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

If `theta=pi/(2^n+1)` , prove that: `2^ncosthetacos2thetacos2^2thetacos2^(n-1)theta=1.`

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In the above result, put `theta=(pi)/(2^(n)+1)`
`rArr RHS.=(sin2^(n)theta)/(2n^(n)sin theta)=(sin((pi)/(2^(n)+1))2^(n))/(2^(n)sin((pi)/(2^(n)+1)))=(sin((2^(n)+1-1)/(2^(n)+1))pi)/(2^(n)sin((pi)/(2^(n)+1)))`
`=(sin(pi-(pi)/(2^(n)+1)))/(2^(n)sin((pi)/(2^(n)+1)))`
`=(sin((pi)/(2^(n)+1)))/(2^(n)sin((pi)/(2^(n)+1)))`
`(1)/(2^(n))`
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