Proving $\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\frac12\sqrt{\frac{13+3\sqrt{13}}2}$ 
Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$

My Attempt
Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}} \implies 16x^4-52x^2+13=0$$
And through some donkey work we can calculate the chebyshev polynomial for $\sin\left(\frac{\pi}{13}\right),\sin\left(\frac{3\pi}{13}\right),\sin\left(\frac{4\pi}{13}\right)$ which will all be the same as $\sin(n\pi)=0,\text{ for all } n \in \mathbb{I}  $, so
$$P(x) = 4096x^{12}-13312x^{10}+16640x^8-9984x^6+2912x^4-364x^2+13$$
where $x = \sin\left(\frac{2i\pi}{13}\right), \text{ from } 1 \le i \le 12 \text{ where } i \in \mathbb{I}$, are the roots of $P(x)$.
Now I am not getting how to connect these two into a possible solution and even it is possible (probably is), its still a pretty donkey method as you need to find the $13^{th}$ chebyshev polynomial, so if possible maybe give some another method of approach to this question.
 A: with $ \; w = \cos \frac{2 \pi}{13}  + i \sin \frac{2 \pi}{13} \; \; \; $  in mind, let
$$ x = -i \left( w - w^{25} +  w^3 - w^{23} +  w^9 - w^{17} \right)  $$
we may calculate polynomials in $x.$   We may then apply the relation $w^{26 }  = 1  $ repeatedly and   express the outcome as  sums of $w^{25}, w^{24}, ..., w^2, w,1 $
I get
$$ x^4 - 13 x^2 + 13 = w^{24} + w^{22} + \cdots w^4 + w^2 + 1 $$
with no odd exponents  and all the coefficients $1$
Multiply by the (nonzero)  $w^2 - 1 $  to get $ x^4 - 13 x^2 + 13 $
The original ideas were due to Gauss, see Cox chapter
Many examples were worked out by Reuschle;
This comes from page 529: Reuschle 1875

A: Let
$$
s=\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)\tag1
$$
If $\alpha=e^{\pi i/13}$, then
$$
\alpha^{13}+1=0\tag2
$$
and
$$
2is=\underbrace{\alpha+\overbrace{\quad\alpha^{12}\quad}^{-\alpha^{-1}}}_{2i\sin\left(\frac{\pi}{13}\right)}+\underbrace{\alpha^3+\overbrace{\quad\alpha^{10}\quad}^{-\alpha^{-3}}}_{2i\sin\left(\frac{3\pi}{13}\right)}+\underbrace{\alpha^4+\overbrace{\quad\alpha^9\quad}^{-\alpha^{-4}}}_{2i\sin\left(\frac{4\pi}{13}\right)}\tag3
$$
Using the approach in this answer, we get that the minimal polynomial of $x=2is$ is
$$
x^5+13x^4+13x=0\tag4
$$
Since we don't want $s=0$, we have
$$
x^4+13x^2+13=0\tag5
$$
which has roots
$$
x\in\left\{\pm i\sqrt{\frac{13\pm3\sqrt{13}}2}\right\}\tag6
$$
Since $\frac2\pi\theta\le\sin(\theta)\le\theta$ for $0\le\theta\le\frac\pi2$, we have $\frac{16}{13}\lt s\lt\frac{8\pi}{13}$. Therefore, we must have
$$
s=\sqrt{\frac{13+3\sqrt{13}}8}\tag7
$$
