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Question Number 115455 by bemath last updated on 26/Sep/20

cos ((π/(65))).cos (((2π)/(65))).cos (((4π)/(65))).cos (((8π)/(65))).cos (((16π)/(65))).cos (((32π)/(65)))=?

$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$

Commented by Adel last updated on 13/Jan/21

cos ((π/(65))).cos (((2π)/(65))).cos (((4π)/(65))).cos (((8π)/(65))).cos (((16π)/(65))).cos (((32π)/(65)))=?

$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{65}}\right).\mathrm{cos}\:\left(\frac{\mathrm{32}\pi}{\mathrm{65}}\right)=? \\ $$

Answered by TANMAY PANACEA last updated on 26/Sep/20

θ=(π/(65)) S=cosθcos2θ....cos32θ 2Ssinθ=(2sinθcosθ)cos2θcos4θcos8θcos16θcos32θ 2^2 Ssinθ=(2sin2θcos2θ)cos4θcos8θcos16θcos32θ 2^3 Ssinθ=(2sin4θcos4θ)cos8θcos16θcos32θ 2^4 Ssinθ=(2sin8θcos8θ)cos16θcos32θ 2^5 Ssinθ=(2sin16θcos16θ)cos32θ 2^6 Ssinθ=(2sin32θcos32θ) 2^6 Ssinθ=sin64θ now sin64θ=sin(65θ−θ)=sin(π−θ)=sinθ so 2^6 Ssinθ=sinθ S=(1/2^6 )=(1/(64))

$$\theta=\frac{\pi}{\mathrm{65}} \\ $$$${S}={cos}\theta{cos}\mathrm{2}\theta....{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}{Ssin}\theta=\left(\mathrm{2}{sin}\theta{cos}\theta\right){cos}\mathrm{2}\theta{cos}\mathrm{4}\theta{cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{2}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{2}\theta{cos}\mathrm{2}\theta\right){cos}\mathrm{4}\theta{cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{3}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{4}\theta{cos}\mathrm{4}\theta\right){cos}\mathrm{8}\theta{cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{4}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{8}\theta{cos}\mathrm{8}\theta\right){cos}\mathrm{16}\theta{cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{5}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{16}\theta{cos}\mathrm{16}\theta\right){cos}\mathrm{32}\theta \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta=\left(\mathrm{2}{sin}\mathrm{32}\theta{cos}\mathrm{32}\theta\right) \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta={sin}\mathrm{64}\theta \\ $$$${now}\:{sin}\mathrm{64}\theta={sin}\left(\mathrm{65}\theta−\theta\right)={sin}\left(\pi−\theta\right)={sin}\theta \\ $$$${so} \\ $$$$\mathrm{2}^{\mathrm{6}} {Ssin}\theta={sin}\theta \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{64}} \\ $$$$ \\ $$

Commented by bemath last updated on 26/Sep/20

santuyy...gave kudos

$${santuyy}...{gave}\:{kudos} \\ $$

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