Sum of trigonometric series $\sum_{m=1}^{N-1} \frac{\sin(4\pi mk/N)}{\sin ^2 (\pi m/N) }$

Question

Anybody has some ideas to prove the following identity?

\begin{equation} \sum_{m=1}^{N-1} \frac{\sin(4\pi mk/N)}{\sin ^2 (\pi m/N) }= 0 \end{equation}

where $N$ is an integer greater than $1$, $k$ could be any integer ranging from $1$ to $N$.

I programmed to show that is true, but how can it be proved?

Answer 1

Consider the function $$f(x) = \frac{\sin 4\pi x}{\sin^2 \pi x}$$ It is odd: $f(-x)=-f(x)$ and $1$-periodic: $f(x+1)=f(x)$. Therefore, $$f(x)+f(1-x)=0\quad \text{ for all } x$$ Your sum consists of the values of this function at the points $$ \frac{1}{N},\frac{2}{N},\frac{3}{N},\dots , 1-\frac{3}{N}, 1-\frac{2}{N}, 1-\frac{1}{N} $$