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.

HINT:
$$cos\theta_1cos\theta_2=\frac{cos(\theta_1 - \theta_2) + cos(\theta_1 + \theta_2)}{2}$$
