I know that the first one of the following identities holds, yet I don't know the identity of the general case as shown in the title and the bottom of the page. Is there anyone who knows the closed form of the product of the above formula? If you know how to prove, I would like you to show some proof. You don't need to consider the case such that the product becomes zero. Thanks. $$\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)=\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$


The first identity's proof is available here. Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$


Proceeding with the complex exponential formula for $\cos$,

$$\prod_{k=1}^{n-1}\cos\left(\theta+\frac{k\pi}{n}\right)=\prod_{k=1}^{n-1}\frac{\zeta ^ke^{i\theta}+\zeta^{-k}e^{-i\theta}}{2}=\frac{1}{2^{n-1}}\left[\prod_{k=1}^{n-1}\zeta^{-k}e^{-i\theta}\right]\prod_{k=1}^{n-1}\left(1+\zeta^{2k}e^{2i\theta}\right).$$

In order to evaluate the last product, consider it a function of $z=e^{2i\theta}$ and further consider

$$P(w)=\prod_{k=1}^{n-1}\left(w-\zeta^{2k}\right)=\begin{cases}\displaystyle\frac{w^n-1}{w-1} & k~~{\rm odd} \\[5pt] \displaystyle\frac{(w^{n/2}-1)^2}{w-1} & k~~{\rm even}.\end{cases}$$

  • $\begingroup$ Thanks. I actually calculated the answer in different method and got $-\dfrac{\sin(\dfrac{n\pi+\theta}{2})}{2^{n-1}}$. I also got similar result using your result. $\endgroup$ – Math.StackExchange Nov 14 '13 at 3:28

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