20. Other useful expressions for the sines and cosines of multiple arcs, may be found thus: Take the sum of the expanded (BA) and sin (BA); that is, add. to. ........ expressions for sin sin (B+ A) = sin B COS A+ COS B Sin A ..sin (B A) sin B COS A COS B Sin A there results, sin (B + A) + sin (BA) = 2 cos a sin B, So again, the sum of the expressions for cos (B+ A) and cos (B A), is COS B+A) +COS (B A) = 2 COS A COS B. COS (B Whence, cos B+A =2 cos a cOS B A). Substituting in these expressions for the sine and cosine of B + A, the successive values of a, 2a, 3a, &c. instead of B, we shall have, sin 2A 2 cos A sin A sin 3A 2 cos A sin 2A sin ᏎᎪ cos 2A= 2 cos A Cos A cos na = 2 cos A Cos (n 21. If the cosine of a be represented by a particular binomial, the formula (K) will be transformed into a class of very elegant and curious theorems: thus then 2 cos 2a = 2 (2 cos 2A − 1) = 2 [} (≈ + 4)* − 1] = x2 + A like substitution in the forms for 2 cos 3A, 2 cos 4a, &o. will give 22. By an inverted process, the sine and cosine of a single arc may be inferred from those of a double arc; or, which comes to the same, those of a half arc from those of a whole arc. Let x = √(1 cos 2A), or x2 2xzz21 cos 2A, and assume x2 + z2 = 1; then 2xz = cos 2a; and, exterminating z, x2 + and 22= cos2 2A = 1. sin 2A, and x = √(1± sin 2a) sin 2a). 3 sin 2A). COS § √(1 Hence, sin A = - (1 + sin 2A) — ¦ √(1 − sin 24 cos A1 + sin 2A) + √(1 3 √(1 cos A = √(1 + sin A) += sin^) (3.) * This remarkable class of formulæ may be of use in the summation of series. Thus, cos a + cos 2A + cos 3A + cos 4A + &c. ..cos na, is equal to the sum of the two series (≈ + z3 + z3 + ployed to determine the sum of all the natural cosines of every minute in the quadrant, gives sin 45 x cos 45° 0′ 30′′ sin 30" 3437 2470374. 23. Otherwise, in order to obtain expressions for the sine and cosine of a half arc, take 1 = cos 2A + sin 2A cos 2A= cos 2A sin 2A (see formulæ D), the sum and difference of these respectively, give 2 cos 2A = 1 + cos 2A, and 2 sin 2A = 1 Changing A and 2A into A and A, we shall have 2 cos2 A = 1 + cos A, and 2 sin2 =1 2 cos 2A. cos A, COS A (N.) Suppose, for example, A 60°, then cos a = sin 30° = 1, and cos 30°√(2 + 1) = { √3 = ·8660254, sin 30°(2-1), as it ought to be: cos 15° = √(2 + √3) = {{√6 + √2) = •9659258 sin 15 = √(2 √3 ) = 1 ( √/6 = /2)=2588190: and so on, by continual bisections, as low as we please. 24. If the formula (c) be divided one by the other, there will result, sin (A+B) COS (A + B) sin Acos B± cos B Sin A COS A COS BF sin A sin B Here dividing the two terms of the second member of the equation by cos a cos B, and recollecting that If A in this formula be 45°, we shall have tan A = 1, and consequently, If A = B, the formula (o) will give for the double arc, While, on the other hand, dividing the first values of sin A, cos A (formula N) one by the other, there will result, and so on, to any proposed extent. √3; 25. Pursuing an analogous process to that by which we obtained tan (A + B), the following may be deduced from formulæ (c): viz. cot (AB): cot A Cot B = 1 cot Bcot A 26. Taking again the equations (c) for the sine of the sum and the difference, there results, by add. sin (A + B) + sin (A — B) by sub. sin (A+B) sin (A 2 sin A COS B B) = 2 sin B COS A. The first of these formulæ divided by the second If, therefore, we suppose, A + B = A', A whence A (A′ + B ́), B = = B = B', (A'B'), we shall have But, it has been seen (formula 2) that The preceding equation, therefore, establishes the truth of the following, viz. 27. Similar computations with the equations (c), combined 2 and 2, would produce a great variety of other formulæ. From these, however, since the ingenious student may investigate them nearly as rapidly as he can write them down, we shall only select the following for insertion here. sin A + sin B2 sin B) (A + B) COS § (A (v.) sin A sin B = 2 sin § (a B) COS A + COS B = 2 cos 3 (a + B) COS § (A — B) 1 + sin B = 2 sin (45° + §B) cos (45° — §B) = 2 sin2 1 (45° + B) sin B = 2 sin2 (45° — }B) = 2 cos2 (45° + }B) (v.) sin B COS A sin B COS A+ COS B =cot (A+B) tan (A-B) =- cot (A+B) cot (A - B) = tan2 (45° + B); 1 + cos A -COS A - sin B 1 - СОЗ В sin2 (45° + B) 1 28. Some useful theorems in the computation of sines, cosines, &c. result from the introduction of the imaginary quantities that arise from the solution of the equations A2 + 1 = 0, and в2 + 10: so that, though these roots, A = ± √− 1, and в = ± √-1, are in |