Prove this $\sin(nx)$ identity

Question

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

Answer 1

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

$$R.H.S=\frac{n\sin(2nx)}{2^{2n-2}}\left(\frac{1}{n2^{2-2n}}\right)=\sin(2nx)=L.H.S$$