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.

Consider the polynomial $x^6+x^5+\dots+x+1$. This has roots at $e^{i2n\pi/7} = cos(2n\pi/7)+ isin(2n\pi/7)$ for $n = 1,2,\dots,7$.
Since $cos(x) = cos(2\pi - x)$ the sum you want is just half of the real part of the sum of the roots of this polynomial.  Now by Vieta's formula the sum of the roots is $-1$, hence the sum is $-1/2$.
