How to prove those “curious identities”?

How to prove $$\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and $$\prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$

• I would have thought that you might have learned from your previous experience here that it's a good idea to say something about where you came across these identities, as it might point the way toward an answer. – Gerry Myerson Oct 6 '11 at 2:25
• +1 for a question that generated a variety of good answers! – lhf Oct 6 '11 at 2:57
• FYI, if you like those kind of identities, those, and some other similar ones, are in "Challenging Mathematical Problems with Elementary Solutions" by Yaglom and Yaglom, volume 2, which is available in an inexpensive Dover edition. – tzs Oct 6 '11 at 4:42
• Related: Similar reasoning as in some of the answers below (Euler's formulas + geometric series) proves the nice but no so widely known multiple-angle formula $$\sin nx = 2^{n-1} \prod_{k=0}^{n-1} \sin(x + \frac{k\pi}{n}).$$ Your first formula can be obtained as a special case after dividing both sides by $\sin x$ and taking the limit as $x\to 0$. – Hans Lundmark Oct 6 '11 at 7:50
• Half-duplicate of math.stackexchange.com/questions/8385/… – Hans Lundmark Mar 20 '15 at 8:04

For the first: $$\lim_{z=1}\frac{z^n-1}{z-1}=n\tag{1a}$$ $$\frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}(z-e^{2\pi ik/n})\tag{1b}$$ $$|1-e^{i2k\pi/n}|=|2\sin(k\pi/n)|\tag{1c}$$ Combining $(1a)$, $(1b)$, and $(1c)$, we get $$2^{n-1}\prod_{k=1}^{n-1}\sin(k\pi/n)=n$$ since everything is positive.

For the second:

If $n$ is even, then $\cos(\frac{\pi}{2})=0$ appears in the product (when $k=n/2$) and $\sin(\frac{n\pi}{2})=0$.

If $n$ is odd, then combining $$\lim_{z=1}\frac{z^n+1}{z+1}=1\tag{2a}$$ $$\frac{z^n+1}{z+1}=\prod_{k=1}^{n-1}(z+e^{2\pi ik/n})\tag{2b}$$ $$1+e^{i2k\pi/n}=2\cos(k\pi/n)e^{ik\pi/n}\tag{2c}$$ and noting that $\displaystyle\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$ so that $\displaystyle\prod_{k=1}^{n-1}e^{ik\pi/n}=(-1)^{(n-1)/2}$ which matches the sign of $\sin(\pi n/2)$, yields $$2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n)=(-1)^{(n-1)/2}=\sin(\pi n/2)$$

• How do you prove (1b)? Can you recommend a page for me to see, or the name of that theorem? Thanks. – D.R. Sep 7 '17 at 1:39
• $\text{(1b)}$ is just noting that the roots of $z^n-1$ are the $n^\text{th}$ roots of unity. Dividing by $z-1$ removes the root at $1$ and so we're left with the other $n-1$ roots $\left\{e^{2\pi ik/n}:1\le k\le n-1\right\}$. – robjohn Sep 7 '17 at 2:04

Denote $w = e^{i \pi/n}$. We have

$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$

Since we have

$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$

Setting $x=1$ yields

$$\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$

So we get

$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \frac{n}{2^{n-1}} \frac{w^{n(n-1)/2}}{i^{n-1}} = \frac{i^{n-1}}{i^{n-1}} \frac{n}{2^{n-1}} = \frac{n}{2^{n-1}}$$

I guess (but did not check) that the same kind of reasoning gives the one with $\cos$.

• similar reasoning for odd $n$ (except you have to watch the sign), but very different, however easy, reasoning for even $n$. – robjohn Oct 6 '11 at 4:03

The second purported identity is equivalent to asking for the constant term of $\dfrac{U_{n-1}(x)}{2^{n-1}}$ (i.e., $\dfrac{U_{n-1}(0)}{2^{n-1}}$), where $U_n(x)$ is the Chebyshev polynomial of the second kind. Since

$$\frac{U_{n-1}(x)}{2^{n-1}}=\frac{\sin(n \arccos\,x)}{2^{n-1}\sqrt{1-x^2}}$$

letting $x=0$ gives your identity.

• This is really clever. – Joel Cohen Oct 6 '11 at 3:13

Define $\zeta_n = e^{2 \pi i/n}$.

Proposition For odd integer $n \geq 1$, \begin{align} \prod_{k = 1}^{n-1}(\zeta_n^{k} - \zeta_n^{-k}) = n. \end{align} and \begin{align} \prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ) = \tfrac{n}{(2 i)^{n-1}}. \end{align} Proof: The claimed identities follow from the identity \begin{align} z^n - 1 = \prod_{ k =0}^{n-1} (z - \zeta_n^{k}) = \prod_{ k =0}^{n-1} (z - \zeta_n^{-2k}). \end{align} Writing $z = x/y$, we have \begin{align} x^n - y^n = \prod_{k = 0}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y). \end{align} Thus, \begin{align} n y^{n-1} = \lim_{x \to y} \frac{x^n - y^n}{x - y} = \lim_{x \to y} \ \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y) = y^{n-1} \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} - \zeta_n^{-k} ). \end{align} For the second identity, let $x =e^{\pi i z}$ and $y = e^{- \pi i z}$ and recall the complex exponential representation of the sine function. This yields \begin{align} n = \lim_{z \to 0} \frac{\sin n \pi z}{\sin z } = (2 i)^{n-1} \lim_{z \to 0} \ \ \prod_{k = 1}^{n-1} \sin( \pi z + \tfrac{2 \pi k }{n} ) = (2 i)^{n-1} \prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ). \end{align}

Similar reasoning works to prove the identities that you mention.