# 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.

###### Theorem 1.

Let $m=\left\lfloor\frac{n}{2}\right\rfloor$. 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 $n>1$ 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

 $\frac{x^{n}-1}{x-1}=1+x+x^{2}+\dots x^{n-1}=\prod_{k=1}^{n-1}(x-\zeta_{n}^{k})$

Substitute $x=1$ to get the result. ∎

###### Proof of Theorem 1.

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

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

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

 $\left\lvert 1-\zeta_{n}^{k}\right\rvert=2\left\lvert\sin\left(\frac{\pi k}{n}% \right)\right\rvert$

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

 $\displaystyle\prod_{k=1}^{m}\sin^{2}\left(\frac{\pi k}{n}\right)$ $\displaystyle=\prod_{k=1}^{m}\sin\left(\frac{\pi k}{n}\right)\sin\left(\pi-% \frac{\pi k}{n}\right)=\prod_{k=1}^{m}\sin\left(\frac{\pi k}{n}\right)\sin% \left(\frac{\pi(n-k)}{n}\right)$ $\displaystyle=\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\prod_{k=1}^{n% -1}\left\lvert\sin\left(\frac{\pi k}{n}\right)\right\rvert$ $\displaystyle=\frac{1}{2^{n-1}}\left\lvert\prod_{k=1}^{n-1}(1-\zeta_{n}^{k})% \right\rvert=\frac{1}{2^{n-1}}\left\lvert n\right\rvert$ $\displaystyle=\frac{n}{2^{n-1}}$

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

Title trigonometric identity involving product of sines of roots of unity TrigonometricIdentityInvolvingProductOfSinesOfRootsOfUnity 2013-03-22 19:00:03 2013-03-22 19:00:03 rm50 (10146) rm50 (10146) 11 rm50 (10146) Theorem msc 26A09 msc 33B10