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.