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If `theta=pi/(2^n+1)`, prove that: `2^ncosthetacos2thetacos2^2 cos2^(n-1)theta=1.` |
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Answer» `L.H.S. = 2^ncosthetacos2theta...cos2^(n-1)theta` `=2costheta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta` `=1/sintheta[2sinthetacostheta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta]` `=1/sintheta[sin2theta*2cos2theta*2cos2^2theta...2cos2^(n-1)theta]` `=1/sintheta[sin2^2theta*2cos2^2theta...2cos2^(n-1)theta]` If we solve it in this way, we finally come to the below solution, `=1/sintheta[2sin2^(n-1)thetacos2^(n-1)theta]` `=sin2^ntheta/sintheta` `=sin(pi-2^ntheta)/sintheta`...As `sin(pi-theta) = sintheta` `=sin(pi-2^n(pi/(2^n+1)))/sintheta` `=sin((2^npi+pi-2^npi)/(2^n+1))/sintheta` `=sin((pi)/(2^n+1))/sintheta` `=sintheta/sintheta` `=1 = R.H.S.` |
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