De Moivre's Formula

Theorem
$$\left(\cos x+i\sin x\right)^n=\cos\left(nx\right)+i\sin\left(nx\right)\,$$, for $$\begin{cases}n=1,2,3,\ldots \\ x\in\mathbb{R}\end{cases}$$

Proof
The formula can be derived from Euler's formula,


 * $$e^{ix} = \cos x + i\sin x\,$$

and laws of exponentials,


 * $$\left( e^{ix} \right)^n = e^{inx} .\,$$

Then, by Euler's formula,


 * $$e^{i(nx)} = \cos(nx) + i\sin(nx)\,$$.

Base Case (n=1)
True since $$(\cos{x}+i\sin{x})^1=\cos(1x)+i\sin(1x)$$

Induction Hypothesis(n=k)
Assume true for $$n=k$$ to get $$ (\cos{x}+i\sin{x})^k=\cos(kx) + i\sin(kx)$$

Show true for n=k+1
$$ \begin{array}{lcl} (\cos{x}+i\sin{x})^{k+1}&=& (\cos{x}+i\sin{x})(\cos{x}+i\sin{x})^k \\ & = & (\cos{x}+i\sin{x})\underbrace{(\cos{kx}+i\sin{kx})}_{\text{by induction hypothesis}} \\ & = & \cos(kx)\cos(x)-\sin(kx)\sin(x)+i[\cos(kx)\sin(x)+\sin(kx)\cos(x)] \\ & = & \cos(kx+x)+i\sin(kx+x) \\ & = & \cos((k+1)x)+i\sin((k+1)x) \end{array} $$

So true for $$n=k+1$$, hence true for all $$n=1,2,3,\ldots$$ by induction.