Question Number 134417 by deleteduser12 last updated on 03/Mar/21
Answered by som(math1967) last updated on 03/Mar/21
$$\frac{\mathrm{2}^{{n}−\mathrm{1}} \mathrm{2}{sin}\theta{cos}\theta.{cos}\mathrm{2}\theta...{cos}\mathrm{2}^{{n}−\mathrm{1}} \theta}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{2}} .\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta.{cos}\mathrm{2}^{\mathrm{2}} \theta.....}{{sin}\theta} \\ $$$$\frac{\mathrm{2}^{{n}−\mathrm{3}} \mathrm{2}{sin}\mathrm{2}^{\mathrm{2}} \theta{cos}\mathrm{2}^{\mathrm{2}} \theta...}{{sin}\theta} \\ $$$$\frac{{sin}\mathrm{2}^{{n}} \theta}{{sin}\theta} \\ $$$${now}\:{given}\:\theta=\frac{\pi}{\mathrm{2}^{{n}} +\mathrm{1}} \\ $$$$\:\mathrm{2}^{{n}} \theta=\pi−\theta \\ $$$${sin}\mathrm{2}^{{n}} \theta={sin}\left(\pi−\theta\right)={sin}\theta \\ $$$$\therefore\frac{{sin}\mathrm{2}^{{n}} \theta}{{sin}\theta}=\frac{{sin}\theta}{{sin}\theta}=\mathrm{1} \\ $$