<s>[INST] 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.
 [/INST] see $ \cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are the 4 roots and go by sum of roots formula i.e. **sum of roots = -b/a **
hence, 
 $ \cos\frac{0\pi}{7}+ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} = \frac{4}{8}$
$  \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{1}{2} -1 $
=$  \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{-1}{2}$
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