggestion : a cos + b sin = a ( cos + - sin Y Put - = tan a 
 
 and get cos (0 or) = , from which the general value of can be 
 
 found. 
 
 35. sin + cos = 1. 
 
 36. sin cos = 1. 
 
 37. sin + cos = Vj. 
 
 38. sin0 + V3 cos0= 1. 
 
 39. V3 sin cos = N/2. 
 
EXAMPLES. VI. 51 
 
 40. sin + 2 cos = -J-. 
 
 41. cos (a + 0) - sin (ex + 0) = V2 cos /5. 
 Construct the graphs of the following expressions : 
 
 42. sin + cos 0. 
 
 43. sin cos 0. 
 
 44. tan0+ctn0. 
 
 45. tan0 ctn0. 
 
 46. sec + esc 0. 
 
 47. sec esc 0. 
 
 48. sec + tan 0. 
 
 49. sec tan 0. 
 
CHAPTER VI. 
 
 A. Right Triangles. 
 
 61. To solve a triangle is to find its unknown from its 
 known parts. It is shown in plane geometry that to deter- 
 mine a triangle three parts, one of which is a side, must be 
 given. In right triangles the right angle counts as one of the 
 three given parts. 
 
 a. Cases. 
 
 62. The following table gives all the cases which can 
 arise in the solution of right triangles. 
 
 Case. 
 
 Given. 
 
 Required. 
 
 1 
 
 *, y 
 
 0, 90 0, r 
 
 2 
 
 r, x or y 
 
 0, y or x, 90 - 
 
 3 
 
 r. or 90 
 
 x, y, 90 or 
 
 4 
 
 x or y, or 90 
 
 r, 90 or 0, y or x 
 
 b. Solutions. 
 
 Since the angles <f> and 90 < are acute, and, therefore, of 
 the first quadrant, all their functions are positive. We will 
 
 Y 
 
 X'- 
 
 M 
 
 Y' 
 
 FIG. 27. 
 
EXAMPLES, VII. 53 
 
 express all solutions in terms of the given parts, but some- 
 times, in order to use logarithmic computation, it will be 
 necessary to express an unknown part in terms of a given 
 part, and a part already found. 
 
 Case 1. By 18, tan < == ctn (90' 
 
 Also, r = +Vx 2 -fy 2 . 
 
 But in order to use logarithms we take r = x sec $ = y esc <. 
 
 Case 2. By 18, cos <f> = sin (90 <) = - 
 
 Also, y = + Vr 2 - x 2 = + V(r + x) (r x). 
 
 Slog y = * [log (r + x) + log (r-x)];. 
 
 If y is given, the acute angles and x can be found in a 
 similar way. 
 
 Case 3. By 18, x = r cos tj>. 
 y = r sin <. 
 
 Solve similarly if the other acute angle is given. 
 
 Case 4. By 18, r x sec <. 
 
 y = x tan <. 
 
 Solve similarly if y is given. 
 
 B. Examples. VII. 
 
 Find the unknown parts of the following right triangles, having given : 
 
 1. x = 273, y= 115. 
 
 2. x = 23, y = 7. 
 
 3. r = 97, x = 16. 
 
 4. r = 42, y = 37. 
 
 5. r = 67, = 43 25'. 
 
 6. r = 19, 90 - = 27 17 X . 
 
 7. x = 17, = 47 32'. 
 
 8. y = 211, 90 - = 47 16'. 
 
54 
 
 PLANE TRIGONOMETRY. 
 
 C. Practical Applications to Problems on Heights 
 and Distances. 
 
 63. The angle of elevation, or de- 
 pression, of a point P is the angle between 
 a horizontal line and the line of vision to 
 the point P, both lines being taken in the 
 same vertical plane. In the accompany- 
 ing figure the eye of the observer is sup- 
 posed to be at 0. In the case of a heav- 
 enly body its angle of elevation is called 
 its altitude. In other cases altitude means 
 the height of a point. Many problems on 
 heights and distances can be made to 
 depend upon the solution of right tri- 
 angles. 
 
 FIG. 28. 
 
 />. Examples. VIII. 
 
 1. When the sun's altitude is 30, the length of the shadow of Bunker 
 Hill monument on the horizontal plane is 381.1 ft. How high is the 
 monument ? 
 
 2. At the distance of 75 ft. from its base, the angle of elevation of the 
 top of a tree standing on a horizontal plain is 63 26'. How high is the 
 tree? 
 
 3. A steeple is 120 ft. high. What is the angle of elevation of its top 
 at a distance of 75 ft. from its base ? 
 
 4. A man on the top of a tower 250 ft. high observes an object whose 
 angle of depression is 53 27'. How far is the object from the base of the 
 tower ? 
 
 5. At two points, 75 yards from each other, in a horizontal line, and 
 in the same vertical plane with the top of a hill, the angles of elevation 
 are observed to be 45 and 60. Find the height of the hill and the 
 distance from the nearer point to the point directly under the top of 
 the hill. 
 
 6. A besieging party comes to a channel at a point directly opposite a 
 fortified castle of the enemy. They measure away from the shore, and 
 in a line with the castle, 500 yards. The angles of elevation of the top 
 of the castle are observed to be 16' 15", and 15' 50" at the extremities of 
 the base line respectively. How high is the castle ? If the maximum 
 
EXAMPLES. VIII. 55 
 
 range of their guns is ten miles, will there be any use in their trying to 
 storm the castle ? (log esc 25" = 3.9105.) 
 
 7. From the top of a tower 175 ft. high two objects on the horizontal 
 plane, in a line with the base of the tower, are observed to have angles of 
 depression of 60 and 70 respectively. What are their distances from 
 the foot of the tower, and from each other ? 
 
 8. At two points, a feet from each other, in a horizontal line, and in 
 the same vertical plane with the top of a tower, the angles of elevation 
 are observed to be a and /3, respectively, where a > /3. Find the height, 
 h, of the tower, and the distance, b, from the nearer point of observation 
 to the base of the tower. 
 
 Ans. h = a sin a sin p esc (a if /3), b = a cos a sin /3 esc (a q= /9), accord- 
 ing as the points are on the same, or on opposite sides of the tower. 
 
 9. Use the formulas obtained in Problem 8 to solve Problems 5 and 6. 
 
 10. From the top of a tower h ft. high two objects in the horizontal 
 plane, in a line with the base of the tower, are observed to have angles of 
 depression of a and /3, respectively, where a > /3. What are their 
 distances from the foot of the tower, and from each other ? 
 
 11. A farmer wishes to make a ditch 100 rds. long on level land. If 
 the inclination of the bottom of the ditch to the horizon is 12', and the 
 depth of the upper end of the ditch is 12 in. , what is the depth of the 
 lower end ? 
 
 12. A winding railroad is built around a cylindrical tower whose 
 radius is 100 ft. and height 500 ft. If the maximum inclination to the 
 horizon which can safely be used is 10, how many times will the road 
 wind around the tower ? If the road is 3 ft. wide, what is the length of 
 a line midway between the rails ? 
 
 13. The length of a road (ascent 1 ft. in 5), from the foot to the top 
 of a hill, is If mi. What will be the length of a zigzag road (ascent 1 ft. 
 in 12) ? 
 
CHAPTER VII. 
 
 A. Some General Formulas for Oblique 
 Triangles. 
 
 64. In treating the oblique triangle we limit the dis- 
 cussion to ^ geometrical triangles. We shall denote the sides 
 of a triangle by a, b, c, and the angles opposite these sides by 
 A, B, C, respectively. 
 
 a. Theorem of Sines. 
 
 65. In any plane triangle the sides are proportional to 
 
 the sines of the opposite 
 angles. 
 
 Let ABC represent any 
 plane triangle. Let p rep- 
 resent a perpendicular let 
 fall from any vertex upon 
 the opposite side. 
 
 By 18, p = b sin A, 
 
 p = a sin B, 
 .'. b sin A = a sin B. 
 
 Similarly, c sin B = b sin C. 
 
 a sin C = c sin A. 
 
 These equations may be written : 
 
 a b c 
 
 sin B ~~ 
 
 or 
 
 sin A sin B sin C 
 a : b : c I : sin A : sin B : sin C. 
 
GENERAL FORMULAS FOR OBLIQUE TRIANGLES. 57 
 
 6. Theorem of Tangents. 
 
 66. From 
 
 sin A 
 sin B 
 
 a+ b _ sin A + sin B _ tan j- (A + B) 
 'a b sin A sin B tan-j-(A B) 
 
 by 60. 
 
 Similarly, 
 
 b ctan(B C) 
 c + a = tan j- (C + A) 
 c a tan $ (C A) 
 
 c. Theorem of Cosines. 
 
 67. Formulas for one side of a triangle, in terms of the 
 other two sides, and their included angle. 
 
 C 
 
 FIG. 30. 
 Whatever allowable value of A is used, 
 
 v 
 
 - = cos A, or x b cos A. 
 b 
 
 Similarly, 
 
 = y 2 + (c - x) 2 
 
 = f + x 2 + c 2 - 2 ex 
 
 = b 2 + c- - 2 be cos A. 
 
 b 2 = c 2 -f- a 2 '2 ca cos B. 
 
 C 2 = a 2 + b 2 2ab cos C 
 
58 
 
 PLANE TRIGONOMETRY. 
 
 d. Formulas for one Side of a Triangle, in Terms of the 
 Adjacent Angles, and the other two Sides. 
 
 FIG. 31. 
 
 68. By 53, 
 
 or 
 
 Similarly, 
 
 a = c cos B + b cos (360 C). 
 a = b cos C + c cos B. 
 b = c cos A + a cos C. 
 c a cos B + b cos A. 
 
 ABC 
 e. Formulas for the Functions of -, , 
 
 . By 67, cos A = 
 
 b 2 + c 2 -; 
 2 be 
 
 sin A = 
 
 1- 
 
 b 2 +c 2 -a 2 
 
 2 be by 58, 
 
 2 
 
 _Ja 2 -(b-e) 2 
 4 be 
 
 _J(a-b + 
 
 b-c). 
 
 4 be 
 
 Put 
 Then, 
 
 s = 
 
 a+b 
 
 -a+b+c 
 
 a- b + c 
 
GENERAL FORMULAS FOR OBLIQUE TRIANGLES. 59 
 a + b-c 
 
 s c 
 
 sin 
 
 _ 2(s-b)2(s-c) 
 y 4bc 
 
 . - J(s-bxr^). 
 
 * be 
 
 Similarly, sin B = + 
 
 By 58, 
 
 2 be 
 
 4 be 
 
 =+v^- 
 
 Similarly, cos B = + A/ S (s ~ b) 
 
 ca 
 
 cos 4- C = 
 
 Since tan A = 
 
 cos A 
 
 Similarly, tan * B = + ($ c) ($ ,^ a) 
 
 s (s b) 
 
60 
 
 PLANE TRIGONOMETRY. 
 
 f. Formulas for the Area of a Triangle, and the Radii of the 
 Inscribed and Circumscribed Circles. 
 
 7O. By geometry, 
 area of the triangle. 
 But, 
 
 By 65, 
 
 represents the 
 
 y = b sin A. 
 = be sin A. 
 a sin B 
 
 b = 
 
 c = 
 
 sin A 
 a sin C 
 
 sin A 
 a 2 sin B sin C , a 2 sin B sin C 
 
GENERAL FORMULAS FOR OBLIQUE TRIANGLES. 
 
 61 
 
 Let be the center of the inscribed circle whose radius is r. 
 By geometry, AM = AP [= a']. 
 
 BM = BN[=b f ]. 
 
 c'+a' = b. 
 a'-fb' = c. 
 
 Adding and dividing by 2, 
 a'+b' + c' = s. 
 
 .-. a'= s a. 
 b'=s-b. 
 c' = s-c. 
 
 A r 
 .*. tan = 
 
 2 s a 
 
 B r 
 
 tan = 
 
 2 s b 
 
 C r 
 
 tan - = 
 
 2 s c 
 
 .-. 4. 
 
 = rs = Vs (s - a) (s - b) (s - c). 
 
 A 
 
 b , 
 
 Let P be the center of the circumscribed circle whose 
 radius is R. By geometry we have the parts as designated 
 in the figure. 
 
62 PLANE TRIGONOMETRY. 
 
 a=B. 
 
 .-.a = 90 A. 
 
 a. 
 
 2 
 
 cos a = sin A = -ir 
 R 
 
 .-. R=|cscA. 
 
 be 
 By 1, esc A = 
 
 5. R= abc 
 
 This demonstration, in connection with 65, gives inci- 
 dentally the interesting relation 
 
 sin A sin B sin C 
 
 The formulas of this chapter enable us to solve any plane 
 triangle when the necessary parts are given, and also to find 
 its area. 
 
 B. Oblique Triangles. 
 a. Cases. 
 
 71. There are four cases. 
 
 1. Given two angles and a side. 
 
 2. Given two sides and an angle opposite one of them. 
 
 3. Given two sides and the included angle. 
 
 4. Given three sides. 
 
OBLIQUE TRIANGLES. 
 
 63 
 
 b. Solutions. 
 
 Case 1. The third angle is the supplement of the sum of 
 the two given angles.* The required sides are given by the 
 formulas of 65. For example, if a is the given side, 
 b = a sin B esc A. c = a sin C esc A. 
 
 Case 2. If a, b, and A are given, 
 
 ,m B = ^^ by 66. 
 
 Since B is found from its sine, and since there are always 
 two angles, each less than 180, which have the same sine, 
 B may have two values, and the triangle may have two 
 solutions. But if A is obtuse, there cannot be more than one 
 solution. The cases may be discriminated as follows : 
 
 (1) A acute. 
 
 (1) Two solutions when 
 b > a > b sin A. 
 
 (2) One solution when 
 a > b, a = b (isosceles), 
 
 a = b sin A (right angled). 
 
 (3) No solution when 
 a < b sin A. 
 
 (#) A obtuse. 
 
 (1) One solution when a > b. 
 
 (2) No solution when b ^> a. 
 Case 8. Given a, b, and C. 
 
 (A-B) = a ~ b 
 
 FIG. 35. 
 
 
 By 66, tan 
 
 . 
 a -)- b 
 
 Since, tan \ (A + B) = ctn % C. 
 
 This gives tan J (A B), and therefore | (A B). 
 Then, i( A+B).+ i(A- B) = A. 
 
 i(A+B)-f(A-B)=B. 
 
64 
 
 PLANE TRIG GNOME TR Y. 
 
 The required side can now be found by 65. 
 
 Case 4' Given a, b, and c. 
 
 By 69 or 70 the angles can be found. 
 
 The tangent formulas should be used in practice. 
 
 FIG. 36. 
 
 C. Examples. IX. 
 
 1. A = 67 24', B = 39 18', a = 15. 
 
 2. C = 25 19', A = 74 30', c = 23. 
 
 3. a = 90, b = 100, A = CO . 
 
 4. b = 65, a = 124, A= 117. 
 
 5. c = 96, b = 45, C = 30. 
 
 6. a = 25, b = 19, C - 59. 
 
 7. c = 43, b = 29, A = 40. 
 
 8. a = 25, b = 33, c = 44. 
 
 9. a = 37, b = 14, c = 29. 
 
 10. a = 29, b = 57, c = 27. 
 
 11. Are the following triangles possible? If impossible, what con- 
 ditions are violated ? " If possible, have they one or two solutions ? 
 
EXAMPLES. IX. 65 
 
 (1) A = 95 27', B = 84, 35', a = 73. 
 
 (2) a = 47, b = 53, A = 91 16'. 
 
 (3) b = 27, c = 54, B = 30. 
 
 (4) a = 65, b = 70, A = 58. 
 
 (5) a = 34, b = 79, c = 44.5. 
 
 12. The angle of elevation of a balloon which is ascending uniformly 
 and vertically, when it is one mile high is observed to be 35 20'; 
 20 minutes later the elevation is observed to be 55 40'. How fast is the 
 balloon moving ? 
 
 13. A base line, AB, 1000 ft. long is measured along the straight bank 
 of a river ; C is an object on the opposite bank ; the angles BAG and CBA 
 are observed to be 65 37' and 53 4' respectively. Find the breadth of 
 the river at C. 
 
 14. A column on a pedestal 20 ft. high subtends an angle 45 to a 
 person on the ground ; on approaching 20 ft. it again subtends an angle 
 45. Find the height of the column. 
 
 15. A and B are two points on opposite sides of a swamp, and C is a 
 point visible from both A and B. AC is 15 rods, and BC 12 rods; the 
 angle ACB is 60. Find the distance AB. 
 
 16. Two mountains, each one mile high, are just visible from each 
 other over the sea. If the radius of the earth is taken equal to 4000 miles, 
 how far are the mountains apart ? 
 
 17. Two men standing at the same point, C, observe the horizontal 
 angle subtended by the line joining two inaccessible objects, A and B ; 
 they then move away, one in the direction AC to D, the other in the 
 direction BC to E, until each observes the horizontal angle to be half 
 what it was before. ACB = 30, CD = 100, CE = 200. Determine AB. 
 
 18. A base line 400 yards long, whose upper end is 8 yards higher than 
 the lower, is measured in the same vertical plane with the top of a hill. 
 The angles of elevation of the top of the hill, measured at the lower and 
 upper ends of the base line, are found to be 5 17' and 3 17' respectively. 
 Find the height of the hill. 
 
 19. A pole 100 ft. high stands in the center of a horizontal equilateral 
 triangle. At the top of the pole each side subtends an angle 60. Find 
 the length of a side of the triangle. 
 
 20. A regular pyramid on a square base has an edge of 150 ft. in 
 length. The side of the base is 200 ft. long. Find the inclination of the 
 face to the base. 
 
 21. The distance between two lighthouses, one of which bears E from 
 the other, is 4 miles. The bearings of the lighthouses from a ship are 
 
66 PLANE TRIGONOMETRY. 
 
 E.b.N. and N.W.b.W. respectively. What is the distance of the ship 
 from each lighthouse ? 
 
 22. Two cruisers start from the same point, and steam at the rate of 
 19 and 23 knots respectively, one S. E., and the other S.|W. How far 
 apart are they in five hours ? 
 
 23. A man standing on a horizontal plain at a distance a feet from a 
 tower observed that a flagstaff on the tower subtends an angle a, and that 
 on walking 2 b feet towards the tower the flagstaff again subtends the 
 same angle. Prove that the height of the flagstaff is 2 (a b) tan a feet. 
 
 24. A tower and a spire on its top subtend equal angles at a point 
 whose distance from the foot of the tower is a. If h is the height of the 
 
 tower, prove that the height of the spire is h 2 _ 
 
 25. A lighthouse is N.b.W. of a ship. After the ship sails 21 miles 
 E.S. E., the bearing of the lighthouse is N.W. What is the distance of 
 the lighthouse from each position of the ship ? 
 
 26. Find the area of a quadilateral field whose sides, AB, BC, CD, are 
 20, 15, and 18 respectively, and whose angles, B and C, are 85 and 116 
 respectively. 
 
SPHERICAL TRIGONOMETRY. 
 
 CHAPTER I. 
 A. Introduction. 
 
 a. Geometrical Principles, 
 
 72. 1. Every tangent to a sphere is perpendicular to the 
 radius drawn to the point of contact, and is also tangent to 
 that great circle arc in whose plane it lies. 
 
 2. Two angles in space which have their sides parallel 
 respectively, and lying in the same or in opposite directions 
 from their vertices, are equal. 
 
 ABC 
 In any spherical triangle : 
 
 clDC 
 
 3. A = 7T-a', B = 9T-b f , C = ir C', 
 
 a = Tr A', b=?r B', C = TT C', 
 
 A'B'C' . f ABC 
 
 where , is the polar triangle 01 
 
 a'b'c' abc 
 
 4. The greater side is opposite the greater angle, and con- 
 versely. 
 
 5. The sum of two sides is greater than the third. 
 That is, a + b > c, b + c > a, c + a > b. 
 
 6. 27T>7T+A>B + C, 
 
 27r>7r+B>C+A, 
 
68 
 
 SPHERICAL TRIGONOMETRY. 
 
 &. Sufficiency of the Convex Spherical Triangle. 
 
 The solution of every spherical triangle may be made to 
 depend upon the solution of a triangle in which the parts are 
 each separately less than TT. This is made evident by the 
 following summary : 
 
 denote that k 
 
 f sides 
 \ angles 
 
 of a spherical triangle are > TT, the possible combinations are 
 16, of which 8 are excluded, and 8 give spherical triangles : 
 
 s i a o 
 
 5. s 2 a 
 
 7. S 3 a 
 
 2. s a 3 
 
 6. S 2 a 3 
 
 . S 3 a 3 
 
 These eight cases are represented graphically by the follow- 
 ing three figures. The numbers accompanying the figures 
 correspond respectively to the cases. 
 
 1 &2 
 
 FIG. 37. 
 
 FIG. 38. 
 
SPHERICAL TRIGONOMETRICAL RELATIONS. 
 
 5 & 6, and 7 &. 8 
 
 \ 
 
 69 
 
 FIG. 39. 
 
 B. General Theory of Spherical Trigonometrical 
 Relations. 
 
 73. It is shown in geometry that a spherical triangle is 
 determined if any three parts, not excluding the three angles, 
 are properly given. In the general case, the three parts are 
 independent of each other, and relations between the parts of 
 the general triangle cannot, therefore, exist unless the parts 
 enter into such relations at least four at a time. For example, 
 if c f(b), or if c = f(a, b), c is determined, and, therefore, 
 three parts are determined as soon as a and b are given in 
 either case ; that is, the triangle is determined by two parts. 
 But if A f(a, b, c), A is not determined until three inde- 
 pendent parts, a, b, and c, have been given. 
 
 74. Let us inquire how many species of relations we 
 may expect to meet. In the relations of parts taken four at 
 a time we shall pick out four and always exclude two parts. 
 The number of species of parts retained will be the same as 
 the number of species of parts excluded. For convenience, it 
 will be easier to consider the latter. The number of combina- 
 
 6X5 
 
 tions of six things taken two at a time is = 15. 
 
 1 X ^ 
 
 The 
 
70 SPHERICAL TRIGONOMETRY. 
 
 six parts of the triangle are a, b. c, A, B, C. These fifteen 
 combinations are ab, ac, aA, aB, aC, be, bA, bB, bC, cA, cB, 
 cC, AB, AC, BC. Excluding those which are repetitions, we 
 find only four distinct species represented by 1. BC, 2. cC, 
 3. cB, 4. be, and these give us, therefore, to enter into rela- 
 tions four at a time the four distinct species where the parts 
 retained are the complementary combinations of the parts 
 excluded, viz.: 
 
 Ketained 1. abcA, 2. abAB, 3. abAC, 4. ABCa. 
 Excluded 1. BC, 2. cC, 3. cB, 4. be. 
 
 75. To find the number of species of parts taken five at 
 a time, we may proceed as above. 
 
 The number of combinations of six things taken one at a 
 time is six. The six combinations are a, b, c, A, B, C. Here 
 are only two distinct cases, 1. C, 2. c, and we have : 
 
 Eetained 1. abcAB, 2. abABC. 
 Excluded 1. C, 2. c. 
 
 When the parts are taken six at a time there can be only 
 one species of relations. 
 
 C. Systems of Equations demanded by the 
 Preceding Theory. 
 
 a. Equations in which the Parts enter Four at a Time. 
 
 These are by 74 of four species. 
 
 76. 1. Species one. In which we have three sides and an 
 angle. 
 
 Let ABC be any spherical triangle. Let be the center of 
 the sphere. OA, OB, OC, are radii of the sphere. In the 
 plane OAB draw DB and OM perpendicular to AO. AT, the 
 tangent to the arc, AB, at A is also perpendicular to OA. 
 Hence AT, DB, and OM are parallel lines. In the plane 
 
SYSTEMS OF EQUATIONS. 
 
 71 
 
 FIG. 40. 
 
 OAC draw PC and ON perpendicular to OA ; then AT, the 
 tangent to the arc AC at A, PC, and ON are parallel. And 
 the angle NOM = the angle T'AT = the angle A of the 
 spherical triangle ABC. 
 
 77. By 53, OB OC = OD OC + DB OC , 
 
 .'. OB cos a = OD cos b -f DB cos COM. 
 
 And 
 
 But 
 
 hence 
 or 
 
 Also 
 
 OC cos MOC = OC cos COM = + PC cos A. 
 
 PC = OC sin b, 
 
 OC cos COM OC sin b cos A, 
 cos COM = sin b cos A. 
 
 OD^OBcosc, 
 DB = OB sine. 
 
 Substituting these values in the equation, 
 
 OB cos a = OD cos b + DB cos COM, 
 
72 SPHERICAL TRIGONOMETRY. 
 
 we get OB cos a = OB cos b cos c -f OB sin b sin c cos A, 
 
 or cos a = cos b cos c + sin b sin c cos A. 
 
 Advancing the letters, cos b = cos c cos a + sin c sin a cos B, 
 cos c = cos a cos b + sin a sin b cos C. 
 
 78. Prom these three equations all other equations what- 
 ever, which are true of any spherical triangle, may be obtained 
 by algebraic transformations, and are therefore dependent or 
 derived equations. For as three parts of a spherical triangle 
 may be taken arbitrarily, but when they are taken the tri- 
 angle is fixed, and three parts must be taken to fix the triangle, 
 there remain three parts unknown. Now the number of 
 simultaneous and independent equations required to deter- 
 mine three unknown quantities is three. As many as three 
 must be given, but not more than three equations can exist 
 independently. If, then, we have a system or systems of 
 simultaneous equations involving three unknown quantities, 
 not more than three of the equations can be taken to be inde- 
 pendent, and as many as three of them must be taken to be 
 independent, if the system is determinately solvable. From 
 this it is evident that we might take any one of the systems 
 of three equations, where the three equations are independent 
 of each other, as the three fundamental equations of spherical 
 trigonometry. But we will choose the preceding three equa- 
 tions of 77, as the fundamental equations. We shall now 
 show how the other systems of equations are derived from 
 them. 
 
 2. Species two. Equations involving two sides and the angles 
 opposite to them. 
 
 79. Here, e.g., we must retain abAB, and exclude cC. 
 By adding and subtracting the first two of the equations of 
 77, we have : 
 
 (cos a + cos b) (1 cos c) = sin c (sin b cos A + sin a cos B). 
 (cos a cos b) (1 + cos c) = sin c (sin b cos A sin a cos B). 
 
SYSTEMS OF EQUATIONS. 73 
 
 Multiplying these two equations together, and dividing by 
 sin 2 c = 1 cos 2 c, which we have a right to do, since c =%= 0, 
 sin c =)= 0, and therefore sin 2 c =(= 0? an( i we g 6 ^ : 
 
 cos 2 a cos 2 b sin 2 b cos 2 A sin 2 a cos 2 B, 
 or expressing in terms of sines, 
 
 sin 2 b sin 2 A sin 2 a sin 2 B, .'. sin b sin A sin a sin B, 
 
 but as all the factors involved in the equation are positive, 
 we must exclude the minus sign, and write : 
 
 sin b sin A = sin a sin B. 
 Advancing the letters, 
 
 sin c sin B = sin b sin C, 
 sin a sin C = sin c sin A. 
 
 These equations may also be put in the forms : 
 
 sin a sin b sin c _ 
 
 sin A ~~ sin B ~~~ sin C ~~ 
 sin a = r sin A, sin b = r sin B, sin c = r sin C. 
 
 3. Species three. Equations involving two sides and an angle 
 opposite to one of them. 
 
 8O. We must, e.g., retain abAC, and exclude cB. 
 
 In the first of the equations of 77, substitute for cos c its 
 value given by the third equation, and for sin c, its value, 
 
 : -r > given by the third one of the equations of 79. 
 
 We get : 
 
 cos a = cos b (cos a cos b + sin a sin b cos C) 
 sin b sin a sin C cos A 
 
 sin A 
 
 cos a (1 cos 2 b) = cos a sin 2 b = cos b sin a sin b cos C 
 + sin b sin a sin C ctn A. 
 
74 SPHERICAL TRIGONOMETRY. 
 
 Dividing both sides of the equation by sin a sin b, since 
 sin b sin c =$= 0, 
 
 ctn a sin b = cos b cos C + sin C ctn A. 
 Advancing the letters, 
 
 ctn b sin c ^ cos c cos A + sin A ctn B, 
 ctn c sin a = cos a cos B + sin B ctn C. 
 
 If instead of retaining abAC, and excluding cB, we had 
 retained acAB, and excluded bC, we should simply have put 
 c in the place of b, and B in the place of C, and instead of 
 the first of the preceding equations, we should have had : 
 
 ctn a sin c = cos c cos B + sin B ctn A. 
 
 Advancing, ctn b sin a = cos a cos C '-{- si n C ctn B, 
 
 ctn c sin b = cos b cos A + sin A ctn C. 
 
 4. Species four. Equations involving three angles and a side. 
 
 81. Here we must, e.g., retain ABCa, and exclude be. 
 If we divide the equation at the beginning of the preceding 
 section by sin b instead of sin a sin b, there results : 
 
 cos a sin b = cos b sin a cos C + sin a sin C ctn A. 
 
 -01 r> i, . sin a sin C 
 
 Keplacmg sin b by r sin B, sm a by r sin A, sm c = : r 
 
 sin A 
 
 by r sin C, and dividing by r, we have : 
 
 cos a sin B = cos b sin A cos C + sin C cos A. 
 
 If cos c is replaced in the second of the equations of 77 
 by its value taken from the third, the result will differ from 
 the preceding in having a changed into b, and A into B, and 
 it will be : 
 
 cos b sin A = cos a sin B cos C + sin C cos B. 
 
 Multiplying the latter equation by cos C, and equating the 
 terms cos b sin A cos C, there results : 
 
SYSTEMS OF EQUATIONS. 75 
 
 cos a siri B sin C cos A = cos a sin B cos 2 C + sin C cos C cos B, 
 or cos a sin B (1 ' cos 2 C) = cos a sin B sin 2 C 
 
 = sin C cos A + sin C cos C cos B. 
 
 Dividing by sin C, since sin C =(= 0, 
 
 cos a sin B sin C = cos A + cos B cos C, 
 or cos A = cos B cos C + si n B sin C cos a. 
 
 Advancing, cos B = cos C cos A + sin C sin A cos b, 
 cos C cos A cos B + sin A sin B cos c. 
 
 b. Equations in which the Parts enter Five at a Time. 
 
 1. Species one. Equations involving three sides and two 
 angles. 
 
 82. We retain here, e.g., abcAB, and exclude C. 
 
 If in the first of the equations of 77, we substitute for 
 cos b, its value as given by the second of those equations, 
 cos a = (cos c cos a + sin c sin a cos B) cos c + sin b sin c cos A, 
 or cos a (1 cos 2 c) = cos a sin 2 C = sin c sin a cos c cos B 
 
 + sin b sin c cos A. 
 
 Dividing by sin c, since sin c =|= 0, 
 
 cos a sin c = cos c sin a cos B + sin b cos A, 
 or as usually written, 
 
 sin b cos A = cos a sin c cos c sin a cos B. 
 
 Advancing, sin c cos B = cos b sin a cos a sin b cos C, 
 sin a cos C = cos c sin b cos b sin c cos A. 
 
 If instead of putting cos b in the first of the equations of 
 77, we had put it in the third, the results would have 
 appeared with a and c, and A and C, interchanged, or 
 
 sin b cos C = cos c sin a cos a sin c cos B, 
 
 and also, sin c cos A = cos a sin b cos b sin a cos C, 
 
 sin a cos B = cos b sin c cos c sin b cos A. 
 
76 SPHERICAL TRIGONOMETRY. 
 
 2. Species tivo. Equations involving three angles and two sides. 
 83. Here we retain, e.g., ABCab, arid exclude c. 
 The first of these equations has already been obtained in 
 81. It is : 
 
 cos a sin B = cos b sin A cos C + sin C cos A. 
 
 Advancing, cos b sin C = cos c sin B cos A + sin A cos B, 
 cos c sin A = cos a sin C cos B + sin B cos C. 
 
 The leading equation of the second division of this set has 
 also been obtained in 81. It is : 
 
 cos b sin A = cos a sin B cos C + sin C cos B. 
 
 Advancing, cos c sin B = cos b sin C cos A -+- sin A cos C, 
 cos a sin C = cos c sin A cos B + sin B cos A. 
 
 c. Equations in which the Parts enter Six at a Time. 
 
 84. Here there is but one species. An example may 
 suffice. From 77 we can obtain, 
 
 cos a cos b cos c (cos b cos c + sin b sin c cos A) 
 (cos c cos a + sine sin a cos B) (cos a cos b-hsinasin bcosC)=0, 
 cos A cos B cos C sin 2 a sin 2 b sin 2 c = (cos a cos b cos c) 
 (cos b cos c cos a) (cos c cos a cos b). 
 
 D. On the Use of the Polar Triangle in Establishing 
 
 Relations. 
 
 85. The relations A = TT a', a = TT A', 
 
 B = 7r-b', b = 7r-B', where ,^,, 
 
 C = 7T-C', C-7T-C', 
 
 is the polar triangle of , , often afford an easy means of 
 
 a b c 
 
 establishing a relation. For example, we may say that the 
 formulas of 77 are transformed into those of 81. This is 
 shown as follows : Express that the formulas of 77 hold 
 for the polar triangle. Then the first one gives : 
 cos a' = cos b' cos c' + sin b' sin c' cos A'. 
 
USE OF THE POLAR TRIANGLE. 77 
 
 Substituting for a', b', c', A', the values given above, and we 
 have : 
 
 cos (TT A) = cos (TT B) cos (TT C) + sin (TT B) 
 
 sin (TT C) cos (TT a), 
 
 or cos A = ( cos B) ( cos C) + sin B sin C ( cos a), 
 
 or cos A/= cos B cos C + sin B sin C cos a. 
 
 The equations of 79 are transformed into themselves ; 
 that is, we get nothing different by using the polar triangle 
 with them. The first half of the formulas of 80 are trans- 
 formed into the second half, and the second half into the first 
 half. Thus the first one of the first half, 
 
 ctn a sin b = cos b cos C + sin C ctn A, 
 is transformed into 
 
 ctn A sin B = cos B cos c sin c ctn a, 
 or ctn a sin c = cos c cos B + sin B ctn A, 
 
 which is the first one of the second half. The formulas of 
 82 are transformed into those of 83, and vice versa, as the 
 student may satisfy himself, by writing out the leading equa- 
 tions and making the substitutions. As this work can be 
 done at a glance when once understood, it is evident that the 
 method affords an easy and powerful means of writing down 
 new formulas without any work at all. It is a method that 
 ought, at least, to be applied to every new formula obtained, 
 for it may enable one to double the number of his formulas 
 without great labor. 
 
 GENERAL NOTE ON SYSTEMS or TRIGONOMETRIC EQUATIONS. The 
 preceding investigations have shown that the number of species of trigo- 
 nometric equations or formulas is exactly determinable. And while 
 complete sets of the various species have been given, it is to be observed 
 that other sets might be derived, and that the number of derived sets of 
 the various species is unlimited. 
 
CHAPTEK II. 
 
 /. Formulas for Spherical Right, and Quadrantal 
 Triangles. 
 
 86. In the formulas of 77-81, making an angle, and a 
 side, as C, and c, successively equal to 90, we have the 
 following formulas for spherical right, and quadrantal 
 triangles : 
 
 a. Spherical Right Trian 
 
 1. cos c = oos a cos b. 
 
 A Silia 
 
 gles. 
 
 77 
 
 79 
 79 
 80 
 80 
 80 
 80 
 81 
 
 81 
 81 
 
 b. Quadrantal Triangle 
 
 1'. cos C = cos A cos B. 
 . . sin A 
 
 JS. 
 
 81 
 79 
 
 79 
 80 
 80 
 80 
 80 
 77 
 
 77 
 77 
 
 25, Sill A . 
 sin c 
 
 -in P Slnb - 
 
 A . bin a . _ 
 sin C 
 
 . . sin B 
 
 Qf Qin h - 
 
 O. &111 D . 
 
 sin c 
 tan b 
 
 sin C 
 tanB 
 
 4. cos A . 
 tan c 
 
 D tan a 
 
 * . CUO d ^ 
 
 tanC 
 tan A 
 
 o. cos b . 
 tan c 
 
 . tan a 
 
 tanC 
 tan A 
 
 o. tan A . , 
 sin b 
 
 7 1 nP tanb - 
 
 u . tan a . -- 
 sin B 
 
 tanB 
 
 4 . Lclll L> . 
 
 sin a 
 cos A 
 
 sin A 
 
 . cos a 
 
 o. cos a . _. * 
 sin B 
 
 cos B 
 
 o . Cub A . . 
 
 sm b 
 cos b 
 
 y. cos D . . 
 sin A 
 
 10. cos c = ctn A ctn B. 
 
 y . cob D . 
 sin a 
 
 10'. cos C = ctn a ctn b. 
 
SPHERICAL RIGHT AND QUADRANTAL TRIANGLES. 79 
 
 c. Attention is called to the Following Deductions 
 from the Formulas of 86. 
 
 87. 1. From 1 and 1' it appears that c will be of the first 
 or second quadrant, and C of the second or first quadrant, 
 according as a and b, or A and B, respectively, are of the 
 same or of different quadrants. 
 
 2. From 6 and 7 it appears that an oblique angle and its 
 opposite side, and from 6' and 7' that a non-quadrantal side, 
 and its opposite angle, are of the same quadrant. 
 
 B. Solution of Spherical Right and Quadrantal 
 Triangles. 
 
 88. In the right spherical triangle, the right angle, and 
 in the quadrantal triangle, the quadrant, counts as one of the 
 three given parts. There are six cases to be treated. 
 
 a. Right Triangle. 
 
 b. Quadrantal Triangle. 
 
 Case. 
 
 Given. 
 
 Re- 
 quired. 
 
 Formulas 
 86. a. 
 
 Check. 
 
 Case. 
 
 Given. 
 
 Re- 
 quired. 
 
 Formulas 
 86. b. 
 
 Check. 
 
 1 
 
 ab 
 
 CAB 
 
 1,6,7 
 
 10 
 
 1 
 
 AB 
 
 Cab 
 
 l',6',7' 
 
 10' 
 
 2 
 
 ac 
 
 bAB 
 
 1,2,5 
 
 9 
 
 2 
 
 AC 
 
 Bab 
 
 l',2',5' 
 
 9' 
 
 3 
 
 aA 
 
 bcB 
 
 6,2,8 
 
 3 
 
 3 
 
 Aa 
 
 BCb 
 
 0', 2', 8' 
 
 3' 
 
 4 
 
 aB 
 
 bcA 
 
 7,5,8 
 
 4 
 
 4 
 
 Ab 
 
 BCa 
 
 7', 6', 8' 
 
 4' 
 
 5 
 
 cA 
 
 abB 
 
 2, 4, 10 
 
 7 
 
 5 
 
 Ca 
 
 ABb 
 
 2', 4', 10' 
 
 r 
 
 6 
 
 AB 
 
 abc 
 
 8, 9, 10 
 
 1 
 
 6 
 
 ab 
 
 ABC 
 
 8', 9', 10' 
 
 V 
 
 89. The remarks of 87, taken in connection with the 
 formulas given in 88 for the solution of the various cases, 
 show that of the six cases, only the third is ambiguous. In 
 this case the required parts must be obtained from their sines. 
 Since eacli required part has two values, supplements of each 
 
80 SPHERICAL TRIGONOMETRY. 
 
 other, there might apparently be eight triangles ; but as b 
 and B (for the right spherical triangle) must be of the same 
 quadrant, only two combinations of a, b, and B are allowable, 
 and when a and b are determined, c is determined, and, there- 
 fore there cannot be more than two solutions. If a differs 
 less than A from 90, the triangle is impossible, since the 
 sine of each required part would be greater than unity. An 
 analogous consideration holds for the corresponding case of 
 the quadrantal triangle. 
 
 9O. A complete set of formulas could be deduced for 
 spherical isosceles triangles from the general formulas of 
 77-81, by assuming that two sides are equal, but as they 
 can be easily solved by the formulas for right triangles, we 
 will not give a special set of formulas. 
 
 C. Some Astronomical Conceptions. 
 
 a. Celestial Sphere. 
 
 1. Definition. The Celestial Sphere is a sphere of indefi- 
 nitely great radius, concentric with the earth, and in whose 
 surface, the heavenly bodies are, for astronomical uses, 
 supposed to be. 
 
 Its axis, called the Celestial Axis, coincides with the axis 
 of the earth, and its ends are the North and South Poles of 
 the celestial sphere. 
 
 2. Great Circles. 
 
 (1) The Sensible Horizon is that small circle of the celes- 
 tial sphere, which is tangent to the earth at the observer's 
 standpoint. 
 
 The Rational Horizon is that great circle of the celestial 
 sphere, which is parallel to the sensible horizon. 
 
 Because the radius of the celestial sphere is indefinitely 
 great, we are justified, in practice, in identifying the sensible 
 
CELESTIAL SPHERE. 81 
 
 and the rational horizon, and the observer's standpoint and 
 the center of the celestial sphere. Accordingly we shall, 
 without distinction, speak only of the horizon. 
 
 The upper and lower poles of the horizon are called the 
 Zenith and Nadir, respectively. 
 
 (2) Vertical Circles are the great circles which pass through 
 the zenith and nadir. 
 
 (3) The Equinoctial, or Celestial Equator, is that great 
 circle of the celestial sphere which is perpendicular to the 
 celestial axis. Obviously its poles are the celestial poles. 
 
 (4) Hour Circles, or Declination Circles, are great circles 
 passing through the celestial poles. 
 
 (5) The Celestial Meridian is that hour circle which passes 
 through the zenith and nadir. It coincides with the terres- 
 trial meridian of the observer's standpoint. 
 
 The intersections of the celestial meridian and horizon are 
 the North and South Points. 
 
 (6) The Prime Vertical is that vertical circle which is 
 perpendicular to the celestial meridian. 
 
 The points of intersection of the prime vertical and the 
 horizon are the East and West Points. 
 
 (7) The Ecliptic is that great circle of the celestial sphere 
 in which the sun appears to move. 
 
 The intersections of the equinoctial and ecliptic are called 
 the Vernal and Autumnal Equinoxes. The sun appears to 
 be at these points on about March 21, and September 21, 
 respectively. 
 
 The angle between the equinoctial and ecliptic is called 
 the Obliquity of the Ecliptic. It is denoted by the letter e. 
 e = 23 27', nearly. 
 
 (8) The hour circle passing through the equinoxes is called 
 the Colure. 
 
 (9) Latitude Circles are the great circles which pass 
 through the poles of the ecliptic. 
 
82 SPHERICAL TRIGONOMETRY. 
 
 (10) That latitude circle which passes through the equi- 
 noxes we shall call the Prime Latitude Circle. 
 
 b. Spherical Coordinates. 
 
 1. Definition. The Spherical Coordinates of a heavenly 
 body are its angular distances from two fixed, mutually per- 
 pendicular, great circles of reference. 
 
 For convenience the great circle arcs which measure the 
 angular distances, are used. 
 
 2. Systems of Spherical Coordinates. 
 
 (1) Using the horizon and meridian as reference circles, 
 we get Altitude and Azimuth, respectively ; altitude taken 
 from the horizon from to 90, and azimuth taken from the 
 south point measured by way of the west from to 360. 
 The complement of the altitude is called the Zenith Distance. 
 
 (2) Using the equinoctial and meridian we get Declination 
 and Hour Angle, respectively ; declination taken from the 
 equinoctial north or south from to 90 or 90, and hour 
 angle taken from the meridian measured by way of the west 
 from at noon to 360, or 24 hrs. The complement of the 
 declination is called the Polar Distance. 
 
 (3) Using the equinoctial and colure we get Declination 
 and Eight Ascension, respectively; right ascension taken 
 from the vernal equinox by way of the east from to 360, 
 or 24 hrs. 
 
 (4) Using the ecliptic and prime latitude circle we get 
 Celestial Latitude and Celestial Longitude, respectively ; 
 celestial latitude taken from the ecliptic from to 90 or 
 90, and celestial longitude taken from the vernal equinox 
 by way of the east from to 360. 
 
 It is important not to confound celestial latitude and longi- 
 tude with terrestrial. 
 
 Pigs. 41 and 42 represent the celestial sphere, with its 
 center at 0. 
 
SPHERICAL CO ORDINA TES. 
 
 83 
 
 In Fig. 41, NWS represents the horizon ; EQR the celestial equator; 
 Z, N a , and P n , P s , their respective poles; V e , the vernal equinox ; and 
 H, a heavenly body. 
 
 AH = altitude denoted by h. 
 
 SWA = azimuth a. 
 
 DH = declination " 5. 
 
 Angle RP n H = Arc RWD = hour angle " t. 
 
 V e D = right ascension " a. 
 
 RZ = NP n = terrestrial latitude of observer's standpoint " 0. 
 ZH = 90 h = zenith distance. 
 p n H = 90 5 = polar distance. 
 
 The triangle P n HZ is called the Astronomical Triangle ; the angle at 
 H, the Position Angle. The student should carefully note the parts of 
 this triangle. 
 
84 
 
 SPHERICAL TRIGONOMETRY. 
 G 
 
 FIG. 42. 
 
 In Fig. 42, QV e R, represents the equinoctial ; EV e C, the ecliptic ; P n , 
 P s , and F, G, their respective poles; V e , the vernal equinox; and H, a 
 heavenly body. 
 
 DH = declination. 
 V e D = right ascension. 
 LH = celestial latitude, denoted by . 
 V e L = celestial longitude, " X. 
 CR = P n G = obliquity of the ecliptic. 
 
 The student should carefully note the parts of the triangle P n HG. 
 
 > 
 
 n. Examples. X. 
 
 1. a = 116, b = 16, C = 90. 
 
 2. A = 60 47', B = 57 16', c = 90. 
 
 3. a = 28, b = 28, B = 79. 
 
EXAMPLES. X. 85 
 
 4. a = 33 40', A = 43 21', C = 90. 
 
 5. A = 105 53', a = 104 54', c = 90. 
 
 6. a = 61, B = 123 40', C = 90. 
 
 7. A = 139, b = 143, c = 90. 
 
 8. a = 69 18', c = 84 27', C = 90. 
 
 9. A = 70 12', C = 106 25', c - 90. 
 
 10. c = 69 25', A = 54 54', C = 90. 
 
 11. C = 134, a = 143, c = 90. 
 
 12. A = 46 59', B = 57 59', C = 90. 
 
 13. a = 174 12', b = 94 08', c = 90. 
 
 For a right spherical triangle prove, when C =.90 : 
 
 14. tan 2 - = tan $ (a + c) tan (c a). 
 
 15. tan 2 (45 - $ A) = tan $ (c - a) ctn $ (c + a). 
 
 16. tanH B = sin (c a) esc (c + a). 
 
 17. tan 2 (45 \ B) = tan (A a) tan -J- (A + a). 
 
 18. tan 2 (45 \ b) = sin (A - a) esc (A + a). 
 
 19. tan 2 (45 - i c) = tan i (A - a) ctn (A + a). 
 
 tan j (90+ B-A) 
 
 21. ctn 2 i b = tan | (90 + A - B) ctn $ (B + A - 90). 
 
 22. ctnHc = cos(B A)sec(B + A): 
 
 23-31. Obtain formulas for the quadrantal triangle corresponding to 
 formulas of problems 14-22. 
 
 32. A spherical square is a spherical quadrilateral which has equal 
 sides and equal angles. The diagonals divide it into four equal spherical 
 right triangles. Given a side a of the spherical square, find an angle A. 
 
 33. A line makes with a plane an angle a ; through the foot of the 
 line and in the plane a line is drawn making an angle ft with the projec- 
 tion of the first line on the plane ; find the angle 7 made by the second 
 line with the first. 
 
 34. Given the number of sides of a regular spherical polygon equal to 
 n and each angle equal to A ; find a side of the polygon and the polar 
 radii of the inscribed and circumscribed circles. 
 
 35. Find the angles between the adjacent faces of each of the five 
 regular polyhedrons. 
 
 36. The base of a regular pyramid is an octagon ; each angle at the 
 vertex of the pyramid is 30; find the angle a between two adjacent 
 lateral faces, and the angle ft between any lateral face and the base. 
 
 37. From the longitude of the sun (A) and the obliquity of the ecliptic 
 (e) find the sun's right ascension (a) and declination (8). 
 
86 SPHERICAL TRIGONOMETRY. 
 
 38. From the latitude (0) of a place on the earth's surface and the 
 declination (5) of the sun on a given day, find the times and places of the 
 rising and setting of the sun and its distance from the zenith at noon. 
 
 When are days and nights equal ? 
 (Neglect the effect of refraction.) 
 
 39. When does the solution of Ex. 38 become impossible? When 
 indeterminate ? And what follows for places so situated on the earth's 
 surface as to give these results ? 
 
 40. When and where does the sun rise in Oberlin (latitude 41 17') on 
 the longest day,, June 21 (5 = +23 27") ? When and where on the 
 shortest day, Dec. 21 (5 = - 23 27') ? 
 
 41. At a place on the earth's surface whose latitude is 0, a rod OA, 
 pointed toward the north pole, makes with a horizontal plane an angle, 
 BOA = 0; let OC be the position of the shadow of the rod at a given 
 time of the day ; find the angle BOC. 
 
 What does the result become for = 41 17', t = 2 hrs. = 30? 
 
 42. Through (Ex. 41) pass a plane perpendicular to OB ; extend 
 the rod through this vertical plane so that its shadow may fall upon it ; 
 find the angle S which the shadow makes with the projection of the rod 
 on the plane at a given time t. Find S for = 41 17', t = 3 hrs. = 45. 
 
 43. At what time of a given day (5 given) at a given place (lat. = 0) 
 is the sun exactly east or west ? 
 
 44. Find the altitude (h) and the azimuth (a) of the sun at a given 
 place and on a given day at 6 A.M. 
 
 45. Find the shortest distance (i.e., length of great circle arc) from 
 Oberlin, long. 82 14', to New Haven, long. 72 55', the latitude of each 
 place being 41 IT, 
 
CHAPTER III. 
 A. Formulas for Spherical Oblique Triangles. 
 
 a. Formulas for Functions of A, B, C. 
 
 91. From the formulas of 77, 
 
 cos a cos b cos c 
 
 sin b sin c 
 
 ^/1-f cos A /cos a (cos b cos c sin b sin c) 
 
 2 * 2 sin b sin c c f 
 
 Vcos a cos (b -f- c) _ /sin ^ (a -J- b + c) sin -j- (b -(- c a) 
 2 sin b sin c ^ sin b sin c 59 
 
 Let 
 
 /sm s sin (s a) 
 
 and cos 4- A = \/ : ; 
 
 ^ sm b sm c 
 
 /sin s sin (s b) 
 Similarly, cos B = V -A L 
 
 * sin r sin a 
 
 /sin s sm (s c) 
 
 ~ \ , ' 
 
 sm a sm b 
 
 92. 
 
 sin b sin c 
 
 58 and 59 
 
 Q . ., , -ID /sin (s c) sin (s a) 
 
 Similarly, sin -J- B = \ 
 
 sm c sm a 
 
 . n Jsin(s-a)sin(s-b) 
 
 ~ 
 
 siri a sin b 
 
88 SPHERICAL TRIGONOMETRY. 
 
 93. From the formulas of 91 and 92, 
 
 . 
 sin s sin (s a) 
 
 sin s sin (s b) 
 
 sin s sin (s c) 
 
 The student may prove that in each formula of the last 
 section the positive square root is to be taken, and that the 
 number under the radical sign in each case is positive. 
 
 94. From the formulas of 81, or from the formulas of 
 the last three sections, by using the properties of polar 
 triangles, the student may prove the following formulas : 
 
 /cos (S B) cos (S C) 
 cos a - - 
 
 = \/ 
 * 
 
 / 
 
 = \ 
 * 
 
 . 
 sin B sin C 
 
 / cos S cos (S A) 
 sm a = - . D . v - L 
 sin B sin C 
 
 j / cos S cos (S A) 
 
 = ^cos(S-B)cos(S-C)' 
 
 Where S = A " [ "^ + C 
 
 The student may write the corresponding formulas for 
 b and c, and prove that in each formula the positive square 
 root is to be taken, and that the number under the radical 
 sign in each case is positive. 
 
 b. Gauss's Equations. 
 95. From 54 and 55 
 
 cos -j- (A . B) = cos % A cos J B rp sin -J- A sin B. 
 Substituting for 
 
 cos \ A, cos \ B, sin J A, and sin B, 
 their values given in 91 and 92 : 
 
SPHERICAL OBLIQUE TRIANGLES. 89 
 
 cos |(A B) = 
 
 7 
 
 sin c * sin a sin b 
 
 s * n ( s ~ c ) /sin (s a) sin (s b) 
 sin c ^ sin a sin b 
 
 sin s rp sin (s c) . C 
 = - ^sm 
 
 sin c 2 
 
 Taking the upper sign, 
 
 c\ _ . /a + b c\ 
 
 cos J (A + B) = ^ ?f ^- sin ^ 
 
 2 sin - cos - 
 
 2 cos i (a + b) sin - p 
 
 ____2 sin C 59 
 
 2 sin - cos - 
 
 c 
 
 cos- 
 
 Taking the lower sign, 
 
 . sin s + sin (s c) . C 
 
 cosi(A-B)=- -- ^sm- 
 
 sin 
 Similarly from 
 
 sin % (A B) = sin % A cos -J B cos |- A sin -J B. 
 Substituting as above, 
 
 sin (s b) sin (s a) C 
 sin i (A B) = - J . ^ z cos - 
 
 sin c 2 
 
 , a-b\ /c a-b 
 
 2 sin - cos - 
 
90 SPHERICAL TRIGONOMETRY. 
 
 . cos J (a b) C 
 .'. sin -J- (A+ B) = * ^cos > 
 
 cos I 
 
 sin % (a b) C 
 and sin (A B) = - * cos 
 
 These formulas may be put in a form in which they may 
 be more easily remembered. They can be written as follows : 
 
 . . . Dx , , Here observe : (1) 
 
 sm-HA+B) cos -J- (a b) ... ,, . _V 
 
 1. v ' z = z All the angles are half 
 
 cos * C cos * c , 
 
 . , /A DX i, .v angles. (2) In the left 
 _ sin i (A B) sin % (a b) 3 , v ' 
 
 2. v _ ^ = =-r-j members of the equa- 
 
 cos i C sin 4- c , . i * iW 
 
 . tions we have A, B, C ; 
 
 _ cos | (A + B) cos (a -f b) . 
 
 3. i /^ ' ^ di ln the right members 
 
 sin j- C cos i c 
 
 i / A D\ i / i u\ a, b, c. (3) In each left 
 cos 4- (A B) sin 4- (a + b) n v ' 
 
 4. . v . _ L = . \ -' member are co-func- 
 
 sm ^ C sm c , . . . . ^ J _. 
 
 tions, in the right, the 
 
 same functions. (4) In each equation to a sign -f~ in the 
 numerator of one member corresponds a cosine in the numer- 
 ator of the other member. And in the same way to a sign 
 corresponds a sine. 
 
 c. Napier's Analogies. 
 
 96. If we divide member by member, 1 of the Gauss 
 equations by 3, 2 by 4, 4 by 3, and 2 by 1, we obtain : 
 
 i /A i D\ i / (1) Here the parts 
 
 tan 4- (A + B) cos 4- (a b) v ' ,, 
 
 1. - 1 , i * = / r~R ' enter nve at a time, and 
 
 ctn -J- C cos ^ (a -f b) . ^. 
 
 . , A DN f / , x m the respective mein- 
 
 _ tan -J- (A B) sm i (a b) , . _ _ 
 
 2. 1 -, ^ L = - H rix * bers we nave A > B, C 
 
 ctn ^ C sin } (a + b) 
 
 i / i . N i /A o\ an a ? b 5 or a ? D, c, and 
 
 tani(a+b) = cos^A-B . g _ ' 
 
 tanjc COS HA + B enter t V e left members 
 
 tan 4- (a b) sm 4- (A B) 
 
 4. -^T L =- ; ;. , ' we have there the co- 
 
 tan -J- c sm -J- (A -f B) 
 
 functions, tangent and 
 
SPHERICAL OBLIQUE TRIANGLES. 91 
 
 cotangent, but the same functions, tangents, when a, b, c, 
 enter the left members. In the right members are the same 
 functions, sines or cosines. (3) The angles are half angles. 
 
 (4) In the right members are sines or cosines according as 
 the signs or + are in the numerators of the left members. 
 
 (5) In the right members the angles are differences in the 
 numerators, sums in the denominators. 
 
 d. 1'Huilier's Formula. 
 
 8 97. If in 1 and 3, 95, A +^+- C ^ be represented 
 
 2i 2i 
 
 byE, 
 
 /C c \ 
 
 cos ( L , , , . 
 \2 I cos j (a b) 
 we get *- L > 
 
 cos - cos - 
 
 sin(- E ) cosHa + b) 
 
 C 
 
 sm- 
 
 cos| 
 
 and 
 
 By division and composition the first of these gives 
 
 cos f E J cos cos -J- (a b) cos - 
 
 (C \ C = c '>sou, 
 
 - E J + cos - cos % (a b) + cos - 
 
 tan (C E) tan ()= tan | (s b) tan [ (s a)], 
 or, tan(C E) tan E tanj- (s b) tan^-(s a). 
 
 In a similar manner the second equation gives, 
 
 ctn (C E) tan $ E = tan s tan i (s c) ; 
 
 multiplying these equations together, "term by term, we have, 
 
 tan 2 !- E = tan % s tan (s a) tan (s b) tan $ (s c). 
 
92 
 
 SPHERICAL TRIGONOMETRY. 
 
 Formula for the Polar Radius of a Circumscribed Circle 
 of a Spherical Triangle. , 
 
 98. The construction is anal- 
 ogous to that for the similar 
 problem in the case of a plane 
 triangle. P, the pole of the 
 required circle is the intersec- 
 tion of perpendicular great circle 
 arcs erected to the sides at their 
 middle points. 
 
 the required radius. 
 
 O A n C D C f* 
 
 In the right spherical triangle LPC, by 86 and 94, 
 tan- 
 
 tan R = 
 
 cos S 
 
 cos (S A) 
 = = tan -J a tan -J b tan -J c, by 
 
 ~~ COS o 
 
 cos (S A) cos (S B) cos (S - C) 
 94. 
 
 /. Formula for the Polar Radius of the Inscribed Circle 
 of a Spherical Triangle. 
 
 99. Bisect the angles of the 
 triangle by great circle arcs. 
 From the point of intersection of 
 the angle bisectors draw perpen- 
 dicular arcs of great circles to the 
 sides of the triangle. Since L, M, 
 and N are points of tan gen cy, 
 BN = BL, AN^AM, CM = CL, 
 a'+b' = c, b'+c' = a, c'-f a' = b, 
 adding and dividing by 2, FIG. 44. 
 
DEDUCTION OF PLANE TRIGONOMETRY. 
 
 93 
 
 a- I b' I c'- a+b + C -c 
 
 2 
 
 ..a' = s-(b'+ c') = (s-a). 
 A tan r 
 
 tan r 
 
 In the right triangle AN P. tan = : . = -; 
 
 2 sm a' sin (s a) 
 
 .'. tan r = tan sin (s a) 
 
 L 
 
 -V" 
 
 sin (s a) sin (s b) sin (s c) 
 
 sin s 
 sin s tan -J A tan -J- B tan -J- C. 
 
 86 
 
 93 
 
 B. Plane Trigonometry a Special Case of Spherical 
 Trigonometry. 
 
 1OO. Let ABC be any spherical triangle of which the 
 sides a, b, and c are measured in some linear unit. Let r be 
 the radius of the sphere, and 
 
 a, /?, and y the angles at the 
 center corresponding to the 
 arcs a, b, and c, respectively. 
 
 Then = 2 p = y = ? 
 by 8, 3. 
 
 If now r increases while a, 
 
 b, and c remain fixed in length, 
 a, (3) and y will diminish ; and 
 if r increases indefinitely, a, /?, 
 and y will approach zero as a 
 limit. And the limit as r in- 
 creases indefinitely of 
 
 lim sin a lim 
 
 FIG. 45. 
 
 lim sin a 
 
 r sin a = _ r a 
 
 a = a 
 
 sm a _ 
 a = a a = 0a 
 
 = a. 
 
 Similarly as r increases indefinitely lim r sin ^ = ?> 
 
94 SPHERICAL TRIGONOMETRY. 
 
 We have by 77 cos a = cos /? cos y -\- sin (3 sin y cos A. 
 cos a = 12 sin 2 - , by 58. Substituting this value for 
 
 cos a and similar values for cos (3 and cos y in the above 
 equation and reducing we have : 
 
 in 2 1 = sin 2 ^ -f sin 2 1 2 sin 2 1 sin 2 1 -J- sin ft sin y cos A. 
 
 Multiplying this equation through by r 2 and taking the limits 
 of both members as r increases indefinitely we have : 
 
 sn 
 
 (r sin \ J= ( r sin |J+ (r sin *J- ? (r sin f)'(r sin * 
 
 -J- (r sin /3) (r sin y) cos A. 
 
 an equation which with two others similar to it, and obtained 
 in the same way, may be considered the fundamental equa- 
 tions of plane trigonometry. This has led us to the broadest 
 generalization of which we are capable at this stage, but the 
 student will learn subsequently that all trigonometry of the 
 sphere and plane is wrapped up in one quaternion equation, 
 an equation which in the quaternion calculus is very elemen- 
 tary. This equation is : 
 
 S (VyoVajS) = a 2 S/?y SyaSa/?, 
 
 and being interpreted with reference to a sphere it gives at 
 once the three equations of 77, which are the fundamental 
 equations of spherical trigonometry, which includes plane 
 trigonometry as a special case. 
 
 C. Spherical Triangles. 
 
 (f. Cases. 
 
 1O1. With any three parts given, a spherical triangle 
 may be solved, and the equations of 77 theoretically deter- 
 
SPHERICAL TRIANGLES. 95 
 
 mine the solution. In practice, however, derived equations 
 adapted to logarithmic computation are used. The three 
 
 given parts may be selected in ' ' = 20 ways, as follows : 
 
 ABC ACa Aarc BCc Cab 
 
 ABa ACb Abe Bab Cac 
 
 ABb ACc BCa Bac Cbc 
 
 ABc Aab BCb Bbc abc 
 
 Of these, there are only six distinct cases. 
 Excluding repetitions, we have : 
 
 1. abc, three sides. 
 
 2. ABC, three angles. 
 
 3. abC, two sides and included angle. 
 
 4. ABc, two angles and included side. 
 
 5. abA, two sides and angle opposite one. 
 
 6. ABa, two angles and side opposite one. 
 
 The number of these cases may again be reduced by one 
 half, by the use of the polar triangle, since, for example, 
 cases 2, 4, and 6 could be made to depend upon cases 1, 3, 
 and 5, respectively. But it is practically better to know 
 how to treat six cases. 
 
 b. Solutions. 
 
 Case 1. Given a, b, and c. In order that the triangle 
 should be possible, the following inequalities must be satisfied : 
 
 a + b>c>0 7r>a>0 
 
 b + c>a>0 7r>b>0 
 
 c + a>b>0 TT>C >0 
 
 If these inequalities are satisfied, the solution is determinate 
 in a single way, and the angles are given by the formulas : 
 
96 SPHERICAL TRIGONOMETRY. 
 
 A tan r B tan r C tan r 
 
 2 sin (s a) 2 sin (s b) '2 sin (s c) 
 
 /sin (s a) sin (s b) sin (s c") 
 
 where tan r = + \ -A z 99 
 
 ^ sin s 
 
 If only a single angle is desired, it may be computed from 
 one of the formulas, 
 
 A /sin (s b) sin (s c) 
 
 tan - = + \l ^ / . v A etc. 93 
 
 2 * sm s sin (s a) 
 
 As a check, the formulas 
 
 sin a sin b sin c 
 
 sin A sin B sin C 
 may be used. 
 
 Case 2. Given A, B, and C. If the following inequalities, 
 
 27T>7T+A>B+C, 7T>A>0, 
 
 27T>7T+B>C+A, 7T>B>0, 
 
 27T>7T+C>A+B, 7T>C>0, 
 
 * 
 
 are satisfied, the three sides may be obtained determinately 
 in a single way, from the formulas : 
 
 tan - = tan R cos (S A), 
 
 tan - = tan R cos (S B), 
 
 2i 
 
 tan = tan R cos (S C), 
 
 V_ pr)c ^ 
 7F K 7^ 5^ 7^ 7^' 98 
 cos (S A) cos (S B) cos (S C) 
 
 The formulas of 79 may be used as a check. If only a 
 single side is desired, one of the formulas of 94 may be 
 used. 
 
SPHERICAL TRIANGLES. 97 
 
 Vase 8. Given a, b, and C. If the inequalities, 
 
 7r>a>0, TT > b >0, TT > C> 0, 
 
 are satisfied, we may solve the triangle, which will be deter- 
 minate in a single way, as follows : A and B may be found 
 by using the formulas : 
 
 a-b 
 A+B COS ~2- C 
 
 COS 
 
 2 
 a-b 96 
 
 A _B 
 
 ~ 
 
 A+ B 
 These give us - = m, 
 
 2 A = m -f n, 
 
 A D whence _ 
 
 A B _ B = m n. 
 
 ~~2~ 
 c may be found from the formula of 96 : 
 
 A + B 
 c S1 "^T- a-b 
 
 As ?, check use 79. 
 
 If the side c alone is desired, it may be found from the 
 formula, 77, 
 
 cos c = cos a cos b 4- sin a sin b cos C, 
 
 which may be adapted to logarithmic computation by the aid 
 of an auxiliary angle, as follows : 
 
 cos c = cos b (cos a + sin a tan b cos C). 
 
 If tan b cos C is put equal to tan x = -- > we have the 
 
 cos x 
 
 two formulas, 
 
 cos b cos (a x) _. 
 
 cos c = - * - * > and x = tan" 1 (tan b cos C), 
 cos x 
 
SPHERICAL TRIG ONOME TR Y. 
 
 or a single formula, 
 
 _ cos b cos [a tan" 1 (tan b cos C)] 
 
 cos tan" 1 (tan b cos C) 
 
 In the same manner, A and B can be computed singly. 
 80, we have 
 
 ctn a sin b = cos b cos C -f- sin C ctn A, 
 ctn b sin a = cos a cos C + sin C ctn B, 
 or sin C ctn A = ctn a (sin b cos b tan a cos C), 
 
 sin C ctn B = ctn b (sin a cos a tan b cos C). 
 If we put tan b cos C tan x, 
 
 tan a cos C = tan y, 
 these formulas become 
 
 _ ctn a sin (b y) _ ctn C sin (b y) 
 ~ sin C cos y sin y 
 
 By 
 
 Ctn D 
 
 or 
 
 ctn b sin (a x) _ ctn C sin (a x) 
 sin C cos x sin x 
 
 ctn C sin [~b tan -1 (tan a cos C)l 
 
 ctn A - * ) 
 
 sin tan -1 (tan a cos C) 
 
 ctn C sin [a tan" 1 (tan b cos C)] 
 
 sin tan -1 (tan b cos C) 
 
 The employment of the auxil- 
 iary angles x and y is equiva- 
 lent to decomposing the triangle 
 
 ABC 
 
 into two right triangles, 
 abc 
 
 For if AD is an arc perpendicular 
 to BC from A, 
 
 tanx' = tanbcosC, 86, 4 
 whence x = x'. 
 
 FIG. 46. 
 
 tan AD = 
 
 Also, cps 
 
 sin x 
 ctn C 
 
 COS X 
 
 86,1 
 86, 6 
 
SPHERICAL TRIANGLES. 99 
 
 And in the triangle ADB we have 
 
 cos b cos (a x) 
 
 cos c = cos AD cos (a x) = - - L > 
 
 cos x 
 
 _ sin (a x) ctn C sin (a x) 
 and ctn b = - . ^ ' = - : 
 tan AD sin x 
 
 In the same way. if B E is a perpendicular arc from B to 
 AC, y = y', and the angle A may be found similarly as the 
 angle B. 
 
 Case 4. Given A, B, and c. If the inequalities TT > A >0, 
 TT > B >0, 7r>c>0, are satisfied, the triangle 'is possible, 
 and determinate in a single way. We may find a, and b by 
 the following formulas : 
 
 A-B 
 a+b COS -T~ c 
 
 cos 
 
 sin 
 
 A-B 
 
 a b 2 c 
 
 tan ^ A , B tan - These give 
 
 ~2~~ P > a = p + q, 
 
 whence 
 
 a b _ b = p q. 
 
 ~T~ ~ q ' 
 
 C may be found from the formula of 96, 
 
 a+ b 
 c sin -y- A _ B 
 
 ctn = tan - Use 79 as a check. 
 
 i a b Z 
 
 sin 
 
 2 
 
 If only the angle C is required, it may be obtained by the 
 formula, 81, 
 
100 SPHERICAL TRIGONOMETRY. 
 
 cos C = cos A cos B -f- sin A sin B cos c 
 = cos B ( cos A -f- sin A tan B cos c) 
 _ cos B sin (A u) 
 
 sin u 
 _ cos B sin [A ctn -1 (tan B cos c)] 
 
 sin ctn -1 (tan B cos c) 
 if ctn u = tan B cos c. 
 
 Similarly the sides a and b may be computed, if in the 
 formulas, 
 
 ctn a sin c = cos c cos B -j- sin B ctn A, 
 ctn b sin c = cos c cos A + sin A ctn B, 
 we put ctn u = tan B cos c, 
 
 ctn v = tan A cos c, 
 
 ctn c cos FB ctn- 1 (tan A cos c)] 
 
 and obtain, ctn a = u r- ; ; : -** > 
 
 cos ctn" 1 (tan A cos c) 
 
 ctn c cos [A ctn- 1 (tan B cos c)] 
 
 cos ctii" 1 (tan B cos c) 
 
 Similarly as in Case 3, it may be shown that the use of the 
 auxiliary angles u and v, is equivalent to decomposing the 
 
 triangle ' into two right triangles. The student may 
 
 8. DC 
 
 that angle u = angle BAD, . 
 prove , & , A _ ' in Fig. 46. 
 
 and angle v = angle ADD, 
 
 Case 5. Given a, b, and A. 
 
 (1) If the inequalities, 7r>a>0, TT > b > 0, 7r>A>0, 
 are satisfied, there will always be one solution, when b differs 
 
 more than a from 
 
 (2) Unless the inequalities : - [= sin B] < 1, 
 
 sin a 
 
 TT > . > , or else > > 0, and TT > b > 0, are satisfied 
 
 r\ +J r\ 
 
 when a differs more than b from , there will be no solu- 
 
SPHERICAL TRIANGLES. 101 
 
 tion ; and when they are satisfied, and a and b thus related, 
 there will always, necessarily, be two solutions. 
 
 These statements may be proved as follows : It is evident 
 that there can be no solution under (2), unless the inequali- 
 ties, TT > ^ > ^ , or else ^ > * > 0, are satisfied under the 
 conditions. For, in any spherical triangle with real parts, if 
 one side differs more than another side from , its opposite 
 angle is of the same quadrant with it. 
 
 This appears from the fact that the parts of the triangle 
 
 cos a cos b cos c __ , 
 
 must satisfy the relation cos A = : ; : > 77, and 
 
 sin b sin c 
 
 if a differs more than b from i cos a > cos b, numerically, 
 
 and if c is real, cos c < 1, numerically, and, therefore, 
 cos a > cos b cos c, numerically, and hence, cos A and cos a 
 have the same sign, and A and a are of the same quadrant. 
 To prove the remaining statements, we consider the equa- 
 cos a cos b cos c 
 
 tion sin c = 
 
 sin b cos A 
 
 cos a cos b , , 
 
 where we let r = m, r r = n, and get the 
 
 sin b cos A sm b cos A 
 
 following equation in sin c : 
 
 (1 + n 2 ) sin 2 c 2 m sin c + m 2 n 2 = 0. 
 
 >2_ n 2 
 
 The product of the roots is 2 " 
 The roots are : 
 
 sin G! 2 (m + n (n 2 + 1 m 2 )^), 
 
 sin c 2 = -^-7- -- (m n (n 2 + 1 m 2 ) 1 ). 
 
IS 
 
 102 SPHERICAL TRIGONOMETRY. 
 
 According as 4 I differs more than \ i from > that i 
 
 according as we have {,<>;}, n' 2 ^ m 2 , and in (2) m > 0. 
 
 I (*) J ^ 
 
 According as we have ! xo\ f *^ e P r duct f the roots, 
 
 m 2 n 2 . ("negative") . 
 
 . > is therefore s > ; and its form shows that 
 
 ir + 1 L positive J 
 
 the factors, i.e., the roots, must be real, and that, there- 
 fore n 2 -fl m 2 >0; and in \ ; ^ V that the roots must be 
 
 iw.J 
 
 f positive and negative "] . 
 
 4 fT, ... .. h And m (2) both roots 
 
 [both positive or both negative j 
 
 must be positive, since, as m > 0, a single root at least must 
 be positive. 
 
 Again, 2 m 2 n 2 (n 2 + 1 - m 2 ) -< m 4 n 4 + (n 2 + 1 m 2 ) 2 , 
 and 
 
 4 m 2 n 2 (n 2 + 1 m 2 ) < m 4 n 4 + 2 m 2 n 2 (n 2 + 1 m 2 ) 
 
 -h(n 2 +l-m 2 ) 2 . 
 
 Taking positive values of m, n, and n 2 -[-l m 2 , 
 2 mn (n 2 + 1 - m 2 )*-< n 2 + 1 - m 2 + m 2 n 2 , 
 .-.2mn(n 2 + l m 2 )^ <n 4 + 2n 2 +l m 2 n 2 (n 2 + l m 2 ), 
 . ' . m 2 + n 2 (n 2 + 1 m 2 ) + 2 mn (n 2 + 1 m 2 )* < (n 2 + I) 2 , 
 . m + n(n 2 +l-m 2 )* 
 li* + 1 
 
 From this it is seen that no one of all the possible values 
 of sin c, in either (1) or (2), can be numerically greater than 
 unity. 
 
 In (1), there is one solution only, for one value of sin c 
 is negative, and B must be of the same quadrant with b. 
 
 In (2), there are two solutions, real and distinct, in sin c, 
 
SPHERICAL TRIANGLES. 103 
 
 unless n (n 2 + 1 m 2 )* = 0, i.e., unless b = , or else cos 2 a 
 
 cos 2 b = sin b cos A, when the two solutions in sin c analyti- 
 cally coincide. For, with this possible exception, we have 
 
 proved that 1 > . 1 > 0, and sin GI 4= sin c 2 . If n = 0, 
 
 smc 2 
 
 I.e., if b = , the two analytically coincident solutions in sin c 
 
 give two real triangles, in which the values of c are supple- 
 mentary, as shown in 89. 
 
 If n 2 -fl m 2 0, i.e., if cos 2 a cos 2 b = sin b cos A, the 
 condition sin B<1, is violated, and the solutions are not 
 real.* 
 
 It might seem as if sin Ci , and sin c 2 , would give two values 
 of G! and of c 2 , which taken with the two values of B, obtained 
 from sin B, would give eight triangles, but the formulas of 
 96, which we use in finding C and c, show that only one 
 value of c can consist with one value of B. (See examples 
 XI. 24, 25, 26.) 
 
 We may now consider that we have 
 
 b 
 
 bc 2 A 
 given 
 
 B 2 C 2 
 required 
 
 where TT > 1 
 
 C 2 
 
 A 
 
 We may find B by the formula, 
 
 sinB = siDAsinb . 79 
 
 sin a 
 
 If two values of B are admissible, they are supplements of 
 each other and may be denoted by B! and B 2 . When B is 
 found, c and C may be obtained from the formulas : 
 
 * It is easily shown in the Theory of Functions that sin 1 x, if x > 1, 
 is a complex angle. 
 
104 SPHERICAL TRIGONOMETRY. 
 
 A+B 
 
 a + b 
 C Sm A-B 
 
 sn 
 
 If there are two solutions we have : 
 
 . A + B, 
 Cl Sm a-b 
 
 sin 
 
 . A+B 2 
 Sm -2- a-b 
 
 with two similar formulas for Ci and C 2 . 
 Case 6. Given A, B, and a. 
 
 (1) If the inequalities, 7r>A>0, 7r>B>0, 7r>a>0, 
 are satisfied, there will always be one solution, when B differs 
 
 more than A from 
 
 LJ 
 
 (2) Unless the inequalities, 
 
 sin a sin B _ . . _. ^ . ^ A ^ TT 
 
 : - [= sm b] < 1, TT > > - > 
 
 sm A L a 2 
 
 or else > > 0, 
 and TT > B > 0, 
 
 are satisfied when A differs more than B from > there will 
 
 be no solution ; and when they are satisfied, and A and B thus 
 related, there will always, necessarily, be two solutions. 
 
SPHERICAL TRIANGLES. 105 
 
 These statements might be proved similarly as in the 
 preceding case, taking the equation, 
 
 . ^ cos A + cos B cos C 
 sin C = - . p - 81 
 
 sin D cos a 
 
 But it is not necessary to do this. If A, B. and a are given, 
 a', b', and A' of the polar triangle are given. If the polar 
 triangle has two solutions, the given triangle will have two 
 solutions, and conversely. The polar triangle will have two 
 
 solutions, when a ' differs more than b' from The given 
 triangle will therefore have two solutions, when IT A differs 
 more than TT B from > i.e., when A differs more than B 
 
 from We may find b by the formula 
 
 L 
 
 sin a sin B 
 
 sin b = -- : T 79 
 
 sm A 
 
 and if two values of b are admissible they are supplements of 
 each other, and may be denoted by b t and b 2 . c and C may 
 be found from the formulas : 
 
 A + B 
 c Sm a-b 
 
 Bn |- 
 
 a + b 96 
 
 Sm ~ A-B 
 
 Similarly if there are two solutions. 
 
106 SPHERICAL TRIGONOMETRY. 
 
 J>. Examples. XL 
 
 Apply the conditions of possibility, and solve the following triangles, 
 having given : 
 
 1. a = 124 12'. 5, b = 54 18', c - 97 12 / .5. 
 
 2. a = 74 26', b = 85 12', c = 29 47'. 
 
 3. A = 20 10', B = 55 52', C = 114 20'. 
 
 4. A = 130, B = 110, C = 80. 
 
 6. a = 35 37', b = 59 12', C = 124 18'. 
 
 6. A = 54 55', b = 69 25', c = 109 46'. 
 
 7. A = 26 59', B = 39 45', c - 154 47'. 
 
 8. A = 57 35', B = 120 48', c = 124 18'. 
 
 9. a = 50, b = 40, A = 80. 
 
 10. a = 150 57', b = 134 16', A = 144 23'. 
 11. A = 113 39', B = 123 40', a = 65 40'. 
 
 12. A = 52 50', B = 66 07', a = 59 28'. 
 
 13. Are the following triangles possible ? If impossible, what condi- 
 tions are violated ? 
 
 (1) a = 160 25', b = 92 23', c = 64 49'. 
 
 (2) A - 137 18', B = 123 15', C = 74 36'. 
 
 (3) a = 2 29', b = 37'. 5, C = 179 48'. 
 
 (4) A = 5 41', B = 15 50', c = 163 34'. 
 
 (5) A = 88 49', a = 16 34', b = 16 30'. 
 
 (6) a = 30 08', b = 54 03', A - 60 01'. 
 
 (7) a = 117 29', A = 174 17', B = 173 25'. 
 
 14. Prove that the polar radii of the inscribed circle of a triangle, and 
 of the circumscribed circle of its polar triangle are complements of each 
 other. 
 
 15. If R, R a , Rb, Re, are the polar radii of the circles circumscribed 
 about the triangles ABC," BCA', CAB', ABC', respectively, where A A', BB', 
 CC', are diameters of the sphere, prove that 
 
 ctn R cos S = ctn R a cos (S A) 
 = ctnRbCos(S B) 
 = ctnR c cos(S C) 
 
 = v cos S cos (S A) cos (S B) cos (S C). 
 [= L, for brevity.] 
 
 16. If r, r a , r b , r c , are, respectively, the polar radii of the circles 
 inscribed in the triangles of the preceding problem, prove that 
 
EXAMPLES. XL 107 
 
 tan r sin s = tan r a sin (s a) = tan r b sin (s b) = tan r c sin (s c) 
 = Vsin s sin (s a) sin (s b) sin (s c). 
 [= I, for brevity.] 
 
 Using the notation of examples 15 and 16, prove the following : 
 
 17. tan r tan r a tan r b tan r c = 1 2 . 
 
 18. ctn R ctn R a ctn Rb ctn R c = L 2 . 
 
 19. tan r a tan r b tan r c = I sin s. 
 
 20. ctn R a ctn Rb ctn R c = L cos S. 
 
 4 L sin S 
 
 21. tan r a + tan r b + tan r c - tan r - 
 
 22. tan R a + tan Rb + tan R c tan R = 2 ctn r. 
 
 23. ctn R ctn R a ctn Rb ctn R c = - j -- 
 
 24. Show that if the inequalities of 101. 6. Case 5. (1), are satisfied, 
 
 sin A sin b r , _, 
 
 - [= sin B] < 1. 
 sin a 
 
 Show that under 101. 6. Case 5. (2), the two analytically coincident 
 solutions in sin c : 
 
 25. Give two triangles whose sum is equal to a lune, if n = 0. 
 
 26. Violate the condition 
 
 sin A sin b r 
 
 for real triangles, if n 2 + 1 m 2 = 0. 
 
 27. Find the volume V of an oblique parallelepiped, having given the 
 three conterminous edges I, m, n, and the three angles cr, 0, 7, which 
 the edges make with each other. 
 
 In the following problems the earth is assumed to be a sphere whose 
 radius is 3963 miles. 
 
 28. Given the latitudes and longitudes of three places on the earth's 
 surface, show how to find the area of the triangle of which the given 
 places are the vertices. 
 
 29. Colorado extends from 37 to 40 N., and from 102 to 107 W. 
 What is its area in statute square miles ? 
 
 30. If Pennsylvania is assumed to extend from 39 43' to 42 N. , and 
 from 75 to 80 35' W., what is its area in. statute square miles ? 
 
 31. If Ohio is assumed to extend from 39 to 41 30' N., and from 
 80 35' to 84 40' W., what is its area in statute square miles ? 
 
 32. Given the latitudes and longitudes of two places on the earth's 
 surface, show how to find the shortest distance between them. 
 
108 SPHERICAL TRIGONOMETRY. 
 
 33. What is the shortest distance in statute miles, from Cambridge, 
 Mass. (lat. 42 23' N., long. 77 08' W.), to Oberlin (lat. 41 17' N., long. 
 82 11' W.)? 
 
 34. What is the shortest distance in statute miles from Philadelphia 
 (lat. 39 57' N., long. 75 10' W.) to Oberlin ? 
 
 35. Given the shortest distance between two places, and their lati- 
 tudes, show how to find the difference in their local time. 
 
 36. The shortest distance between Sandy Hook (lat. 40 28' N.) and 
 Cork Harbor (lat. 51 47' N.) is 2726 geographical miles. When it is 
 6 A.M. at Sandy Hook, what time is it at Cork Harbor ? 
 
 37. Given 0, 5, and t. What is the formula for the position angle of 
 the astronomical triangle ? 
 
 38. At Oberlin Observatory, the declination of a star is 54 25', its 
 hour angle 10 h. 20 m. What is its position angle ? 
 
 39. Find the distance between the sun and moon at Greenwich mean 
 noon on June 20, 1896, when their respective right ascensions are 5 h. 
 58 m. 11.09s. and 13 h. 58 m. 21.65 s., and their declinations 23 27' 14" 
 and 17 06' 15". 3. 
 
 40. Find the distance between Uranus and Neptune at Greenwich 
 mean noon on March 31, 1896, when their respective right ascensions are 
 15 h. 26 m. 59.06 s. and 4 h. 58 m. 8s., and their declinations 18 30' 34" 
 and 21 16' 40". 4. 
 
 41. Find the distance between Venus and Jupiter at Greenwich mean 
 noon on Sept. 10, 1896, when their respective right ascensions are 12 h. 
 21 m. 38.16 s. and 9 h. 55 m. 29.76 s., and their declinations 1 09' 19". 9 
 and 13 27' 06". 3. 
 
 42. Find the moon's right ascension and declination at Greenwich 
 mean midnight Oct. 29, 1896, when her true longitude is 124 53' 30". 7, 
 and the obliquity of the ecliptic is 23 27' 16". 78. 
 
 43. Find the declination of a star whose longitude is 47 25' and 
 latitude 51 36'. (Take e = 23 27'.) 
 
 44. Find the latitude of a star whose longitude is 27 19', and 
 declination 32 48'. 
 
ADVERTISEMENTS 
 
MATHEMATICS. 93 
 
 The Method of Least Squares. 
 
 With Numerical Examples of its Application. By GEORGE C. COM- 
 STOCK, Professor of Astronomy in the University of Wisconsin, and 
 Director of the Washburri Observatory. 8vo. Cloth, viii + 68 pages. 
 Mailing price, $1.05 ; for introduction, $1.00. 
 
 College Requirements in Algebra: A Final Review. 
 
 By GEORGE PARSONS TIBBETS, A.M., Instructor in Mathematics, 
 Williston Seminary. 12mo. Cloth. 46 pages. By mail, 55 cents ; to 
 teachers and for introduction, 50 cents. 
 
 Peirce's Elements of Logarithms. 
 
 With an explanation of the author's Three and Four Place Tables. By 
 Professor JAMES MILLS PEIRCE, of Harvard University. 12mo. Cloth. 
 80 pages. Mailing price, 55 cents : for introduction, 50 cents. 
 
 Mathematical Tables Chiefly to Four Figures. 
 
 With full explanations. By Professor JAMES MILLS PEIRCE, of Har- 
 vard University. 12ino. Cloth. Mailing price, 45 cents; for intro- 
 duction, 40 cents. 
 
 Elements of the Differential Calculus. 
 
 With numerous Examples and Applications. Designed for Use as a 
 College Text-Book. By W. E. BYERLY, Professor of Mathematics, 
 Harvard University. 8vo. 273 pages. Mailing price, $2.15; for intro- 
 duction, $2.00. 
 
 rpHE peculiarities of this treatise are the rigorous use of the 
 Doctrine of Limits, as a foundation of the subject, and as 
 preliminary to the adoption of the more direct and practically 
 convenient infinitesimal notation and nomenclature ; the early 
 introduction of a few simple formulas and methods for integrat- 
 ing ; a rather elaborate treatment of the use of infinitesimals in 
 pure geometry ; and the attempt to excite and keep up the interest 
 of the student by bringing in throughout the whole book, and not 
 merely at the end, numerous applications to practical problems in 
 geometry and mechanics. 
 
 E. H. Moore, Professor of Mathe- 1 ablest text-book on the calculus yet 
 matics, University of Chicago : The I written by an American. 
 
94 MATHEMATICS. 
 
 Elements of the Integral Calculus. 
 
 Second Edition, revised and enlarged. By W. E. BYERLY, Professor 
 of Mathematics in Harvard University. 8vo. xvi + 383 pages. Mail- 
 ing price, $2.15; for introduction, $2.00. 
 
 work contains, in addition to the subjects usually treated 
 in a text-book on the Integral Calculus, an introduction to. 
 Elliptic Integrals and Elliptic Functions ; the Elements of the 
 Theory of Functions ; a Key to the Solution of Differential Equa- 
 tions ; and a Table of Integrals. 
 
 John E. Clark, Prof, of Mathe- 
 matics, Sheffield Scientific School of 
 Yale University: The additions to 
 the present edition seem to me most 
 judicious and to greatly enhance its 
 
 value for the purposes of university 
 instruction, for which in several im- 
 portant respects it seems to me hetter 
 adapted than any other American 
 text-book on the subject. 
 
 An Elementary Treatise on Fourier's Series, 
 
 and Spherical, Cylindrical, and Ellipsoidal Harmonics, with Appli- 
 cations to Problems in Mathematical Physics. 
 
 By WILLIAM E. BYERLY, Ph.D., Professor of Mathematics, Harvard 
 University. 8vo. Cloth, x + 288 pages. Mailing price, $3.15 ; for 
 introduction, $3.00. 
 
 HHHIS book is intended as an introduction to the treatment of 
 some of the important Linear Partial Differential Equations 
 which lie at the foundation of modern theories in physics, and 
 deals mainly with the methods of building up solutions of a 
 differential equation from easily obtained particular solutions, in 
 such a manner as to satisfy given initial conditions. 
 
 John Perry, Technical College, 
 Finfsbury, London, England: Byer- 
 ly's book is one of the most useful 
 books in existence. I have read it 
 with great delight and I am happy 
 
 to say that although it seemed to be 
 written expressly for me, one of my 
 friends who is a great mathemati- 
 cian, seems as delighted with it as I 
 am myself. 
 
 A Short Table Of Integrals. Revised and Enlarged. 
 
 By B. O. PEIRCE, Prof. Math., Harvard Univ. 32 pages. Mailing 
 price, 15 cents. Bound also with Byerly's Calculus. 
 
 B LI er Iy f s Syllabi. 
 
 Each, 8 or 12 pages, 10 cents. The series includes, Plane Trigonom- 
 etry, Plane Analytical Geometry, Plane Analytic Geometry (Advanced 
 Course), Analytical Geometry of Three Dimensions, Modern Methods 
 in Analytic Geometry, the Theory of Equations. 
 
MATHEMATICS. 95 
 
 Directional Calculus. 
 
 By E. W. HYDE, Professor of Mathematics in the University of Cincin- 
 nati. 8vo. Cloth. xii + 247 pages, with blank leaves for notes. Price 
 by mail, $2.15; for introduction, $2.00. 
 
 HHHIS work follows, in the main, the methods of Grassmann's 
 Ausdehnungslehre, but deals only with space of two and three 
 dimensions. The first two chapters which give the theory and 
 fundamental ideas and processes of his method, will enable students 
 to master the remaining chapters, containing applications to Plane 
 and Solid Geometry and Mechanics ; or to read Grassmann's original 
 works. A very elementary knowledge of Trigonometry, the Differ- 
 ential Calculus and Determinants, will be sufficient as a preparation 
 for reading this book. 
 
 Daniel Carhart, Prof, of Mathe- 
 matics, Western University of Penn- 
 sylvania : I am pleased to note the 
 success which has attended Professor 
 
 Hyde's efforts to bring into more 
 popular form a branch of mathemat- 
 ics which is at once so abbreviated in 
 form and so comprehensive in results. 
 
 Elements of the Differential and Integral Calculus. 
 
 With Examples and Applications. By J. M. TAYLOR, Professor of 
 Mathematics in Colgate University. 8vo. Cloth. 249 pages. Mailing 
 price, $1.95; for introduction, $1.80. 
 
 rpHE aim of this treatise is to present simply and concisely the 
 fundamental problems of the Calculus, their solution, and more 
 common applications. 
 
 Many theorems are proved both by the method of rates and that 
 of limits, and thus each is made to throw light upon the other. 
 The chapter on differentiation is followed by one on direct integra- 
 tion and its more important applications. Throughout the work 
 there are numerous practical problems in Geometry and Mechanics, 
 which serve to exhibit the power and use of the science, and to 
 excite and keep alive the interest of the student. In February, 1891, 
 Taylor's Calculus was found to be in use in about sixty colleges. 
 
 The Nation, New York : In the 
 first place, it is evidently a most 
 carefully written book. . . . We are 
 acquainted with no text-book of the 
 Calculus which compresses so much 
 matter into so few pages, and at the 
 same time leaves the impression that 
 
 all that is necessary has been said. 
 In the second place, the number of 
 carefully selected examples, both of 
 those worked out in full in illustra- 
 tion of the text, and of those left for 
 the student to work out for himself, 
 is extraordinary. 
 
MATHEMATICS. 
 
 Elements of Solid Geometry. 
 
 By ARTHUR LATHAM BAKER, Professor of Mathematics, University of 
 Rochester. 12mo. Cloth, xii + 126 pages. Mailing price, 90 cents; 
 for introduction, 80 cents. 
 
 distinctive features of this work are improved notation, 
 tending to simplify the text and figures ; improved diagrams, 
 particular attention being paid to the perspective of the figures ; 
 clear presentation, each part of the discussion being presented 
 under a distinct heading ; generalized conceptions, which is the 
 principal feature of the work, the general theorems for the frustum 
 of a pyramid being first worked out, and then the pyramid, cone, 
 prism, and cylinder discussed as special cases of the pyramidal 
 frustum and of the prismatoid. The essential unity of the sub- 
 ject is constantly impressed upon the reader. 
 
 Benjamin G. Brown, Professor of 
 Mathematics in Tufts College : It is 
 a most excellent book. I have never 
 
 used a book for the first time with 
 greater satisfaction. 
 
 Elementary Co-ordinate Geometry. 
 
 By W. B. SMITH, Professor of Mathematics, Missouri State University. 
 8vo. Cloth. 312 pages. Mailing price, $2.15; for introduction, $2.00. 
 
 book is spoken of as the most exhaustive work on the 
 subject yet issued in America ; and in colleges where an easier 
 text-book is required for the regular course, this will be found of 
 great value for post-graduate study. 
 
 Wm. G. Peck, late Prof, of Math- 
 ematics and Astronomy, Columbia 
 College: Its well compacted pages 
 
 contain an immense amount of mat' 
 ter, most admirably arranged. It is 
 an excellent book. 
 
 Theory of the Newtonian Potential Functions. 
 
 By B. O. PEIRCE, Professor of Mathematics and Physics, in Harvard 
 Univ. 8vo. Cloth. 154 pages. Mailing price, $1.60; for introd. $1.50. 
 
 book gives as briefly as is consistent with clearness so 
 much of that theory as is needed before the study of standard 
 works on Physics can be taken up with advantage. A brief treat- 
 ment of Electrokinematics and many problems are included. 
 
MATHEMATICS. 
 
 97 
 
 Academic Trigonometry: pi ane and spherical. 
 
 By T. M. BLAKSLEE, 
 
 Moines College, Iowa. 12mo. Cloth. 
 
 Ph.D. (Yale), Professor of Mathematics in Des 
 . 12mo. Cloth. 33 pages. Mailing price, 30 
 cents; for introduction, 25 cents. 
 
 fPHE Plane and Spherical portions are arranged on opposite pages. 
 
 The memory is aided by analogies, and it is believed that the 
 
 entire subject can be mastered in less time than is usually given to 
 
 Plane Trigonometry alone, as the work contains but 29 pages of text 
 
 The Plane portion is compact, and complete in itself. 
 
 Examples of Differential Equations. 
 
 By GEORGE A. OSBORNE, Professor of Mathematics in the Massachu- 
 setts Institute of Technology, Boston. 12mo. Cloth, vii + 50 pages. 
 Mailing Price, 60 cents; for introduction, 50 cents. 
 
 A SERIES of nearly three hundred examples with answers, sys- 
 "^ tematically arranged and grouped under the different cases, 
 and accompanied by concise rules for the solution of each case. 
 
 Selden J. Coffin, Prof, of Astron- 1 ance is most timely, and it supplies 
 omy, Lafayette College : Its appear- | a manifest want. 
 
 Determinants. 
 
 The Theory of Determinants: an Elementary Treatise. By PAUL H. 
 HANUS, B.S., recently Professor of Mathematics in the University of 
 Colorado, now Assistant Professor, Harvard University. 8vo. Cloth. 
 viii + 217 pages. Mailing price, $1.90 ; for introduction, $1.80. 
 
 book is written especially for those who have had no pre- 
 vious knowledge of the subject, and is therefore adapted to 
 self-instruction as well as to the needs of the class-room. The 
 subject is at first presented in a very simple manner. As the 
 reader advances, less and less attention is given to details. 
 Throughout the entire work it is the constant aim to arouse 
 and enliven the reader's interest, by first showing how the various 
 concepts have arisen naturally, and by giving such applications as 
 can be presented without exceeding the limits of the treatise. 
 
 William G. Peck, late Prof, of 
 Mathematics, Columbia College, 
 N. Y. : A hasty glance convinces me 
 that it is an improvement on Muir. 
 
 T. W. Wright, Prof, of Mathemat- 
 ics, Union Univ., Schenectady, N.Y.: 
 It fills admirably a vacancy in our 
 mathematical literature, and is a 
 very welcome addition indeed. 
 
98 MATHEMATICS. 
 
 Analytic Geometry. 
 
 By A. S. HARDY, Ph.D., recently Professor of Mathematics in Dart- 
 mouth College, and author of Elements of Quaternions. 8vo. Cloth. 
 xiv + 239 pages. Mailing price, $1.60; for introduction, $1.50. 
 
 work is designed for the student, not for the teacher. It 
 is hoped that it will prove to be a text-book which the teacher 
 will wish to use in his class-room, rather than a book of reference 
 to be placed on his study shelf. 
 
 Oren Root, Professor of Mathe- 
 matics, Hamilton College : It meets 
 quite fully my notion of a text for 
 our classes. 
 
 John E. Clark, Professor of Mathe- 
 
 matics, Sheffield Scientific School of 
 Yale College : I need not hesitate to 
 say, after even a cursory examina- 
 tion, that it seems to me a very at- 
 tractive book. 
 
 Elements of Quaternions. 
 
 By A. S. HARDY, Ph.D., recently Professor of Mathematics in Dart- 
 mouth College. Second edition revised. Crown. 8vo. Cloth, viii 
 + 234 pages. Mailing price, $2.15 ; for introduction, $2.00. 
 
 HPHE chief aim has been to meet the wants of beginners in the 
 class-room. 
 
 Elements of the Calculus. 
 
 By A. S. HARDY, Ph.D., recently Professor of Mathematics in Dart- 
 mouth College. 8vo. Cloth. xi + 239 pages. Mailing price, $1.60; 
 for introduction, $1.50. 
 
 Part I., Differential Calculus, occupies 166 pages. Part II., Integral 
 Calculus, 73 pages. 
 
 HPHIS text-book is based upon the method of rates. From the 
 author's experience in presenting the Calculus to beginners, 
 the method of rates gives the student a more intelligent, that is, a 
 less mechanical, grasp of the problems within its scope than any 
 other. No comparison has been made between this method and 
 those of limits and of infinitesimals. This larger view of the 
 Calculus is for special or advanced students, for whom this work 
 is not intended. 
 
 Ellen Hayes, Professor of Mathe- 
 matics, Wellesley College : I have 
 found it a pleasure to examine the 
 book. It must commend itself in 
 many respects to teachers of Cal- 
 culus. 
 
 J. B. Coit, Professor of Mathe- 
 matics, Boston University : The 
 treatment of the first principles of 
 Calculus by the method of rates is 
 eminently clear. 
 
14 DAY USE 
 
 RETURN TO DESK FROM WHICH BORROWED 
 
 LOAN DEPT. 
 
 This book is d, 
 
 ( 
 Renewed 
 
 on 1 le 
 bo.ks 
 
 REcfre-f-rT 
 
 JM31'fi4-?IH 
 
 ^ 
 
 Rgc'D 
 
 SIB12'8A-b 
 
 i^^r'6SVS 
 
 REC'D 
 
 flPfl 6'65-m 
 
 ^IMTOB** 
 
 LD 21A-40m-4,'63 
 (D6471slO)476B 
 
 General Library 
 
 University of California 
 
 Berkeley 
 
 LD 2lA-50m-8,'61 
 (Cl795slO)476B 
 
10 i 
 
 v-ai&n&'&A 
 
 - r_*t * *" **** * / 
 
 
 
 THE UNIVERSITY OF CALIFORNIA ^IBRARY