How to prove an identity (Trigonometry Angles--Pi/13) In this page http://mathworld.wolfram.com/TrigonometryAnglesPi13.html
I found equation (11) and (12).
$$\cos^2\frac{\pi}{13}+\cos^2\frac{3\pi}{13}+\cos^2\frac{4\pi}{13}=\frac{11+\sqrt{13}}{8}$$
$$\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\sqrt{\frac{13+3\sqrt{13}}{8}}$$
How to prove it ?
Thanks in advances

For the first one, let $\zeta=e^{2\pi i/13}$.  This is a thirteenth root of $1$, so $1+\zeta+\zeta^2+...+\zeta^{12}=0$.
\begin{align}
A &=\cos^2(\pi/13)+\cos^2(3\pi/13)+\cos^2(4\pi/13)\\
B &=2A-3=\cos(2\pi/13)+\cos(6\pi/13)+\cos(8\pi/13)\\
2B &=\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^4+\zeta^{-4}\\
C &=-1-2B=\zeta^2+\zeta^{-2}+\zeta^5+\zeta^{-5}+\zeta^6+\zeta^{-6}\\
2BC &=3(\zeta+\zeta^2+...+\zeta^{12})=-3\\
& 2B(2B+1) *=3
\end{align}
