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.

Here is one way to do it.  It doesn't use your previous work, though.
$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\\
\frac {\sin\frac \pi7(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})}{\sin
 \frac \pi7}\\
\frac {\sin\frac \pi7\cos\frac{2\pi}{7}+\sin\frac \pi7\cos\frac{4\pi}{7}+\sin\frac \pi7\cos\frac{6\pi}{7})}{\sin
 \frac \pi7}\\
\sin A\cos B = \frac12 \sin(A+B) - \frac12\sin(B-A)\\
\frac {\sin\frac {3\pi}7 - \sin \frac {\pi}{7}+\sin\frac {5\pi}7 - \sin\frac{3\pi}{7} +\sin\frac {7\pi}7 - \sin\frac {5\pi}{7}}{2\sin
 \frac \pi7}\\
\frac {\sin \pi - \sin \frac {\pi}{7}}{2\sin
 \frac \pi7} = -\frac 12\\
$
Using the info from part 1.
Let $x = \cos \theta$
if $\theta = \frac{2n\pi}{7}$
$8x^4-4x^3 -8x^2+3x+1 = 0$
and $1, \cos \frac{2\pi}{7}, \cos \frac{4\pi}{7}, \cos \frac{6\pi}{7}$ are roots of the polynomial.
$8(x - 1)(x - \cos{2\pi}{7})(x - \cos{4\pi}{7})(x - \cos{6\pi}{7}) = 8x^4-4x^3 -8x^2+3x+1$
By Viteta's rules
$8(1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7}) = 4\\
1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = \frac 12\\
\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = -\frac 12$
