TITLE: How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$?
QUESTION [7 upvotes]: How prove 
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$ 
for $n \in N$?

REPLY [3 votes]: Using De Moivre's formula for odd $n=2m+1$,
and writing $\cos x=c,\sin x=s$
$$i\sin(2m+1)x=(i s)^{2m+1}+\binom{2m+1}2(i s)^{2m-1}c^2+\binom{2m+1}4(i s)^{2m-3}c^4+\cdots+\binom{2m+1}{2m}(is)c^{2m}$$
$$=i^{2m+1}[s^{2m+1}-\binom{2m+1}2s^{2m-1}(1-s^2)+\binom{2m+1}4s^{2m-3}(1-s^2)^2+\cdots+\binom{2m+1}{2m}(-1)^ms(1-s^2)^m]$$
$$\iff s^{2m+1}\left[1+\binom{2m+1}2+\binom{2m+1}4+\cdots+\binom{2m+1}{2m}\right]+\cdots= (-1)^m\sin(2m+1)x$$
$$\iff s^{2m+1}\left[1+1\right]^{2m+1-1}- (-1)^m\sin(2m+1)x=0$$
Now if $\sin(2m+1)x=\sin(2m+1)a,$
$(2m+1)x=2r\pi+(2m+1)a, x=a+\dfrac{2r\pi}{2m+1}$ where $-m\le r\le m$
$\implies2^{2m}\prod_{r=-m}^m\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=-(-1)^m\sin(2m+1)a$
Now for $-m\le r<0, r=-t$(say) $\implies m\ge t>0$
$\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=\sin\left(a-\dfrac{2t\pi}{2m+1}\right)$
$=-\sin\left(a-\dfrac{2t\pi}{2m+1}+\pi\right)=-\sin\left(a+\dfrac{(2m+1-2t)\pi}{2m+1}\right)$
As $m\ge t>0,-2m\le -2t<0\iff1\le2m+1-2t<2m+1$
$\implies2^{2m}\prod_{t=0}^{2m}\sin\left(a+\dfrac{t\pi}{2m+1}\right)(-1)^m=-(-1)^m\sin(2m+1)a$
Similarly for the even $n=2m$(say)