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Question Number 192985 by MM42 last updated on 01/Jun/23

$$proveit:$$$${lim}_{n\to \mathrm{\infty }}\underset{i=1}{\prod ^n}cos\frac{\theta }{2^i}=\frac{sin\theta }{\theta }$$$$thenshow:$$$${im}_{n\to \mathrm{\infty }}cos\frac{\pi }{4}cos\frac{\pi }{8}...cos\frac{\pi }{2^{n+1}}=\frac{2}{\pi }$$$$$$

Answered by witcher3 last updated on 02/Jun/23

$$\cos \left(\frac{\theta }{2^{\mathrm{i}}}\right)=\frac{\sin \left(\frac{\theta }{2^{\mathrm{i}-1}}\right)}{2\sin \left(\frac{\theta }{2^{\mathrm{i}}}\right)}$$$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{i}=1}{\prod ^{\mathrm{n}}}\cos \left(\frac{\theta }{2^{\mathrm{i}}}\right)=\underset{1}{\prod ^{\mathrm{n}}}\frac{\sin \left(\frac{\theta }{2^{\mathrm{i}-1}}\right)}{2\sin \left(\frac{\theta }{2^{\mathrm{i}}}\right)}=\frac{\sin \left(\theta \right)}{2^{\mathrm{n}}\sin \left(\frac{\theta }{2^{\mathrm{n}}}\right)}$$$$\text{Double subscripts: use braces to clarify}$$$$\left(2\right)\theta =\frac{\pi }{2}$$

Commented by MM42 last updated on 02/Jun/23

$$verygood$$

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