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Question

Prove that: cosθ. cos2θ. cos22θ. cos23θ cos2n1θ= sin2nθ2n sinθ

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Solution

sin2x=2sinx cosx [double angle formula]
sin4x=sin(2(2x))=2sin2x cos2x=4sinx cosx cos2x
Similarly
sin16x=2sin(8x)cos(8x)
=2.2sin(4x)cos(4x)cos(8x)
=2.2.2.sin(2x)cos(2x)cos(4x)cos(8x)
=2.2.2.2.cos(x)sin(x)cos(2x)cos(4x)cos(8x)
Now similarly of we generalise this
sin(2.2.2....(n+1)times)=2cos(2.2....(n)times)sin(2.2...ntimes)
22cos(2nx)sin(2n1x)cos(2n1x)
2ncos(2nx)cos(2n1x)......cos(4x)cos(2x)sin(2x)
2n+1cos(2nx)cos(2n1x).......cos(2x)cos(x)sin(x)
cos(x)cos(2x)cos(4x)cos(8x).......cos(2nx)=sin(2n+1x)/(2n+1sin(x))




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