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Question Number 170872 by mathlove last updated on 02/Jun/22

Answered by aleks041103 last updated on 04/Jun/22

$$\underset{k=1}{\prod ^{n-1}}sin\left(\frac{k\pi }{n}\right)=\underset{k=1}{\prod ^{n-1}}\frac{e^{\frac{k\pi i}{n}}-e^{-\frac{k\pi i}{n}}}{2i}=$$$$=\frac{\left(\underset{k=1}{\prod ^{n-1}}{e}^{\frac{k\pi i}{n}}\right)\left(\underset{k=1}{\prod ^{n-1}}(1-e^{-\frac{2k\pi i}{n}})\right)}{{\left(2i\right)}^{n-1}}=$$$$=\frac{e^{\frac{\pi i}{n}\underset{k=1}{\sum ^{n-1}}k}}{2^{n-1}{\left(e^{\frac{\pi i}{2}}\right)}^{n-1}}\underset{k=1}{\prod ^{n-1}}(1-e^{-\frac{2k\pi i}{n}})=$$$$=\frac{e^{\frac{\pi i}{n}\frac{n(n-1)}{2}-\frac{(n-1)\pi i}{2}}}{2^{n-1}}\underset{k=1}{\prod ^{n-1}}(1-e^{-\frac{2k\pi i}{n}})=$$$$=\frac{1}{2^{n-1}}f\left(1\right)$$$$wheref\left(x\right)=\underset{k=1}{\prod ^{n-1}}(x-e^{-\frac{2k\pi i}{n}})$$$$f\left(x\right)=\underset{k=1}{\prod ^{n-1}}(x-e^{-\frac{2k\pi i}{n}})=\frac{\underset{k=0}{\prod ^{n-1}}(x-e^{-\frac{2k\pi i}{n}})}{x-1}$$$$but{e}^{-\frac{2k\pi i}{n}}fork=0,...,(n-1)arethe$$$$n-throotsofunity.$$$$\Rightarrow \underset{k=0}{\prod ^{n-1}}(x-e^{-\frac{2k\pi i}{n}})=x^n-1$$$$\Rightarrow f\left(x\right)=\frac{x^n-1}{x-1}=1+x+x^2+...+x^{n-1}$$$$f\left(1\right)=1+1+...+1=n$$$$\Rightarrow \underset{k=1}{\prod ^{n-1}}sin\left(\frac{k\pi }{n}\right)=\frac{n}{2^{n-1}}$$

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