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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.

Nick Jones
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1 Answers1

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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.

Nick Jones
  • 31