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

1 Answer
Abhishek K.
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#

Related questions
Impact of this question
6259 views around the world
You can reuse this answer
Creative Commons License