Prove that $\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}=4\sqrt{3}$ I'm trying to calculate the expression: $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ and show that it is equal $4\sqrt{3}$. 
I was trying to group the summands and calculate sums of $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}} \hspace{0.5cm}\text{and} \hspace{0.5cm} -\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ where we get $$\frac{2\cos\frac{2\pi}{15}+1}{\sin\frac{2\pi}{15}}-\frac{2\cos\frac{4\pi}{15}-1}{\sin\frac{8\pi}{15}}$$ but unfortunately this sum is not simplified.
How to prove this equality?

Well, this is not a full solution (I will not do the calculations), but I guess this is one of the ways to reach the end ...

*

*$\sin\left(\frac{\pi}{15}\right) = \sin\left(\frac{\pi}{6} - \frac{\pi}{10}\right)$

*$\sin\left(\frac{\pi}{10}\right) = \sin\left(\frac{1}{2}\cdot\frac{\pi}{5}\right)$

*for the $\frac{\pi}{5}$ use the identity: $\sin(5x) = 5\sin(x)-20\sin^3(x) + 16\sin^5(x)$. It has a straightforward solution

Rest of the angles are possible to calculate with the double angle formula.
99,9% not indended to be solved this way, but ...
