<s>[INST] 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.
 [/INST] 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 $$  \sqrt{\frac{13 + 3 \sqrt{13}}{8}   } $$
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