Prove the identity: $\displaystyle \prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{2N} \right) = N$

Question

Context: trying to diagonalize a sparse matrix here $$ \begin{pmatrix} 2 \eta^2 & -\eta & & & -\eta \\ -\eta & 2 & -1 & & \\ & & \ddots & & \\ & & -1 & 2 & -1 \\ -\eta & & & -1 & 2 \\ \end{pmatrix} $$

By using the constant term of the characteristic polynomial, I got the identity below.

$$\prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{2N} \right) = N$$

If anyone has a nice proof for it, do share!

E.g. of numerical verification WolframAlpha: $N=101$

Answer 1

Note that $$\sin^2 \frac{k\pi}{2N}=\frac{1-\cos\frac{k\pi}{N}}2,~~~\cos^2 \frac{k\pi}{2N}=\frac{1+\cos\frac{k\pi}{N}}2$$ we have:

$$\cos\frac{k\pi}N=-\cos\frac{(N-k)\pi}N$$

hence, $$1-\cos\frac{k\pi}N=1+\cos\frac{(N-k)\pi}N$$ From symmetry, we have $$\prod_{k=1}^{N-1} \left( 1-\cos\frac{k\pi}{N} \right) =\prod_{k=1}^{N-1} \left( 1+\cos\frac{(N-k)\pi}N\right)= \prod_{k=1}^{N-1} \left( 1+\cos\frac{k\pi}{N} \right)$$

Therefore, we get

$$\prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{2N} \right) = \prod_{k=1}^{N-1} \left( 4 \cos^2 \frac{k\pi}{2N} \right)=P$$

Next, multiply them, and use $\sin2\theta=2\sin\theta\cos\theta$

$$P^2=\prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{2N} \right) \prod_{k=1}^{N-1} \left( 4 \cos^2 \frac{k\pi}{2N} \right)=\prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{N} \right)=\left[\prod_{k=1}^{N-1} \left( 2 \sin \frac{k\pi}{N} \right)\right]^2$$

We know: $\displaystyle \prod_{k=1}^{N-1}\left(2\sin\frac{k \pi}{N} \right)= N$, hence

$$P^2=N^2$$

Finally,

$$\boxed{\prod_{k=1}^{N-1} \left( 4 \sin^2 \frac{k\pi}{2N} \right) = \prod_{k=1}^{N-1} \left( 4 \cos^2 \frac{k\pi}{2N} \right)=N}$$

Answer 2

Note that the product is a real positive number $P$. Using euler's theorem, the product equals $$P=\prod_{k=1}^{n-1}(-(e^{ik\pi/2n}-e^{-ik\pi/2n})^2)$$ $$=\prod_{k=1}^{n-1}-e^{-ik\pi/n}(e^{ik\pi/n}-1)^2$$ $$=\prod_{k=1}^{n-1}-e^{-ik\pi/n}(e^{ik\pi/n}-1)^2$$ $$=(-1)^{n-1}e^{i\pi(n+1)/2}\prod_{k=1}^{n-1}(e^{ik\pi/n}-1)^2$$ $$=\prod_{k=1}^{n-1}(e^{ik\pi/n}-1)^2$$ Now we compute $P\overline{P}$ $$P\overline{P}=\left(\prod_{k=1}^{n-1}(e^{ik\pi/n}-1)^2\right)\left(\prod_{k=1}^{n-1}(e^{-ik\pi/n}-1)^2\right)$$ $$=\frac{1}{4}\left(\prod_{k=1}^{2n-1}(e^{ik\pi/n}-1)^2\right)$$ $$=\frac{1}{4}\left(\prod_{k=1}^{2n-1}(1-e^{ik\pi/n})^2\right)$$ If $f(x)$ is the polynomial with roots $e^{ik\pi/n}$ for all $k\in [2n-1]$, we know $f(x)=1+x+\cdots+x^{2n-1}$, hence the product is equal to $$=\frac{1}{4}f(1)^2$$ $$=\frac{1}{4}(2n)^2$$ $$=n^2$$ Since $P$ is real and positive, we must have $P=n$.