Question Number 134417 by faysal last updated on 03/Mar/21
![](https://www.tinkutara.com/question/19205.png)
Answered by som(math1967) last updated on 03/Mar/21
![((2^(n−1) 2sinθcosθ.cos2θ...cos2^(n−1) θ)/(sinθ)) ((2^(n−2) .2sin2θcos2θ.cos2^2 θ.....)/(sinθ)) ((2^(n−3) 2sin2^2 θcos2^2 θ...)/(sinθ)) ((sin2^n θ)/(sinθ)) now given θ=(π/(2^n +1)) 2^n θ=π−θ sin2^n θ=sin(π−θ)=sinθ ∴((sin2^n θ)/(sinθ))=((sinθ)/(sinθ))=1](https://www.tinkutara.com/question/Q134425.png)
$$\frac{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{2}{sin}\theta{cos}\theta.{cos}\mathrm{2}\theta…{cos}\mathrm{2}^{{n}−\mathrm{1}} \theta}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{2}} .\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta.{cos}\mathrm{2}^{\mathrm{2}} \theta…..}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{3}} \mathrm{2}{sin}\mathrm{2}^{\mathrm{2}} \theta{cos}\mathrm{2}^{\mathrm{2}} \theta…}{{sin}\theta} \\ $$$$\frac{{sin}\mathrm{2}^{{n}} \theta}{{sin}\theta} \\ $$$${now}\:{given}\:\theta=\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}} \\ $$$$\:\mathrm{2}^{{n}} \theta=\pi−\theta \\ $$$${sin}\mathrm{2}^{{n}} \theta={sin}\left(\pi−\theta\right)={sin}\theta \\ $$$$\therefore\frac{{sin}\mathrm{2}^{{n}} \theta}{{sin}\theta}=\frac{{sin}\theta}{{sin}\theta}=\mathrm{1} \\ $$