Prove that $\sum_{l=0}^{N-1}\frac{\sin^2(\pi x)}{\sin^2(\frac{\pi}{N}(l-k+x))}=N^2$ [closed]

Question

When I numerically compute the sum below it is always $1$. How can I prove this? $N$ is an integer number and $k$ is an integer number between $0$ to $N-1$ and $x$ is real number between $0$ and $0.5$ $${\frac{1}{N^2}\sum_{l=0}^{N-1}\frac{\sin^2(\pi x)}{\sin^2(\frac{\pi}{N}(l-k+x))}}$$

Answer 1

It is sufficient to handle the case $k=0$, i.e. to show that $$\sum_{n=0}^{N-1}\frac1{\sin^2\frac\pi N(n+x)}=\frac{N^2}{\sin^2\pi x},$$ since the general case follows if you put $l=(n+k)\bmod N$.

Recall that $\cos Nt$ is a polynomial in $\cos t$ (of degree $N$). Hence $$\sin^2 Nt-\sin^2\pi x=P_N(\sin^2 t)$$ with a polynomial $P_N$ of degree $N$ (whose coefficients may depend on $x$).

Let $t_n=(n+x)\pi/N$ for $0\leqslant n<N$. Then $\sin^2 Nt_n=\sin^2\pi x$, hence $y_n=\sin^2 t_n$ are roots of $P_N(y)$, and in fact different ones (because of $0<x<1/2$, and more generally $2x\notin\mathbb{Z}$). Thus $$\sin^2 Nt-\sin^2\pi x=A(x)\prod_{n=0}^{N-1}\left(\sin^2 t-\sin^2\frac\pi N(n+x)\right),$$ where $A(x)$ depends on $x$ only. Taking $\partial/\partial t$, we get $$\frac{N\sin 2Nt}{\sin^2 Nt-\sin^2\pi x}=\sum_{n=0}^{N-1}\frac{\sin 2t}{\sin^2 t-\sin^2\frac\pi N(n+x)}.$$

It remains to divide this by $\sin 2t$ and take $t\to 0$.