Prove this $\sin(nx)$ identity Without induction How to prove that $$\sin(2nx)=2n\sin x \cos x \prod_{k=1}^{n-1}\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)$$ 
for Natural n
I Tried by several way and the last try is to use euler formula which lead me to
$$\sin(2nx)=\sin x \cos^{2m+1} x \sum_{k=0}^{n-1} \left( (-1)^k\binom{2n}{2k+1}(\cos x \sin x )^{2k}\right)$$ 
and no idea how to continue or if this method lead to the answer
 A: Finally it solved ^_^ :
first by use: $\sin^2x=\frac{1-\cos 2x}{2}$
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=\left(\frac{\cos 2x - \cos{\frac{k\pi}{n}}}{2\sin^2\frac{k\pi}{2n}}\right)$$ 
then by use :$\cos x - \cos y $ formula 
I got 
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=-\left(\frac{\sin(x+\frac{k\pi}{2n})\sin(x-\frac{k\pi}{2n})}{\sin^2\frac{k\pi}{2n}}\right)$$ 
so 
$$R.H.S=2n\sin x \cos x \prod_{k=1}^{n-1}\left(-\frac{\sin(x+\frac{k\pi}{2n})\sin(x-\frac{k\pi}{2n})}{\sin^2\frac{k\pi}{2n}}\right)$$
but 
$$\prod_{k=1}^{n-1}\sin(\frac{k\pi}{2n}-x)= \prod_{k=1}^{n-1}\cos(x+\frac{k\pi}{2n})$$
so
$$R.H.S=2n\sin x \cos x \prod_{k=1}^{n-1}\left(\frac{\sin(2x+\frac{k\pi}{n})}{2\sin^2\frac{k\pi}{2n}}\right)$$
and by This LINK i've got
$$\frac{\sin(2nx)}{2\sin x \cos x 2^{n-1}}=\prod_{k=1}^{n-1}\sin(2x+\frac{k\pi}{n})$$
so 
$$R.H.S=\frac{n\sin(2nx)}{2^{2n-2}}\prod_{k=1}^{n-1}\left(\frac{1}{\sin^2\frac{k\pi}{2n}}\right)$$
and Finally from this
$$R.H.S=\frac{n\sin(2nx)}{2^{2n-2}}\left(\frac{1}{n2^{2-2n}}\right)=\sin(2nx)=L.H.S$$
