<s>[INST] 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
$$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
 [/INST] 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.
</s>