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

#### Solutions Collecting From Web of "How to prove those “curious identities”?"

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
$$2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n)=(-1)^{(n-1)/2}=\sin(\pi n/2)$$

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 $\cos$.

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.

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.