If we suppose n equal to 2, 3, 4, etc., in succession, we have the We may make the results more uniform by substituting for powers of 8 their values in powers of c, thus: s Making these substitutions, and reducing the numerical coefficients, we find Next equating the coefficients of the imaginary terms of (17) to i sin no and dividing by i, we find Supposing n = 2, 3, 4, 5, etc., this form gives sin 2x = 28c; sin 3x = 8{3c — s2} ; sin 4x84c-4c8'}; sin 5x = s {5c* — 10 c2 s2 + s1 } ; sin 686c20c's+Gcs'}; (20) etc. etc. Substituting for the even powers of s their expressions (18), we find Instead of substituting for the powers of s their expressions (18) in terms of the powers of c, we might have expressed the powers of c in terms of s, and by substituting them in (20) have developed the multiple sines in powers of s = sin x. 82. Expression for powers of the cosine. The reverse problem, to express the powers of the sine or cosine of an angle in terms of simple sines and cosines of multiples of the angle, is of yet more frequent application. Let us take the first equation (15), 2 cos x = exi + e-xi, and raise it to the nth power by the binomial theorem. We shall then have To understand this general form let the student take the special cases n = 4 and n = 5. Then Supposing n to be a positive integer, we see (1) that the coefficients of terms equally distant from the two ends of the series are equal, and (2) that the exponents of e in such terms are equal and of opposite signs. Also, if n is even, the middle term does not contain x; but if n is odd, the terms on each side of the middle will contain x. Therefore by joining the first and last terms, the term after the first and that before the last, etc., the development (21) may be put into the form But by the general equations (15') we have, putting na for whatever be the value of n. Hence, substituting this value of the exponential terms, 2n cos" x = cos nx + (笑) (11) cos (n 2)x+(2) cos (n − 4)x+ etc. + cos (− nx) + (†) cos (2 − n) x + (1⁄2 2 ) cos (4 (22) the terms in the third line forming the end of the series, which is doubled over so that the end comes under the beginning. Giving to n the successive values 2, 3, 4, etc., we find 2a cos3 x = cos 2x + 2 + cos (-2x); 23 cos3 x = cos 3x + (†) cos x + (3) cos (−x)+(3) cos (— 3x); 2' cos x = cos 4x+(4) cos 2x + ()+(1) cos (-2x) +(4) cos (4x); 2° cos3 x = cos 5x + (§) cos 3x + (§) cos x +(§) cos (— 5x) + (†) cos (− 3x) + (§) cos (− x). (23) We have extended the series in this form that the student may see the law of its formation, which is as follows: The successive coefficients are the binomial coefficients. The coefficient of x in the first term is n, and it diminishes by 2 in each following term, so as to become n in the last term. will enable us to combine every pair of terms equally distant from the extremes into a single one. For instance, we have 6.5.4.3 6.5 /6 Combining them thus and dividing each equation by 2, we find A careful study of these forms will show the general law of formation, and enable us to carry the powers forward to any extent. 83. Powers of the sine. To find the powers of the sine we take the second of equations (15), 2i sin x = evi — e − xi, and raise it to the nth power by the binomial theorem. We then have The bottom line gives the last terms of the series, arranged in the reverse order, so that terms equally distant from the extremes are under each other. The upper signs are to be used in the bottom line when n is even; the lower ones when n is odd. This will be seen by forming the developments for n = 5 and n = 6. We first take the case of n even. The nth power of i is then +1 or 1 according as n is even or odd. Adding the terms of (25) and substituting for the exponential functions their values in terms of cosines from (15'), we find, when n is even, When n is odd, the sum of each pair of corresponding terms in (25) will give rise to a sine. Making the substitution and dividing by i, we shall have Giving n the successive values 2, 3, 4, etc., and applying the forms (26) and (26') alternately, and changing all the signs in (26) when in is odd, we find 10 2o sin' x = cos 4x + (4) cos (— 4x), 4 cos 2x +3. 2° sina sin 5x (4) sin 3x + sin x - sin (- 5x) 2' sin x = sin 5x or 2° sinR x = 5 sin 3x + 10 sin x. cos 6x+() cos 4x (§) cos 2x + (3) 84. We might have obtained the above results for cos x, cos x, cos3 x, etc., and sin x, sin3 x, etc., by consecutive multiplication, and substitution of sines or cosines of sums for products, by § 43. Thus, by § 44, eq. 16', 2 cos x = cos 2x + 1; × 2 cos x, 2' cos3 x = 2 cos x cos 2x + 2 cos x. |
