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 MANUAL 
 
 OF 
 
 TRIGONOMETRY. 
 
 FOR THE USE 
 
 OF 
 
 HIGH SCHOOLS. ACADEMIES AND COLLEGES. 
 
 BY 
 
 J. B. CLARKE, Ph. B. 
 
 ASSISTANT PROFESSOR OF MATHEMATICS 
 
 IN THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
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 [AUTHOa'S EDITION.] 
 

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 F»ACIF^IC PRESS, 
 
 PRINTERS, ELECTROTYPERS, AND BINDERS. 
 
PREFACE. 
 
 In the following pages the author has endeavored to 
 present a brief, but thorough and complete, course in 
 Trigonometry. While non-essentials have been excluded, 
 considerable attention has been given to some matters 
 not discussed in any of the usual text-books. It is dis- 
 couraging even to a bright student, to meet, in his work 
 in the Integral Calculus, difficulties that should have 
 been surmounted in his Trigonometrical course. The 
 chapter on Algebraic Trigonometry will, it is hoped, 
 lighten the labor of the student preparing for advanced 
 work and broaden the views of the reader whose Trigo- 
 nometry is merely a minor part of a symmetrical course 
 of liberal ytudy. The text, taken in connection with the 
 exercises, (all of which should be worked out), will afford 
 ample preparation for any University Mathematical 
 course. 
 
 The author desires to express his sense of obligation 
 to Prof. Irving Stringham for encouraging and valuable 
 hints and suggestions. 
 
 J. B. CLARKE. 
 
 Department of Mathematics, 
 University of California, 
 February, 1888. 
 
 183630 
 
TABLH OK CONTKNTS. 
 
 Paoe. 
 
 Chapter I. — Preliminary Definitions. — Signs and Limiting Values 
 
 of Functions. — Fundamental Relations. — Geometric Repre- 
 sentation of Functions. — Trigonometric Tables 9 
 
 Chapter II. —Solution of Right Triangles. — Solution of Oblique 
 
 Triangles by means of Right Triangles. — Examples 20 
 
 Chapter III. — Functions of the Sum and of the Difference. — 
 Sums and Differences of Functions. — Miscellaneous Form- 
 ulae 26 
 
 Chapter IV. — Additional Formulae for Plane Triangles. — Special 
 
 Formulae 30 
 
 Chapter V. — Spherical Trigonometey. — Napier's Rules. — De- 
 termination of Species. — Solution of Right Triangles and 
 Solution of Oblique Triangles by means of Right Triangles . . 33 
 
 Section II. — Additional Formulae for Oblique Triangles. — Special 
 
 Formulae, — Napier's and Delambre's Analogies 41 
 
 Chapter VI. — Algebraic Trigonometry. — Trigonometric De- 
 velopments. — Exponential Functions. — Circular and Hyper- 
 bolic Functions. — Properties and Relations of the Functions. — 
 Geometric Representation of the Exponential Functions 47 
 
 Exercises. — 
 
OF THE \ 
 
 IVERSfTT ^ 
 
 . OF 
 
 TRIGONOMETRY. 
 
 CHAPTER I. 
 
 1. Trigonometry is a branch of Mathematics, in which 
 are discussed the properties, relations, and applications of cer- 
 tain functions of angles. These functions, to be hereafter 
 defined, are called the Trigonometric functions, and are practi- 
 cally useful mainly in the solution of triangles. 
 
 2. An angle, AOB (Fig. 1), 
 is considered as generated by 
 the revolution of a line about 
 the vertex, O, from the posi- 
 tion OA to the position OB. 
 Ois called the Angular Centre; 
 OA, the Initial, and OB, the 
 Tei-minal dine. We call the 
 revolution, and the angle gen- 
 erated, joo^i^ive, when the motion 
 is opposite in direction to that of 
 fig- 1- a watch-hand ; negative, when 
 
 the motion is in the direction of that of a watch-hand. Angles 
 are assumed to be positive unless they are explicitly stated to 
 be negative. Angular magnitude, thus considered, evidently 
 admits of infinite extension ; AOD may be generated by a 
 simple passage of the moving line from OA to OD, or by a 
 movement in which this line reaches OD after passing around 
 O any number of times. This requires, of course, that the 
 magnitude of the measuring arcs be similarly considered ; 
 hence the occurrence of angles of 540°, 630°, etc. (i. e., the 
 
 (9) 
 
 HHkSBSUHE 
 
10 DEFINITIONS. 
 
 angle as ordinarily taken, + any multiple of 360°). The 
 plane space about O is, for convenience, divided into four equal 
 parts, called quadrants, by two lines, viz., the Initial line, as- 
 sumed horizontal and unlimited in length, and a perpendicular 
 thereto through O. These quadrants are numbered as in the 
 figure, and an angle lies in that quadrant in which is found its 
 terminal line. 
 
 3. An angle is usually denoted by a single symbol, thus : x, 
 y, A. In general this symbol will denote the radial measure 
 of the angle, except when we deal with triangles. The radial 
 measure (Ven. IV, 19) of an angle is the arc intercepted (by 
 the initial and terminal lines) divided by the radius with which 
 it is described from the vertex as a centre. The unit is the 
 angle whose sides intercept on the circumference an arc equal 
 in length to the radius, for, in the case of this angle, arc : ra- 
 dius = 1. As 2-R = 360°, R = 360 ^ 2r = 180° -r 3.1415926 = 
 57°.29578, or about 57°.3 (57° 17' 45'^ nearly). To pass 
 from the " degree " measure of an angle to its " radial " meas- 
 ure, we therefore divide by 57°.3, and conversely. 
 
 The complement of an angle (or arc) is a right angle (or 
 90°), minus the angle (or arc); the supplement is the difference 
 between the angle (or arc) and two right angles (180°). 
 
 4. Angles are dealt with by means of certain Trigonometric 
 Functions. If a perpendicular be dropped from any point of 
 the Terminal line on the (unlimited) Initial line, a right triangle 
 is formed called the Triangle of Reference. The following ratios, 
 involving its sides, are the Trigonometric Functions of the angle. 
 
 (1.) The perpendicular (p) divided by the hypothenuse, is 
 the SINE (sin). Thus (Fig. 1), BM:OB = sin AOB. 
 
 (2.) The base (6) divided by the hypothenuse, is the cosine 
 (cos). Thus, cos AOB = OM : OB ; cos AOD = OL : OD. 
 
 (3.) The perpendicular (p) divided by the base (h) is the 
 TANGENT (tan). Thus, tan AOB = BM : OM. 
 
 (4.) The base (h) divided by the perpendicular (p) is the 
 COTANGENT (cot). Thus, cot AOC = OK : KG. 
 
5/: 
 
 TRIGONOMETRY. 11 
 
 (5.) The reciprocal of the sine is called the co-secant (csc). 
 Thus, CSC AOB = OB : BM. 
 
 (6.) The reciprocal of the cosine is called the Secant (sec). 
 Thus, sec AOD = OD:OL. 
 
 Unity minus the cosine is called the versed-sine (yersin) ; 
 thus versin AOB =1 - OM : OB. 
 
 Unity minus the sine is called the coversed-sine (coversin); 
 thus coversin AOC = 1 - CK : CO 
 
 It will be noticed: — 
 
 (1.) That each ratio can have but one value for any one 
 angle. For if the perpendicular were dropped from any other 
 point of OB than B (e. g., B', C, D'), the new triangles of 
 reference would be similar to OBM, and hence the ratios of 
 their sides would equal the corresponding ratios in OBM. 
 
 (2.) Neither the sine nor the cosine can exceed 1. 
 
 (3.) The secant and cosecant must both exceed 1. 
 
 (4.) Neither the versed-sine nor the coversed-sine can ex- 
 ceed 2. 
 
 5. The functions (4) may evidently be grouped in two triads, 
 viz.: Sine, Tangent, Secant, and Co-sine, Co-tangent, Co-secant. 
 Fig. 2 will render clear the reason of this nomenclature. Let 
 AOB be any angle, BOQ its com- 
 plement, BOM and KOL the tri- 
 angles of reference. As BOM is sim- 
 ilar to KOL, 
 
 OM : OB = KL : OK, or co-sine 
 AOB = sin (90° -AOB), 
 
 OM : BM = KL : OL, or co-tangent 
 AOB = tan (90° - AOB), 
 
 OB:BM = OK:OL, or co-secant 
 AOB = sec (90° - AOB), ,^'»- ^' 
 
 that is, the co-sine of an angle is the sine of the complement, 
 the co-tangent is the tangent of the complement, and the co- 
 secant is the secant of the complement. 
 
12 SIGNS — LIMITING VALUES. 
 
 ^ 6, While each function has but one value for any one angle, 
 the converse is not true. To a single numerical value of a 
 function may correspond four different angles, each less than 
 360°. Any confusion that might result from this fact is 
 avoided by the adoption of the following convention (Cf. Alg. 
 §§8, 46, 55, 429, 432) :~ 
 
 Vertical distances are estimated from the initial line XOX' 
 (Fig. 1) ; those measured upward are said to he positive; hence 
 those measured downward must be taken as negative. Hori- 
 zontal distances are estimated from the vertical line YOY'; 
 those laid off to the right are said to he positive; hence those 
 measured to the left are negative. The hypothenuses of the 
 triangles of reference being, in general, neither vertical nor 
 horizontal, may be considered signless. 
 
 Consequently all functions of angles in the first quadrant 
 are positive. For angles in the second quadrant the base of 
 the triangles of reference are negative, and the perpendiculars 
 positive ; hence the sine and cosecant are positive^ and the other 
 functions negative. In the third quadrant both base and per- 
 pendicular are negative; hence the tangent and cotangent are 
 positive, and the other functions negative. Finally, for angles 
 in the fourth quadrant the cosine and secant are positive, and 
 the other functions negative. 
 
 7. Limiting Values of Functions. Tlie Sine. At 0° 
 p in the triangle of reference is evidently 0, and hence sin 0° = 
 : /i = 0. It will be convenient, though not necessary, to assume 
 that the length of h remains constant as the angle increases 
 from 0° to 360°; let the changing angle be denoted by x. As 
 X increases from 0° onward, p increases, and hence sin x in- 
 creases, rapidly at first, but slowly as x approaches 90°. 
 When X is nearly 90°, p and h are nearly equal, and, finally, 
 when X = 90°, p and h coincide, and p : /*= 1, .-. sin 90° = 1. 
 As X increases,^ (and hence p : h) decreases, slowly at first, but 
 
TRIGONOMETRY. 13 
 
 more rapidly as x approaches 180°. When x is very near 
 180°,/) (and hence ^: h) is very small; finally when re = 180°, 
 p: h = 0. A peculiar circumstance here presents itself. As 
 180° = 180° - 0°, and also 180° + 0° (Alg. Ch. VIII) the angle 
 of 180° may be considered as ending in either the second 
 quadrant or the third. In the first case sin 180° is +, and in 
 the second case-. Obliged to recognize both contingencies 
 we say that sin 180° =±0. Here, then, sin x changes its 
 sign, becoming negative. As x increases,^, and therefore p : h, 
 increases numerically until, when a; = 270°, p and h coin- 
 cide, and p: h= -1. An x increases, p (and hence p : Ji) de- 
 creases numerically, until, when x = 360°, sin x = + according 
 as X is taken as 360° -0° or 360° + 0°. If the angle gen- 
 eratrix continue to revolve, the same values will evidently 
 recur in the same order. 
 
 The Tangent When a; = 0°,^ = 0, and .-. p:b = 0, i. e., tan 0° 
 = 0. As a; increases p increases and b decreases at the same 
 time; hence tan x increases very rapidly. When x is nearly 
 90°, tan X is very great, and, when x = 90°, tan x =p ; = oo . 
 As X may at this stage be considered as ending in the first 
 (90° -0°), or in the second (90° + 0°), quadrant, and as the 
 tangent is positive in the first and negative in the second, we 
 say that tan 90° = ±od . Here the tangent changes sign by 
 passing through qo . As a: increases toward 180° tan x rapidly 
 diminishes (numerically), since p diminishes and b increases, 
 until, when a; = 180°, tan x^T 0. Tan x, changing its sign 
 by passing through 0, increases rapidly as x increases toward 
 270°. When a; = 270°, tan .t= tec , and as x increases, tan x, 
 changing its sign by passing through co , decreases, numeri- 
 cally, with great rapidity, until, when a; = 360° tan a;=+ 0. 
 
 A consideration of the changes in all the functions as x 
 changes from 0° to 360° leads to the results given in the fol- 
 lowing 
 
14 "FUNDAMENTAL RELATIONS." 
 
 ►: Table of Limiting Values of the Functions. 
 
 Function. 
 
 0° to 90° 
 
 90° to 180° 
 
 180° to 270° 
 
 270° to 360° 
 
 Sine 
 
 Oto+1 
 + 1 to±0 
 
 to±oo 
 4- ooto±0 
 
 1 t0±00 
 
 C0t04l 
 
 4-1 to±0 
 ±0 to-1 
 ±ooto+0 
 ±0 toqrx 
 ±x to-1 
 + 1 to±oo 
 
 ±0 to-1 
 -1 to + 
 + to±c» 
 T CO to±0 
 
 - 1 toqroo 
 ± 00 to - 1 
 
 -1 to + 
 
 Cosine 
 
 + to 4-1 
 
 Tangent 
 
 Cotangent 
 
 Secant 
 
 ±00 to + 
 ±0 to+co 
 
 Cosecant 
 
 -1 to^co 
 
 8. Negative Angles. By comparing the functions of 
 negative angles (2) with the functions of numerically equal 
 positive angles, it will be seen that, A being any angle, 
 sin ( - J.) = - sin J. ; esc ( - J.) = - esc A; cos ( - J.) = cos J. ; 
 sec ( — A)= sec A; tan ( - J.) = - tan A; cot ( - ^) = - cot A. 
 Hence changing the sign of an angle changes the sign of the 
 SINE, COSECANT, TANGENT, and COTANGENT, but does not af- 
 fect the sign of the cosine and secant. 
 
 9. Fundamental Relations.* If A represent any angle, 
 (1.) Sin' A + cos'^ ^ = 1. (4.) Tan ^ cot JL = 1. 
 
 sin A (5.) Sin A esc A = 1. 
 
 cos A (6.) Cos A sec A = 1. 
 
 i cos A (7.) Sec' A = l+ tan' A. 
 
 ~" "■ • (8.) Csc^ 
 
 (2.) Tan A 
 (3.) Cot 
 
 sin^ • (8.) Csc'^ = l + cot'X 
 
 Of these, (2), (3), (4), (5), (6), follow immediately from the 
 
 P ^ 
 
 definitions. To prove (1): Sin A = -/ cos A = - hence sin' A + 
 
 « h h 
 
 p' 6' p'+¥ h' , . M , 
 
 cos' A = — -{- — = = -- = 1. (7) and (8) may be similarly 
 
 /t' /i' /r h 
 
 established. 
 
 10. Anti-functions. If ab=.m, we write b = a-hn. By 
 
 analogy we usually say that if tan A=m, J. = tan~^m,. al- 
 
 *The student must memorize these, and be thoroughly familiar with them. 
 
TRIGONOMETRY. 
 
 15 
 
 though the word tangent cannot be considered a factor. 
 Though this notation can be justified, it is inconvenient and 
 may produce confusion, as powers of functions are indicated 
 by placing the exponent above and a little to the right of the 
 abbreviation for the function ; thus (tan A)^ is usually written 
 tan' A ; so that tan""^ A might mean "t^" as well as " the angle 
 whose tangent is A." As it is useless now to attempt to 
 change the notation, the student must be on his guard to avoid 
 confusion. While sin~^ A, cot~^ A, etc., are sometimes read " anti- 
 sine of tI," " inverse sine of J.," etc., it is better to read, in 
 full, " the angle whose sine is A," etc. 
 
 11. The trigonometric functions may be represented by lines, 
 if it be assumed that all angles are formed about the centre of 
 a circle of unit radius, and that in every triangle of reference 
 one side is equal to the radius. In this case, however, the 
 angles are treated by means of the measuring arcs, and it is 
 better to consider the functions as functions immediately of the 
 arcs. Arcs considered directly are generally supposed to be- 
 gin at the right hand extremity of a horizontal diameter (and 
 we shall assume that they do so begin); this point is called the 
 primary origin, or simply, the origin. The complements of 
 arcs are in this case estimated from the upper extremity of a 
 vertical diameter ; this point is called the secondary origin. 
 
 The sine is the perpendicular 
 dropped from the termination of 
 the arc on the diameter through 
 the origin. Thus sin AP (Fig. 
 3) is PM. The cosine of 
 an arc is the sine of the 
 complement; thus cos AP = 
 PL; cos AD = KD. The tan- 
 gent is that part of the geo- 
 metric tangent a( the origin 
 which is limited by the diame- 
 ^^' '• ter through the termination of 
 
 r. 
 
1(5 TRIGONOIk^ETRIC LINES. 
 
 the arc; thus tan AP = tan ABF = AT. The cotangent of an 
 arc is the tangent of the complement ; thus cot AD = BE. 
 The secant is the line connecting the centre of the circle with 
 the extremity of the tangent ; thus sec ABD = OS. The co- 
 secant of an arc is the secant of the complement ; thus esc AP 
 = 0V. The versed-sine is the radius minus the cosine, and 
 the co-versed-sine is the radius minus the sine ; thus versin AP 
 = MA ; coversin AD = KB. 
 
 Lines are always measured /ro?^ A A' and BB'; those meas- 
 ured upward and those measured to the right of BB' are pos- 
 itive ; all other horizontal or vertical lines are negative. The 
 secant and cosecant follow in ^ign their reciprocals, the cosine 
 and sine, respectively.* 
 
 If the radius be different from unity the lines indicated 
 above must be divided by the radius, in order to get the various 
 functions. As these definitions agree in substance with those 
 given in (4) we may use angles or their measuring arcs in- 
 differently, and likewise the trigonometric functions or their 
 line representatives. 
 
 12. Functions of SO'' and of ^5°. 
 
 Let AOB (Fig. 4) = 30°. Then (Ven. 
 
 36 p. 51) OB = 2 AB, .-. sin AOB = 
 
 AB : OB = ^. As sin' 30° + cos' 30^ 
 
 = 1, cos'' 30° = 1 .-. cos 30° = ^ Vs. 
 
 Fig. 4^ T,an_30°=i--- 
 
 J V^3 = Jl/'S; cot 30° = 1/3; sec 30° = 
 
 f 1/3 ; CSC 30° = 2. 
 
 Again let AOC (Fig. 5) = 45°. As 
 OC is the diagonal of a square, AC : 
 OC=sin 45° = il/2=cos 45°. As OA ^ig. 5. ^ 
 
 = AC, tan 45° ^ cot 45° = 1 . Finally, sec 45° = esc 45° =i/2. 
 13. Trigonometric Tables. — To enable us to deal with 
 angles by means of their functions tables are computed, giving 
 the values of these functions for all angles from 0° to 90°. 
 
 *The versed-sine and coversed-sine are measured from the foot of the sine and 
 cosine respectively, and therefore are never negative. 
 
TRIGONOMETRY. 17 
 
 The angles are usually taken at intervals of V, though the in- 
 terval may be as large as 10' in tables for ordinary use, 
 and is frequently as small as 1" in tables used for calculations 
 requiring great accuracy. The functions of an angle lying 
 between any two angles of the table are found by interpola- 
 tion (Alg. 491 and Appendix III). 
 
 14. The Tables of Natural Functions give the actual val- 
 ues of the functions ; the Tables of Logarithmic Functions 
 give the common logarithms of these values. As all sines and 
 cosines, tangents of angles between 0° and 4^°, and cotangents 
 of angles between 45° and 90°, are " proper " fractions, the 
 characteristics of their logarithms are negative. To avoid 
 these negative characteristics, 10 is added to the characteristic 
 of the logarithm of each of these functions, and, in many tables, 
 for uniformity, 10 is added to every characteristic. 
 
 15, Elementary Method of Computing a Table of 
 Natural Functions. — We will consider the line representa- 
 tives of functions, i. e., (11) the functions of the arc (radius 
 being unity). Though an arc exceeds its sine, the inequality 
 diminishes as the arc decreases, the ratio (sin x) : x having for 
 its limit 1. This inequality is so small for an arc of V that 
 there is no error, within the limits of ordinary tables, in 
 assuming arc l' = sin 1'. In a circle of radius 1, the semi- 
 circumference = 10800' = TT = 3.14159265; hence arc 1' = 
 3.14159265^10800 = .000290888-sinr. cos V = ^/l~mi'V 
 
 = V (1+sin l')(l-sin T) =-/l.000290888 x .999709112 
 = .999999958 
 It will be shown (37) that if A and B be any two angles, 
 
 (I) sin (^ + J5) - 2 sin ^ cos J5 - sin {A - B). 
 
 (II) cos {A + B) = 2 cos ^ cos i> - cos {A - B), 
 
 Let B= V, and let A become successively, V, 2', 3', etc. 
 From (I), sin (1' + 1') = sin 2' = 2 sin 1' cos V 
 Sin (2' + V) = sin 3' = 2 sin 2' cos 1' - sin V 
 sin 4'= 2 sin 3' cos 1' - sin 2', 
 and so on. 
 
18 TRIGONOMETRIC TABLES. 
 
 . From (II) COS 2' = 2 cos'^ 1' - / 
 
 cos 3'-- 2 cos 2' cos V - cos 1' 
 cos 4'= 2 cos 3' cos V -cos 2' 
 and so on. 
 
 As 2 cos 1'= 1.999999154 = 2 -.000000846 is a constant 
 factor in the first term of the second member, these equations 
 may be written 
 
 sin 2'- 2 sin 1' - .000000846 sin V = .000581776 
 
 sin 3' = 2 sin 2' - .000000846 sin 2' -sin 1' = . 000872665 
 sin 4' = 2 sin 3' - .000000846 sin 3' - sin 2' = .001163553 
 and similarly for the cosines. After computing the functions 
 of angles up to 30° by these formulae, we let A (I) remain 
 equal to 30°, while B changes from V successively at minute 
 intervals up to 15°. Then, as sin 30° = ^, 2 sin 30°- 1, and 
 
 we have 
 
 sin (30° + 1') = cos 1' - sin 29° 59' 
 
 sin (30° + 2') = cos 2' - sin 29° 58' 
 sin (30° + 3') = cos 3' - sin 29° 57' 
 and so on up to 45°, the only operation required in the second 
 members being mere subtraction. 
 
 Instead of (II) we now use, (37), the formula 
 
 cos (A + B)^ - 2 sin ^ sin ^ + cos (A - B). 
 Letting A remain equal to 30°, we have 
 
 cos (30° + ,1') = - sin 1' + cos 29° 59' 
 cos (30° + 2') = - sin 2' + cos 29° 58' 
 and so on, up to 45° inclusive. 
 
 We need not compute beyond 45°, as the sine and cosine of 
 any angle between 45° and 90° are respectively the cosine and 
 sine of the complement, i. e., of an angle between 45° and 0° 
 (of which the cosine and sine have been computed). 
 
 The tangents, cotangents, secants, and cosecants may be 
 obtained from the sines and cosines by means of the funda- 
 mental relations (9). Functions of angles greater than 90° 
 may be obtained from the table by means of the relations 
 suggested in (5, 6). 
 
TRIGONO]NIETRY. 19 
 
 16. The Table of Logarithmic Functions may be obtained 
 by simply tabulating the common logarithms of the natural 
 functions and adding 10 to each characteristic. Other and 
 better methods will be indicated later. This table is usually 
 employed in preference to the table of natural functions, as 
 logarithms are utilized whenever they serve to facilitate com- 
 putation, as in multiplications, divisions, involution, and evolu- 
 tion. 
 
 17. Tables are usually so arranged that for angles from 0^ 
 to 45° the degrees and the names of the functions are to be 
 taken from the top of the page, and the minutes from the left- 
 hand column, while for angles between 45° and 90° the 
 degrees and the names of the functions are to be taken from 
 the bottom of the page, and the minutes from the right-hand 
 column. 
 
 18. Interpolation is effected on the hypothesis that as t]ie 
 arc changes uniformly, the function changes uniformly. The 
 error involved in this assumption being greatest for cosines of 
 very small angles and sines of angles near 90° ^ we avoid com- 
 puting small angles by means of the cosine, and angles near 
 90° by means of the sine, unless we have tables specially cal- 
 culated for the purpose. It is best in general to compute 
 angles by means of the tangent or cotangent. 
 
 19. To guard against the errors likely to creep into the 
 extended computations of (15) we check the results of those 
 computations by comparing them with the values of functions 
 of various angles, determined independently by formulae that 
 might in this connection be termed verification formulse.. 
 Numerous examples of these will be given later. 
 
 I 
 
CHAPTER II. 
 
 SOLUTION" OF TRIANGLES. 
 
 The parts of a triangle, trigonometrically considered, are 
 its sides and angles. Plane Trigonometry treats in detail the 
 problem : — " Given three parts of a triangle, to compute the 
 other three," or, more generally, "From data sufficient to 
 determine a plane triangle, to compute the unknown parts." 
 Of given parts one at least must fix the length of a line; angles 
 only can determine nothing more than the shape of a tri- 
 angle : — the ratios of the sides to each other. 
 
 In any triangle, ABC, the angles will be denoted by 
 A, B, C, and the sides opposite by a, 6, c, respectively. 
 
 20. Right Triangles. — The problems that arise in the 
 solution of right triangles may be grouped thus : — 
 I. Given the hypothenuse and another side. 
 II. Given the two sides about the right angle. 
 
 III. Given the hypothenuse and an acute angle. 
 
 IV. Given an acute angle and a side, opposite or adjacent. 
 The definitions (4) may, (Fig. 6) 
 
 be stated thus: — 
 
 (1) sin C = c^ a 
 
 (2) cos C=h^a 
 
 (3) tanC = c-^ h 
 
 (4) cot Q=h^G 
 These suffice for the solution of all 
 
 right triangles. ^'°" ^' 
 
 I. 1°. Given a, h, (2) gives cos C, whence is obtained at 
 once from the tables, c'^ = d^ - h^ ; hence c = V {a->rb) (a- h) 
 and the solution is complete, c might also be computed from 
 (1) after C has been found; but, (i) any error in the computa- 
 tion of would vitiate this computation, and (2) if C be near 
 90° its sine cannot be accurately determined from the usual 
 (20) 
 
TRIGONOMETRY. 21 
 
 tables. Required parts should always be determined directly 
 (if possible) from given parts. Unused relations may then 
 serve as " checks " on the work. 
 
 2°. Given a, c. Solution similar to that just indicated. 
 
 II. Given h, c. From (3) and the Tables, C and hence B 
 are determined at once. To determine a we have the two for- 
 mulae a=l/6^ + c^ and a = c-rsin C; the former is direct, but 
 not adapted to logarithmic computation ; the latter, the simpler, 
 has the disadvantage already indicated. 
 
 III. Given b, B. C = 90° - B. (2) gives a, and (3) deter- 
 mines c. When c, C are given the solution is similar. 
 
 21. Oblique TrianorleS. — As every oblique triangle may 
 be divided into two right triangles, all oblique triangles may 
 be solved by (20). 
 
 Fig. 7. Fig. 8. 
 
 I. Given two sides and the included angle, e. g., c, h, A. 
 Draw from the extremity of one known side, (6), CD J_ c, the 
 other known side. Solve ACD (20, HI), thus determining 
 k and p; if ^ >• c, CD falls without the triangle (Fig. 7). 
 BD = G— k or k- c, according as CD falls within or without. 
 Knowing p and m, solve the triangle BDC, thus obtaining a 
 and BCD. ACB = ACD + BCD or ACD -BCD according 
 as CD falls within or without ACB. 
 
 II. Given two angles and a side, e. g.. A, B, h. All the 
 angles are known, and the side 6. Draw, from the extremity 
 of this side, CD J_ c. Solve ACD (I). Then, knowing p 
 and the angle DCB, solve DCB. If CD falls mthin the tri- 
 
22 
 
 SOLUTION OF TRIANGLES. 
 
 angle, i. e., if ACD < C (as given), AB = AD + DB. If CD 
 falls without, i. e., if ACD > C, AB = AD- DB. 
 
 Fig. 9. Fi?. 10. 
 
 III. Given two sides and the angle opposite one of them, 
 e. g., h, a A. From the extremity of a known side, h, draw 
 CD _L c, the unknown side. Solve ACD, as before. Know- 
 ing |) and a, solve DCB (20, 1). As in this case two triangles 
 may fulfill the given conditions, it is called the "ambiguous 
 case" Thus if a < 6, a may take either one of the positions 
 CB or CB\ cutting c at equal distances from the foot of the 
 p3rpendicular, and both ACB and ACB' fulfill the require- 
 msnts. If b = a, ACB' shrinks to a line, and there is but one 
 solution. Again if a >■ 6, a can fall on only one side of AC 
 and there can be but one solution. If a=p, ACB' and ACB 
 coincide. Finally, if a < ]:> there is no solution. 
 
 IV. Given the sildi a, b, c. Drop the parpendicular from 
 any vertex, as C. Then (Fig. 7), 
 
 p'=.b'-k' = a^-^m' 
 .'. 6'— x^ = Ic^ - m\ whence 
 (6 - a)(6 + a) = (^ - ?n)(^ -f »i), i. e., 
 b + a: k + m: : k-m: b-a (1); 
 ^ (6 + «)(6-a) 
 
 or k 
 
 Determining k-m from (2), we have 
 
 A; = J(A; -H m) -f J(^ - 7rt), and 
 
 m = ^(k -f- m) - \{k - m). 
 
TRIGONOMETRY. 23 
 
 Knowing k and m, we solve ACD and DCB, and thus de- 
 termine A, B, C. 
 
 22. The solutions, I, III, (21) may be abbreviated by 
 noting that " the sides of a plane triangle are proportional to 
 the sines of the opposite angles" Thus (Figs. 9, 10), 
 p = bsin.A = a sin B, whence a: 6 = sin A : sin B. 
 
 This proposition may be used to " check " solutions of (2.1). 
 
 The constant ratio = — = is the modulus of 
 
 sni A sin B sin C 
 
 the triangle, and is numerically equal to the diameter of 
 
 the circumscribing circle. 
 
 NUMERICAL EXAMPLES. 
 
 1. Right Triangles.— 1. Let a = 256, 5 = 172. From 
 (2) cos C = sin B = 172 -^ 256 = .67188. The table of Natural 
 Functions gives C = 47° 47' 14", and hence B = 42° 12' 46". 
 € = V {ct + b)(a-h)= l/428x84 = l/35952 = 189.6. (1) now 
 serves as a " check," to test the correctness of the computation. 
 Thus sin 47° 47' 14" = .74065, .-. c = 256 x .74065 = 189.6 as 
 before. 
 
 2. Let 6 = .0795; c = .0386 [Logarithmic solution]. From 
 (3), tan C = c-7-6, whence log tan C = log c-log 6 + 10 = 
 2.58659- 2.90037 + 10= 9.68622, whence C= 25° 53' 26". 
 From (1), a = c + sin C, .*. log a = log c-log sin C + 10 = 
 2.58659 - 9.64025 + 10 = 2.94634 /. a = 0.08838. The rela- 
 tion a"^= b'^ + c"^ may be used as a " check." 
 
 3. Let a =27.125, C= 37° 42' 18". From (1), 
 €=a sin C 
 
 log a=- 1.43337 Again, 6 = a sin B; log a = 1.43337 
 log sin C= 9.78647 log sin B = 9.89827 
 
 log c= 1.21984 log ^> = 1.33164 
 
 c= 16.589 6=21.46 
 
 IL Oblique Triangle8.-l°.Given A=56°41', B=74° 28' 
 
24 NUMERICAL EXAMPLES. 
 
 ^and 6=1326. C=180°- (A + B) = 48° 51'. InACD(Fig.7) 
 AD= b cos A and CD= 6 sin A. 
 
 log 1326=3.12254 log 1326= 3.12254 
 
 log cos 56° 41' = 9.73978 log sin 56° 41' = 9.92202 
 
 log AD = 2.86232 log CD = 3.04456 
 
 .-.AD =728.31 
 In CDB, CB = CD -^ sin B, and DB = CD.cotB. 
 
 log CD = 3.04456 log CD - 3.04456 
 
 log sin 74° 28' = 9.98384 log cot 74° 28' = 9.44397 
 
 log CB = 3.02840 log DB = 2.48853 
 
 a= 1067.59 DB = 307.99 
 c= AD + DB= 1036.3 
 
 We check our results by the law of sines ; thus : 
 sin A -^ a should equal sin B -^ 6. 
 
 log sin B- 9.98384 log sin A= 9.87679 
 
 log ^>= 3.12254 log a= 3.01549 
 
 6.86130 6.86130 
 
 Thus the logarithms of the two ratios agree exactly. 
 
 2. Given a= 1200, b= 758, A= 72° 14'. As a> 5 there 
 is but one solution. 
 
 In ACD, CD= 5 sin A, and AD= b cos A. 
 log sin 72° 14'= 9.97878 log cos 72° 14'= 9.48450 
 log 758= 2.879(37 =2.87967 
 
 log CD= 2.85845 log AD =T3G4r7 
 
 CD= 721.8 AD= 231.3 
 
 In CDB, BD= |/(a + CD)(a - CD) = v^ 1921.8 x 478.2 
 log 1921.8= 3.28371 log sin B= log CD - log a + 10 
 log 478. 2=2.67961 log CD= 2.85845 
 
 2 I 5.96332 log a = 3.07918 
 
 log BD= 2.98166 log sin B= 9.77927 
 
 BD= 958.6 B= 36° 58' 49" 
 
 ACB= 180° - (A + B) = 70° 47' 11". 
 
 3. Given a = 12.76; 6 = 26.58, c=31.6. Then, (Fig. 7), 
 
TRIGONOMETRY. 25 
 
 , 39.34x13.82 ,^^, i oa 4 no 
 
 k-m= — = 17.21 .-. ^'=24.4; w^7.2. 
 
 ol.D 
 
 log COS B=log 7.2 - log 12.76 + 10 
 
 =.85733 - 1.10585 + 10 = 9.75148.-. B=55°38' 57". 
 log cos A=log 24.4 - log 26.58 4- 10 
 
 =1.38739 - 1.42455 + 10= 9.96284 .-. A= 23° 22' 
 C= 180° - (A + B) = 100° 59' 3". As a check we may use 
 (22) ; thus :— log b = 1.42455 log c= 1.49969 
 
 log sin B =9.91677 log sin C = 9.99197 
 
 log modulus= 1.50778 or, 1.50772 
 
 results agreeing to the fourth decimal place. 
 4. Given 6=.0079, c= .0124, A= 37° 22' 17". 
 In ACD (Fig. 7), 
 
 log sin A = 9.78318 log CD = 3.68081 
 
 log .0079 = 3.89763 log DB = 3.78675 
 
 log CD = 3.68081 log tan B = 9.89406 
 
 .-. B = 38°4'48" 
 log cos A = 9.90021 log CD = 3.68081 
 
 log .0079 = 3.89763 log sin B = 9.79012 
 
 log AD = 3179784 log a = 3.89069 
 
 AD= .00628 a= .00777 
 
 .-. DB= .00612 
 We-may again check by use of (22), thus: — 
 
 loga= 3.89069 log b= 3.89763 
 
 log sin A =9.78318 log sin B =9.79012 
 
 log modulus= 2.10751 log modulus =2. 107 51 
 
 results agreeing exactly. 
 
CHAPTEK III. 
 
 FUNCTIONS OF THE SUM AND OF THE DIFFERENCE 
 OF TWO ANGLES. MISCELLANEOUS FORMUL.F. 
 
 23. Sine and Cosine of the Sum. — To show that 
 
 (1) sin (^x + 2/) = sin x cosy + cos x sin y 
 
 (2) cos ^y')^cosx cos y-sinx sin y. 
 
 Let OA, OB, and OC be any three lines whatever, forming 
 the angles AOB and BOC (Figs. 11, 12). Let AOB- x and 
 BOC=- y. From any point of OC, as F, drop FE _|_ OA and 
 FP j^ OB, and from P drop PD j_ OA and PG _[_ EF. 
 
 Fig. 11. Fig. 12. 
 
 Then PD :0P = sin x =PG: FP (similar triangles), and 
 OD : 0P= cos a;= FG : FP ; sin y^ FP : OF, cos y=0¥: OF, 
 sin (x + y) = FE : OF, and cos (a; + y) = OE : OF. 
 From the figures, EF : EG + GF = PD + GF, i. e., 
 
 EF^PD PO FG FP . 
 OF~ PO • OF "^ FP * OF ' ^* ^'' 
 
 8m(x + y) = sin X cos y + cos X sin y 
 Again, (Fig. 11), OE = OD-GP, whence 
 OE^OD OPGP FP . 
 OF~OP*OF FPOF'*'^' 
 cos(a; + y) = cos x cosy- sin x sin y 
 (26) 
 
 (I). 
 
 (II). 
 
TRIGONOMETRY. 27 
 
 The same formulae may be deduced from figure 12, if it be 
 noted that in this case OE is negative. 
 24. Sine and Cosine of the Difference. — To show that 
 
 (1) sin (x-y)^ sin x cos y - cos x sin y 
 
 (2) cos (x-y') = cos x cos y + sin a sin y. 
 
 Fig. 13. Fig. U. 
 
 Let AOB (Fig. 13) be x, and BOC (laid off backward, i. e., 
 in a negative direction), be - y, so that AOC is x-y. From 
 any point, D, of OC. draw DH _l OA, and DE _l OB. 
 Draw EG i OA, and DF x EG. Then EG : OE = FD : ED 
 = di\x, OG:OE = EF:ED = cosa;, DE:0D = sin2/, OE:OD 
 = cos y, DH : OD = sin (x - y), and OH : OD = cos (x - y). 
 From the figure we see that DH = EG - EF, whence 
 
 DHEG OEEF ED . 
 
 OD"OE'OD EDOD'"'^' 
 
 sin (a; -y)= sin x cosy- cos x sin y (HI)* 
 
 Again, OH = OG + FD, whence 
 
 ohog oe fd ed . 
 od~oeod"^ed'od''-^" 
 
 cos (a; -y) = cos x cos y + smx sin y • (I^« 
 
 In Fig. 14, both x and x-y are obtuse. 
 
 mil f Of^ -I- 7/1 
 
 25. Tanirents. — Since tan(a; + w)= — \ ^, we obtain, 
 
 cos {x + y) 
 
 by substitution from (31), tan (x + y') = — — - 
 
 cos X cos y -s\n X sm y, 
 
 Tlie student should notice that no supposition has been made concerning the 
 direction of OB and OC, and that, therefore, the formulae are true for all angles. 
 
28 
 
 MISCELLANEOUS FORMULAE. 
 
 or, dividing both numerator and denominator- by 'CO&^c-cosy, 
 
 (remembering that = tan x, etc.), 
 
 cos a; 
 
 tan {x + y) 
 
 _ tan X + tan y 
 1 - tan X tan % 
 
 (V). 
 
 Bya similarprocess we^obtain tan (x - y) =~.^ ~ (VT)- 
 
 -' ^ \ :/y 1-f tana:tan y ^ ^ 
 
 cos \ X "4" \l^ 
 
 34. Cotangents. — Since cot(a;4-v) =— r— ^ — =^, weob- 
 
 ^ sm(a; + 2/) 
 
 * • V u i-i. i.- /otN x/ N COS a; COS w - sin a; sin y 
 
 tam,- by substitution- («>1)> cot(a; + v) = -; ; — ^• 
 
 sin a; cos y + cos x sin y* 
 
 or, dividing both terms of the fraction by sin x sin y, 
 
 cot a; cot 2^ - 1 
 
 cot (a3H-i/) = - 
 
 cot 2/ + cot a; 
 
 Similarly we obtain cot (a^ - y) = 
 
 cot a; cot 3/ - 1 
 
 (VII). 
 (VIII). 
 
 cot 2/ - cot a; 
 
 35. Exercise.— Find (31 - 34) the functions of (90±a:), 
 (180°±a;), (270°±a;), (360°±a;). The results are given in Table 
 II. Observe that if the number of quadrants involved be 
 odd the function changes name, and not otherwise. Thus^ 
 tan(270° + a;)= -cota;; but tan(180° + aj) =tanaj. The sign 
 is determined by the quadrant in which the angle would lie, 
 if a; were less than 90°. 
 
 Table II. 
 
 Function. 
 
 90^ +x 
 
 90* -X 
 
 180°+a; 
 
 ISC-x 
 
 270°+x 
 
 270°— X 
 
 360° +x 
 
 360°-x 
 
 Sink. 
 
 cos X 
 
 cos X 
 
 —sin X 
 
 sin X 
 
 —cos X 
 
 —cos X 
 —sin X 
 
 sin X 
 
 —sin X 
 
 Cosine. 
 
 -sin X 
 
 sin X 
 
 —cos X 
 
 —cos X 
 
 sin X 
 
 cos X 
 
 cos X 
 
 Tangent. 
 
 —cot X 
 
 cot X 
 
 tan X 
 
 —tan X 
 
 —cot X 
 —tan X 
 
 cot X 
 
 tanx 
 
 —tan X 
 
 Cotangent. 
 
 —tan X 
 
 tan X 
 
 cot X 
 
 —cot X 
 
 tan X 
 
 cot X 
 
 —cot X 
 
 Secant. 
 
 —CSC X 
 
 CSC X 
 
 —sec X 
 
 —sec X 
 
 CSC X 
 
 —esc X 
 
 sec X 
 
 sec X 
 
 Cosecant. 
 
 sec X 
 
 sec X 
 
 —CSC X 
 
 CSC « 
 
 -sec X —sec x 
 
 CSC X 
 
 --CSC X 
 
 36. Functions of Double Angles and Half Angles. 
 
 If in (31-4) we make y= x, we obtain, 
 
TRIGONOMETRY. 29 
 
 sin t^+"^) =sin (x + x)= sin 2 :r= 2 sin x cos x (A) ' 
 
 cos (x + y) --cos (.i; + ^) = cos 2x= cos'^ a; - siu^ x 
 
 =2cos''a;-l=l-2sin2a; (B) 
 
 / N ^ o 2 tan ^ ^i-,N 
 
 tan (x + x)= tan 2 a;=- ■ — {yj 
 
 1 - tan'' X 
 
 cot^ic-1 ^■n>. 
 
 cot (x + x)^ cot z a;= — r K^J 
 
 ^ ^ 2 cot a; 
 
 Again, from (B), we have 1 - cos2a:= 2 sin^a;. Let a; 
 
 = -^, whence 2 x=a. Then 2 sin' -= 1 - cos a, whence sin ^ 
 2 2 2 
 
 = ^"1(1 ~ ^o^ ^0 (^')- From (B) we have, also, 2 cos''' x= 
 
 1 + cos 2 X, or, if a;= -^, 2 cos'^ ^=1 4- cos a, whence cos— = 
 2 2 -^ 
 
 l/j(l +COS a) (B'). From (A') and (B') we derive, at once, 
 
 a /l - cos a .^,. , , ^ /l + cos a ^-p.,. 
 
 tan - =^/- (C); and cot ^ =-y/- (D ). 
 
 2 \ 1 -f-cos a ^ \ 1 - cos a 
 
 These are formulae of great importance. 
 
 28. Adding (I) and (III), member to member, we obtain 
 sin {x-\-y)-\- sin (a? - ?/) = 2 sin x cos y (1). Let x-\-y = m and 
 a; - 2/ = ^, whence ic = |^(?/i + /:) ; 2/ = i(^'i'- ^)- Then (1) be- 
 comes sin m + sin ^• = 2 sin \{in + A;) cos ^(m - ^) (2); 
 
 Similarly, 
 
 sin m - sin ^ = 2 cos \{iin + ^) sin |(?/i - U) («J); 
 
 cos m 4- cos ^ = 2 cos |^(m + ^) cos \{m - k') (4); 
 
 cos m - cos ^ = - 2 sin ^(r^i + U) sin J(m - ^) (5). 
 
 29. From (2) -(4), (28) we deduce, by direct division, 
 
 sin m + sin k _2 sin ^(m + k) cos \{m - k~) _ tan ^(m + ^ ) ,^ ^ 
 
 sin m- sin ^ 2cos^(m + ^) sin J(m-A;) tan^(m-^) ' 
 
 and similarly, 
 
 cos?)i-cosA; ^ 1/ . 7^x i> 7N r9^. 
 
 — i = - tan *(wr+ A^) tan -J^(m - A;) K.'^jf 
 
 cos ?7i + cos A; "^ ^ 
 
 sinm + sinA; ^ ,, zn /on sinm-sin k i\/a\ 
 
 = tan Urn + k) (3): j- =-cot|(m— A;) (4). 
 
 cosm + cosA; ^ oosm-cosA; 
 
 30. The formulae of (28, 29) enable us to adapt formulae 
 involving sums or differences of functions to logarithmic calcu- 
 lation. 
 
CHAPTER lY. 
 
 ADDITIONAL FORMULA FOR PLANE TRIANGLES. 
 
 40, A7iy side of a plane triangle equals the sum of the prod- 
 ucts of the remaining sides into the cosines of the respective 
 angles included between them and the first side. 
 
 Thus (in Fig. 7), if CD i AB, AD =6 cos A, and BD = 
 a cos B, whence c = AD + DB = b cos A + a cos B. If (as in 
 Fig. 8), CD fall without the triangle, c = AD - DB. But DB = 
 a cos CBD = a cos (180° - B) = - a cos B 
 and we have as before, c= 6 cos A + a cos B (1) 
 
 Similarly 6 = c cos A + a cos C (2) 
 
 a = ^ cos C + c cos B (3) 
 
 41, The square of any side of a triangle equals the sum of 
 the squares of the other two sides, minus twice the product of these 
 sides into the cosine of their included angle. 
 
 From (40), (2), by transposing and squaring, we obtain 
 d^ cos"^ C = 5^ - 2 6c cos A + c^ cos'^A 
 
 From (22 ) a'^sin^C = c^sin^A 
 
 Adding, a' = b'' -2 be cos A + c' (1) 
 
 Similarly, * h' = a' + G'-2ac cos B (2) 
 
 c'=b'' + d'-2abcosO (3) 
 
 42, From (1) - (3), (41) we deduce 
 
 cosA = 
 
 cosC = ^^±^^'(3) 
 2 be 
 
 These formulae, not being adapted for logarithmic computa- 
 tion, are usually transformed as follows : — 
 (30) 
 
 2 be ^^ 
 
TRIGONOMETRY. 31 
 
 Subtracting each member of (1) from luiity, we have 
 
 , , b^ + c'-a:' 2cb-b'-c' + ci:' ci'-(b-cy 
 l-eosA=.l ^^— = ^^ ==-^^- ; or, 
 
 since l-cosA^2sm-^4' 2sin^^i = (.^zA±£^^ 
 
 Let 2 8=a + b + c; then s—b^^(a + c- b), and 8-c = ^(a + b-c'), 
 so that 
 
 sm ~^- ^X-e) 
 
 . A Ks-bX' 
 
 ■^ (2); VR^ rtrr ^Hft- r 
 
 similarly, «n- = y -^^ (2); WB) r^ 
 
 ^^"2 = V-^;i— ^^>j 
 
 Adding both members of the equations (A) to unity we 
 obtain, by similar reductions, formulse (C). 
 
 cos 
 
 cos 
 
 |B = ^!(i2_*) (2); 1(C) ^-^«^ '^f'^ 
 
 /.■^ (.s - c) 
 ab 
 
 cosiC = y^^l^^ (3). 
 
 From (B) and (C) we obtain, by direct division, (D). 
 tan 
 
 tan 
 
 
 tan^C=^/^l-;i^^^ (3). 
 
 (s-a)(s- 
 
 Of these three sets, (B), (C), (D), the first should be used 
 only when the angles are small; the second should never be used 
 
32 
 
 SPECIAL FORMULA. 
 
 with small angles. The formulse (D) 
 are the best for general itse, being 
 accurate for all angles. These form- 
 ulae furnish the means of solving Case 
 IV (21). 
 
 34. Radius of Inscribed Circle 
 (r). Area.— Let ABC be any triangle. 
 The bisectrices of the angles meet in 
 the centre of the inscribed circle. 
 Also AE = s-CB, BD = s-AC, etc. 
 Hence, in the right triangle AOE, 
 
 tanOAE = tan^A--=-^ = 
 s -a 
 
 - a 
 
 = (33, D), 
 
 Us-b)(is-c) _ 1 / 
 \ sijs — a) s-a\ 
 
 (s—a)(s- bXs- c) 
 
 = tan ^A.{s-a 
 
 )-^/ o^-«x.^- 
 
 bXs-c) 
 
 Again, as the area of a triangle equals one-half the peri- 
 meter multiplied by the radius of the inscribed circle, the area 
 of ABC 
 
 V(s—a')(s--b)(s — c) /—. ^, TTT : 
 ^^: <-^ ^-^ ^ = -^ s {s — a){s—b){s ~- g). 
 
 35. Another expression for the area is "one-half the prod- 
 uct of any two sides into the sine of the included angle." 
 
 Thus (Figs. 7, 8), area ABC = ^ AB . CD = J AB . 6 sin A = 
 ■| be sin A. This is equivalent to be sin ^ A cos J A. 
 
 36 . The sum of any two sides of a triangle is to their dif- 
 ference as the tangent of half the sum of the opposite angles is 
 to the tangent of half their difference. 
 
 Let ABC be the triangle. By (22), a:b :: sin A: sin B. 
 By composition and division, a + b:a-b: : sin A + sin B : 
 
 sin A - sin B : : (29), tan i(A + B) : tan |(A - B). Q. E. D. 
 
 Case II (21) may be solved by means of this proposition. 
 
CHAPTER V. 
 
 SPHERICAL TRIGONOMETRY. 
 
 36. Spherical Trigonometry treats of the solution of 
 spherical triangles, or the relations between the face-angles 
 and diedrals of the triedrals, whose common vertex is the 
 centre of the sphere, and which intercept these triangles on 
 the surface. As the sides of these triangles are expressed in 
 degrees, or in radial measure, the solution of the triangle gives, 
 not the absolute lengths of the sides, but only the ratios ex- 
 isting between them. 
 
 To determine the absolute lengths w^e must know the radius. 
 This premised, any three parts of a spherical triangle suffice 
 to determine the other three. Two parts are of the same 
 species when they lie in the same quadrant, and of diflferent 
 species w^hen they lie in different quadrants. We consider 
 only triangles in which no part exceeds 180°; this convention, 
 though not absolutely necessary, simplifies our calculations 
 without causing any material loss of generality. 
 
 37. Right Triangles. Napier's Parts. Napier's 
 Rules. — ^To express, in two general rules, the various formulae 
 required in solving a right triangle, Napier adopted the follow- 
 ing convention : In any right triangle the two sides about the 
 right angle, the complements of the two angles opposite them, 
 and the complement of the hypothenuse, are called the five 
 circular parts. If any three of these be considered, one 
 will be adjacent to both the others, or one will be separated 
 from each of the others by the intervention of a circular part. 
 A part adjacent to two others is called a middle part, and the 
 other two are called adjacent extremes. A part separated 
 from two others is called a middle, and the other two are called 
 
 opposite extremes. 
 
 (33) 
 
34 
 
 NAPIER S EULES. 
 
 Napier's Rules —I. The 
 
 SINE of the middle part 
 equals the product ,of the 
 
 COSINES of the OPPOSITE 
 
 extremes. 
 
 Taking the circular parts 
 successively as middles, we 
 Fig. 15. have the following equa- 
 
 tions to establish : — 
 
 (1) sin J = sin a sin B 
 
 (2) sin c = sin a sin C 
 
 (3) cos a = cos h cos c 
 
 (4) cos B = cos h sin C 
 
 (5) cos C = cos c sin B 
 
 Let ABC be the right triangle, A being the right angle. 
 Connect the vertices with the centre (O) of the sphere. From 
 any point D, of OC, drop DE ± O A, and from E draw EF x OB. 
 Then DF must be perpendicular to OB, and DFE, the recti- 
 lineal angle of the diedral, OB, will equal the spherical angle 
 B. DEO, EOF, DOF, and DFE, are respectively triangles 
 of reference for h, c, a, and B. 
 
 (1) As sin i = DE :0D, sin a = DF : OD, and sin B = DE :DF, 
 we have, at once, 
 
 DE_DF DE 
 0D~uI3 DF' 
 sin h = sin a sin B. Q. E. D. 
 The lettering of the figure being purely arbitrary, so far as 
 B and C (and therefore b and c), are concerned, (2) follows at 
 once from (1). 
 
 OF^OE OF 
 
 OD ODOE' ' 
 
 we have, at once, cos a = cos h cos c. Q. E. D. 
 
 (4) To r.how that cos B = cos 6 sin C. We cannot get sin C 
 
 sin c 
 directly fro::a the %ure, so we replace it by -: — (2), as sin c and 
 
 or 
 
 (3) As 
 
TRIGONOMETRY. 35 
 
 sin a can be obtained directly. We must prove, then, that 
 
 „ sin c . 
 
 cos ±>= cos 0—. ; 1. e., cos i3= cos o sin c esc a. 
 
 , sin a 
 
 ^ ^ , ,. , EF OE EF OD . 
 But we have, directly, j^^q^ " OE ' FD ' ^' ^•' 
 
 cos B = cos h sin c esc a = cos h sin C. Q. E. D. 
 
 II. The SINE of the middle part equals the product of the 
 TANGENTS o/ the ADJACENT extremes. 
 
 We are to prove that : — 
 
 (1) sin 6 = tanc cotC 
 
 (2) sin c= tan h cot B 
 
 (3) cos B = tan c cot a 
 
 (4) cos C= tan h cot a 
 
 (5) cos a = cot B cot C 
 
 (1) From (I, (1)) we have, sin 6 = sin a sinB; from,-(I, (2)) 
 
 siaa= -. — ;^,and from (1,(5)) sinB= ; whence 
 
 sm C ^ ^ ^^ cos c 
 
 . , sin c cos C sin c cos C /-. ^ -r^ t^ 
 
 sm b = ■-:—r- . = -: — — = tan -c cot C Q, E. D. 
 
 sm O cos c cos c sm C 
 
 (2) Similarly sin c = tan h cot B. 
 
 (3) From (1,(4)) cos B= cos 6 sin C, and as 
 
 , cos a - . „ sin c , 
 
 cos 0= , and sm C= -. — ,we have 
 
 cos c sin a 
 
 ^ cos a sin c ^ ^ >^ -r^ t^ 
 
 cos B = : — =tan c cot a. Q. E. D. 
 
 cos G sin a 
 
 (4) Similarly, cos C= tan h cot a. 
 
 cos B 
 
 (5) Finally, as (I) cos a= cos 6 cos c, and cos h=—. — ^, 
 
 sin v^ 
 
 , cos C cos B cos C , T^ ^ ^ T^ T^ 
 
 and cos c^-. — =^, cosa=-^ — -,.-^ — ^=cotB cotC. Q. E. D. 
 sm B sm C sin B ^ 
 
 88. Napier's Rules suffice for the solution of all problems 
 involving only right triangles. Slight difficulties may, however, 
 present themselves in actual practice, when we wish to decide 
 whether an angle (arc) whose functions are numerically deter- 
 mined, lies in the first, or in the second, quadrant. 
 
36 DETERMINATION OP SPECIES. 
 
 38. Determination of Species.— If, in any investiga- 
 tion, a part be determined by means of the cosine, tangent, or 
 cotangent, the sign of the function will indicate the qu^-drant 
 in which the part is to be found. If, however, the part be 
 determined by means of the sine, an ambiguity presents itself, 
 as the sine is positive in both the first and^ second quadrants. 
 In this event we note that : — 
 
 I. An oblique angle and the side opposite are always of the 
 same species. 
 
 From (37)sinC= p as sinC is necessarily positive, 
 
 cos B and cos b must agree in sign ; i. e., B and b must lie in 
 the same quadrant. 
 
 II. If the hypothenuse exceed 90°, the remaining sides, and 
 consequently the oblique angles, are of different species. 
 
 From (37), cos a = cos b cos c. Hence, if a> 90°, cos a is 
 negative, and therefore cos b and cos c are of different signs ; 
 i. e., b and c, and therefore, by (I), B and C, lie in different 
 quadrants. If, however, a < 90°, cos a is positive, and hence 
 cos b and cos c agree in sign, and therefore b and c (and like- 
 wise B and C), lie in the same quadrant. Hence 
 
 III. If the hypothenuse be less thayi 90°, the remaining sides, 
 and therefore the oblique angles, are of the same species. 
 
 IV. If a side {other than the hypothenuse'), and the angle 
 opposite be given, there will be two solutions, if the sine of the 
 side be less than the sine of the angle; one solution if the sine 
 of the side equal the sine of the angle, and no solution if the 
 sine of the side exceed the sine of the angle. 
 
 From (87), sin a=-^— — -; hence (4) sin b cannot exceed 
 ^ ^ smB . ^ 
 
 sin B. If sin 6 = sinB, sin a = \, a = 90°, and the triangle is 
 
 birectangular. If sin b < sin B, sin a < 1, but positive, and 
 
 a may lie either in the first or in the second quadrant. 
 
 40. Quadrantal Triangles, i. e., triangles in which at 
 
 least one side is a quadrant, may be solved by Napier's Rules, 
 
TRIGONOMETRY. 37 
 
 as the polar of every quadrautal triangle must be a right 
 triangle. 
 
 41 i Example. 
 
 In a right triangle, 6= 38° 16'; c= 120° 35'; solve. 
 
 From (37) cos a=cos h cos c, whence 
 
 log cos a= log cos 6 + log cos c - 10 
 
 = 9.89495 + 9.73121 - 10= 9.62616. 
 h <90° and c> 90°; .-. cos 6 is +, and cos c is -, so that, 
 cos a is negative, and a> 90°; .*. a= 155° 0'47". To avoid. 
 using a we compute C by (37, II, 1), which gives 
 log cot C= log sin 6 - log tan c 
 
 = 9.79192 -10.19442 + 10 = . 5975. 
 e > 90°; C > 90°, (89) /. C= 111° 35' 40". 
 Finally, log cot B= log sin c - log tan b 
 
 = 9.92563- 9.89697 =.02866 
 whence B= 43° 06' 40". 
 
 42. The formulae of Spherical Trigonometry correspond 
 closely to those of Plane Trigonometry ; differences arise from 
 the fact that the sides of plane triangles are dealt with directly, 
 while the sides of spherical triangles are dealt with only by 
 means of their functions. Analogies deserve attention so far 
 as they aid in fixing formulae in mind. Table III suggests 
 analogies between the formulae for solving Plane Right trian- 
 gles and those employed in solving Spherical Right triangles. 
 More striking analogies will present themselves in the case of 
 oblique triangles. 
 
 Table III. 
 
 Higfit Plane Triangles. JUgJa Spherical Triangles. 
 
 sin B= - ; sin C = 
 
 a* 
 
 c 
 
 a ' 
 
 • T) sm * 
 
 sm B= ; 
 
 sm a 
 
 . ^ sm c 
 
 smC= -: . 
 
 sm a 
 
 cosB=-; cosC = 
 
 b 
 
 a ' 
 
 ^ tane 
 
 cosB = - : 
 
 tan a 
 
 ^ tan 5 
 
 C08C=-T . 
 
 tana 
 
 tanB = -; tanC = 
 c 
 
 c 
 
 b' 
 
 ^ T> tan 6 
 tanB=-^ — ■; 
 sm c 
 
 ^ ^ tan c 
 
 tanC= -T— ,. 
 
 sm6 
 
 COtB = ^; cotC = 
 
 b 
 
 c ' 
 
 ^ -r, sin c 
 
 cotB== -• 
 
 tan 6 
 
 ^ ^ sin b 
 
 cotO=-- . 
 
 tan c 
 
38 
 
 OBLIQUE TRIANGLES. 
 
 Fig. 17. 
 
 '" 43. Oblique Triangles. — I. Given two sides and the in- 
 cluded angle, e. g., 6, c, A. From the extremity of one known 
 side drop BD, perpendicular to 
 the other. Solve (37) ABD. If 
 AD •< b, the perpendicular falls 
 within the triangle ; if AD >• b, 
 the perpendicular falls without. In 
 either case, p and m being known, 
 BCD may be solved. Finally, 
 the angle ABC = ABD + DBC, or 
 ABD - DBC, according as DB 
 falls within or without the triangle. 
 
 II, Given two angles and the in- 
 cluded side, e. g., B, A, c. Drop 
 the perpendicular, BD, from one 
 extremity of the known side, to b. Solve the triangle ABD 
 as in I. If ABD >• B the perpendicular falls without ; if 
 ABD < B, the perpendicular falls within the triangle, p 
 and DBC being known, the triangle DBC may be solved. 
 Finally, AC = k + m, or k-m, according as BD falls within or 
 without ABC. 
 
 III. Given two sides, and an angle opposite one of them; 
 
 e. g., a, c, A. Drop BD i. b, 
 the unknown side. Solve ABD 
 (37); in DBC the sides a and 
 BD will then be known, and 
 DBC may be solved. But as 
 two equal arcs may be drawn 
 from a point to an arc of a great 
 circle, a may occupy either of 
 two positions, one on the right, 
 and the other on the left, of BD. 
 Hence (compare (21) III), there 
 
 ^^s-'^s. ujay \yQ t^YQ solutions. There 
 
TRIGONOMETRY. 39 
 
 will evidently be two solutions if a be intermediate in value 
 between jo and c, and aho between p and 180° - c. 
 
 I. If A < 90°, a>p, and < c, but a > 180° - c, there can 
 be but one solution, as ABC"; for, though a might occupy a 
 second position BC", as far to the right as BC" is to the left, 
 of BD, C" would fall beyond A', since BC" > BA', and the 
 arc AC" would exceed 180°, so that we could not (36) con- 
 sider the triangle ABC"'. 
 
 II. If A<90°, a>j9, and a> c, but ^«<180°-c, a 
 must fall to the right of the perpendicular, and there will be 
 but one solution. 
 
 III. If A > 90°, the conditions I, II, must of course be 
 changed by reversing the signs of inequality. 
 
 IV. If o=i?, ABC and ABC coincide with ABD. 
 
 V. If a <.pj A •< 90°, there will be no solution, as, when 
 A < 90°, the perpendicular is the shortest arc (of a great 
 circle) from B to ADA'. 
 
 VI. If a >• JO, A >• 90°, there will be no solution, for if 
 A > 90° the perpendicular is the longest arc of a great circle 
 that can be drawn from B to ADA'. 
 
 The number of solutions should be determined as soon as 
 BD is computed. 
 
 IV • Given two angles and a side opposite one of them, e. g., 
 A, C, a. The polar or supplemental triangle may be solved 
 a? in III. The angles and sides of ABC may then be ob- 
 tained by taking the supplements respectively of the sides and 
 angles of the polar. 
 
 V. Given the sides : a, b, c. From any vertex, as B, draw 
 
40 OBLIQUE TRIANGLES. 
 
 ^'BJ),J.b. Let AD = w, and CT>=k. Then, (87), 
 cos c=cosm cos^ (1), and cosa= cos A; cos^ (2), whence 
 
 cos c : cos a^ cos m : cos k, and, by division and composition, 
 
 cos c - cos a : cos c + cos a^ cos m - cos k : cos m + cos k, or 
 (29,2), tan i(c + a).tan i(c-a) - tan ^(m + k) . tan |(m - yt)(3). 
 From (3) either m+k ovm-k may be determined if the other 
 be known. If BD fall within the triangle, in + k=b; if BD 
 fall without, m-k=h (numerically). The symmetry of (3) 
 shows that either ^(in + k) or \(m— k) may be assumed equal 
 to ^ 6 in the first instance, the other being 
 t^j^-i tanKc + ft) tanK^-g) 
 
 tan ^6 ^ ^' 
 
 Adding this latter value, (4), to ^b, we obtain the greater seg- 
 ment ; subtracting (4) from ^b we obtain the other segment. 
 If either segment exceed b, the perpendicular must fall with- 
 out, and b must be the numerical difference of the segments. 
 
 The signs of the functions in (3) are important ; 
 tan |(c -H a) will be + if a + c < 180°, and - if c + a > 180° ; 
 tan ^(c - a) is + if e'>' a, and - if c << a ; for c - a cannot ex- 
 exceed 1 80°. tan J6 is + . 
 
 If the tangent (4) be negative it is the tangent of half the nu- 
 merical difference of the segments, and m < k. For tan ^(_,n + k) 
 is always + . For, the feet of the two perpendiculars that can 
 be dropped from B to AC... are 180° apart, and if they both 
 fall without the triangle we take D as the foot of the nearer. 
 Hence k + m is always less than 180°. 
 
 k and m being determined, ABD and CBD may be solved, 
 (37), and thus all parts of ABC obtained. 
 
 From (3) we derive the following important theorem (com- 
 pare 21, IV):— 
 
 If from any vertex of a spherical triangle, a perpendicular 
 be dropped to the opposite side, the tangent of half the sum of 
 the segments of this side is to the tangent of half the sum of the 
 other tivo sides as the tangent of half the difference of the sides 
 is to the tangent of half the difference of the segments. 
 
TRIGONOMETRY. 41 
 
 VI. Given the angles A, B, C. Solve the polar by Y. The 
 supplements of its angles are the sides of the original triangle. 
 
 The solutions III, IT, may be facilitated by use of the the- 
 orem, " The sines of the sides of a spherical triangle are propor- 
 tional to the sines of the opposite angles." 
 
 Thus (Figs. 19, 20), sin j9= sin a sin C ; also, in ABD, sin^ = 
 sin c sin A, whence sin a sin C— sin c sin A or, 
 sin a : sin c : : sin A : sin C. 
 
 This theorem may also be used as a " check " on the other 
 solutions of this article. 
 
 SECTION II.— ADDITIONAL FORMULA. 
 
 4:4:, Right Triangles. — When B or C is small, computa- 
 tion by means of the cosine should (18) be avoided. We may 
 use the following formula : — 
 
 , B 1 - cos B 1 - tan c cot a tan a - tan c sin (a - c), 
 2 1 + cos B 1 + tan c cot a tan a + tan c sin (a + c) 
 
 whence tan 4 B = -i / -; — 7 ^ (I). 
 
 ^ V sni (a + c) ^ ^ 
 
 Again, if a be near 90°, we may use the formula 
 
 tan ^ = y - ^^g^gt^jx [proof similar to proof of (I)] (II). 
 
 If 6 or c be near 90°, it will be better not to compute by 
 means of the sine. 
 
 45. Oblique Triangles. — In any spherical triangle. 
 
 The cosine of any side equals the product of the cosines of the 
 other two, plus the product of the sines of these sides into the 
 cosine of the included angle. 
 
 To prove (Figs. 16, 17) cos b = cosa cos c + sin a sin c cos B. (1) 
 By (C7) cos a = cos m cos p; cos c = cos k cos p, \ ,on 
 
 sin a = sin ^ ^ sin C ; sin c = sin /> -f^ sin A ) ^ ^ 
 
 Again, (37) cos B = cos ABD cos DBG - sin ABD sin DBC. 
 By (37) cos ABD = sin A cos ^ ; sin ABD = sin ^ -^ sin c ; 
 cos DBC = sin C cos m ; sin DBC = sin m -f sin a 
 
42 ADDITIONAL FORMULAE. 
 
 XT T> J ... ^ sin m sink ^ 
 
 xlence cos B = cos m cos k sin A sin C - -; : — . (Z) 
 
 sin a sin c ^ "^ 
 
 Substituting the values (2),(3) in the second member-of (1), 
 we obtain 
 
 cosm cos k cos^p + cos m cos k sin^jy - sin k sin m, i. e., 
 cos m cos k - sin w sin k, which equals cos b, or, 
 cos a cos c -f sin a sin c cos B = cos b. Q, E. D. 
 Il> fall without, (Fig. 17), B = ABD - DBC, and 6 = ^ - m, 
 and the reduction is similar to that just indicated. For a and c 
 we have 
 
 cos a = cos b cos c + sin b sin c cos J. (4), 
 cos c = cos a cos b + sin a sin 6 cos C (5). 
 46. -From (1), (4), (5), we deduce 
 cos a - cos b cos c 
 
 cos A = 
 cos B = 
 cos C = 
 
 sin 6 sin c 
 cos 6 - cos a cos c 
 
 sin a sin c 
 cos c - cos a cos 6 
 
 sin a sin b 
 
 These formulae, like their analogues, (33) are ill adapted 
 for logarithmic coitiputation. We therefore transform them, 
 as follows: From (1) we deduce 
 
 cos a - cos b cos c sin b sin c + cos 6 cos c - cos a 
 
 1 - cos A = 1 
 
 or, sin""^ ^ A = 
 
 sin b sine sin 6 sine 
 
 cos (6 - e) - cos a 2 sin ^(a + b—c) sin ^(a + c—b) 
 2 sin 6 sin e 2 sin 6 sin e * 
 
 1 • n T . i 1 A sin(s— c)sin (s-ft) 
 
 or, placing 2s = a + b + c, sm^ -J- A = — ~.~-,-,— \ 
 
 sin b sm c 
 
 \ sin 6 sin c 
 
 sinularlv, sin J B ^ /«i" (^ 7 «) sin (^I) (2). 1(B) 
 \ sin a sine | 
 
 sinJC = ^AMH5fl£5E5 (3). 
 V sin a sin A 
 
TRIGONOMETRY. 43 
 
 4:7. Adding both members of (1), (2), (3), (A), to unity 
 and reducing in a manner similar to that just shown, we obtain 
 
 cos-^ ^ — ' ^ ^ 
 
 I \ _ /sin s sin {s — a] 
 \ sin b sin c 
 
 T -D /sin « si 
 V smi 
 
 sin (s — b) 
 
 b sine 
 
 1 r^ /sin«sin(s - c) 
 
 cos i C= ^ — —-4 ^^ 
 
 \ sni 6 sm c 
 
 The tabulated cosines are inaccurate for small angles, and 
 the sines are unsatisfactory for angles near 90°. We therefore 
 obtain, by direct division, from (B) and (C), the tangents, (D), 
 which may be satisfactorily employed for all angles. 
 
 tan|A = ..M"(^-"^^'"^\-^> (1); 
 \ sm«sm(«— a) 
 
 tanJB=J!l^5IS5E5 (2); 1(D) 
 
 taniC= j!i^l!HllIlfEJ) (3). 
 V sin.'?sin(«~c) 
 
 If A'B'C be the polar of ABC, 
 
 A' = 180° - a, B' = 180° - 6, C = 180° - c. ) 
 a' = 180° - A, b' = 180° - B, c' = 180° - C. j 
 In this polar then we have (45) 
 cos a' = cos b' cos c' - sin b' sin c' cos A', etc., or, by substitution, 
 
 - cos A = cos B cos C - sin B sin C cos a (1); | 
 
 - cos B = cos A cos C - sin A sin C cos b (2); > 
 
 - cos C = cos' A cosB - sin A sin B cos c (3). ) 
 
 From (1;, (2), (3), by methods similar to those employed 
 in (45-6) we obtain the formulae (A'), and, adapting these to 
 logarithmic computation, the sets (B^ and (€')> ^^^^ which by 
 direct division we obtain (D')- 
 
 In all these formula s = ^(A + B + C). (B'). (C), (DO, may 
 also be obtained from (B), (C), (B), by passing directly to the 
 
44 
 
 ADDITIONAL FORMULAE. 
 
 v; 
 
 polar triangle, as in the deduction of (A') from (A). The 
 general remarks on (A), (B), (C), (D), are applicable also to 
 
 (AO, (BO, (CO, (DO- 
 
 cos a = 
 
 cos b = 
 
 cos c = 
 
 cosB cos C- cos A, 
 
 sin B sin C 
 cos A cos C - cos B^ 
 
 sin A sin C 
 cos A cos B - cos C 
 
 sin A sin B 
 
 cosScos(S- A)^ 
 
 sin-|-a = -*/ — _, . V 
 
 \ sniBsinC 
 
 • 1 7. / - c^s S cos (S - B) 
 \ sin A sui C 
 
 sin^c 
 
 cos 
 
 cos 
 
 -V 
 
 -cosScos(8-C) 
 sin A sin B 
 
 cos(S-B)cos(S-C). 
 sin B sin C 
 
 ^ \ cos S cos OS -A) 
 
 i.^ 1. /- cos (8 - A) cos (8 - C) 
 
 cot i- 6 = -• / !i-- ^-- — -^^- '' 
 
 ^ V cosScos(S-B) 
 
 \ cos 8 cos (8 - C) 
 
 .(A') 
 
 (B') 
 
 / cos(S-A)co.(S-C X 1 
 V sin A sin C j v^ / 
 
 lc= / cos (8 -A) cos (8 -B ) 
 \ sin A sin B 
 
 (D') 
 
 48. The analogies obtaining between the formulae of Spheri- 
 cal Trigonometry and those of Plane Trigonometry are no- 
 where more interesting than at this part of our work. We 
 deduced in (34) an expression for the radius of the inscribed 
 
TRIGONOMETRY. 
 
 45 
 
 circle (r) in terms of the sides of a 
 plane triangle. We have an analo- 
 gous expression in the spherical tri- 
 angle. Let O, (Fig. 21), be the pole 
 of the inscribed circle of ABC. 
 AOF, FOB, etc., are right triangles; 
 OAF = JA, OBD = |B, etc., and 
 AE = AF, BF = BD, CD = CE. 
 Hence (36) 
 
 ^ , tan OF tan r 
 tan ^ A 
 
 Fiff. 21. 
 
 tan h A 
 
 sin AF 
 1 
 
 sin(«— a) 
 
 i 
 sin {s—a)\ 
 
 whence tan r = -♦ 
 
 V 
 
 sin (s -6) sin (s-c) 
 
 sin 8 sin (s - a) 
 
 or 
 
 /sin (s - b) sin (s - c) sin {s - a) tan r 
 
 sin (s - a) sin (.s* - b) sin (« 
 
 sin (s - a) 
 -\ in which r is 
 inscribed 
 
 Fig. 22. 
 
 cotR^ 
 
 sin 5 
 the arcual radius of the 
 circle, and s = ^ (a + 6 + c). 
 
 Again, let R = O'A represent (Fig. 
 22) the arcual radius of the cir- 
 cumscribed circle of ABC. In any 
 right triangle, as AO'E, we have, 
 cos O' AE = cot R tan AE. By con- 
 sideration of the isosceles 
 AO'C, etc., we find that 
 S-Cand AE=-|C;hence 
 cos (S — C) 
 
 triangles 
 0'AE = 
 
 tan -^ C 
 
 = (47,D') 
 
 cos (S 
 
 V 
 
 -cos(S-A)cos(S-B ) _ 
 cos 8 cos (S - C) 
 
 - cos (S - A) cos (S - B) cos (S - C) 
 
 49. Napier's Analogies.- 
 
 JVapier',9 Analogies, viz. : — 
 
 cosS 
 -From (46-7) we readily deduce 
 
46 napier's and delambre's analogies. 
 
 tan-J(A+B)_cos^(a- 6) tan ^{a+b) eos^(A-B). 
 
 cotiC ~cosK«+^) tan^c ""cosiCA + B) ^ '^' 
 
 tan i( A-B) ^ sin |(a - 6) tan |(a- ^) ^ sin |(A - B) 
 
 cot J (J sini(a+6) ^ ^' tanjc sini(A + B)^ ^' 
 
 To establish (1):— 
 
 cos J(a - b)= cos (|-a - 1^6) = cos |- a cos |^ 6 + sin ^ a sin ^ 6; 
 substituting for cos^a, etc., their values (46-7), we obtain 
 
 . , , ^ cos (S - C) - cos S /cos (8 - B) cos (S - A) 
 
 cosi(a-6)= ^ — ;— ^^ -x — - . -p . .^ ^ 
 
 sinC V sinBsmA 
 
 cos (S - C) - cos S 1 ,j-,^ , X 
 
 = sinC •'''''*'' ^*^) ^''^• 
 
 «-.. -1 1 w ,. cos(S-C) + cosS ^ ,-^ 
 
 Similarly, cos ^(a + b} = ^ — ^~F^~ ^^* 2 ^ W* 
 
 By division, 
 cos -|(a- ^ j cos (S - C) -cos S sin (S-^C) sin^C 
 cos i(a + 6) "~ cos (S - C) + cos S ~ cos (S - -|^ C) cos ^ C 
 As S = i( A + B + C), S - ^ C = J( A + B), ancTwe have 
 cos^(ct-6) _ tanyA + B ) q j. -p 
 cos -J(a + b) cot 1^ C 
 
 The other equations may be similarly established. 
 50 . In a similar manner we may also establish the truth of 
 Delambre's Analogies, otherwise known as Gaiiss's Equations, 
 viz. : — 
 
 ^ ^ cos^C " cos^c 
 
 sinKA-B) _ sinK^-5) . 
 ^ ^ cos^C sin^c 
 
 cosKA + B) _ cos|(a + 6) , 
 ^ ^ sin^C cos^c 
 
 cosKA-B) _ sinK« + 6) 
 ^ -^ sin^^C sin^c 
 
 These formulae afford solutions of cases I, II (43); the species 
 of the parts should be carefully determined from the signs of 
 the functions. 
 
 Napier's Analogies may be readily deduced, by division, 
 from Delambre's. 
 
i UNIVERSITY 
 
 CHAPTEIi yi, 
 
 ALGEBRAIC TRIGONOMETEY. 
 
 51. The Trigonometric functions have thus far been treated 
 in a somewhat restricted way, from a point of view ahnost 
 purely geometric. We now propose to show that the relation 
 between the arc (or angle) and the function may be expressed 
 algebraically, and thus that Trigonometry may be treated as a 
 branch of pure Algebra. We will also extend our discussions 
 to other transcendental functions of the same general form as 
 those with which we have been dealing, and in properties and 
 relations closely analogous to them. 
 
 52. To develop sinx and cosx into series arranged accord- 
 ing to the powers of x, i. e., to express algebraically the relation 
 between x and sin x and cos x respectively. 
 
 Whatever the relation between x and sin x, let it be expressed 
 by the identity 
 
 sina; = M + Na; + Pa;' + Qic' + Ra^* + . . . (1) 
 
 in which M, N, P, Q, etc., are independent of x, and are to be 
 determined. After we have determined them we must further 
 show that the series is convergent. 
 
 As sin = 0, the second member of (1) nmst vanish when 
 x = 0, and hence can contain no term independent of x ; .*. M = 0. 
 Again, as sin(-£c)= -sin a*, the second member of (1) must 
 change its sign (without changing its numerical value), when 
 - X is substituted for x. Hence it can contain no even power 
 of X. Further, the coefficient of the first term must be 1, 
 since for an infinitely small arc the ratio (sin .t): .r= 1. We 
 must therefore modify the form of (1) by assuming a develop- 
 ment of the form, 
 
 smx = x + qx' + Sx' + \Jx' + (2). 
 
 (47) 
 
48 TRIGONOMETRIC DEVELOPMENTS. 
 
 In the development of cos x no term of an odd degree can 
 occur, for the development must remain unchanged when for x 
 we substitute -x. Again, the development must contain a 
 term independent of x and equal to unity, since, when a; = 
 we must have cos £c = 1 (7). We assume then, 
 
 co^x= 1 + RV+ TV+ W + (3). 
 
 R', T', V, etc., being independent of x, and to be determined. 
 Since (2) is an identity we may substitute 2x for x, thus : — 
 
 sin2a; = 2sina; cos£c = 2x + 8Qa;=* + 32S:c' + 128Ux' + (4). 
 
 or, sin x qo?,x= x+ A Qx^ + 16 Sa:^ + 64 JJx' + (5). 
 
 Multiplying (2) by (3), member by member, we obtain, 
 sin xQ.o^x = x+((^ + W)x^ + (S + T' + QR'jic' 
 
 + (U + V + QT' + SR')a:^ + . . . . (6). 
 Equating the second members of (5) and (6), we obtain, 
 a; + (Q + Wy + (S + T' + QR')^' + (U + V + QT' + m!)x' + 
 
 = a; + 4Qx^ + 16Sa;' + 64LV + 
 
 Equating coefficients, (Alg. 462), we obtain 
 
 Q + R'^4Q (a); 8 + T' + QR' = 16S (6); ) ,^. 
 U + V' + QT' + SR' = 64U(c);etc. P ^ 
 
 Again, sin^a; + cos^a; = 1; substituting for sin x, cos x, their 
 values, (2), (3), we obtain 
 
 1 + (1 + 2'K')x' + (R' 2 + 2Q + 2T)x' 
 + (Q"^ + 2S + 2T'R' + 2\')x^ 
 
 + (T'=^ + 2QS + 2U + 2V'R' + 2WK+ = 1. (8). 
 
 As this is an identity w^e have, 
 
 1 + 2R' = (a'); R' '^ + 2Q + 2T' = (h'y, | 
 
 Q-^ + 2S + 2T'R' + 2V' = (cO; etc. J ^^^ 
 
 From (A) and (B) we obtain Q, R', S, T', etc., thus:— 
 
 1 + 2R' = 0, .-. R'=-i- 
 
 From (a), R' = 3Q, .-. Q = |r' = - g- = - ^j 
 From(6'),l-i- + 2T' = 0, .-. T' = -i-4= \ 
 From (6), 8+1 + 1 = 168, .-. S = jl= ^ 
 
TRIGONOMETRY. 49 
 
 etc., etc. 
 Substituting the values of these coefficients in (2) and (3) 
 we obtain 
 
 „3 „5 „7 ^9 
 
 (A). 
 
 «• x' 
 
 x' x' 
 
 " 7! ^ 9! 
 
 ^2 ^4 
 
 cos^ = l--+-- 
 
 x' x' 
 -6!'' 8!"- 
 
 (B). 
 
 The series (A) and (B) are convergent. For, the ratios of 
 
 xi^ x^ x^ 
 the terms in successive pairs in (A) are — -^ -— , — -, and so 
 
 ^.o 4.0 D.7 
 
 on, and in (B), — , — , ^ , and on on. Whatever the 
 ^ ^ 3.4' 6.7 9.10 
 
 value of X we may continue these fractions until the denomina- 
 tor is, say, greater than 3x^ Then the sum of the remaining 
 terms in each series will be less than the sum of a geometric 
 progression of which the ratio is ^ ; hence (Alg. Ch. IX) each 
 series is convergent. The convergence may also be shown for 
 all ordinary values of x by (Alg. 500). On account of the 
 rapid convergence, these series give a shorter method of com- 
 puting a table of natural sines and cosines than that indicated 
 in Chapter I.* 
 
 53. From (A) and (B) we may deduce, by the relations of 
 (9) 
 
 (C). 
 
 (E). 
 
 • (F). 
 
 ^Elegant methods of computing Tables of Natural Functions, and also methods 
 of computing directly Tables of Logarithmic Functions, which would be out of place, 
 however, in a work'^of this character, may be found in Caliet's "Tables de Loga- 
 rithiues" Paris, 1879. 
 
 tancc 
 
 — X ■ 
 
 o 
 
 ^3.5 
 
 3^5.7 
 
 3^5.7.9 
 
 + •• 
 
 cot a; 
 
 1 
 
 X 
 
 X 
 
 "3 
 
 x' 
 5.9 
 
 2x' 
 5.7.9 
 
 x' 
 
 
 7.25.27 
 
 
 secic 
 
 = 1 
 
 x' 
 ^2 
 
 6x' 
 ^2^3 
 
 
 ' 277x« 
 5 ^ 2^3^.7 
 
 + 
 
 CSC a; 
 
 = x~ 
 
 ■^ 4-- 
 
 X 
 
 W2 
 
 7x' 
 
 Six' 
 2\^\b.l ^ 
 
 127a;^ 
 
 2^3^5^7 ^ 
 
50 
 
 THE EXPONENTIAL SERIES. 
 
 54, The relations expressed in (A) .... (F) might evidently 
 be taken as definitions, replacing those given in (4). The 
 trigonometric functions might then be treated precisely as any 
 other algebraic functions involving infkiite converging series. 
 We proceed to the purely algebraic treatment of these func- 
 tions, and to some important extensions of Trigonometry, 
 viewed as a department of pure algebra. 
 
 55. The Exponential Series. — Let us first expand a' 
 into a series arranged according to the ascending powers of x. 
 Assume 
 
 a-=P + Qa; + Rx^ + Sa;^-f-Tx* + Va;^ + 
 
 This equation, being identical, must be true for every value 
 of X. Making ic=0 we have a°=P; i. e., P=l. Hence 
 
 a^=l+Qa; + Ric2-fS.'«' + Ta:* + Va;^ + 
 
 Squaring both members of (1) we obtain 
 
 
 + 2QR 
 
 x^+ R^ 
 
 + 2QS 
 
 + T 
 
 (1). 
 
 (2). 
 
 Again, since (1) is an identity, we may substitute- 2x for a;; 
 so doing we obtain 
 
 a^= 1 + 2Qir + 4Ra;^ + 8Sx^ + IGToj* + 32Vx^ + . . . (3). 
 
 Equating the second members of (2) and (3), we obtain the 
 identity 
 
 l + 2Q£c + 2R 
 
 a;^ + 2S 
 4-2QR 
 
 x^ + . 
 
 + R^ 
 
 + 2QS 
 + T 
 
 1 + 2Qa; + AUx' + 8Sx^ + 16Ta;V .-.-(4> 
 Equating coefiicients in (4), we obtain, 
 
 2R + Q2= 4R, whence R = -|-' 
 2S + 2QR = 8S, whence S= ^' --= %; 
 
 24 4!* 
 etc., etc. 
 Substituting the values thus found in (1), we obtain, 
 
 R' + 2QS + T= 16T, whence T= 
 
TRIGONOMETRY. 51 
 
 «'=l + Qx + |!x^+|V + ^jX' + ..... (5), 
 
 the required series, in which Q, independent of a, depends 
 only on the value of a. Since x may have any value what- 
 ever, let ic=l : Q. Then 
 
 a^ = a^=<^ = l + l+i + ^j + ^ + (6). 
 
 The second member of (6) is e, the Natural Base, (x\lg. 
 511); hence a}'-^== e, or e^=a; i. e., Q=lna, so that 
 
 a^ =1 + (In a)x + (In a)^|^+ (In a)^|^+ (In a)*^J + . . . . (7). 
 
 56. In (7), (55), let a=e ; we then have 
 
 „^ X OC X X r-i^ 
 
 e«=l + «+- + -+^+.-+ ^^> 
 
 As (1) is an identity, xV - 1 may be substituted for x. 
 Making this substitution, denoting V - 1 by i, we obtain 
 
 .^ , XXtXXlilL /■c\^ 
 
 ^=^+"'-2-- 4! +3!+ 5! -6!- ^^^^ 
 
 _ x"^ X*' x^ x^ 
 "2""^4!~6!'^8!"**' 
 a? x"" x' 
 
 or, by (52), 
 
 e^' = cos cc + i sin x (3). 
 
 Similarly we may show that 
 
 e~^* = cos x — i sin x (4). 
 
 Combining (3) and (4) we obtain, 
 
 e-- + e— ■ = 2 cos aj ; .-. cos x = ^(e'' + e'^O (5). 
 
 e"' _ e-- = 2i sin .r ; /. sin x = lie"' - e""') (6). 
 
 These relations may also be expressed thus : — 
 
 ^xi _^ ^-xi = cos ic + COS ( - x'). (7). 
 
 ^xl _ ^-xi ^ ^ gjjj ^ _ ^' gii^ (^ _ x). (8). 
 
 (7) and (8) are called "Euler's Equations," 
 
 57. The trigonometric functions are, then, merely expo- 
 
52 FUNDAMENTAL RELATIONS. LIMITING VALUES. 
 
 nential fanctions of the arc, and may, therefore, be thus 
 defined : — 
 
 8inx=l-.(e''-0 (l);tan.= .^j,7;"l^ (3); 
 
 , cosa! = i(e- + e-'') (2); -cotx = lg^^^ (4); 
 
 2 2i 
 
 ^'''^=-pTP''' ^^^' ^^^^=^^rr^ (6). 
 
 All the formulae established with the old definitions may 
 now be established analytically with these, so that Trigonome- 
 try may be considered merely a department of Algebra. 
 
 58. Fundamental Relations. — Multiplying (3) and (4) 
 (56), we obtain 
 
 1 = cos'^ic + sin^a;. 
 
 This may also be obtained by direct substitution. Again, 
 squaring the fractions defined as tan x and sec a;, and subtract- 
 ing the first square from the second, we obtain 
 
 sec^a; - tan^a; = 1 . Similarly, 
 
 csc'^a; — cot'^a; = 1. 
 The other relations are mere restatements of definitions. 
 59 . Limiting Values. — By substituting, successively, 
 |-, r,|r, 2', etc., for x, in (1) . . . (6), (57), we find the same 
 set of limiting values as given in the Table on page 14. 
 
 60. Again, by substituting for x in (5), (6), (56), ^tt - x, re- 
 membering that, by equations (3), (4), when a;=i-, e^ = i, 
 and e~^^ = -i, we shall obtain, 
 
 sin a;= cos (^-—x); and similarly, 
 
 tan aj= cot (J- —x), and esc x= sec (|-- - x), as in (5). 
 
 61. The addition and subtraction formulae of Chapter III 
 may be established by direct substitution of the values of sin x^ 
 cos a;, etc., as given in the definitions of (57). Or, we may 
 proceed thus : — 
 
 e(.x+y)i _ COS (ix + y) + i sin (x + xj) (a), by (3) (56). 
 
 But 
 
 (Xi+yi ^ ^i^yi ^ ^^.^g x-\-i^\\\ x) (cos y + i siu y) (Jb). 
 
 Equating the values (a) and (6) we obtain 
 
TRIGONOMETRY. 63 
 
 (cos X + { sin X) (cos y-\-i siny) = cos (a; + y) 4- i sin (ic + y), 
 or, expanding in the first member, 
 
 cos cc cos y - sin a; sin y + % (sin a; cos y + cos ic sin y)= 
 cos (a; + 2/) + i sin {x + ?/). 
 
 As the real parts of the two members of the identity must 
 be equal, 
 
 cos (a; + y) = cos a; cos y - sin a; sin 2/ (1). 
 
 As the coefficients of i in the two members must be equal, 
 
 sin (a; + y) = sin x cos y + cos a; sin y (2). 
 
 Since y may be negative in (1) and (2), the formulae for 
 cos {x - y) and sin (a; - y) are established. From these we 
 derive at once the formulse for tan (x±y'), cot (x±y\ and the 
 remaining formulae of Chapter III. 
 
 62. Demoivre's Theorem.— In (3), (4), of (56), sub- 
 stitute nx for X ; there results 
 
 f,nxi ^ (.Qg nx + i sin nx ; e~"^ = cos nx-i sin nx. 
 As e"^ = (/'')"', and e~"^ = (e~^^y, we obtain, on substituting 
 for e^^ and e~^ their values, (56), 
 
 (cos x±i sin a;)" = cos nx±i sin nXy Demoivre's Theorem. 
 
 63 . It is clear that in the functions we have been consider- 
 ing X is not necessarily real. On the other hand it is plain 
 that we may define and discuss a series of exponential func- 
 tions, analogous to those already treated, but having the 
 exponents real. From these we may develop a theory as 
 complete and interesting as that of the ordinary trigonometric 
 functions. These latter have those relations to the circular arc 
 of length X, which we have already discussed, (11), whence 
 they are usually called circular functions. The functions, 
 the treatment of which we are about to undertake, have anal- 
 ogous relations to the arc of the equiangular hyperbola, whence 
 they are called hyperbolic functions. Taking again the natu- 
 ral base, e, we define as follows : — 
 
 the hyperbolic sine of x, (sinhx) = o(^ - 6~^) O-y^ 
 
54 DEFINITIONS THE HYPERBOLIC FUNCTIONS. 
 
 % 1 
 
 the hyperbolic cosine of x, (coshx)^ -^(e^ + e~^) (2) 
 
 the hyperbolic tangent of x, (tanh x) = ^^ (3) 
 
 e^ -T €■ 
 
 the hyperbolic cotangent of x, (coth a?) = (4) 
 
 2 
 
 the hyperbolic secant of x, (sech x') = — ■ (5) 
 
 2 
 
 the hyperbolic cosecant of x,(cschx)= — - (6). 
 
 From these definitions we derive at once series and relations 
 analogous to those of (53, 55). Thus, 
 
 e^ = cosh X + sinh x (7); 
 
 e~^ — cosh x - sinh x (8). 
 
 Whence, by using the exponential series, we deduce 
 
 q? rJ' ly^ 
 
 sinha; = ^ + ^j+^ + ^j + (9); 
 
 cosha; = l+| + | + | + (10): 
 
 and from these we may derive, by mere division, series for 
 tanh x^ coth x, sech x and csch x. The convergence of these 
 series may be shown as in the case of the analogous series for 
 the circular functions. 
 
 64. Fundamental Relations. —These, for the hyperbolic 
 functions, are : — 
 
 cosPo; — sinh^a; =1 (1) 
 
 - sinh X .^. 
 
 tanh a; = — -, — (2) 
 cosh a; 
 
 coshaj 
 
 coth a; = -.— .; — (d): 
 
 smn£c 
 
 tanh X . coth x = l (4) 
 
 sinh X csch x =1 (5); 
 cosh :c sech a; =1 (6); 
 
 tanh^oj + sech'^x- = 1 (7); 
 
 coth^ic- csch^a? = l (8). 
 
 (2), (3), (4), (5), (6) are evident from equations (l)-(6), (63). 
 To establish (1) we may multiply (7) by (8), (63), member 
 by member. To establish (7) and (8) we may substitute at 
 once the exponential values of tanh x, etc., from (l)-(G) (63). 
 
TRIGONOMETRY. 55 
 
 65. Limiting Talues. — Sinhx. AVhen ^- = sinha; = 
 •|(1 - 1) = 0. As a; increases sinh x rapidly increases, becom- 
 ing infinite when x — cc. If a;, diminishing from 0, becomes 
 negative, sinh x becomes negative, and numerically increases, 
 until, when aj = - qo , sinh a; = - oo . It will be observed that, 
 as in the case of sin Xy changing the sign of x changes the 
 sign of sinh x. 
 
 Cosh X. When x = 0, cosh a; = 1 , and this is the smallest 
 value of cosh x; as x increases cosh x increases very rapidly, 
 until, when a; = qo , cosh a: = oo . Since, however, 
 
 cosh'-'a; — sinh^ic = 1, 
 cosh X always exceeds sinh x. Again, cosh x = cosh ( — x); 
 i. e., changing the sign of x does not affect the sign of cosh x. 
 
 Tank X. Tanh =^ 0, and as x increases tanh x increases, 
 from onward, but very slowly. As x approaches qo , tanh x 
 approaches 1 ; when x =oo , tanh a: = 1. For, 
 
 e^ _ e--' 1 - e-^ 1 e-^^' 1 1 
 
 = 1, when 03= 00. 
 
 If X, passing through 0, becomes negative, tanh x becomes 
 negative, and passes through the same numerical changes as in 
 the case of positive x. 
 
 Coth X. AVhen ic = 0, coth aj = l:0=oc;as x increases 
 coth X decreases, but remains the reciprocal of tanh x, and, 
 when £c = QO , coth a: = 1 : 1 = 1. 
 
 For negative arcs the numerical changes in the coth x are 
 the same as in the case of positive arcs. 
 
 Seek X and Csch x. These functions are the reciprocals 
 respectively of the cosh x and sinh x. Hence, as x passes 
 from to oo , sech x passes from 1 to 0,.and as x passes from 
 to — QO , sech X again passes from 1 to 0. As x passes from to 
 oo , csch X passes from cc to 0, and as x passes from to - go , 
 csch X passes from — oo to 0. 
 
 66. Addition and Subtraction Formulae . — By direct 
 
ba 
 
 PERIODICITY OF THE FUNCTIONS. 
 
 substitution of the values (63) of the functions we may readily 
 
 show that : — 
 
 sinh (ic + 1/) = sinh x cosh y + cosh x sinh 2/(1) 
 sinh {x-y) = sinh x cosh y - cosh x sinh y (2) : 
 cosh {x + y) = cosh x cosh y + sinh x sinh i/ (3) 
 cosh {x-y) = cosh £c cosh y - sinh cc sinh y (4) 
 tanh a:±tanh?/ 
 
 tanh {x±y') = 
 coth (a3±t/) = 
 
 l±tanhaj tanhi/ 
 coth X coth 2/± 1 
 
 (5); 
 
 (6). 
 
 coth 2/±coth X 
 
 We may also derive these from the corresponding formulae 
 for circular functions (and vice versa), by using the following 
 relations, which are evident from the definitions, viz. : — 
 
 sin x = - sinh ix (1); tan x = - tanh ix (3) ; 
 I I 
 
 cos X =i cosh ix (2) ; cot x — i coth ix (4). 
 
 67. Periodicity of tlie Functions. — The circular func- 
 tions are periodic; i. e., as x passes from to oo the same 
 values of the functions recur at intervals of 2-. The functions 
 of ic'are the same as those of {x-\-2n7i), n being any integer. 
 The values of cos x, sin x, sec x, esc x, recur at intervals of 
 2- ; the values of tan x and cot x at intervals of -. The re- 
 lations just expressed indicate a similar periodicity in the 
 hyperbolic, functions. But while the circular functions are 
 periodic for real values of x, and continuous for imaginary 
 Talues, the hyperbolic functions, on the other hand, are con- 
 tinuous for real values of x (65), and periodic for imaginary 
 values. As 
 
 cosh 2nrd = cos "In- = 1, and sinh 2nr:i = i sin 2??- = 0, 
 we have the following formulae, analogous in pairs : — 
 
 CIRCULAR FUNCTIONS. 
 
 sin {x + 2?i-) = sin x ; 
 cos {x + 2/1-) = cos a; ; 
 tan {x -H nrr) = tan x ; 
 cot {x + W-) = cot X ; 
 
 HYPERBOLIC FUNCTIONS. 
 
 sinh (x -f- 2n-i) = sinh x ; 
 cosh (x- 4- 27iTd) = cosh x ; 
 tanh (a; + nrA) = tanh x ; 
 coth (x + 7irA) = coth X ; 
 
TRIGONOMETRY. 
 
 57 
 
 The student should substitute, successively, Jri, -ni, |ri, and 
 2rt" for y in all the formulae of (66), and thus construct a 
 table for hyperbolic functions analogous to that on page 14. 
 
 68. By processes analogous to those of (27, 28) we may 
 deduce expressions for the hyperbolic functions of 2x and of 
 ^x, and also for cosh a^cosh i, sinh a±sinh b, etc. Or, we 
 may derive these formulae directly from the definitions. Thus, 
 
 as 
 
 (e^+ib ^ g-ia-i6>)^gia-i6 ^ g-t«+i6) = e« -f e"^ + g" + e-\ 
 
 we have 
 
 cosh a + cosh 6 = 2 cosh J(a 4- b) cosh ^(a - 6). 
 
 Similarly for the other formulae. 
 
 69. The exponential character of the hy}:)erbolic and circu- 
 lar functions suggests important relations that obtain between 
 them and the ordinary logarithmic functions. For example, 
 let cosh x^^z, so that sinh a;= Vz'^ - 1, then, ((7), (8), 63), 
 
 a;= In (2; + V^z' - 1), and - x= In (z - Vz^ - 1). 
 Similarly, if z = sinh x, 
 
 x^ In (2 4- Vz' + l), and - re- In (l^zN^ - z). 
 
 * . . l+tanha; e^ ,_ .^^ •/. ^ 1 
 Agam, smce - — - — -- =—, = e'^ (6 3 j; if « = tanh x, 
 1 - tanh X e~^ 
 
 2a; = In ^ , or aj = ^ In q- — = In-v/ t^ 
 
 1 -z 2 1-z \ 1 
 
 + z 
 
 -z 
 
 Further, if iv= coth x, x-. 
 
 V 1 - w 
 
 The corresponding relations for the circular functions are 
 evidently, 
 
 xi:= In (Vl^' + zi)= - In (VY^'-zi) (1); 
 
 70. This connection of the Trigonometric functions with 
 logarithms indicates that we are not confined to the natural 
 base, e, in our definitions. Thus, if a be the base, instead of e, 
 
58 GEOMETRIC INTERPRETATION OF HYPERBOLIC FUNCTIONS. 
 
 since a^=e^^°«, if, in all discussions we replace a bye, the 
 only change required in the functions will be the change of the 
 exponent from x to x In a, which will 
 evidently not affect the nature or the 
 results of our investigations. 
 
 71. Geometric Interpretation 
 of the Hyperbolic Functions. — In 
 (4) we dealt with the relations of the 
 Fig. 23. circular functions to the sides of tri- 
 
 angles of reference. Let us now investigate the analogous re- 
 lations of the hyperbolic functions. If BAG, (Fig. 23), be 
 right-angled at A, we have 
 
 sin C = c : a ; cos C = 6 : a ; tan C = c:h. 
 Kow let BA revolve about B, the angle C remaining con- 
 stant, and let x be an arc, some function of the measuring arc 
 of C, such that 
 
 sinha; = c:a; cosha;=6:a; tanhic = c:6. 
 We are to determine the proper shape of the triangle, and 
 the relation betwaen C and x. Since, (22), 
 sin C c^ sin B 6, sin C c 
 sin A a sinA~a sinB 6' 
 and since cosh^a; - sinh^aj = 1 (or); we obtain, on substituting in 
 (or) for cosh X and sinh x their values, 
 
 sin'B - sin'C = sin^^A = sin"' (C + A) - sin'C ; 
 or, since sinV - sin^^/ = sin (x + y) sin (x + y) 
 sin (2C + A) sin A = sin^A; whence 
 sin (2C + A) = sin A (1), or sin A = (2). 
 The latter value of sin A being inadmissible, we have, from (1), 
 2C + A = --A, orC + A = j7:; 
 whence B = ^r; L e., BA ± CB. 
 This determines the shape of the triangle. The relation be- 
 tween C and X is evident; for 
 
 sinh a? = tan C ; tanh a? = sin C; 
 cosh a; = sec C ; coth x = esc C. 
 72. If, in any circle, (Fig. 24), a perpendicular, PM, be 
 dropped from the termination, (M), of any arc on the diame- 
 
TRIGONOMETRY. 
 
 59 
 
 ter through the origin thereof, and if PM = y, OM ic, and 
 radius = a, then evidently, 
 
 Hence we may at once write : — 
 
 ic = a cos ^; y = a sin ^, if 6 denote the radial measure of 
 AOP. This convention gives 
 us the circular functions as de- 
 fined geometrically in Chapter 
 I. \a^d is the area of the sector 
 AOP, and hence, if a = 1, d 
 represents double this area. 
 
 The hyperbolic functions are 
 connected in an analogous way 
 with the rectangular hyperbola. 
 Thus, (Fig. 25), if PM be the 
 perpendicular dropped from any 
 point P, of a rectangular hyper- ^*^' ^^' 
 
 bola on the diameter through the origin, (A), of the hyperbolic 
 arc, and if, as before, PM = y, OM = x, and the radius of the 
 auxiliary circle be a. we have, from the well-known property 
 of the curve. 
 
 Hence we may define the hyperbolic functions geometrically 
 as follows: — 
 
 (1) The hyperbolic sine is the perpendicular dropped from 
 the termination of the hyperbolic arc on the diameter through 
 the origin, divided by the radius of the auxiliary circle. 
 This radius is to the hyperbola what the radius is to the circle, 
 viz., the semi-axis of the curve. 
 
 (2) The hyperbolic cosine is the distance from the centre to 
 the foot of the hyperbolic sine (the base of the hyperbolic 
 triangle of reference), divided by the semi-axis (a). 
 
 In a circle every diameter is an axis; hence any semi-diameter (or radius) may be 
 taken as a semi-axis. In the hyperbola, however," tliis is not the fact; hence we take 
 the only semi-diameter that is also a semi-axis, as the denominator of the hyperbolic 
 functions. 
 
60 GEOMETRIC INTERPRETATION OF HYPERBOLIC FUNCTIONS. 
 
 The definitions of the geometric representatives of the other 
 hyperbolic functions are at once evident (63). 
 
 73. As will be shown later, the area of the hyperbolic 
 sector AOP is ^a\ (u being the hyperbolic arc AP), and 
 hence, just as in the case of the circle, if a = 1, i* is double the 
 area of AOP. 
 
 From M draw ML, tangent to the auxiliary circle (of radius 
 a), at L, and denote the angle (arc) AOL by 6. Then 
 a sec ^ = OM = a cosh u ; 
 a tan ^ = LM = a sinh u = PM. 
 
 When 6 is thus connected with u it 
 is called the Gudermannian of t*,* 
 ordinarily abbreviated gd u. A tri- 
 angle of reference for any hyperbolic 
 function of u may be readily con- 
 structed ; the calculation of any hy- 
 perbolic function may thus be reduced 
 to the calculation of a circular func- 
 Fig. 25. tion. For example, lay off 
 
 MK = a, and draw PK ; tan MKP = PM : a = sinh u. 
 Similarly for the other functions. 
 
 74. Hyperbolic and circular functions are of great impor- 
 tance in the Integral Calculus, where it will be shown that they 
 are special forms of certain more general functions, known as 
 Elliptic Functions. Any attempt at a discussion of these latter 
 would, however, be entirely beyond the scope of this work. 
 
 Interesting examples of the methods of employing these 
 functions in the solution of equations will be found in Tod- 
 hunter's and Burnside and Panton's treatises on the Theory of 
 Equations, and also in the various mathematical journals, 
 among which Crelle's is perhaps the best. These discussions 
 are not, however, deemed of sufficient practical importance to 
 warrant their introduction into the present work. 
 
 *cf . Orelle, " Journal fUr die reine und an^'ewandte Mathematik," Band 6. 
 
EXERCISES- 
 
 CHAPTER L 
 
 1. Construct 
 
 sin-Y.3); tan-^e^; cof^C - 4); sec-'( - 3); coS"\ - . 6); 
 
 sin~'j - |; tan~^7r; csc~V —1 |; cos~M - J. 
 
 Determine (with a protractor) these angles, to within 1°; 
 I determine also their radial measure. 
 
 2. Determine, with protractor and scale, the functions of 
 5°, 10°, 15°, etc., to 180° inclusive. 
 
 3. Determine geometrically the functions of (90° ±a;), 
 (180°±a:), (270°tx), and (360°±ic), in terms of functions 
 of X. 
 
 4. Find the functions of 
 
 sin-^'- ; cos"'^ ; tan-^c ; cot-^( - 2); csc~\ - m). 
 5 
 
 5. Express successively all the functions of x in terms of 
 each. 
 
 6. State the signs of the functions of 
 
 (2?i + l)180°±A; 2?ir±A; (2?i. - l)r- ^^^ rr. 
 
 7. Prove that 
 sec X + CSC X 1 + cot X tan ^ + 1 
 
 (1) 
 
 secic — cscx 1-cota; tanic-1 
 
 .^^ sec X + CSC X - sec x esc x sin a: cos x sec x esc x 
 
 sec X CSC X sec x + esc x + sec x esc x 
 
 .^^ 1 H-tan'a; 
 
 (^) A/r" -^=tanic; tan a; + cot x = ^ sec a; esc a;. 
 
 ^ \ 1+COt'iC 
 
 ^]:r^<^- ^ (1) 
 
EXERCISES. 
 
 (4) — - = - ^ r— .tan^aj. 
 
 CSC a; - 1 1 + sina; coversm x 
 
 (5) tan-^^ + cot-X2aJ + l) = ^ 
 
 (6) tan A + cot A = sec A esc A. 
 
 1 
 
 (7) secB + V'sec=^B-l = 
 
 sec B - -y/sec^B - 1 
 
 (8) ] 'l^'f = l + 2tan^A-2secAtaiiA. 
 1 + sin A 
 
 .(.-. tan A + tan B , . ^ _, 
 
 (9) =:=tanAtanB. 
 
 ^ cot A + cot B 
 
 (10) sec-^Acsc'A-(sec'A + csc'A) = 0. 
 
 (11) sinA + cosA= /?5^A^!:^A±^. 
 \ sec A CSC A 
 
 8. Solve, analytically and graphically, the following equa- 
 tions : — 
 
 (1) 2 cos'^a; = tan x ; (2) esc a; = 4 sin x ; (3) sin cc = 2 cos x, 
 (4) sin a; + cos ic = CSC ic ; (5) 5 sec'^ic — 12 sin^a; = 4 tan x. 
 
 (6) sin X cos a; + tan a; cot x = . 
 
 sec a; esc x 
 
 (7) 2 (sin X + cos a;) + sec a; esc a; - cot x = S. 
 
 (8) 3 sec X esc x tan x = cot^a; + 1 . 
 
 (9) 3sina; + 5cosa; = 8; (10) 2 tan aj + 3cscaj = 6. 
 (11) cot aj - 4 sin x --=.7 ; (12) sin-'a; + 7 cos a; = 1. 
 
 (13) 3 cos^a; + 6 sin^a; cos x= 1. 
 
 (14) secaj— tanaj= 4'v/sina;-2. 
 
 (15) a cos x + b sin x= c. 
 
 9. Show that, for any angle, 
 
 2 (sin*A + cos*A + 2 sin^A cos'^A) = 
 
 (tan A + cot A)^ - tan^A — cot''^A= constant ; 
 
 cos A + COS B sin A + sin B 
 
 sin A - sin B cos B - cos A 
 
 10. Solve the equations r sin x=a (1) 
 
 r cos a; =6 (2), 
 
CHAPTER II. 
 
 1. Solve the right triangle ABC, given 
 
 (1) 6= 3;c=4 ; (4) a=625 B^52°13'; 
 
 (2) a=29;6=20 ; (5) 6=.00216 B== 87°26'; 
 
 (3) a= 30.126 ; C=27° 50' ; (6) c=^. 00326 C=: 2° 15'. 
 
 2. When the sun's altitude is 36° 19' the shadow of a flag, 
 pole is 175 ft. What is the height of the pole? What is the 
 length of the shadow when the sun's altitude increases to 
 40° 16' 27"? 
 
 3. A winding road 3 miles long has a rise of 1,196 feet. 
 Required the angle of (uniform) slope. 
 
 4. The angles of elevation of the top of an inaccessible crag 
 are 32° 46', and 24° 17', as observed from two points, 200 
 yards apart, in the same vertical plane with the top. If the 
 theodolite stands four feet from the ground, what is the height 
 of the crag? 
 
 5. Two towers stand on opposite banks of a river. From 
 the top of one, 168 feet high, the angles of depression of the 
 top and bottom of the other are observed to be 13° 28', and 
 28° 16', respectively. Required the breadth of the river and 
 the height of the second tower. 
 
 6. What is the angle of depression of a tangent to the 
 earth's surface as observed from the top of a mountain 18,260 
 feet high ? What is the distance from the foot of the mountain 
 at which the top must cease to be visible, the average slope 
 
 being 28° 17'? 
 
 7. Let E represent 
 
 the earth and M the 
 
 moon. The angle EMS 
 
 = 57', (moon's paral- 
 
 Fij:. i6. lax); the radius of the 
 
 earthis 3,956 miles. Required thegeocentric distance of the moon. 
 
 8. The angle MET (the moon's apparent semi-diameter), is 
 
 31' 20" What is the diameter of the moon ? 
 
 (3) 
 
4 EXERCISES. 
 
 9. To find the point at which a perpendicular from an in- 
 accessible point will meet a given line. 
 
 10. In a right triangle are given R and r. To solve. 
 
 11. Find the volume of a cone of revolution whose vertical 
 angle is 75° 16' 23", and whose slant height is 2a 136. 
 
 1 2. To solve an isosceles triangle, when the base and r are 
 given. 
 
 13. Each side of the base of a regular heptagonal pyramid 
 is 10.307, each lateral edge makes with the base an angle of 
 39° 46' 19". Required the volume and surface of the pyramid. 
 
 14. To solve a right triangle, given R and B, or R and h. 
 
 CHAPTER HI. 
 
 1. Establish geometrically the formulae of (27-8-9). 
 
 2. Show that:— 
 
 (1) sin (30° + A) + sin (30° - A) =cos (60° + A)cos(60° - A) 
 
 = cos A ; 
 
 x^x -i A cotiA + tanA . cotiA + taniA 
 
 (2) sec4A= ^-— r-i ; secA=— -f-i — r — ?-*-: 
 
 ^ ^ ^ cotJA cot^A-tan^A' 
 
 (3) sin 3a7= sin x (4 cos'^cc— 1) = 3 sin a; - 4 sin^ 
 sin 4ic = sin £c ( .8 qos^x - 4 cos .^•), 
 
 sin bx= sin x { \% cos*a; - 12 co^^x + 1), 
 
 sin 6ic= sin x ( 32 cos^:^ - 32 cos^an- 6 cos a;), 
 
 sin 7a; = sin a; ( 64 cos^'a; - 80 cos*a; + 24 cosmic — 1), 
 
 sin 80;= sin x (128 cos'a;- 192 cos^it'H- 80 cos^a?-8cosa;); 
 
 (4) cos 3a; = 4 cos^a; - 3 cos x, 
 
 cos 4a: = 8cos*a;- 8cos^a;+ 1, 
 
 cos bx^ 16 cos^a; - 20 cos*.r + 5 cos x, 
 
 cos Qx= 32 cos^a; - 48 cos*a; + 18 cos^a; - 1, 
 
 cos 7a;= 64 cos'a; - 1 1 2 cos^'c + 56 cos^a; - 7 cos x, 
 
 cos 8a;- 128 cos^a; - 256 cos^a; + 160 cos*a; - 32 coslr + 1. 
 
EXERCISES. 5 
 
 Establish also the formulae for the tangents and cotangents 
 of multiples of x, 
 
 Gos^k=^-^ cos 2^ + — . 
 
 1 3 
 
 cos^^= -J- cos 3/: + — cos h, 
 
 cos*^= -r- COS 4^ + -^ cos 2^ + -^ , 
 
 L o 
 
 C0S^^=-Tr77 COS ^h-V—t COS 3^ + -5- COS h, 
 
 lb lb o 
 
 C0S^^=-:r7r COS 6^ + :j7; COS 4^ + r^ COS 2^ + :^ , 
 
 oz lb Iz lb 
 
 cos^^=-rr cos Ih + 7,-: cos 5^ + ^ : cos 3^ + tt. cos h, 
 b4 b4 b4 b4 
 
 1 1 7 7 35 
 
 - C0S^^=r7T^ cos %h + — COS GA; + — - COS 4^ + 77^ COS 'Ih + zrr^-- . 
 
 Izo lb oZ lb Izo 
 
 3. Show that : — 
 
 tan.x±tanw= — ^-^^ (1); tana; + cotw= ^-t^^- (3j; 
 
 cosxcosy cos^smi/ 
 
 ^ , ^ sin (w+x) ,^^ / , X l+tanrt-tan^ 
 
 cota;±cotw=^ — ^T (2); cos(a;± w) = --:- ^ (4^ 
 
 sm x sin 1/ ^ ^ ^ sec xaecy ^ ^* 
 
 What do (1), (2), (3), become when 
 
 x = a-^-, and ?/= a + 1-- ? 
 
 (5) sin (aj + y) sin {x-y)-= sin^cc - sin"^^/^ cos^i/ - cos'^x ; 
 
 (6) cos (aj + 2/) cos {x -y) = cos'^a^ - siii^y = cos^^/ ~ sin^^ • 
 From (6) deduce the value of cos 2x. 
 
 (7) (tan'a: - tsLii'y){cofx - coi^y)= ? 
 
 J au ^u ^ ^ 1 1-cosaj , ^, 1 + cosa; 
 
 4. fenow that tanj.7;= — : , and cot Jo; = 
 
 smic 
 
 Prove that tan2ic= — ^ — ; also that 
 
 COSiC + COSdx' 
 
 ^ sin 2x + sin 4x 
 tan 3a; = 
 
 tan 4x = 
 
 cos 2x + COS 4x ' 
 
 sin X + sin 3a; + sin 5a; + sin 7 
 
 cos X + cos 3a; + cos ox + cos 7a; 
 
 X 
 
tan 5x = 
 
 EXERCISES. 
 
 sin 2x -h bin 4x + sin Qx + sin 8a: 
 
 cos 2x 4- cos 4£P + cos 6x 4- cos Sx ' 
 Deduce the corresponding expression for tan nx. 
 Show that cot x= tan x + 2 tan 20? + 4 tan 4ir + 8 tan Sx. 
 
 5. Prove that : — 
 
 (1 ) cos (x + y) + sin (x - 2/) = -/ 2 sin ( Jrr + x) cos ( J-r ± ?/) ; 
 
 (2) cos i/= cos (x + 2/) cos ;c + sin {x ■\-y)smx; 
 
 (3) tan-^^ + cot-V2a; + 1)= 7- ; 
 ^ ^ ic + 1 4 
 
 (4) 2cot-'3 + cot-7 =-r; 
 
 /rx . 1 . 1 . A-x-y-xy TT 
 
 (5) tan-U- + tan-V + tan-^^; ^^ = 7 • 
 
 ^^ ^ l+x + y-\-xy4 
 
 6. Given sin 2.r = cos Sx ; find x and its functions. 
 
 7. From the functions of 30°, 45°, and 18°, find the func- 
 tions of all those angles between 0° and 90°, the expressions for 
 which involve no fraction less than 0.°5, and the calculation 
 of whose functions requires no other data. 
 
 8. Show that:— 
 
 W. ^ 2 tan x 2 r. r^ cos^a- + sin'ic 
 sin 2x = ~ = = 2-2 : — : 
 1 + tan-a; tan x + cot x cos x + sin x 
 
 cot'a;-! ^ . cot ccipl 
 
 (2) cos 2x= -— ~ = (l±sin2a:) -V ; 
 
 ^ ^ cot'a3 + l ^ ^cot^±l 
 
 X 1+sin^ + cosiP 
 
 , (3) cot -. = :; ; ^ 
 
 \ 2 1 + sin 0? - cos a; 
 
 i 9. Express, as the sum or difference of two functions, 
 
 (1), 2 sin 3a; cos 2 2/; (2) sin 3.r cos 5a; ; (3) sin 20° sin 30°. 
 
 ^. -.„ sin 3a;±sin 5a; cos 4;i;±cos 6.^' 
 
 bimpliiy =— ; — — : — -— . 
 
 cos 3a;±cos 5a; sm 4a;+sni bx 
 
 Prove that tan 5a; - tan 3a; - tan 2a; = tan 5x tan 3a; tan 2a;. 
 
 1 . Solve the following equations : — 
 
 (1) cos X - sin X = (cos x + sin a;)(2 cos^a; - 1); 
 
 (2) cos (a; + -5-) + 2 sin a; + 3 tan (x + —-) = cot'-^a; + esc x ; 
 
 (3) 2 tan a; + tan (a- a;) = tan (6 4- a;); 
 
EXERCISES. 
 
 (4). 2a sec X = (a -\- b^O- - tan x + sec x) ; 
 
 (5). ^/smx - CSC a; - l/ 1 - esc a; = esc x (sin x - 1). 
 
 11. Show that :— 
 
 sin A + sinB + sin (A + B) A B ... 
 
 r— :5 ^-)-r ^ =COt^ cot - (1): 
 
 sin A + sin B - sm ( A + B) 2 2 ^ ^ 
 
 tan 3A cot A (2 cos 2A - 1) = 2 cos 2A + 1 (2) 
 
 12. Sum the series, sinic + asin2ic + rt'^sin Sx + 
 
 Assuming that a? = cos m + V - 2 sin m, sum x + x'^ + x^ + ,, . 
 
 13. Show that, if C = 90°, 
 
 a + h a — b 4 sin^A 
 
 -2. 
 
 1 
 
 COSOJ 
 
 COS 2/ 
 
 
 
 COS a; 
 
 1 
 
 cos 2 
 
 
 
 cosy 
 
 COS 2! 
 
 1 
 
 COS w 
 
 
 
 
 
 COS 11^ 
 
 1 
 
 a - 6 a + 6 COS 2B 
 
 14. COS (x + y + z) + cos (^x + y — z) + cos (^x— y -^ z) + 
 
 _ cos(z + y-x') = ^ y__ 
 
 15. Show that if ^^^^^^tkavv-ao^. 
 
 = 0, 
 
 sin'''a5 cos'^w; = 1 - cost's - cos'-'y ~ cos'^o; + 2 cos ic cos y cos z. 
 
 16. Evaluate: — 
 
 1, 1, 1, since, sin 2/, sin 2, 
 sin aj, sin y, sin 2, cos a;, cosy, cos 2, 
 cos x, cos y, cos 2, , sin 2a;, sin 2y, sin 22, 
 
 1 7. Given, (1), sin £c - J sin 2x = m; 
 
 (2), cos x-^ cos 2a; = ^ : find a;. 
 18' Eliminate x and 1/ from the groups of equations: — 
 r cos X sec y-- (r: a) esc (a; -f y) 
 = [(t : 6) - sin x sec y] sec (^x + y) = tan 3/ (1)^ 
 
 r cos (x — 37/) = 2a cos^i/ 5 ^ sin (a; — 3?/) = 2a sin^y C^)- 
 19- Show that two geometrical progressions may be formed 
 
 firom the roots of 
 
 x^ - 2x cos a + 1 = 0, 
 x'- - 2x cos 2a + 1 = 0, 
 
 sin x, sin ?/, sin 2, 
 sin 2x, sin 2y, sin 22, 
 sin 4a;, sin 4y, sin 42, 
 
 or* - 2a; cos Tia + 1 = 0; 
 
EXERCISES. 
 
 the product of the two series being sin^ I — jcsc^ ( o I 
 20. Show that:— 
 
 a" 
 6 sin A 
 c sin A 
 
 
 h 
 c 
 
 21. 
 
 6 sin A 
 1 
 
 cos A 
 h 
 
 1 
 cos A 
 Show that 
 
 cosi (B-C) 
 cos 4 (B + C) 
 sin 4 (B + C) 
 
 c sin A 
 
 
 1 
 
 cos A 
 
 =0; 
 
 1 
 
 
 c 
 
 
 
 COS A 
 
 - 
 
 6- 
 
 1 
 
 
 
 e- 
 
 -1 
 
 cosC 
 
 cosC 
 
 -1 
 
 cosB 
 
 cosA 
 
 a 
 
 h 
 
 ccosA 
 
 a 
 
 h cos A 
 
 
 
 cosB 
 
 cos A 
 -1 
 
 0; 
 
 = a^ 
 
 equals 
 
 22, 
 
 cosi (C-A) cos^ (A-B) 
 cos|(C + A) cos|(A + B) 
 sinJ(C + A) sin| (A + B) 
 
 sin (B - C) + sin (C - A) + sin (A - B). 
 If A, B, C, be the angles of a triangle : — 
 cos A cos B cos C ^ 
 
 -: h -: ; h -; : ' = 2 
 
 sin B sin C sin A sin C sin B sin C 
 sin^C = cos"B + cos'^A + 2 cos A cos B cos C 
 sin A + sin B + sin C = (semi-perimeter) : R 
 e+h:a::e-h:c cos B - 5 cos C 
 & cos ( X - C) + c cos (X + B) = 6 cos (X + C) + c cos (X ■ 
 he + 4R^ = cos A + cos B cos C = 2 (Area) sin A + a^ 
 2R sin C = 6 cos A + \/{a' — b'^ sin'^A) 
 23. What is the shape of a triangle in which 
 
 B)? 
 
 (1); 
 
 (2); 
 
 (3); 
 
 (4); 
 
 B)(5); 
 
 (6); 
 (7). 
 
 a' + h' : ce - h' = sin ( A + B) : sin (A 
 
 24. In any right triangle, (A - 90°): - 
 
 sin 2B + sin 2C = 4 a^ + c^ (1); 
 
 sin 2B - sin 2C = cos 2B + cos 2C = (2); 
 
 sin^C + sin^B - (sin'^JB + sin'^C) = {h + c>+ 2a (3). 
 
 Express (3) in terms of cosines. 
 
 25. In any triangle, ABC, 
 
 6 - c : a : : sin ^(B - C) : cos ^A : : cos C - cos B : 1 + cos A; 
 6 + c : a : : cos J(B - C) : sin J A : : cos C + cos B : 1 - cos A; 
 a + i -I- c = (6 + c) cos A + (c + a) cos B + (a + 6) cos C ; 
 (cos A -f 6 sin C) + (cos B + c sin A) + (cos C -r a sin B) = 
 2 (sinA + 2sinB)-r-(a + 2i»). 
 
EXERCISES. 9 
 
 26. If i,j, h be the altitudes of a triangle, 
 
 a sin A + i sin B + c sin C = 2(icos A +j cos B + ^cos C) (1); 
 sinA = sin(2A4-B + C) = -cosK3A + B + C) (2); 
 
 tan (B - 2C) = - tan (A + 2C) (3); 
 
 csc(P-C) = sec(iA + C);tan(A + 2B + 3C)=tan(A-C)(4); 
 cot ^(B + C) cot ^(C + A) -f cot J(C + A) cot i(A + B) 
 
 + cot |(B + C) cot ^(C + A) = 1 (5). 
 
 27. Extend to quadrilaterals (and to polygons in general), 
 the relation a^= fe^ + c'^ — 26c cos A. Extend also the relation 
 
 a = b cos C + c cos B. 
 
 28. The roots of a cubic are the tangents of the angles of 
 a triangle. Required the form of the cubic. Investigate also 
 the cubic whose roots are sines and cosines. 
 
 EXERCISES IN SOLVING PLANE TRIANGLES. 
 
 1. To solve a triangle ABC, given: — 
 
 (1) a = 237.062, h = 127.268, A = 27° 32' 16". 
 
 (2) a = 156.82. 6 = 284.072, c= 192.007. 
 
 2. To solve a triangle ABC, given: — 
 
 _ (1) a, 6,A±B 
 
 U^(2) A,a,6^±c^ 
 
 (3) A, C, a±6: 
 
 Jk- (4) c,A,a'^±6^: 
 
 (5) The altitudes; 
 
 (6) The medians ; 
 
 (7) The bisectrices of the angles ; ^^ 
 
 (8) The sides of the Pedal triangle ; 
 (9) The perimeter, an altitude and a side ; 
 
 (10) The radius of the inscribed circle, a±h, and c; 
 
 (11) The radius of the inscribed circle, the radius of the 
 circumscribed circle, and one altitude. 
 
 3. The distance between two buoys, A and B, being required, 
 a base line, CD, of 300 yards is laid oif, and the angles, CDA, 
 ( = 28° 19'), ADB( = 62°16'), ACB ( = 57°38'), and BCD 
 ( = 38° 52') are measured. Required, AB, and the angle CBD. 
 
 4*. Deduce expressions for the side, apothem and area of 
 each regular polygon inscribed in a circle whose radius is 10, 
 up to and including, the dodecagon. 
 
10 EXERCISES. 
 
 5. An aeronaut observes the angles of depression of two 
 towers on opposite shores of a lake, and 12 miles apart, to be 
 47° 58' and 33° 25', respectively. What is his altitude? 
 
 6. An engine passes a station, C, that is 8 miles distant, 
 and that beai-s S 32° 17' W, at 10 A. m.; moving continuously, 
 it passes B, N 68° 26' W, and 5 miles away, at 10 : 20 a. m. If 
 the road BC is straight, what is the average rate of speed of 
 the engine ? 
 
 « n, 1 , /T^- rf ON ^-r^ c sin A sin B 
 
 7. Show that, (Fig. 7, 8), CD= — ^— ^- — ^— ; express the 
 
 area of ABC in terms of c. A, B. 
 
 8. The area of the largest circle that can be cut from a 
 triangular block, is 1312. The area of the block is 2018, and 
 the angle at one corner is 57° 10'. Required the dimensions of 
 the block. 
 
 9. If ABC, DEF, have a = d,b = e, and C = 180°- F, they 
 are equivalent. Generalize this statement. 
 
 10. To determine the distance from D to an invisible, inac- 
 cessible point, B. 
 
 11. Show that in any triangle, ABC : — 
 cot^A-rCot JB + cot|C = s-^r; 
 
 sin A + sin B + sin C = s-^ R = sin A sin B sin C • 2 R-7-r ; 
 tan A + tan B+tan C =^ tan A tan B tan C ; 
 cos^A + cos^B + cos'^C = 1 - 2 cos A cos B cos C. 
 
 12. A line AD is drawn from A to BC. Compare the 
 ratios D AC : DAB, and DB : DC. 
 
 1 3. To find the area of an inaccessible field, knowing a base 
 line KL, and the angles subtended at K and L by the sides of 
 the field. 
 
 17. Express the area of a quadrilateral in terms of the 
 diagonals and their included angle. 
 
 18. Determine a trapezoid, knowing the sides a, b, c, d; a 
 being parallel to c. 
 
 19. B is due south of A. An observer at C notes the re- 
 spective intervals (6 seconds,. 9 seconds), between the flashes of 
 
EXERCISES. 11 
 
 the sunset guns at B and A, and the times at which the reports 
 reach him. AB subtends to him an angle of 68° 29'. How 
 far is A from B ? 
 
 20. A circular arch, 98 feet in (curvilinear) length, rests on 
 piers 16 feet high and 82 feet apart (from centre to centre). 
 
 ,What is the height of the highest point of the arch? 
 
 21. Three circles, of radii 8, 12, and 20, respectively, are 
 tangent, each to the other two. Required the perimeter and 
 area of the enclosed space. 
 
 22. Find the distance in space described in one day, in con- 
 sequence of the rotation of the earth, by the Great Equatorial 
 at the Lick Observatory of the University of California, at ^It. 
 Hamilton, (Lat. 37° 20' 22".51 K Long. 121° 38' 35".75 W). 
 
 23. Find the distance between the orthocentre and the cen- 
 tre of the circumscribed circle, (of ABC), in terms of the 
 radius of the latter and the angles of ABC. 
 
 24. Express, in terms of the radii, the distance between the 
 centres of the inscribed and circumscribed circles. Express, 
 (in terms of a, h, c), the distances between the centres of the 
 escribed circles. 
 
 25. Show that the area of ABC = 4Rr cos |A cos JB cos |C. 
 20. Circles passing through the centre of the circumscribed 
 
 circle, and having a, h, c as chords, have for radii k, I, m). 
 Prove that {(i^k) + {h^l) + {c-^ m) = 8 sin A sin B sin C. 
 
 27. Show how to determine the height of a cloud from its 
 angle of elevation and the angle of depression of its reflection 
 in a lake h feet below the level of the observer. 
 
 28. A man sees two spires, the tops of which seem to be in 
 a straight line whose angular elevation is m ; the angles of de- 
 pression of the reflections in a lake h feet below him are B and 
 C. Determine the distance between the spires, and the height 
 of each. 
 
 29. In any triangle, E = 
 
 J(a cot A CSC B CSC C+ 6 cot B esc A esc C+c cot C esc B esc A). 
 
12 • EXERCISES. 
 
 BO. Determine the radii (/,/',/'' )of the-escribed circles, in 
 terms of the sides. Hence show that 
 
 - + —+-777 = (1); area=-/rrVV" (2); 
 
 r r' r area ^ ' 
 
 / + /' +r'"-r = 4i^(3). 
 
 31. A, walking along a straight road, observes that the 
 greatest angle that two distant spires subtend at his eye is a\ 
 walking c rods farther he observes that they are in a line mak- 
 ing an angle ft with the road. Find the distance between the 
 spires. 
 
 32. A spire S is observed from A and B. If the angles 
 SAB and SBA, the distance AB, and the angle of elevation of 
 the spire at B be known, find the distance of the spire from A 
 and its height above the level road AB. 
 
 33- The angular elevation of a hill, as observed from a 
 point due north of it, is A; as observed from a point due west 
 of the first and m miles distant across a lake, the angle of ele- 
 vation is B. How high is the hill ? 
 
 34. An observer at sea takes the angles subtended at his 
 position, (P), by the known shore distances, AB, AD, DB. 
 Find AP, DP, BP, when APD = 42° 17', BPD = 36° 25', AD = 
 2100, BD = 1672, AB = 2867. 
 
 35. To determine the geocentric distance of a heavenly 
 body M, knowing the zenith distances of M at two stations of 
 which the latitudes are known, and which are situated on the 
 same meridian. 
 
 36. To determine the area of the greatest pentagonal star 
 that can be cut from a circular disc, four inches in diameter. 
 
 37 . The area of the base of a regular quadrangular pyra- 
 mid is 289 square feet. Find the area of a section inclined at 
 an angle of 26° 37' to the base, one of the sides of the section 
 being parallel to the base, and at a distance therefrom equal to 
 two-thirds of the altitude. 
 
 38 . The sides of an inscribed quadrilateral are a, h, c, d. 
 What is the radius of the circumscribing circle? 
 
EXERCISES. 13 
 
 |>9. If a triangle be inscribed in ABC, its perimeter must 
 lie between a-\-b + G and a cos A -f ^ cos B + c cos C, ( = 4R sin A 
 sin B sin C). 
 
 CHAPTER V. 
 
 1. Solve the spherical triangle ABC, given: — 
 
 (1) a =106° 17' 22"; ft = 120° 08' 20"; A= 39° 52' ; 
 
 (2) A = 129° 05' 28" ; B = 140° 25' ; C = 110° 27' ; 
 
 (3) a= 70° 32'; b= 38° 25' ; C= 51° 17' 24"; 
 (4)A=49°37'; B=126°; ft=110°17'; 
 (5) C= 50° 32'; B= 136° 15' 28" ; • a= 68° 35' . 
 
 2 . To solve a spherical triangle, given 
 
 c, B, and A + C, or c, B, and a + c. 
 
 3. Establish the propositions concerning spherical triangles 
 usually given in the text-books in Geometry, by means of the 
 formulse of chapter V. 
 
 4. Prove directly from a figure that 
 
 sin a : sin c : : sin A : siaC. 
 
 5. Show that the values given (47) for sin|^, tan ^a, etc., 
 are real. 
 
 6. Show that sin a : sin A : : 
 
 sin b sin ciV^l - cos'^a - cos'^b - cos'^c + 2 cos a cos b Cos c. 
 
 7. Sliow that the sum of any two angles of a spherical tri- 
 angle is less than 180° + the third. AYhat property of a right 
 triangle is suggested hereby ? 
 
 8. If OX, OY, OZ, be mutually at right angles, and any 
 two lines be drawn through O, 
 
 (1) The sum of the squares of the cosines of the angles that 
 each makes with OX, OY, and OZ, respectively, will equal 
 unity. 
 
 (2) The cosine of the angle between the two lines will equal 
 the sum of the products of tlie cosines of the angles made by 
 the two lines, with OX, with OY, and with OZ, respectively. 
 
14 EXERCISES. 
 
 9. If CD be a median in ABC, 
 
 cos AC + cos CB = 2 cos |AB cos CD. 
 
 10. If two angles of a spherical triangle be equal to the 
 opposite sides respectively, the remaining angle will be equal or 
 supplementary to the third side. 
 
 11. D is any point in AB ; show that 
 
 cos AD sin BC = cos AB sin DC + cos AC sin DB. 
 
 12. If arcs be drawn from the vertices of any spherical 
 triangle through any point, the product of the sines of one set 
 of alternate segments of the sides will equal the product of the 
 sines of the other set ; and conversely. 
 
 IB. The last proposition proves the concurrence of what 
 sets of angle-transversals of a spherical triangle ? 
 
 1 4 . If an arc cuts the sides of ABC, in D, E, and F, re- 
 spectively, what relation obtains between the two sets of alter- 
 nate segments ? 
 
 15. If ^, /, m, be the altitudes of a spherical triangle, 
 
 sin k sin a = sin Z sin b = sin m sin c ; 
 also, cos m = esc c V cos^a + cos^ft - 2 cos a cos b cos c. 
 
 1 6. San Francisco is 122° 25' 40." TCd W., 37° AT 22."55 N., 
 and Yokohama is 139° 40' 27" K, 35° 26' 50" N.; what is the 
 shortest sailing distance between the cities? What is the 
 shortest distance on the surface of the earth between Mount 
 Hamilton and Annapolis', Md.? (See Ex. 22). 
 
 17. Show that in any right spherical triangle (A = 90°). 
 
 sin^Ja = sin"-J6 cos'^-Jc + cos%6 sin^Jc. 
 
 18. If ^, /?, y^ be the lengths of the angle-bisectrices of 
 ABC, 
 
 cot a cos ^A + cot f3 cos JB + cot y cos ^C = cot a + cot b 4- cot c. 
 
 19. Determine the arcual radii (polar distances), of the 
 circumscribed, inscribed, and escribed circles, of the triangles 
 formed by the intersection of any three arcs (of great circles), 
 on the sphere. 
 
EXERCISES. 15 
 
 20. ABC is equilateral, P the pole of its circumscribing 
 circle, and Q any point on the sphere. To show that 
 
 cos QA + cos QB + cos QC = 3 cos PA cos PQ. 
 What is the analogous formula for the plane triangle ? 
 
 21. The angle of elevation of the top of a cliff, A, is 
 23° 27' 12" ; the angle of elevation of a cliff, B, is 10° 28' 53" ; 
 the angle subtended at the point of observation by AB is 
 73° 47' 18". What is the horizontal angle between A and B ? * 
 
 22. ABC is trirectangular ; P is the pole of its circum- 
 scribed circle, and Q any point on the sphere. Show that 
 
 (cos AQ + cos BQ + cos CQ)"'= 3 cosTQ. 
 2 B. In any triangle, ^(a + c) and J(A + C) agree in species. 
 
 24. If r>, E, F, be the feet of the altitudes of ABC, 
 tan BD . tan CE . tan AF --= tan DC . tan E A . tan FB. 
 
 25. Through any point, in the side of a spherical triangle, 
 to draw an arc (of a great circle), that shall cut off a given 
 fracti(>n of the area. 
 
 2 6 . Three circles are described in a triangle, each angle of 
 which is 120°, so that each circle touches the two others and 
 two sides of the triangle. Find the arcual radii. 
 
 27. To find the angle formed by two adjacent faces of a 
 regular polyedron. 
 
 28. To find, for any regular polyedron, the radii of the in- 
 scribed and circumscribed spheres. 
 
 29. To find, for any tetraedron the radii of the -circum- 
 scribed, the inscribed, and the escribed, spheres. 
 
 30. To express the volume of any tetraedron, (and hence 
 of any parallelopiped), in terms of the edges. 
 
 31. If, wath the vertices of any parallelopiped as centres, 
 equal spheres be described, the sum of the intercepted portions 
 of the parallelopiped will equal the volume of any one of the 
 spheres. 
 
 *The solution of examples of this character is called "reducing angles to the 
 horizon." 
 
CHAPTER VI. 
 
 1. Show that : — 
 
 (1) cosh^cc - cosh'-i/ = sinh'^a? - sinh'''2/= sinh (x + y) sinh {x - y) 
 
 (2) cosh^^ 4- sinh^o; = cosh^a; -f sinh^y = ^^sh (x + y) cosh (x - y) ; 
 
 ^^, , X sinha? . . i ^* r^ , 
 
 (d) tanh ^=- :; i — = coth a; - cscn x^ coth r - 2 csch a; ; 
 
 ^ ' 2 1+ cosh X 2 
 
 1 + tanh X tanh y 
 
 (4) cosh(a; + y) •= -. ^; 
 
 ^ ' ^ ^' sech re- sech y 
 
 (5) cosh (n + 2)^ = 2 cosh k cosh (?i + 1)A; — cosh nk. 
 
 What are the corresponding formulse for circular functions? 
 
 2. Prove that 
 
 \ a (p + qx^) y p{a + hx') 
 
 What is the corresponding formulse for inverse hyperbolic 
 functions ? 
 
 3. Show that 
 
 "/ (a + 1) cos J- a? + V{ct - h) sin J x 
 
 In ———-—- = 
 
 V {a + ^) cos -J a; - V{a - h) sin ^ x 
 
 2 tanh~^ \ , tan |- x 
 (Va + 6 
 
 Find corresponding logarithmic equivalents for 
 
 tan~^ -! tan \x\- . and tanh~^ -^ . tanh -^ x 
 
 (l^a + 6 y (l/a+6 
 
 4. Show that : — 
 
 ( 1 ) cosh (In ?/) + sinh (1 n ?/) = 2/ ; 
 
 (2) tanh (In y)^ {if - 1) + {f + 1). Thence show that 
 
 hi2=6.9;jl4718 + . 
 
 (3)co^.„^'[±|)=^;tanh(.n^[±i)=, 
 
 (4) If f=- 1w - 1, tanh (In t/) = l-w-\ 
 
 (5) cosh-^(sec x) = tanh-^(sin x) = 2 tanh-^(tan \ x) 
 
 (16) 
 
 , 1 + tan -A- a; - , / - \ 
 
 = ln - — - — f— Intan^ 13 + ^ • 
 1-tan^^ "\ 2 / 
 
EXERCISES. 17 
 
 (6) If tan-Jcc =2, "^ 
 
 sinli a:= .j ; cosh x = . 
 
 1-2^ 1 - z^ 
 
 (7) OO.I.-- f-+^- = sinh-- l aV + -2abx + ac 
 
 v¥ -ac \ 6- - ac 
 
 (8) In tan ^ . 
 
 a — a? a 
 
 (9)ln^:±^^^=tanh-^. 
 
 a;-^_^-/2 + l 1 + cc^ 
 
 5. Find the hyperbolic functions of J-i, ^-i, and y\7~i. 
 
 6. Calculate directly sinh 1 and cosh 1 ; thence show that 
 
 e = 2.71828182845904523536 + . 
 Show also that 
 
 sin 1 = 0.841470984807896 + ; 
 cos 1 = 0.540302305868039 + . 
 
 7. Apply De Moivre's theorem to hyperbolic functions so as 
 to deduce, for the powers of sinh k and cosh k formulae analo- 
 gous to those given, for circular functions, on pages 4, 5. 
 
 8. Show that :— 
 \nsmhk=^k-\n2-e-''-ie-''-ie-'^-ie-'^-.. . . .; 
 lncoshk = k-\n2 + e-''-ie-'^ + le-^-ie-'' + 
 
 Thence derive the development of In tanh k. 
 
 9. Show that 
 
 cosh (£c + ^) = cosh k + sinh kj + cosh k— + sinh.^- .57 + .... ; 
 
 sinh (a; + ^) = sinh k + cosh h~-\- sinh ^ t^ + cosh X;— + 
 
 What are the corresponding developments for the circular 
 functions ? 
 
 10. Prove the truth of the following equations : — 
 cosh 2aj = 2 cosh-oj - 1, 
 
 cosh 3ic = 4 cosh^o; - 3 cosh x, 
 
 cosh 4:c = 8 cosh% - 8 cosh^a; + 1 , 
 
 cosh 5^ = 16 cosh^rc — 20 cosh^.?; + 5 cosh x. 
 
18 EXERCISES. 
 
 cosh6ir= 32 cosher- 48cosh*a;+ IBcosh^a;- 1, 
 cosh 7x = 64 cosh'x - 1 1 2 cosh^ic + 56 cosh^a? - 7 cosh x, 
 cosh So; = 128 cosh'o; - 256 cosh'^ + 1 60 cosh*a; - 32 cosh=^:c + 1. 
 Establish the corresponding formulae for the hyperbolic sine, 
 and tangents. Express also sinh^o:, sinh^ic, cosh^:c, cosli^a;, etc., 
 in terms of functions of x. Compare Ex. 4-, page 5. 
 
 11. If ?i = 2r, .T = a + ftcot<9, 
 
 (A + I'B) {x - a + hiy + (A - iB)(a; - g - 6i) » _ 
 \{x-ay + b'\^' ~ 
 
 A cos n^' - B sin ?i^ , 
 
 \ix-ay + hY 
 
 12. Show that, as 
 
 In (cos k + i sin k) ^ ki, 
 
 the logarithm of unity has an infinite number of values, of 
 which all but one are imaginary. Thence show that any 
 number has an infinite number of logarithms for any given 
 base. 
 
 1 3. Show that 
 
 ."-(.-;)(.-,4).... 
 
 This may be shown l)y proving, first, that the equation 
 sin ^ = can have no imaginary roots, and then showing that 
 the series (see page 49) for sin 6 is exactly divisible by 6±n-, 
 n being any integer. Thus sin 8 is shown to be a product of 
 the form 
 
 a^(^+r)(^-r)(/9+2rr)(^-2ir)(^+3T:)(<9-3-) 
 
 in which a cannot contain ^, since the equation sin^ = has 
 no imaginary roots. Finally it may be shown that a = l, 
 whence we may factor sin 6 as shown, cos 6 may, of course, 
 be treated similarly. 
 
 From these expressions for sin ^ and cos ^ we may deduce 
 interesting forms of algebraic statement for the other expo- 
 nential functions. 
 
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