first of these equa ference of the sines; it follows, that the tions converted into an analogy, becomes sin (A — B) : sin A-sin B; sin A + sin B That is to say, the sine of the difference angles, is to the difference of their sines, as the sum of those sines is to the sine of their sum. sin (A + B) (XX.) of any two arcs or If A and B be to each other as n + 1 to n, then the preceding proportion will be converted into sin A: sin (n + 1) ▲ — sin na:: sin (n + 1) A + sin na sin (2n + 1) A (XXI.) These two proportions are highly useful in computing a table of sines: as will be shown in the practical examples at the end of this chapter. = - 25. Let us suppose A + B = ▲', and a B = B'; then the half sum and the half difference of these equations will give respectively A = (A+B), and B (A'B'). Putting these values of A and B. in the expressions of sinA COSB, sin B. Cos A, COS A. COS B, sin A sin B, obtained in arts. 21, 22, 23, there would arise the following formulæ : = sin(A+B). COSA - B sin(A+B). sin བ། R(Sin Asin B'), R(COS A+COS B), R(COS B - COS A'). Dividing the second of these formulæ by the first, there will be had = sin(A-B') cos (A+B) sin(A-B') cos (A+B) sina-sins' sin A+B) cos(A-B ́) (a'—B ́)` sin}( a' + B) Sina+sinв' But since = it follows that the two sin = tan and R factors of the first member of this equation, are This equation is readily converted into a very useful proportion, viz. The sum of the sines of two arcs or angles, is to their difference, as the tangent of half the sum of those arcs or angles, is to the tangent of half their difference. 26. Operating with the third and fourth formulæ of the preceding article, as we have already done with the first and second, we shall obtain (A'B') B') COS B -COS A tan (A Making B = 0, in one or other of these expressions, there tan A tan2 A 1 + cos A 1-cos A These theorems will find their application in some of the investigations of spherical trigonometry. 27 Once more, dividing the expression for sin (A ± B) by that for cos (A+B), there results sin A sin (A + B) : then dividing both numerator and denominator of the second fraction, by cos A cos B, and recollecting that shall thus obtain sin COS (XXIII.) R2 tan A tan B which, after a little reduction, becomes tan Atan B cot A. Cot BR3 cot (AB)== .... (XXIV.) cot Bcot A 28. We might now proceed to deduce expressions for the tangents, cotangents, secants, &c. of multiple arcs, as well as some of the usual formulæ of verification in the construction of tables, such as sin(540+A)+sin(54°-A)-sin(180+A)-sin(180-A)=sin(90°-A); Sina+sin(36"-a)+sin(72o+a)=sin(36°+^) + sin(72o — A). &c &c. But, as these enquiries would extend this chapter to too great a length, we shall pass them by; and merely investigate a few properties where more than two arcs or angles are concerned, and which may be of use in some subsequent part of this volume. 29. Let 29. Let A, B, C, be in any three arcs or angles, and suppose radius to be unity; then sin A. sin c + sin B. sin (A + B + C). sin(A+B) sin (B+c)=For, by equa. v, sin (A+B+C) = sin A COS (B+c) + cos a . sin (BC), which, (putting cos B COS C-sin B. sin c for cos (B+c)), is = sin A .cos B. cos c-sin A.Sin B. sin c+ COS A. Sin (B+c); and, multiplying by sin B, and adding sin A sin c, there results sin a. sin c + sin B. sin (A+B+C) = sin A. COS B cos c. sin B + sin A sin c. cos2 B + COS A. sin ■ . sin (B+c)= sin A. cOS B. (sin в . cos C+COS B. sin c) +cos A. sin B. sin (B+c) = (sin A. cos B + COS ▲ · Sin B) × sin (BC) sin (A+B). sin (B + c). Consequently, by dividing by sin (A + B), we obtain the expression above given. In a similar manner it may be shown, that sin (B - c = 30 If A, B, C, D, represent four arcs or angles, then writing C+D for c in the preceding investigation, there will result, sin A. sin(c+D) + sin B. sin(A+B+C+D) sin (A+B) sin (B+C+D)= A like process for five arcs or angles will give sin A sin (C+D+E) + Sia B. Sin(A+B+C+D+E) (A+B) sin(B+C+D+E)= And for any number, A, B, C, &c. to L, sin (B++....L)= sinA.SIN(C+D+..... L)+S:NB.SIN(A+B+C+ ..L) sin (A+B) COS B, 31. Taking again the three A, B, C, we have sin A. COS B-Sin B COS A. Multiplying the first of these equations by sin a, the second by sin B, the third by sin c; then adding together the equations thus transformed, and reducing; there will result, sin A. sin (B-C)+sin B sin (ca) + sin c sin (A— B) =0, COS A. Sin (B-C)+ cos B. sin (c~A) + cos c. sin (A—B)=0. These two equations obtaining for any three angles whatever, apply evidently to the three angles of any triangles 32. Let the series of arcs or angles A, B, C D . contemplated, then we have (art. 24), .... E, be sin sin (L + A). sin (L-A) sin? L- sin2 A. If all these equations be added together, the second member of the equation will vanish, and of consequence we shall have sin (A+B) sin (A-B) + sin (B+c) sin (B-C) + &c... + sin (L+A) + sin (L - - A) = 0. Proceeding in a similar manner with sin (A-B), COS (A + B), sin (B-C), Cos (B+c), &c. there will at length be obtained COS (A+B) sin(A — B) + CUS (B+C). sin (в - c) + &c ... ..... A +cos (L+A). sin (L—A) = 0. 33. If the arcs A, B, C, &c .... L form an arithmetical progression, of which the first term is 0, the common difference D', and the last term L any number n of circumferences; D', B + C = 3D′, then will B-A=D', C—B = D', &C. A + B &c and dividing the whole by sin D', the preceding equations will become = (XXV.) sin D' + sin 3D + sin 5D' + &c: = 0, cos D' + cos 3D' + cos 5D' + &c. If E' were equal 2Dʻ, these equations would become sin D'+ sin (D'+E') +sin (D'+ 2e') + sin (D'+3E')+ &c= 0, COS D'+ COS (D'+E') + cos (D'+2E') + cos (D'+3E')+ &c = 0. 34. The last equation, however, only shows the sums of sines and cosines of arcs or angles in arithmetical progression, when the common difference is to the first term in the ration of 2 to 1. To investigate a general expression for an infinite series of this kind, let s + sin a + sin (A + B) + sin (A + 2B) sin (A +3B) + &c. Then, since this series is a recurring series, whose scale of relation is 2 cos B 1, it will arise from the developement of 22. cos + z2, making a fraction whose denominator is 1 z = 1. Sina+[sin(A+B) − 2 sina. cos B). 1-22.COS B + 22 Now this fraction will be = 2-2 COS B ; and this,because 2 sin A. COS B Sin (A+B) + sin (A-B) (art. 21), is equal to 2(1-CUS B) sin COS B sin (B), by art. 25, it follows, that sin A-sin (A-1)= 2 cos (AE) sin B; besides which, we have 1 2 sin2 B. Consequently the preceding expression becomes sin Asin (A+B) + sin (A +28) + sin (A + 3B) + &c. ad infinitum = COS (A-3B) .... 2 sin 42 (XXVI.) 35. To find the sum of n + 1 terms of this series, we have simply to consider that the sun of the terms past the (n+1)th, that is, the sum of sin [▲ + (n + 1) B} + sin ( a + (n + 2}B} + sin (A+(+3) B)+ &c. ad infinitum, is, by the preceding theorem,= cos [A+(2+1)B]. Deducting this, therefore, from 2 sin B the former expression, there will remain, sin a + sin (a + B) + sin (A + 2B) + sin (a + SB) + . . . sin (A † NB) = Cos(AR)-CsA+(n+1)B]_sin(+B)sin(n+1)B 2 sin B sin B (XXVII.) By like means it will be found, that the sums of the cosines of arcs or angles in arithmetical progression will be COS A + COS (A + B) + COS (A + 2B) + cos (A + 3B) + &C.. sin (A - 3B) ad infinitum COS A † COS (A + B) ..... (cos A † NB) = 2 sin B Also, ... (XXVII.) + CƠS (A + 2B) + cos (a + 3B) + = sin B (XXIX.) 36. With regard to the tangents of more than two arcs, the following property (the only one we shall here deduce) is a very curious one, which has not yet been inserted in works of Trigonometry, though it has been long known so mathematicians. Let the three arcs A, B, C, together make up the whole circumference, O: then since tan (A + B) = R2 (tan A+ tan B), ·(by equa.xxIII), we have R2 x(tan A+tanв+ tán c) = R3 X [tan A + tan в-tan (A + B)]=R2 × (tan A+ The result therefore is, that the sum of the tangents of any three arcs which together constitute a circle, multiplied by the square of the radius, is equal to the product of those tangents. . (XXX.) Since both arcs in the second and fourth quadrants have their tangents considered negative, the above property will apply to arcs any way trisecting a semicircle; and it will thereVOL. II. fore |