A nice trignometric identity How to prove that:

$$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=\dfrac{\sqrt{13}-1}{4} $$

I have a solution but its quite lengthy, I would like to see some elegant solutions. Thanks!

I add another answer although it is not different in principle from some of the others. 
We have 
$$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}$$
it is natural to then consider the other even divisions, 
$$s=\cos \frac{4\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}$$
Now 
$$t+s=\cos \frac{2\pi}{13}+\cos \frac{4\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}+\cos \frac{10\pi}{13}+\cos \frac{12\pi}{13}=\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}$$
$$\frac{\cos\frac{7\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}}=\frac{1}{2}\frac{\sin\frac{13\pi}{13}-\sin \frac{\pi}{13}}{\sin\frac{\pi}{13}}=-\frac{1}{2}$$
Next we calculate $st$ using $\cos A \cos B=\frac{1}{2}(\cos (A+B)+\cos(A-B))$ we find $st=\frac{3}{2}(s+t)=-\frac{3}{4}$
So $s$ and $t$ are solutions to $4x^2+2x-3=0$ whose roots are $\frac{\sqrt{13}-1}{4}$ and 
$\frac{-\sqrt{13}-1}{4}$
Note that $$t=\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}-\cos \frac{5\pi}{13}$$
and since $\cos \frac{5\pi}{13}<\cos \frac{6\pi}{13}$ we have that $t>0$ and thus
$$\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$$
