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