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The value of cosx2x.cos3x…..cos 999x whe...

The value of `cosx2x.cos3x…..cos 999x` where `x=(2pi)/1999` is `p` then `2^(1000)p` is equal to __________

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The correct Answer is:
2
Let `P=cosxcos2x.cos3x……..cos 999x` and `Q=sinxsin2xsin3x……sin999x`
`PQ.2^(99)=2sin2xsin4xsin6xx….sin1998x`
`= (sin2x.sin4x……sin998x)(-sin(2pi-1000x)(-sin(2pi-1002x)…..[-sin(2pi-1998x)]`
`=sin2x.sin4x.sin998x.sin999x.sin997x…sinx=Q`
`P=1/(2^(99))(Q!=0)`
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