I recently came across a sum (whose closed-form solution I was able to verify via Wolfram Alpha) but have no idea how to get there.
$$\sum\limits_{k=1}^{N-1}\left[\frac{\sin\left(\frac{\pi km}{N}\right)}{\sin\left(\frac{\pi k}{N}\right)}\right]^{2}=m(N-m)$$
I'm pretty sure it's valid for any $m\in 0,1,2\ldots N-1$ and $N\geq2$
My only ideas were to make exponential substitutions or to bring an integral into the mix like this one $$\frac{\pi^{2}}{\sin^{2}(\pi s)}=\int_{0}^{\infty}\frac{x^{s-1}}{1-x}\ln\left(\frac{1}{x}\right)\mathrm dx$$
I couldn't get very far. I figure there's a better way to simplify the fraction of sine functions since they differ only by a dilation of $m$. Let me know what you guys think!