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. 
 A: Let $\mu=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$ and so $\mu^7=1$. What we want is $\text{Im}(\mu+\mu^2+\mu^4)$. Consider
\begin{align*}
(\mu+\mu^2+\mu^4)(\mu^3+\mu^5+\mu^6)&=3+\mu+\mu^2+\mu^3+\mu^4+\mu^5+\mu^6\\
&=3-1\\
&=2
\end{align*}
Let $x=\mu+\mu^2+\mu^4$. Then $\mu^3+\mu^5+\mu^6=-1-x$. We then have
\begin{eqnarray}x(-1-x)=2\end{eqnarray}
Thus $\displaystyle x=\frac{-1\pm\sqrt{7}i}{2}$. Note that $\text{Im}(x)=\text{Im}(\mu+\mu^2+\mu^4)$ is positive, therefore the answer must be $\displaystyle\frac{\sqrt{7}}{2}$
A: Starting from ${\sin ^2}\frac{{2\pi }}{7} + {\sin ^2}\frac{{4\pi }}{7} + {\sin ^2}\frac{{8\pi }}{7} = \frac{7}{4}$, we only need to prove
$$-\cos\frac{4\pi}{7}-\cos\frac{8\pi}{7}-\cos\frac{16\pi}{7}=\frac12,$$
or equivalently
$$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{1}{2}.$$
Multiplying $2\sin\frac{\pi}7$ on both sides, 
$$\sin\frac{2\pi}{7}+\left(\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}\right)+\left(\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}\right)=\sin\frac\pi7.$$
This is obviously true.
A: 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 $m=3$
