- B') sin B', cos B', their values, there will result sin (A sin A sin B COS A COS B. Hence, because B′ = 10-B, we have sin (A - B') = sin (A + B = 10) sin [(A+B) 40]=sin [1Ò − (A + B) = -COS (A+B)]. This value of sin (A B') being substituted for it in the equation above, it becomes cos (A+B) COS A COS B sin A sin B. 16. If B in this latter equation be made subtractive, sin B will become sin B, while cos B will not change (art. 3, 4). The equation will consequently be transformed into this: viz. COS (AB) COS A COS B + sin A sin B. If A and B be regarded as arcs instead of the angles which they measure, the results will be equally conclusive and correct. They may be expressed generally for the sines and cosines of the sums or differences of any two arcs or angles, by these two equations, viz. 1} (0.) sin (AB) sin A COS B sin B COS A COS (AB) COS A COS B sin a sin 17. The actual value of the sine, cosine, &c. depends, obviously, not only upon the magnitude of the arc, but also upon that of the assumed radius. In the preceding investigation we have supposed it to be unity. If we wish to make the above or any other formule, applicable to cases where the radius has another value R, we for sin A, have only to substitute in the expression, COS A R tan A sin A for tan A, and so on. R Or, gene for cos A, rally, we must so distribute the several powers of R, as to make all the terms homogeneous, as to the number of lines multiplied: this is effected by multiplying each term by such a power of R as shall make it of the same dimension, as that term in the equation which has the highest dimension. Thus the expression, sin 3A 3 sin A 4 sin 3A (rad = 1) when radius is assumed = R, becomes 18. Supposing that in formula (c) A = E, and taking the superior sign, we shall have -- 2 cos 2A sin 2a = 2 sin A cos a = 2 sin a √(1 COS 2A. sin 2A) 1 (D) cos 2A for sin 2▲, and 1 sin 2A for So, again, by supposing в to be successively equal to 2A, 3A, 4A, &c. we shall find sin 3A sin A cos 2A + cos A sin 2A sin 4A sin A cos 3A + cos A sin 3A (E.) sin 5A = sin A cos 4A + cos A sin 4A Or, for the successive cosines, 19. If in the above expressions for the several multiples of the sines, we introduce for the cosines their values in terms of the sines, and in those for the multiples of the cosines the values of the sines in terms of the cosines, the following expressions will be obtained, in which each quantity is expressed in terms of its own kind. 20. Other useful expressions for the sines and cosines of multiple arcs, may be found thus: Take the sum of the expanded expressions for sin (B+ A) and sin (B - A); that is, to - add. .......... 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 (B — A) =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. Whence, cos (B+ A) =2 cos a coS B COS (BA). 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, cos na cos 0 (=1) COS A cos 2A 2 cos a cos (n − 1) a (x.) sin (n-2) A (K.) cos (n 2) A 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 [3 (z + ¦-)* − 1] A like substitution in the forms for 2 cos 3A, 2 cos 4A, &c. 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 2x2 + 22 = 1 then 2xz = cos 2a; Z= √(1 - cos 2A), or x2 cos 2A, and assume x2 + 22: = 1; and, exterminating z, x2 + 4.x2 cos2 24 = 1. and z2 sinA = (1 + sin A) — } √(1 A) A) S *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 (≈ + 22 + z3 + (-1)); which, after due reduction, comes n+ 1 n A COS 2 2 A) sin A. This theorem em 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 2A1 + cos 2A, and 2 sin 'A = 1 - cos 2A. Changing A and 2A into A and A, we shall have 2 cos2 A1+ cos A, and 2 sin2 A ..cos A = or cos A = sinA (2-2 cos A) 1. cos A, - COSA (N.) Suppose, for example, a = = 60°, then cos A = sin 30° and cos 30° } √(2 + 1) }√3 = ·8660254, sin 30° cos 15° 1) = , as it ought to be: }√(2 + √3) = {( √6 + √2) =⚫9659258 √3) = 1 (√6 sin 15° = √(2 - /2)=2588190: and so on, by continual bisections, as low as we please. 24. If the formulæ (c) be divided one by the other, there will result, 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 (0) will give for the double arc, |