<s>[INST] Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$ 
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$

I tried showing the equation, but my attempts did not get the result.
I already showed that
$$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})=\frac18$$ $$\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=-\frac12$$
Using in particular the trigonometric formulas:
$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ , $\sin(\pi+\alpha)=-\sin(\alpha)$ , $\sin(\pi-\alpha)=\sin(\alpha)$ , $\cos(\pi-\alpha)=-\cos(\alpha)$ , $\sin(-\alpha)=-\sin(\alpha)$ , and $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$
Any hint would be helpful. Thanks in advance.
 [/INST] $$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; \;  2 \cos \left( \frac{4 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$
are the roots of 
$$ x^3 + x^2 - 2x - 1  $$
so
$$  \cos \left( \frac{2 \pi}{7} \right) \; , \; \;   \cos \left( \frac{4 \pi}{7} \right) \; , \; \;  \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$
are the roots 
$$ 8 x^3 + 4 x^2 - 4x - 1 $$
or
$$ x^3 + \frac{1}{2}x^2 - \frac{1}{2} x - \frac{1}{8}  $$
The sum of products of two roots ate a time is $-\frac{1}{2}$ 
The original claim is from a method due to Gauss, this time we take
$$ \omega = e^{2 \pi i / 7} , $$ 
next take
$$  \eta = \omega + \frac{1}{\omega}, $$
then look for a cubic (in $\eta$) that vanishes because of the relation
$$ \omega^6 +  \omega^5 +  \omega^4 +  \omega^3 + \omega^2 +  \omega + 1=0$$
Here we find
$$
 \eta^3 + \eta^2 - 2\eta - 1 =
\frac{ \omega^6 +  \omega^5 +  \omega^4 +  \omega^3 + \omega^2 +  \omega + 1}{\omega^3}
$$
</s>