How to calculate $\sin\frac{2\pi}{13}-\sin\frac{5\pi}{13}+\sin\frac{6\pi}{13}$ 
How to calculate this trigonometric function?
$$\sin\frac{2\pi}{13}-\sin\frac{5\pi}{13}+\sin\frac{6\pi}{13}$$

I think this function is related to $x^{26}=-1$.
This Problem is provided by Tieba(Chinese facebook) users.
 A: Your number is $$  \sqrt{\frac{13 - 3 \sqrt{13}}{8}   } $$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
If
$$ \alpha = e^{2 \pi i/13}  $$
we can confirm, without tears, that
$$  x = \alpha + \alpha^3 + \alpha^9$$
satisfies
$$  x^4 + x^3 + 2 x^2 - 4x +3 = 0$$
see page 13 of Reuschle (1875)

Next factored
$$ \left(x^2 + \frac{1+ \sqrt{13}}{2 } x +  \frac{5+ \sqrt{13}}{2 }\right)  \left(x^2 + \frac{1- \sqrt{13}}{2 } x +  \frac{5- \sqrt{13}}{2 }\right)   $$
while your  number is the imaginary part of $x$
Your number is $$  \sqrt{\frac{13 - 3 \sqrt{13}}{8}   } $$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
Recalling
$ \alpha = e^{2 \pi i/13}  ,$
$$ y = a^2 + a^5 + a^6 $$ also satisfies
$$ y^4 + y^3 + 2 y^2 - 4y+3 = 0 $$
The imaginary part of $y$ is $$  \sqrt{\frac{13 + 3 \sqrt{13}}{8}   } $$
A: With the shorthands $a=\frac\pi{13}$,
$$p=\cos2a+\cos6a+\cos8a, \>\>\>\>\>
q=\cos4a+\cos10a+\cos12a$$
and the fact
$$p+q=\sum_{k=1}^6 \cos2ka = Re \sum_{k=1}^6 e^{i2ka} =-\frac12$$
it is straightforward to verify that $pq =\frac32(p+q)=-\frac34
$, which lead to
$$p= \frac14(\sqrt{13}-1),\>\>\>\>\>q= -\frac14(\sqrt{13}+1)$$
Then, use $2\sin x \sin y= \cos(x-y)-\cos(x+y)$ to express
$$\left( \sin2a-\sin5a+\sin6a\right)^2= \frac32 +\frac12q -p
$$
and plug in the values for $p$ and $q$ to obtain
$$\sin2a-\sin5a+\sin6a = \sqrt{\frac{13-3\sqrt{13}}8}
$$
