\hspace{-16}\bf{\mathbb{P}}$rove that $\bf{\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)=\frac{\sin \left(\frac{n\pi}{2}\right)}{2^{n-1}}}$
can anyone help me for solving the Given Question.
I have Got \hspace{-16}$\displaystyle \bf{\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)=\frac{1}{2^{n-1}}}$
I did not Understand How can I get \hspace{-16}\bf{\sin\left(\frac{n\pi}{2}\right)}
Where I have make Mistake.
Thanks
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2 Answers
21It is easily proved that over the polynomial ring \mathbb{C[X]} we have \prod_{i=1}^{n-1}\left ( X^2 - 2X \cos \frac{i \pi}{n} +1 \right ) = \frac{X^{2n} - 1}{X^2 - 1}
Divide both sides by X^{n-1} to get that \prod_{i=1}^{n-1}\left ( X - 2 \cos \frac{i \pi}{n} + \frac{1}{X} \right ) = \frac{X^{n} - \frac{1}{X^{n}}}{X - \frac{1}{X}}
Put X = \cos \theta + i \sin \theta to get
2^{n-1} \prod_{i=1}^{n-1}\left ( \cos \theta - \cos \frac{i \pi}{n} \right ) = \frac{\sin n \theta}{ \sin \theta}
Put
\theta = \frac{\pi}{2} to get
\prod_{i=1}^{n-1}\left ( \cos \frac{i \pi}{n} \right ) = (-1)^{n-1}\frac{\sin \frac{n \pi}{2} }{2^{n-1} } = \frac{\sin \frac{n \pi}{2} }{2^{n-1} }
