Is there an elementary way to understand this trigonometric identity? Let $\{k_r\}$ be the $N$ solutions between $0$ and $\pi$ of $\cos(N k) = h$, where $\lvert h \rvert <1$.
I have come across the following identity in a physics research problem:
\begin{align}
 \Bigg(\prod_{r=1}^N 2 \sin(k_r) \prod_{1\leq r<s\leq N}  
\Bigg(\frac{\sin^2 (\frac{k_r-k_s}{2})}{\sin^2 (\frac{k_r+k_s}{2})} \Bigg)^{(-1)^{r+s+1}}\Bigg)^{1/N}=\frac{2\sqrt{1-h^2}}{N}.
\end{align}
I am able to prove it by calculating the same quantity in two different ways, but I was wondering if there is an elementary way to understand it? If anyone knows a reference that contains such an identity I would be interested.
 A: This trigonometric identity does have an elementary proof. The physical context and the argument below is given in arxiv:2202.11728.
The identity in the end comes down to the simple identity $\sin(N z) = 2^{N-1} \prod_{k=0}^{N-1} \sin(z+\frac{k \pi}{N})$ (this is DLMF Eq. 4.21.35) which can be derived by writing $(z^{2n}-1) = \prod_{j=0}^{2n-1} (z-e^{i \pi j/n})$.
Let us consider $\cos(Nk)= h = \cos(\alpha)$, for $0<\alpha<\pi$. Then we have solutions between $0$ and $\pi$ as follows:
\begin{align}
k_{2n+1} = \frac{\alpha}{N} +\frac{2\pi n}{N} \qquad& n=0,\dots, \lceil N/2\rceil -1 \\
k_{2n} = - \frac{\alpha}{N} +\frac{2\pi n}{N} \qquad& n=1,\dots, \lfloor N/2\rfloor.
\end{align}
Another way to write the identity in the question is:
\begin{align}
 \Bigg(\prod_{r=1}^N 2 \sin(k_r) \prod_{1\leq r<s\leq N}  
\Bigg(\frac{\sin^2 (\frac{k_r+k_s}{2})}{\sin^2 (\frac{k_r-k_s}{2})} \Bigg)^{(-1)^{r+s}}\Bigg)^{1/N}=\frac{2\sin(\alpha)}{N}.
\end{align}
Inserting the explicit form for $k_n$ given above and simplifying, this is equivalent to:
\begin{align}
&\sin \big({\alpha}/{N}\big)\prod_{s=1}^{N-1}\Bigg(\frac{
{\sin \big(\frac{\alpha}{N}+ \frac{\pi s}{N}\big)}}{{\sin \big(\frac{\pi s}{N}\big)}}\Bigg)= \frac{\sin(\alpha)}{N},\end{align}
which in turn follows from the simple identity given above.
