Question Number 115455 by bemath last updated on 26/Sep/20
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$
Commented by Adel last updated on 13/Jan/21
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$
Answered by TANMAY PANACEA last updated on 26/Sep/20
$$\theta=\frac{\pi}{\mathrm{65}} \\ $$$${S}={cos}\theta{cos}\mathrm{2}\theta….{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}{Ssin}\theta=\left(\mathrm{2}{sin}\theta{cos}\theta\right){cos}\mathrm{2}\theta{cos}\mathrm{4}\theta{cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{2}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta\right){cos}\mathrm{4}\theta{cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{3}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{4}\theta{cos}\mathrm{4}\theta\right){cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{4}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{8}\theta{cos}\mathrm{8}\theta\right){cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{5}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{16}\theta{cos}\mathrm{16}\theta\right){cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{32}\theta{cos}\mathrm{32}\theta\right) \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta={sin}\mathrm{64}\theta \\ $$$${now}\:{sin}\mathrm{64}\theta={sin}\left(\mathrm{65}\theta−\theta\right)={sin}\left(\pi−\theta\right)={sin}\theta \\ $$$${so} \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta={sin}\theta \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$ \\ $$
Commented by bemath last updated on 26/Sep/20
$${santuyy}…{gave}\:{kudos} \\ $$