|
:<math>
\begin{alignat}{2}
\left(\cos x+i\sin x\right)^{k+1} & = \left(\cos x+i\sin x\right)^{k} \left(\cos x+i\sin x\right)\\
& = \left[\cos\left(kx\right) + i\sin\left(kx\right)\right] \left(\cos x+i\sin x\right) &&\qquad \mbox{by the induction hypothesis}\\
& = \cos \left(kx\right) \cos x - \sin \left(kx\right) \sin x + i \left[\cos \left(kx\right) \sin x + \sin \left(kx\right) \cos x\right]\\
& = \cos \left[ \left(k+1\right) x \right] + i\sin \left[ \left(k+1\right) x \right] &&\qquad \mbox{by the trigonometric identities}
\end{alignat}
</math>
:<math>
\begin{align}
\left(\cos x + i\sin x\right)^{n} & = \left(\cos x + i\sin x\right)^{-m}\\
& = \frac{1}{\left(\cos x + i\sin x\right)^{m}}\\
& = \frac{1}{\left(\cos mx + i\sin mx\right)}\\
& = \cos\left(mx\right) - i\sin\left(mx\right)\\
& = \cos\left(-mx\right) + i\sin\left(-mx\right)\\
& = \cos\left(nx\right) + i\sin\left(nx\right).
\end{align}
</math>
Being an equality of [[complex number]]s, one necessarily has equality both of the [[real part]]s and of the [[imaginary part]]s of both members of the equation. If ''x'', and therefore also <math>\cos x</math> and <math>\sin x</math>, are [[real number]]s, then the identity of these parts can be written (interchanging sides) as
:<math>\begin{alignat}2
\cos(nx)&=\sum_{k=0}^{\lfloor n/2\rfloor}{\tbinom{n}{2k}}(-1)^k(\cos{x})^{n-2k}(\sin{x})^{2k}& &=\sum_{k=0}^{\lfloor n/2\rfloor}{\tbinom{n}{2k}}(\cos{x})^{n-2k}((\cos{x})^2-1)^k\\
\sin(nx)&=\sum_{k=0}^{\lfloor (n-1)/2\rfloor}{\tbinom{n}{2k+1}}(-1)^k(\cos{x})^{n-2k-1}(\sin{x})^{2k+1}& &=(\sin{x})\sum_{k=0}^{\lfloor(n-1)/2\rfloor}{\tbinom{n}{2k+1}}(\cos{x})^{n-2k-1}((\cos{x})^2-1)^k.\\
\end{alignat}</math>
These equations are in fact even valid for complex values of ''x'', because both sides are [[holomorphic function]]s of ''x'', and two such functions that coincide on the real axis necessarily coincide on the whole [[complex plane]]. Here are the concretre instances of these equations for <math>n=2</math> and <math>n=3</math>:
:<math>\begin{alignat}2
\cos(2x) &= (\cos{x})^2 +((\cos{x})^2-1) &&= 2(\cos{x})^2-1\\
\sin(2x) &= 2(\sin{x})(\cos{x})\\
\cos(3x) &= (\cos{x})^3 +3\cos{x}((\cos{x})^2-1) &&= 4(\cos{x})^3-3\cos{x}\\
\sin(3x) &= 3(\cos{x})^2(\sin{x})-(\sin{x})^3 &&= 3\sin{x}-4(\sin{x})^3.\\
\end{alignat}</math>
The right hand side of the formula for <math>\cos(nx)</math> is in fact the value <math>T_n(\cos x)</math> of the [[Chebyshev polynomial]] <math>T_n</math> at <math>\cos x.</math>
: <math>
z^{{}^{\frac{1}{n}}}= \left[ r\left(\cos x+i\sin x \right) \right]^ {{}^{\frac{1}{n}}}= r^{{}^{\frac{1}{n}}} \left[ \cos \left(\frac{x+2k\pi}{n} \right) + i\sin \left(\frac{x+2k\pi}{n} \right) \right]
</math>
| 