Respuesta :

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.

Answer:

the answer is goin to be D

sin^2(x) = 1- cos(2x)/2

cos^2(x) = 1 + cos(2x) /2

2sin^2xcos^2x  =  2 * 1- cos(2x)/2 * 1 + cos(2x) /2

(1- cos(2x) * 1+cos(2x))/2

(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

1-cos(4x) /4

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