How to prove this trigonometric identity? $$\sum_{k=1}^n\frac{1}{\displaystyle \sin^4\left(\frac{k\pi}{2n+1}\right)}=\frac8{45}n(n+1)(n^2+n+3)$$
I think it maybe use 
$\cos(nx)+i\sin(nx) = \displaystyle\sum_{k=0}^n\binom nki^k(\sin^kx)(\cos^{n-k}x)$.
 But I can't .
 A: From this, $$\sin(2n+1)x$$
$$=(2n+1)s-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^3+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^5+\cdots+(-1)^n2^{2n}s^{2n+1}$$
If $\displaystyle\sin(2n+1)x=0,(2n+1)x=k\pi$ where $k$ is any integer
$\displaystyle\implies x=\frac{k\pi}{2n+1}$ where $-n\le k\le n$
So,  $\displaystyle\sin\frac{k\pi}{2n+1},-n\le k\le n$ are the roots of $$(2n+1)s-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^3+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^5+\cdots+(-1)^n2^{2n}s^{2n+1}=0$$
So,  $\displaystyle\sin\frac{k\pi}{2n+1},-n\le k\le n,k\ne0$ are the roots of $$(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}s^2+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}s^4+\cdots+(-1)^n2^{2n}s^{2n}=0$$
If we set $\displaystyle\sin^2\frac{k\pi}{2n+1}=\frac1{y_k},1\le k\le n$
$y_k$ will be roots of $$(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}\frac1t+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}\frac1{t^2}+\cdots+(-1)^n2^{2n}\frac1{t^n}=0$$
$$\iff t^n(2n+1)-\frac{(2n+1)\{(2n+1)^2-1^2\}}{3!}t^{n-1}+\frac{(2n+1)\{(2n+1)^2-1^2\}\{(2n+1)^2-3^2\}}{5!}t^{n-2}+\cdots+(-1)^n2^{2n}=0$$
Now, we need $\displaystyle\sum_{k=1}^ny_k^2=\left(\sum_{k=1}^n y_k\right)^2-2\sum_{i,j=1,i\ne j}^ny_iy_j$ which can be safely managed by Vieta's Formula
