Finding a closed form of $\cos^n(x)$ in terms of $\cos$ 
By definition we have $$\cos(x):= \frac{1}{2}\left(e^{ix}+e^{-ix} \right) $$

Using this definition and applying the binomial expansion to the right hand side I would get $$\cos^n(x)= \left(\frac{1}{2}\right)^n\sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{ixk} e^{(-ix)(n-k)} = \left(\frac{1}{2} \right)^n \sum_{k=0}^n\left(\begin{matrix} n \\ k\end{matrix}\right) e^{(2k-n)ix}$$
which looks promising, but I can't continue from there, apparently to this Wikipedia entry here there is a much more elegant form. I do of course with the above already obtain the $\cos(2k-n)$ part but there is still the $i\sin(2k-n)$ part present. Namely: $$\cos^n(x) = \frac{1}{2^n}\sum_{k=0}^n \left(\begin{matrix} n \\ k\end{matrix}\right)\left(\cos((2k-n)x)+i \sin((2k-n)x)\right) $$
It feels like an invalid argument to say that $\cos^n(x)$ is real and therefore the $i \sin(2k-n)$ part will vanish. 
 A: For the power of trigonometric function we have the following identities:
\begin{eqnarray}
\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] \tag 1\\
\cos^{2n+1}x &=& \frac{1}{4^n}\sum_{k=0}^{n}\binom{2n+1}{k}\cos[(2n+1-2k)x]\tag 2\\
\sin^{2n}x &=& \frac{1}{2^{2n}}\binom{2n}{n}+\frac{(-1)^n}{2^{2n-1}}\sum_{k=0}^{n}(-1)^k\binom{2n}{k}\cos[2(n-k)x] \tag 3
\\
\sin^{2n+1}x &=& \frac{(-1)^n}{4^n}\sum_{k=0}^{n-1}\binom{2n+1}{k}\sin[(2n+1-2k)x]  \tag 4\\
\end{eqnarray}
We prove the identity (1). We have 
$$
\cos x=\frac{(\cos x+i\sin x)+(\cos x-i\sin x)}{2}.
$$
Put $\cos x+i\sin x=z$ so that $\cos x-i\sin x=z^{-1}$. Then we have
$$
\cos^{2n}x=\left (\frac{z+z^{-1}}{2}\right)^{2n}=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}z^{2n-k}.
$$
Further
$$
2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}z^{2(n-k)}+\binom{2n}{n}+\sum_{k=n+1}^{2n}\binom{2n}{k}z^{2(n-k)}.
$$
In the second sum put $n-k=-(n-k')$. Then this sum is rewritten as
$$
\sum_{k'=n-1}^{0}\binom{2n}{2n-k'}z^{-2(n-k')}=\sum_{k=0}^{n-1}\binom{2n}{k}z^{-2(n-k)}
$$
and so
$$
2^{2n}\cos^{2n}x=\sum_{k=0}^{n-1}\binom{2n}{k}\left[z^{2(n-k)}+z^{-2(n-k)}\right]+\binom{2n}{n}.
$$
However, $z^{2(n-k)}+z^{-2(n-k)}=2\cos[2(n-k)x]$, and therefore
$$
\cos^{2n}x =\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos[2(n-k)x].
$$
Replacing in (1) $x$ by $\frac{\pi}{2}-x$, we get formula (2). Formulas (3) and (4) are deduced as (1) and (2).
