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Question Number 162589 by mkam last updated on 30/Dec/21

Answered by Ar Brandon last updated on 30/Dec/21

$$T=\cos \vartheta \cos 2\vartheta \cos 4\vartheta ...\cos \left(2^{n-1}\vartheta \right)$$$$2\sin \vartheta T=\sin 2\vartheta \cos 2\vartheta \cos 4\vartheta ...\cos \left(2^{n-1}\vartheta \right)$$$$4\sin \vartheta T=\sin 4\vartheta ...\cos \left(2^{n-1}\vartheta \right)$$$$2^n\sin \vartheta T=\sin \left(2^n\vartheta \right)$$$$T=\frac{1}{2^n}\cdot \frac{\sin \left(2^n\vartheta \right)}{\sin \vartheta },\vartheta =\frac{\pi }{2^n+1}$$$$T=\frac{1}{2^n}\cdot \frac{\sin \left(\frac{2^n}{2^n+1}\pi \right)}{\sin \left(\frac{1}{2^n+1}\pi \right)}=\frac{1}{2^n}\cdot \frac{\sin (\pi -\frac{\pi }{2^n+1})}{\sin \left(\frac{\pi }{2^n+1}\right)}$$$$T=\frac{1}{2^n}\cdot \frac{\sin \left(\frac{\pi }{2^n+1}\right)}{\sin \left(\frac{\pi }{2^n+1}\right)}=\frac{1}{2^n},\sin (\pi -t)=\sin \left(t\right)$$

Commented by peter frank last updated on 31/Dec/21

$$\mathrm{thank}\mathrm{you}$$

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