TITLE: Finding $\prod_{k=1}^{n-1}\cos\frac{2k\pi}n$
QUESTION [5 upvotes]: Finding $$\mu=\prod_{k=1}^{n-1}\cos\frac{2k\pi}n$$
I thought $$z^n=1=e^{i2\pi}\implies z=\cos\frac{2k\pi}n+i\sin\frac{2k\pi}n\quad k\in\{1,2,...,n-1\}$$
Now we have:
$$\begin{align}\mu&=\prod_{k=1}^{n-1}\frac{z+\bar z}{2}\\&=\prod_{k=1}^{n-1}\frac{e^{i\theta}+ e^{-i\theta}}{2}\\&=\frac1{2^{n-1}}\prod_{k=1}^{n-1}e^{-i\theta}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\
\\&=\frac1{2^{n-1}}e^{-i\frac{2\pi}n(1+2+...(n-1))}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\
\\&=\frac1{2^{n-1}}e^{-i(n-1)\pi}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\
\\&=\frac1{2^{n-1}}(-1)^{n-1}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\end{align}$$
Now since $(e^{2i\theta})^{n/2}=1$:
$$\Theta(z)=z^{n/2}-1=\prod_{k=1}^{n-1}(z-e^{2i\theta})$$
At $(-1)$:
$$\Theta(-1)=(-1)^{n-1}\prod_{k=1}^{n-1}(1+e^{2i\theta})=(-1)^{n/2}-1=e^{in\pi/2}-1=i\sin n\pi/2$$
It turns out something imaginary (but it's not clearly) also I would like to work on:
$$\nu=\prod_{k=1}^{n-1}\sin\frac{2k\pi}n=\prod_{k=1}^{n-1}\frac{e^{i\theta}- e^{-i\theta}}{2i}$$
Similarly:
$$\nu=\frac1{(2i)^{n-1}}(-1)^{n-1}\prod_{k=1}^n(e^{2i\theta}-1)=\frac1{(2i)^{n-1}}(-1)^{n-1}(-1)^{n-1}\Theta(1)=\frac1{2^{n-1}i\sin (n-1)\pi/2}0=0$$
But this is also not zero?
The earlier $\eta$ is obviously:
$$\prod_{k=1}^{n-1}\tan\frac{2k\pi}n=\eta=\frac{\mu}{\nu}$$

REPLY [7 votes]: $$
\begin{align}
2^{n-1}\prod_{k=1}^{n-1}\cos\left(\frac{2k\pi}n\right)
&=\prod_{k=1}^{n-1}\left(1+e^{4\pi ik/n}\right)e^{-2\pi ik/n}\\
&=(-1)^{n-1}\left[\prod_{k=1}^{n-1}\left(-z^2+e^{4\pi ik/n}\right)\right]_{\large z=i}\\
&=\left[\prod_{k=1}^{n-1}\left(z^2-e^{4\pi ik/n}\right)\right]_{\large z=i}\\
&=\left[\prod_{k=1}^{n-1}\left(z-e^{2\pi ik/n}\right)\prod_{k=1}^{n-1}\left(z+e^{2\pi ik/n}\right)\right]_{\large z=i}\\[3pt]
&=\left[\frac{z^n-1}{z-1}\frac{z^n-(-1)^n}{z+1}\right]_{\large z=i}\\[6pt]
&=\frac{(i^n-1)(i^n-(-1)^n)}{-2}\\[9pt]
&=\frac{2(-1)^n-i^n(1+(-1)^n)}{-2}\\[12pt]
&=\cos(n\pi/2)-(-1)^n
\end{align}
$$
Therefore,
$$
\begin{align}
\prod_{k=1}^{n-1}\cos\left(\frac{2k\pi}n\right)
&=\frac{\cos(n\pi/2)-(-1)^n}{2^{n-1}}\\[3pt]
&=\frac{\cos(n\pi/2)-\cos(n\pi)}{2^{n-1}}\\[6pt]
&=\frac{\sin(n\pi/4)\sin(3n\pi/4)}{2^{n-2}}
\end{align}
$$