Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$ 
Prove the identity 
  $$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce, 
  $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$

My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since 
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$

I cannot prove the final part. 

Please help me. Thanks in advance.

$\cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are distinct roots of the fourth order polynomial $$P(x)=8x^4-4x^3-8x^2+3x+1$$
So $P(x)$ can be re-written $$P(x)=8\left(x-\cos\frac{0\pi}{7}\right)\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{6\pi}{7}\right)$$
Therefore looking at $x^3$ coefficient gives $$\cos\frac{0\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{4}{8}=\frac{1}{2}$$
Actually, for any $n>=2$, $$P_n(x)=T_n(x)-T_{n-1}(x)=2^{n-1}x^n-2^{n-2}x^{n-1}+...$$ 
Where $T_n$ is the nth Chebyshev polynomial. 
So $$P_n(cos(x))=cos(nx)-cos((n-1)x)$$And Likewise, $$\sum_{k=0}^{n-1} cos\frac{2k\pi}{2n-1} = \frac{1}{2}$$
