How your verify 1/8[3+4cos2x+cos4x]=cos^ 4x?

1 Answer
May 14, 2018

RHS=cos^4x

=[(2cos^2x)/2]^2

=1/4[1+cos2x]^2

=2/(4*2)[1+2cos2x+cos^2(2x)]

=1/8[2+4cos2x+2cos^2(2x)]

=1/8[2+4cos2x+1+cos4x]=1/8[3+4cos2x+cos4x]=LHS