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Question-162589




Question Number 162589 by mkam last updated on 30/Dec/21
Answered by Ar Brandon last updated on 30/Dec/21
T=cosϑcos2ϑcos4ϑ...cos(2^(n−1) ϑ)  2sinϑT=sin2ϑcos2ϑcos4ϑ...cos(2^(n−1) ϑ)  4sinϑT=sin4ϑ...cos(2^(n−1) ϑ)  2^n sinϑT=sin(2^n ϑ)  T=(1/2^n )∙((sin(2^n ϑ))/(sinϑ)) , ϑ=(π/(2^n +1))  T=(1/2^n )∙((sin((2^n /(2^n +1))π))/(sin((1/(2^n +1))π)))=(1/2^n )∙((sin(π−(π/(2^n +1))))/(sin((π/(2^n +1)))))  T=(1/2^n )∙((sin((π/(2^n +1))))/(sin((π/(2^n +1)))))=(1/2^n )  , sin(π−t)=sin(t)
$${T}=\mathrm{cos}\vartheta\mathrm{cos2}\vartheta\mathrm{cos4}\vartheta…\mathrm{cos}\left(\mathrm{2}^{{n}−\mathrm{1}} \vartheta\right) \\ $$$$\mathrm{2sin}\vartheta{T}=\mathrm{sin2}\vartheta\mathrm{cos2}\vartheta\mathrm{cos4}\vartheta…\mathrm{cos}\left(\mathrm{2}^{{n}−\mathrm{1}} \vartheta\right) \\ $$$$\mathrm{4sin}\vartheta{T}=\mathrm{sin4}\vartheta…\mathrm{cos}\left(\mathrm{2}^{{n}−\mathrm{1}} \vartheta\right) \\ $$$$\mathrm{2}^{{n}} \mathrm{sin}\vartheta{T}=\mathrm{sin}\left(\mathrm{2}^{{n}} \vartheta\right) \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\centerdot\frac{\mathrm{sin}\left(\mathrm{2}^{{n}} \vartheta\right)}{\mathrm{sin}\vartheta}\:,\:\vartheta=\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\centerdot\frac{\mathrm{sin}\left(\frac{\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} +\mathrm{1}}\pi\right)}{\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} +\mathrm{1}}\pi\right)}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\centerdot\frac{\mathrm{sin}\left(\pi−\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\right)} \\ $$$${T}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\centerdot\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}}\right)}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\:,\:\mathrm{sin}\left(\pi−{t}\right)=\mathrm{sin}\left({t}\right) \\ $$
Commented by peter frank last updated on 31/Dec/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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