The product cos(π/N)·cos(2π/N)···cos(nπ/N) with N = 2n + 1

Idea. Doubling the angle turns the cosine product into a sine product that can be compared with the original sine product.

Step 1. Let $S=\prod_{k=1}^n\sin\frac{k\pi}N>0$ and $C=\prod_{k=1}^n\cos\frac{k\pi}N$. Then
$$\prod_{k=1}^n\sin\frac{2k\pi}N=2^nSC.$$

Step 2: the left side equals $S$. For each $k\in\{1,\dots,n\}$, either $2k\le n$, then $\sin\frac{2k\pi}N$ is already a factor of $S$; or $2k>n$, then
$$\sin\frac{2k\pi}N=\sin\left(\pi-\frac{2k\pi}N\right)=\sin\frac{(N-2k)\pi}N,\qquad N-2k=2(n-k)+1\in\{1,3,\dots\}.$$
Here $j=N-2k$ satisfies $1\le j\le n$. The indices $2k$ (for $2k\le n$) are the even numbers in $\{1,\dots,n\}$; the indices $N-2k$ (for $2k>n$) are the odd numbers in $\{1,\dots,n\}$. Each integer in $\{1,\dots,n\}$ appears exactly once, so the left side equals $S$.

Step 3. Therefore $2^nSC=S$, and since $S\ne0$, $C=2^{-n}$. $\blacksquare$

Check. $n=3$ ($N=7$): $\cos\frac\pi7\cos\frac{2\pi}7\cos\frac{3\pi}7=\frac18$ ✓.

Lesson. To evaluate a product of trigonometric values at equally spaced angles, double the angles and show that the new product is a permutation of the old one.