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?



An equality about sin function?
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 $$\prod\_{s=1}^{2n}
\frac{\omega^s-1}{\omega^s+1}=\frac
{P(1)}{P(-1)}=2n+1$$


