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

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

Commented by Adel last updated on 13/Jan/21

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

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

$$\theta =\frac{\pi }{65}$$$$S=cos\theta cos2\theta ....cos32\theta$$$$2Ssin\theta =\left(2sin\theta cos\theta \right)cos2\theta cos4\theta cos8\theta cos16\theta cos32\theta$$$$2^{2}Ssin\theta =\left(2sin2\theta cos2\theta \right)cos4\theta cos8\theta cos16\theta cos32\theta$$$$2^{3}Ssin\theta =\left(2sin4\theta cos4\theta \right)cos8\theta cos16\theta cos32\theta$$$$2^{4}Ssin\theta =\left(2sin8\theta cos8\theta \right)cos16\theta cos32\theta$$$$2^{5}Ssin\theta =\left(2sin16\theta cos16\theta \right)cos32\theta$$$$2^{6}Ssin\theta =\left(2sin32\theta cos32\theta \right)$$$$2^{6}Ssin\theta =sin64\theta$$$$nowsin64\theta =sin(65\theta -\theta )=sin(\pi -\theta )=sin\theta$$$$so$$$$2^{6}Ssin\theta =sin\theta$$$$S=\frac{1}{2^6}=\frac{1}{64}$$$$$$

Commented by bemath last updated on 26/Sep/20

$$santuyy...gavekudos$$

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