<s>[INST] Sum of the reciprocal of sine squared I encountered an interesting identity when doing physics homework, that is,
$$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$
How is this identity derived? Are there any more related identities?
 [/INST] Here it is another approach, just for fun. We may notice that $\cos\frac{\pi}{n+1},\cos\frac{2\pi}{n+1},\ldots,\cos\frac{n\pi}{n+1}$ are the roots of the polynomial
$$ U_n(x) = \sin((n+1)\arccos x)/\sqrt{1-x^2} $$
hence
$$ S_n=\sum_{k=1}^{n-1}\frac{1}{\sin^2\frac{\pi k}{n}}=\frac{1}{2\pi i}\oint \frac{1}{1-z^2}\cdot\frac{U_{n-1}'(z)}{U_{n-1}(z)}\,dx $$
where the integral is performed along a simple contour enclosing $[-1+\varepsilon,1-\varepsilon]$. By enforcing the substitution $z=\cos\theta$ we are left with
$$ S_n = -\frac{1}{2\pi i}\oint \frac{\cot\theta-n\cot(n\theta)}{\sin^2\theta}\,d\theta $$
where the integral is performed along a simple contour enclosing $[\varepsilon,\pi-\varepsilon]$. Due to the periodicity of the integrand function and the fact that the sum of the residues of such meromorphic function has to be zero we have
$$ S_n = \operatorname*{Res}_{z=0}\frac{\cot z-n\cot(n z)}{\sin^2 z} = [z^{2}]\frac{z\cot z-nz\cot(n z)}{\left(\frac{\sin z}{z}\right)^2}=(1-n^2)\cdot[z^2]z\cot z$$
and we may finish be considering that in a neighbourhood of the origin, up to a term $O(z^3)$,
$$ z\cot z = \frac{z\cos z}{\sin z} \equiv \frac{z\left(1-\frac{z^2}{2}\right)}{z\left(1-\frac{z^2}{6}\right)}\equiv 1-\frac{z^2}{3} $$
such that $S_n = \frac{n^2-1}{3}$.
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