Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ 
Prove that 
  $$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$

My attempt 
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.

Recall the identity $$\sin \alpha \cos \beta = \frac{1}{2} \left( \sin(\alpha+\beta) + \sin(\alpha - \beta) \right).$$  Then
$$\begin{align*} \sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right)
&= \frac{1}{2} \left( \sin \frac{4\pi}{7} +  \sin \frac{6\pi}{7} + \sin \frac{-2\pi}{7} + \sin \frac{10\pi}{7} + \sin \frac{-6\pi}{7} \right), \\
\end{align*}$$
and observing that $$\sin \frac{10\pi}{7} = \sin \left(2\pi - \frac{4\pi}{7}\right) = \sin \frac{-4\pi}{7} = -\sin \frac{4\pi}{7}, \\
\sin \frac{-2\pi}{7} = - \sin \frac{2\pi}{7}, \\
\sin \frac{-6\pi}{7} = - \sin \frac{6\pi}{7},$$
we have
$$\sin \frac{2\pi}{7} \left( \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} \right) = -\frac{1}{2} \sin \frac{2\pi}{7},$$ where upon dividing both sides by $\sin 2\pi/7$, we obtain the desired result.
