Question Number 192985 by MM42 last updated on 01/Jun/23
$${prove}\:{it}\:: \\ $$$${lim}_{{n}\rightarrow\infty} \:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{\theta}{\mathrm{2}^{{i}} }=\frac{{sin}\theta}{\theta} \\ $$$${then}\:{show}\:: \\ $$$${im}_{{n}\rightarrow\infty} \:{cos}\frac{\pi}{\mathrm{4}}{cos}\frac{\pi}{\mathrm{8}}...{cos}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\frac{\mathrm{2}}{\pi} \\ $$$$ \\ $$
Answered by witcher3 last updated on 02/Jun/23
$$\mathrm{cos}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}} }\right)=\frac{\mathrm{sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}} }\right)} \\ $$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{cos}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}} }\right)=\underset{\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\frac{\mathrm{sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}−\mathrm{1}} }\right)}{\mathrm{2sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{i}} }\right)}=\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{2}^{\mathrm{n}} \mathrm{sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{n}} }\right)} \\ $$$$\mathrm{sin}\left(\frac{\theta}{\mathrm{2}^{\mathrm{n}} }\right)\sim\frac{\theta}{\mathrm{2}^{\mathrm{n}} }\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}A}_{\mathrm{n}} =\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{2}^{\mathrm{n}} .\frac{\theta}{\mathrm{2}^{\mathrm{n}} }}=\frac{\mathrm{sin}\left(\theta\right)}{\theta} \\ $$$$\left(\mathrm{2}\right)\theta=\frac{\pi}{\mathrm{2}} \\ $$
Commented by MM42 last updated on 02/Jun/23
$${very}\:{good} \\ $$