1.

If `theta=pi/(2^n+1)`, prove that: `2^ncosthetacos2thetacos2^2 cos2^(n-1)theta=1.`

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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