Prove that 32sin^4x.cos^2x=cos6x-2cos4x-cos 2x+2?

1 Answer
May 3, 2018

RHS=cos6x-2cos4x-cos2x+2

=cos6x-cos2x+2(1-cos4x)

=-2sin((6x+2x)/2)*sin((6x-2x)/2)+2*2sin^2(2x)

=4sin^2(2x)-2sin4x*sin2x=4sin^2(2x)-2*2*sin2x*cos2x*sin2x

=4sin^2(2x)-4sin^2(2x)*cos2x

=4sin^2(2x)[1-cos2x]

=4*(2sinx*cosx)^2*2sin^2x

=4*4sin^2x*cos^2x*2sin^2x=32sin^4x*cos^2x=LHS