Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
\begin{equation*}
\prod_{s=1}^{2n}\sum_{k=1}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=(-1)^n\frac{2n+1}{2^n},
\end{equation*}
where $i=\sqrt{-1}$.
Is it a known equality? If it is true, would you please give me some insights on how to derive this equality?
We have $$ \sum_{k=0}^{2n}(-i)^k\sin\frac{sk\pi}{2n+1}=\frac{h(s)-h(-s)}{2i},\quad\text{where}\\ h(s)=\sum_{k=0}^{2n}e^{i(-\pi/2+\frac{\pi s}{2n+1})k}=\frac{1-e^{-i\pi(2n+1)/2+i\pi s}}{1-e^{i(-\pi/2+\frac{\pi s}{2n+1})}}=\frac{1+i(-1)^{n+s}}{1+ie^{i\frac{\pi s}{2n+1}}}. $$ The numerators for $s$ and $-s$ are the same, and $$ \frac1{1+ie^{i\theta}}- \frac1{1+ie^{-i\theta}}=\frac{2\sin\theta}{2i\cos \theta}=-i\tan\theta, $$ so the product reads as $$ 2^{-2n}\prod_{s=1}^{2n} (1+i(-1)^{n+s})\tan \frac{s\pi}{2n+1}. $$ The product of $(1+i(-1)^{n+s})$ equals $2^n$, since the product of two consecutive guys equals 2. It remains to prove that $$ \prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=(-1)^n(2n+1). $$ This should be well known, and in any case it is standard: using the formula $$ i\tan \theta=\frac{e^{2i\theta}-1}{e^{2i\theta}+1} $$ we get $$ (-1)^n\prod_{s=1}^{2n}\tan \frac{s\pi}{2n+1}=\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1},\quad\text{where}\, \omega=e^{2\pi i/(2n+1)}. $$ We have $\prod_{s=1}^{2n}(z-\omega^s)=1+z+\ldots+z^{2n}=:P(z)$, therefore $$\prod_{s=1}^{2n} \frac{\omega^s-1}{\omega^s+1}=\frac {P(1)}{P(-1)}=2n+1$$
