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!
 A: As @Tunk-Fey said in the comment, we only need to show $4x^2 + 2x - 3 =0$ with $x = \cos\frac{2\pi}{13} + \cos\frac{6\pi}{13} + \cos\frac{8\pi}{13} $ since we know $x$ should be positive.
Take $\theta = \frac{2\pi}{13}$, $w = \exp(i\theta)$ and $$A = w + w^3 + w^4$$
$$B = w^{-1} + w^{-3} + w^{-4}$$
we have $A + B = 2x$, thus we only need to show $(A+B)^2 + A+B - 3 = 0$
We will need the following facts:
since we have $w^{13} - 1 = (w-1)\sum_{k=0}^{12}w^k=0$ thus $\sum_{k=0}^{12}w^k = 0$, dividing by $w^6$($\neq 0$) gives
$$\sum_{k=1}^6(w^k + w^{-k}) + 1 =0$$
and we have also $$w^5 + w^{-5} = w^8 + w^{-8}$$  $$w^6 + w^{-6} = w^7 + w^{-7}$$
Simple computation gives 
\begin{align}
(A+B)^2 =&w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6\\
&+2(w + w^{-1} + w^2 + w^{-2} + w^3 + w^{-3} + w^4 + w^{-4} + w^5 + w^{-5} + w^{7}+w^{-7}) \\
\end{align}
replacing $w^{7}+w^{-7}$ by $w^{6}+w^{-6}$ gives
\begin{align}
(A+B)^2 = &w^2 + w^{-2} + w^{6} + w^{-6} + w^8 + w^{-8} + 6+2(-1) \\
 =& w^2 + w^{-2} + w^5 + w^{-5} + w^6 + w^{-6} + 4
\end{align}
Now adding together $(A+B)^2$, $A+B$ and $-3$ allows to conclude.
A: Using complex numbers:
$$\sum_{k=0}^{12}e^{2ik\pi/13}=0$$
So,
$$1+e^{2i\pi/13}+e^{4i\pi/13}..e^{24i\pi/13}=0$$
For Real part,
$$1+\cos(2\pi/13)+\cos(4\pi/13)...\cos(24\pi/13)=0$$
$$\cos(2\pi/13)+\cos(4\pi/13)...\cos(12\pi/13)=-1/2$$
$$\cos\dfrac{2\pi}{13}+\cos\dfrac{6\pi}{13}+\cos\dfrac{8\pi}{13}=-1/2-\cos\dfrac{4\pi}{13}+\cos\dfrac{10\pi}{13}+\cos\dfrac{12\pi}{13}$$
I think you can do it from here.
A: I shall finish it based on what Aditya did, and all the credit goes to him. If you simply write $a_{2k}$ to denote $\cos(2k\pi/13)$, the identity you have is:
$a_0+a_2+a_4+\dots+a_{24}=0$. You also need the basic identities:
$$a_{2k}=a_{(26-2k)},\quad a_{2k}=2a_{k}^2-1,\quad 2a_{2k}a_{2j}=a_{2(k+j)}+a_{2(k-j)}$$
Now 
$$\begin{split}
0&=a_0+a_2+a_4+\dots+a_{24}\\
&=a_0+(a_2+a_6+a_8)+(a_{24}+a_{20}+a_{18})+\\
&(a_4+a_{12}+a_{16})+(a_{22}+a_{14}+a_{10})\\
&=1+2(a_2+a_6+a_8)+2(a_4+a_{12}+a_{16})
\end{split}
$$
What you want is $x:=a_2+a_6+a_8$. Lets denote $y:=a_4+a_{12}+a_{16}$. So you have:
$$1+2x+2y=0$$
But we also have:
$$x^2=\frac{3}{2}+x+\frac{3}{2}y$$
That is, your
$4x^2+2x-3=0$
A: 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
$$\cos \frac{2\pi}{13}+\cos \frac{6\pi}{13}+\cos \frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$$
A: Consider the polynomial $P=X^{13}-1$, and discuss their    roots 
