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Question Number 27153 by abdo imad last updated on 02/Jan/18

$$findthevalueof\prod _{k=1}^{n-1}sin\left(\frac{k\pi }{2n}\right).$$

Commented by abdo imad last updated on 03/Jan/18

$$letintroducethepolynomialp\left(x\right)=x^{2n}-1therootsofp\left(x\right)$$$$arethecomplex{z}_k=e^{i\frac{2k}{n}}andk\in \left[[0,2n-1]\right]and$$$$p\left(x\right)=\lambda \prod _{k=0}^{n-1}(x-z_k)itsclearthat\lambda =1and$$$$p\left(x\right)=\prod _{k=0}^{n-1}(x-z_k)wehave{z}_0=1,z_1=e^{i\frac{\pi }{n}},z_2=e^{i\frac{2\pi }{n}}$$$$z_{n-1}=e^{i\frac{(n-1)\pi }{n}},z_n=-1,z_{n+1}=e^{i\frac{(n+1)\pi }{n}},z_{2n-1}=e^{i\frac{(2n-1)\pi }{2n}}$$$$weseethat{z}_{2n-1}=z_1^{-},z_{2n-2}=z_2^{-},z_{n+1}=z_{n-1}^{-}$$$$\Rightarrow p\left(x\right)=(x^2-1)\prod _{k=1}^{n-1}(x-z_k)(x-z_k^{-})$$$$=(x^2-1)\prod _{k=1}^{n-1}(x^2-2cos\left(\frac{k\pi }{n}\right)x+1)andfor{x}^2\ne 1$$$$\frac{p\left(x\right)}{x^2-1}=\prod _{k=1}^{n-1}(x^2-2cos\left(\frac{k\pi }{n}\right)x+1)andbyusinghospitaltheoem$$$${lim}_{x->1}\prod _{k=1}^{n-1}(x^2-2cos\left(\frac{k\pi }{n}\right)x+1)={lim}_{x->1}\frac{p^{{}^{\prime }}\left(x\right)}{2x}$$$$\prod _{k=1}^{n-1}2(1-cos\left(\frac{k\pi }{n}\right))={lim}_{x->1}\frac{2{nx}^{2n-1}}{2x}=n$$$$\Rightarrow 4^{n-1}\prod _{k=1}^{k=n-1}{sin}^2\left(\frac{k\pi }{2n}\right)=n$$$$\Rightarrow \prod _{k=1}^{n-1}{sin}^2\left(\frac{k\pi }{2n}\right)=\frac{n}{4^{n-1}}$$$$\Rightarrow \prod _{k=1}^{n-1}sin\left(\frac{k\pi }{2n}\right)=\frac{\sqrt{n}}{2^{n-1}}(n⩾2)$$

Commented by Tinkutara last updated on 03/Jan/18

I have already asked this a long time back. See this at Q 19292.

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