Prove that $$\prod_{r=1}^m \sin \left( \frac {r\pi}{2m+1}\right) =\frac {\sqrt {2m+1}}{2^m}$$

My try: $$\prod_{r=1}^m \sin \left( \frac {r\pi}{2m+1}\right) =\prod_{r=1}^m \left(\frac {e^{\frac {ir\pi}{2m+1}} -e^{\frac {-ir\pi}{2m+1}}}{2i}\right) =\frac {1}{2^mi^m}\left(\prod_{r=1}^m e^{\frac {-ir\pi}{2m+1}}\right) \prod_{r=1}^m \left(e^{\frac {2ir\pi}{2m+1}} -1\right)$$

But I can't get a clue to tackle $$\prod_{r=1}^m \left(e^{\frac {2ir\pi}{2m+1}} -1\right)$$ . I tried to use it's relation with roots of unity but couldn't proceed much.

I also tried writing the $\sin$ using Euler's reflection formula that for $z\in (0,1)$ $$\sin (\pi z)=\frac {\pi}{\Gamma(z)\Gamma(1-z)}$$ where in our case $z=\frac {r}{2m+1}$. But that too didn't last long.

Any help is greatly appreciated. Thanks.

  • 3
    $\begingroup$ Check out the answer here. $\endgroup$ – MisterRiemann Dec 8 '18 at 12:03



$$ P = \prod_{r=1}^m \sin \left( \frac {r\pi}{2m+1}\right) $$ then

$$ P^2 = \prod_{r=1}^{2m} \sin \left( \frac {r\pi}{2m+1}\right) $$


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