Xjyuk

Empirical evidence suggests that, for each positive integer $n$, the following equality holds:
beginequation*
prod_s=1^2nsum_k=1^2n(-i)^ksinfracskpi2n+1=(-1)^nfrac2n+12^n,
endequation*
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
$$
sum_k=0^2n(-i)^ksinfracskpi2n+1=frach(s)-h(-s)2i,quadtextwhere\
h(s)=sum_k=0^2ne^i(-pi/2+fracpi s2n+1)k=frac1-e^-ipi(2n+1)/2+ipi s1-e^i(-pi/2+fracpi s2n+1)=frac1+i(-1)^n+s1+ie^ifracpi s2n+1.
$$
The numerators for $s$ and $-s$ are the same, and
$$
frac11+ie^itheta-
frac11+ie^-itheta=frac2sintheta2icos theta=-itantheta,
$$
so the product reads as
$$
2^-2nprod_s=1^2n (1+i(-1)^n+s)tan fracspi2n+1.
$$
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
$$
prod_s=1^2ntan fracspi2n+1=(-1)^n(2n+1). quadquad (ast)
$$
This should be well known, and in any case it is standard:
using the formula
$$
itan theta=frace^2itheta-1e^2itheta+1
$$
we get
$$
(-1)^nprod_s=1^2ntan fracspi2n+1=prod_s=1^2n
fracomega^s-1omega^s+1,quadtextwhere, omega=e^2pi i/(2n+1).
$$
We have $prod_s=1^2n(z-omega^s)=1+z+ldots+z^2n=:P(z)$, therefore $$prod_s=1^2n
fracomega^s-1omega^s+1=frac
P(1)P(-1)=2n+1$$