Do somebody know the closed form of the following sum (m is an integer)

$$f(\beta)=\sum _{k=1}^{2m+1} \sin^{2 m+1}\left[\frac{-\beta+k \pi }{1+2 m}\right]$$

If instead of $n=1+2m$ we put $n=2m$, then the above sum do not depends on $\beta$ and is equal to $\frac{2 \Gamma(1/2+m)}{\sqrt{\pi} \Gamma(m)}$.

I need the maximum of $f(\beta)$, it seems that it is achieved in $0$ or in $\pi/2$ but I don't have the proof.

Indeed I need the maximum M_{n,q} of this integral (at least for q=1) $$F_k(\beta)=\int_0^{2\pi }\left(1- \cos\left(\frac{2\beta}{k+1}+s\right)\right)^{\frac{(k+1)q-2}{2}}\left| \cos\frac{(1+k) (s-\pi)}{2}\right|^q ds.$$

It is related to the inequality $$|f^{(k)}(z)|\le (1-|z|^2)^{1-q-kq}M_{k,q}\|\Re f\|_{p}$$ where $f$ is an analytic function in the unit disk and the corresponding norm is the Hardy norm. For $n=1+k$ an even integer I have the proof...