Question 889804
cos2A=2cos^2(A)-1
cos^(A)=(cos2A+1)/2
so cos^2(4x) can be written as cos^2(4x)=(1+cos8x)/2
cos2A=1-2sin^2(A)
sin^2(A)=(1-cos2A)/2
sin^2(4x)=(1-cos8x)/2
substituting these in eqn sin^2(4x)cos^2(4x) we get ((1-cos8x)/2)((1+cos8x)/2)=(1-cos^2(8x))/4
                                               =(1-(cos16x+1)/2)/4
                                               =1/4-(1/8*cos16x)-1/8
                                               =1/8-(1/8*cos16x)
                                               =1/8*(1-cos16x)
answer to the question
 sin^2(4x)cos^2(4x) = 1/8*(1-cos16x)