2sin^2xcos^2x
Using trig functions we know:
sin^2(x) = 1- cos(2x)/2
cos^2(x) = 1 + cos(2x) /2
Now we have:
2sin^2xcos^2x = 2 * 1- cos(2x)/2 * 1 + cos(2x) /2
Simplify to: (1- cos(2x) * 1+cos(2x))/2
Difference of squares is (a-b) (a+b) = a^2 -b^2
(1- cos(2x) * 1+cos(2x))/2 = 1^2 -cos^2(x)/2 *1^2 +cos^2(x) /2
Multiply to get 1-cos(4x) /4
The answer is D.