Proving a trigonometric identity: 
Prove that $\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} + \sin
 \frac{{8\pi }}{7} = \frac{{\sqrt 7 }}{2}$.

I have tried to square both side and got ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$. But I cannot proceed further. Any help would be appreciated. 

If $7y=(2n+1)\pi$ where $7\nmid(2n+1)$
Using  Prosthaphaeresis  & Double angle Formulas
$$F=\sin2y+\sin4y+\sin8y$$
$$=2\sin\dfrac{2y+8y}2\cos\dfrac{8y-2y}2+2\sin2y\cos2y$$
$$=2\sin5y\cos3y+2\sin2y\cos2y$$
As $\cos3y=\cdots=-\cos4y,\sin5y=\sin2y,$
$$F=2\sin2y(\cos2y-\cos4y)=2\sin2y(2\sin3y\sin y)$$
Using Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$, 
can you prove $$\prod_{k=1}^m\sin\dfrac{k\pi}{2m+1}=+\dfrac{\sqrt{2m+1}}{2^m}$$
Here $m=3$
