How does one begin to show (natural $n$): $$\cos^{2n}(x) =\frac{1}{2^{2n}} \binom{2n}{n}+ \frac{1}{2^{2n-1}} \sum_{k=0}^{n-1} \binom{2n}{k} \cos[2(n-k)x]$$ $$\cos^{2n+1}(x) =\frac{1}{4^{n}} \sum_{k=0}^{n} \binom{2n+1}{k} \cos[(2n + 1 - 2k)x]$$
Alternately, how can I prove that that the l.h.s of the second is a linear combinations of all $\cos{kx}$, where $k$ are the odd $k \leq 2n+1$?
This leads to the intriguing fact that $cos^m{x}$ is orthogonal to $\cos nx$ for all $m < n$..
