TITLE: Proving formulas for $\cos(nx)$ and $\sin(nx)$
QUESTION [6 upvotes]: How do I prove the following formulas?

Let $n \in \mathbb{N}, x \in \mathbb{R}$. Prove that:
$$\cos(nx)=\sum_{j=0}^{[n/2]} (-1)^j {n \choose 2j} (\cos x)^{n-2j} (\sin x)^{2j}$$
$$\sin(nx)=\sum_{j=0}^{[(n-1)/2]} (-1)^j {n \choose 2j+1} (\cos x)^{n-2j-1} (\sin x)^{2j+1}$$

REPLY [9 votes]: Since $n\in\mathbb{N}$ and $x\in\mathbb{R}$, the we can start a derivation for both at once using DeMoivre's formula and Binomial Theorem as @André Nicolas says:
$$
\cos(nx)+i\sin(nx)
=e^{inx}
=(\cos x+i\sin x)^n
=\sum_{k=0}^{n}i^k\binom{n}{k}\cos^{n-k}x\sin^{k}x
$$
Next we must observe that even and odd $k\in\{0,1,\dots,n\}$ give rise to the real and imaginary part, respectively, and can be parametrized by $k=2j$ for $0 \leq j \leq [\frac{n}{2}]$ and by $k=2j+1$ for $0 \leq j \leq [\frac{n-1}{2}]$.