Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$
My friend and I have tested many values of $k > 1$, and in each case this summation is equal to 
$$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$
The paper asserts this as well.
We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically? 
TL;DR

Is this true, and if so, why?
  $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$


This is not an answer but it is too long for a comment.
Considering $$u_k=\sum_{n=0}^k \cos\left(x + \frac{n \,\pi}{k+1}\right)^{2k}$$ and computing the first terms (with some minor trigonometric simplifications) 
$$\left(
\begin{array}{ccc}
k & u_k & \text{value}\\
 0 & 1 & 1 \\
 1 & \cos ^2(x)+\sin ^2(x) & 1 \\
 2 & \cos ^4(x)+\sin ^4\left(x-\frac{\pi }{6}\right)+\sin ^4\left(x+\frac{\pi
   }{6}\right) & \frac{9}{8} \\
 3 & \cos ^6\left(x-\frac{\pi }{4}\right)+\cos ^6(x)+\cos ^6\left(x+\frac{\pi
   }{4}\right)+\sin ^6(x) & \frac{5}{4} \\
 4 & \cos ^8\left(x-\frac{\pi }{5}\right)+\cos ^8(x)+\cos ^8\left(x+\frac{\pi
   }{5}\right)+\sin ^8\left(x-\frac{\pi }{10}\right)+\sin ^8\left(x+\frac{\pi
   }{10}\right) & \frac{175}{128} 
\end{array}
\right)$$ Expanding the terms (this is quite tedious) leads to the constants.
I admire the fact that you have been able to identify that the numbers are such that $$u_k=\frac{ (k+1)\, (2 k)!}{2^{2 k} \,(k!)^2}$$ Could you tell how you did arrive to such an identification ?
