<s>[INST] 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. 
 [/INST] 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.
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