trigonometric identity involving product of sines of roots of unity

Let $n>1$ be a positive integer, and ${\zeta }_n=e^{2i\pi /n}$, a primitive $n^{\mathrm{th}}$ root of unity.

The purpose of this article is to prove

Theorem 1.

Let $m\mathrm{=}\mathrm{\lfloor }\frac{n}{\mathrm{2}}\mathrm{\rfloor }$. Then

$$\prod _{k=1}^m{\sin }^2\left(\frac{\pi k}{n}\right)=\prod _{k=1}^{n-1}\sin \left(\frac{\pi k}{n}\right)=\frac{n}{2^{n-1}}$$(1)

The theorem follows easily from the following simple lemma:

Lemma 2.

Let $n\mathrm{>}\mathrm{1}$ be a positive integer. Then

$$\prod _{k=1}^{n-1}(1-{\zeta }_n^k)=n$$

Proof.

We have $x^n-1={\prod }_{k=1}^n(x-{\zeta }_n^k)$. Dividing both sides by $x-1$ gives

$$\frac{x^n-1}{x-1}=1+x+x^2+\mathrm{\dots }{x}^{n-1}=\prod _{k=1}^{n-1}(x-{\zeta }_n^k)$$

Substitute $x=1$ to get the result.∎

Proof of Theorem 1.

Using the definition of ${\zeta }_n$ and the half-angle formulas, we have

$1-{\zeta }_n^k$$=1-\cos \left({\displaystyle \frac{2\pi k}{n}}\right)-i\sin \left({\displaystyle \frac{2\pi k}{n}}\right)$

$=2{\sin }^2\left({\displaystyle \frac{\pi k}{n}}\right)-2i\sin \left({\displaystyle \frac{\pi k}{n}}\right)\cos \left({\displaystyle \frac{\pi k}{n}}\right)$

$=2\sin \left({\displaystyle \frac{\pi k}{n}}\right)\left(\sin \left({\displaystyle \frac{\pi k}{n}}\right)-i\cos \left({\displaystyle \frac{\pi k}{n}}\right)\right)$

Note that $|\sin \theta -i\cos \theta |={\sin }^2\theta +{\cos }^2\theta =1$, so taking absolute values, we get

$$\left|1-{\zeta }_n^k\right|=2\left|\sin \left(\frac{\pi k}{n}\right)\right|$$

Now, for $1\le k\le n-1$, $\sin \left(\frac{\pi k}{n}\right)>0$ so is equal to its absolute value. Thus (using, for $n$ even, the fact that $\sin \frac{\pi }{2}=1$),

${\displaystyle \prod _{k=1}^m}{\sin }^2\left({\displaystyle \frac{\pi k}{n}}\right)$$={\displaystyle \prod _{k=1}^m}\sin \left({\displaystyle \frac{\pi k}{n}}\right)\sin \left(\pi -{\displaystyle \frac{\pi k}{n}}\right)={\displaystyle \prod _{k=1}^m}\sin \left({\displaystyle \frac{\pi k}{n}}\right)\sin \left({\displaystyle \frac{\pi (n-k)}{n}}\right)$

$={\displaystyle \prod _{k=1}^{n-1}}\sin \left({\displaystyle \frac{\pi k}{n}}\right)={\displaystyle \prod _{k=1}^{n-1}}\left|\sin \left({\displaystyle \frac{\pi k}{n}}\right)\right|$

$={\displaystyle \frac{1}{2^{n-1}}}\left|{\displaystyle \prod _{k=1}^{n-1}}(1-{\zeta }_n^k)\right|={\displaystyle \frac{1}{2^{n-1}}}\left|n\right|$

$={\displaystyle \frac{n}{2^{n-1}}}$

∎

(Thanks to dh2718 for greatly simplifying the original proof.)