Binomial theorem
Multiple-angle identities

For the complex numbers the binomial theorem can be combined with de Moivre's formula to yield multiple-angle formulas for the sine and cosine. According to De Moivre's formula, cos (n x) + i sin (n x) = (cos x + i sin x) n. {\displaystyle \cos \left(nx\right)+i\sin \left(nx\right)=\left(\cos x+i\sin x\right)^{n}.}

Using the binomial theorem, the expression on the right can be expanded, and then the real and imaginary parts can be taken to yield formulas for cos(nx) and sin(nx). For example, since (cos x + i sin x) 2 = cos 2 x + 2 i cos x sin x − sin 2 x = (cos 2 x − sin 2 x) + i (2 cos x sin x), {\displaystyle \left(\cos x+i\sin x\right)^{2}=\cos ^{2}x+2i\cos x\sin x-\sin ^{2}x=(\cos ^{2}x-\sin ^{2}x)+i(2\cos x\sin x),} But De Moivre's formula identifies the left side with (cos x + i sin x) 2 = cos (2 x) + i sin (2 x) {\displaystyle (\cos x+i\sin x)^{2}=\cos(2x)+i\sin(2x)}, so cos (2 x) = cos 2 x − sin 2 x and sin (2 x) = 2 cos x sin x, {\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x\quad {\text{and}}\quad \sin(2x)=2\cos x\sin x,} which are the usual double-angle identities. Similarly, since (cos x + i sin x) 3 = cos 3 x + 3 i cos 2 x sin x − 3 cos x sin 2 x − i sin 3 x, {\displaystyle \left(\cos x+i\sin x\right)^{3}=\cos ^{3}x+3i\cos ^{2}x\sin x-3\cos x\sin ^{2}x-i\sin ^{3}x,} De Moivre's formula yields cos (3 x) = cos 3 x − 3 cos x sin 2 x and sin (3 x) = 3 cos 2 x sin x − sin 3 x. {\displaystyle \cos(3x)=\cos ^{3}x-3\cos x\sin ^{2}x\quad {\text{and}}\quad \sin(3x)=3\cos ^{2}x\sin x-\sin ^{3}x.} In general, cos (n x) = ∑ k even (− 1) k / 2 (n k) cos n − k x sin k x {\displaystyle \cos(nx)=\sum _{k{\text{ even}}}(-1)^{k/2}{n \choose k}\cos ^{n-k}x\sin ^{k}x} and sin (n x) = ∑ k odd (− 1) (k − 1) / 2 (n k) cos n − k x sin k x. {\displaystyle \sin(nx)=\sum _{k{\text{ odd}}}(-1)^{(k-1)/2}{n \choose k}\cos ^{n-k}x\sin ^{k}x.} There are also similar formulas using Chebyshev polynomials.
combination:two_responses