Editing Binomial theorem (section)

=== Multiple-angle identities ===
For the [[complex numbers]] the binomial theorem can be combined with [[de Moivre's formula]] to yield [[List of trigonometric identities#Multiple-angle formulae|multiple-angle formulas]] for the [[sine]] and [[cosine]]. According to De Moivre's formula,
<math display="block">\cos\left(nx\right)+i\sin\left(nx\right) = \left(\cos x+i\sin x\right)^n.</math>

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 {{math|cos(''nx'')}} and {{math|sin(''nx'')}}. For example, since
<math display="block">\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),</math>
But De Moivre's formula identifies the left side with <math>(\cos x+i\sin x)^2 = \cos(2x)+i\sin(2x)</math>, so
<math display="block">\cos(2x) = \cos^2 x - \sin^2 x \quad\text{and}\quad\sin(2x) = 2 \cos x \sin x,</math>
which are the usual double-angle identities. Similarly, since
<math display="block">\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,</math>
De Moivre's formula yields
<math display="block">\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.</math>
In general,
<math display="block">\cos(nx) = \sum_{k\text{ even}} (-1)^{k/2} {n \choose k}\cos^{n-k} x \sin^k x</math>
and
<math display="block">\sin(nx) = \sum_{k\text{ odd}} (-1)^{(k-1)/2} {n \choose k}\cos^{n-k} x \sin^k x.</math>There are also similar formulas using [[Chebyshev polynomials]].