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Question Number 142698 by Willson last updated on 04/Jun/21

$$\mathrm{Prove}\mathrm{that}\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }$$$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\sin \left(\frac{\mathrm{k}\pi }{2\mathrm{n}}\right)=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\cos \left(\frac{\mathrm{k}\pi }{2\mathrm{n}}\right)$$

Answered by Ar Brandon last updated on 04/Jun/21

$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\sin \left(\frac{\mathrm{k}\pi }{2\mathrm{n}}\right)=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\cos (\frac{\pi }{2}-\frac{\mathrm{k}\pi }{2\mathrm{n}})=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\cos \left(\frac{(\mathrm{n}-\mathrm{k})\pi }{2\mathrm{n}}\right)$$$$=\cos \left(\frac{\mathrm{n}-1}{2\mathrm{n}}\pi \right)\cos \left(\frac{\mathrm{n}-2}{2\mathrm{n}}\pi \right)\cdot \cdot \cdot \cos \left(\frac{2}{2\mathrm{n}}\pi \right)\cos \left(\frac{1}{2\mathrm{n}}\pi \right)$$$$=\cos \left(\frac{1}{2\mathrm{n}}\pi \right)\cos \left(\frac{2}{2\mathrm{n}}\pi \right)\cdot \cdot \cdot \cos \left(\frac{\mathrm{n}-2}{2\mathrm{n}}\pi \right)\cos \left(\frac{\mathrm{n}-1}{2\mathrm{n}}\pi \right)$$$$=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\cos \left(\frac{\mathrm{k}}{2\mathrm{n}}\pi \right)$$

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