TITLE: Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: $$
\begin{align}
&1+\color{#C00}{2\cos\left(\frac{2\pi}7\right)}+\color{#090}{2\cos\left(\frac{4\pi}7\right)}+\color{#00F}{2\cos\left(\frac{6\pi}7\right)}\\
&=1+\color{#C00}{\cos\left(\frac{2\pi}7\right)}+\color{#090}{\cos\left(\frac{4\pi}7\right)}+\color{#00F}{\cos\left(\frac{6\pi}7\right)}+\color{#00F}{\cos\left(\frac{8\pi}7\right)}+\color{#090}{\cos\left(\frac{10\pi}7\right)}+\color{#C00}{\cos\left(\frac{12\pi}7\right)}\\
&=\operatorname{Re}\left(1+e^{2\pi i/7}+e^{4\pi i/7}+e^{6\pi i/7}+e^{8\pi i/7}+e^{10\pi i/7}+e^{12\pi i/7}\right)\\
&=\operatorname{Re}\left(\frac{e^{14\pi i/7}-1}{e^{2\pi i/7}-1}\right)\\
&=0
\end{align}
$$
Therefore,
$$
\cos\left(\frac{2\pi}7\right)+\cos\left(\frac{4\pi}7\right)+\cos\left(\frac{6\pi}7\right)=-\frac12
$$

REPLY [2 votes]: 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
$$\begin{align}\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)&=\frac{\omega+\omega^{-1}}{2}+\frac{\omega^2+\omega^{-2}}{2}+\frac{\omega^{3}+\omega^{-3}}{2}\\&=\frac{(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1)-\omega^3}{2\omega^3}=-\frac12\,.\end{align}$$