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.
 A: $\cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ 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, $$\sum_{k=0}^{n-1} cos\frac{2k\pi}{2n-1} = \frac{1}{2}$$
A: $$
\begin{align}
&1+\color{#C00}{2\cos\left(\frac{2\pi}7\right)}+\color{#090}{2\cos\left(\frac{4\pi}7\right)}+\color{#00F}{2\cos\left(\frac{6\pi}7\right)}\\
&=1+\color{#C00}{\cos\left(\frac{2\pi}7\right)}+\color{#090}{\cos\left(\frac{4\pi}7\right)}+\color{#00F}{\cos\left(\frac{6\pi}7\right)}+\color{#00F}{\cos\left(\frac{8\pi}7\right)}+\color{#090}{\cos\left(\frac{10\pi}7\right)}+\color{#C00}{\cos\left(\frac{12\pi}7\right)}\\
&=\operatorname{Re}\left(1+e^{2\pi i/7}+e^{4\pi i/7}+e^{6\pi i/7}+e^{8\pi i/7}+e^{10\pi i/7}+e^{12\pi i/7}\right)\\
&=\operatorname{Re}\left(\frac{e^{14\pi i/7}-1}{e^{2\pi i/7}-1}\right)\\
&=0
\end{align}
$$
Therefore,
$$
\cos\left(\frac{2\pi}7\right)+\cos\left(\frac{4\pi}7\right)+\cos\left(\frac{6\pi}7\right)=-\frac12
$$
A: 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
$8(1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7}) = 4\\
1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = \frac 12\\
\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = -\frac 12$
A: Since you have obtained sufficient information regarding the route you take to solve the problem, I am giving a different solution.  It is a sketch.
Let $\omega:=\exp\left(\dfrac{2\pi\text{i}}{7}\right)$.  Show that $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$.  This implies
$$\begin{align}\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)&=\frac{\omega+\omega^{-1}}{2}+\frac{\omega^2+\omega^{-2}}{2}+\frac{\omega^{3}+\omega^{-3}}{2}\\&=\frac{(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1)-\omega^3}{2\omega^3}=-\frac12\,.\end{align}$$
A: 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}$
