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. 

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 $\displaystyle\frac{\sqrt{7}}{2}$
