General expression for the k-th derivative of $(\cos(x))^n$ Is there a general expression for the higher-order derivatives of $(\cos(x))^n$ evaluated at the origin? 
The odd derivatives are zero due to the symmetry, but what about the even derivatives?
 A: To expand @NotAloner's answer above,
$$\cos^n x = \left( { e^{ix} + e^{-ix} \over 2} \right)^n = {1 \over 2^n} \sum_{j=0}^n {n \choose j} e^{(n-2j)ix}$$
Hence
$$D^k \cos^n x = {1 \over 2^n} \sum_{j=0}^n {n \choose j} i^k(n-2j)^k e^{(n-2j)ix}$$
and evaluated at $x = 0$
$$c(n,k) = {1 \over 2^n} \sum_{j=0}^n {n \choose j} i^k(n-2j)^k$$
As commented in the OP, by symmetry considerations $c(n,k) = 0$ for all odd $k$. For even $k$, writing $k = 2m$,
$$c(n,2m) = { (-1)^m \over 2^n } \sum_{j=0}^n {n \choose j} (n-2j)^{2m}$$
It's not clear to me how to simplify this further (WolframAlpha also draws a blank).
A: $(\cos x)^n = \left(\dfrac{e^{ix}+e^{-ix}}{2}\right)^n$. I think some pattern should emerge when taking derivatives of higher order like the $k$th one.
You edited your question so my answer would not be useful anymore...fine.
As for the edited question, you have $f(x) = \cos x$, taking derivatives:
$f'(x) = -\sin x, f''(x) = -\cos x, f'''(x) = \sin x, f^{(4)}(x) = \cos x, f^{(5)}(x) = - \sin x, f^{(6)}(x) = -\cos x$. Thus we can sum it up what we have found here:
$f^{(k)}(0) = \begin{cases} 0 , k = 2n+1 \\ -1, k = 4n+2, n \geq 0 \\ 1, k = 4n, n\geq 1\end{cases}$
A: i don't have a general expression but a sum. we have 
$$\begin{align} 2^{n-1}\cos^n x &= {n \choose 0} \cos nx + {n \choose 1} \cos(n-2)x +  {n \choose 1} \cos(n-2)x  + \cdots\end{align} $$ taking derivatives $k$ times and use the fact that $\frac{d^k}{dx^k}\cos(x) = \cos(x + k\pi/2),$  we have 
$$\begin{align} 2^{n-1}\frac{d^k}{dx^k}\left( \cos^n x\right) &= {n \choose 0} n^k\cos( nx+k\pi/2) + {n \choose 1} (n-2)^k\cos((n-2)x+k\pi/2) \\
& +  {n \choose 1} (n-4)^k\cos((n-4)x +k\pi/2) + \cdots\end{align} $$ 
i don't know if there is a closed form for this.

added later:
if you evaluate it at the origin, then you have 
$$2^{n-1}\frac{d^k}{dx^k}\left( \cos^n x\right) \Big|_{x=0}= \cos(k\pi/2) \left({n \choose 0} n^k+ {n \choose 1} (n-2)^k
+  {n \choose 1} (n-4)^k + \cdots \right)$$ 
A: \begin{align}
f(x)&=\cos^n{x}\\
f'(x)&=n\cos^{n-1}{x}\cdot(-\sin{x})\\
f''(x)&=n(n-1)\cos^{n-2}{x}\cdot(-\sin{x})^2-n\cos^n{x}\\
f'''(x)&=n(n-1)(n-2)\cos^{n-3}{x}\cdot(-\sin{x})^3+2n(n-1)\sin{x}\cdot\cos^{n-1}(x)-nf'(x)\\
\ldots\\
f^{(k)}(x)&=n\binom{k-n}{k}\sum_{i=0}^k\left(\dfrac{(-1)^i\binom{k}{i}}{n-i}\cdot\cos^{n-i}(x)\cdot\dfrac{\partial^k}{\partial x^k}\left[\cos^i{x}\right]\right)
\end{align}
Thus for $k$ odd, $\dfrac{\partial^k}{\partial x^k}\left[\cos^i{x}\right]$ involving a $\sin{x}$ term, $f^{(k)}(0)=0$.
And for $k$ even, we have:
$$f^{(k)}(0)=n\binom{k-n}{k}\sum_{i=0}^k\dfrac{(-1)^i\binom{k}{i}}{n-i}\neq 0$$

If you have Mathematica, you can test the above formula and check that the values are correct:
Manipulate[{D[(Cos[x])^n,{x,k}]/.{x->0},n*Binomial[k-n,k]*Sum[(-1)^i*Binomial[k,i]/(n-i)*D[(Cos[x])^i,{x,k}]/.{x->0},{i,0, k}]},{n,1,20,1},{k,0,20,1}]
