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If theta=pi/(2^n+1) , prove that: 2^ncos...

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

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Given,
`costhetacos2thetacos2^2 cos2^(n-1)theta`
`=>1/(2sinA)(sinthetacosthetacos2thetacos2^2 cos2^(n-1)theta)`
`=>1/(4sintheta)(sin2thetacos2thetacos2^2 cos2^(n-1)theta)`
expanding serialy,
`costhetacos2thetacos2^2 cos2^(n-1)theta=(sin2^ntheta)/(2^nsintheta)`
or,
...
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