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`sin2x+2sin4x+sin6x= 4cos^(2)xsin4x`

Answer» LHS `=sin2x+2sin4x+sin6x`
`=sin4x+[2sin(6x+2x)/(2)cos(6x-2x)/(2)]`
`=2sin4x+2sin4xcos2x=2sin4x[1+cos2x]`
`=2sin4x[1+2cos^(2)x-1]=2sin4x.2cos^(2)x`
`=4cos^(2)x.sin4x`=RHS Hence Proved.


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