How do you simplify cos^2(t)*sin^2(t)?

Question

Pascal...how did you go from
2 sin² (2θ)/8
(sin² (2θ) + 1 - cos² (2θ))/8?
Thank you so much for answering

Answer 1

cos² θ sin² θ
(cos θ sin θ)²
(1/2 sin (2θ))²
sin² (2θ)/4

This is probably the most useful form, unless you're trying to integrate, in which case you would proceed to:

2 sin² (2θ)/8
(sin² (2θ) + 1 - cos² (2θ))/8
(1 - (cos² (2θ) - sin² (2θ)))/8
(1 - cos (4θ))/8

Edit: "Pascal...how did you go from
2 sin² (2θ)/8
(sin² (2θ) + 1 - cos² (2θ))/8?"

By the Pythagorean theorem, sin² (2θ) + cos² (2θ) = 1, so sin² (2θ) = 1 - cos² (2θ). Thus 2 sin² (2θ)/8 = (sin² (2θ) + sin² (2θ))/8 = (sin² (2θ) + 1 - cos² (2θ))/8.

Does that answer your question?

Answer 2

It already looks "simplified" to me, becuase you have things down to products of 2 terms that don't cancel each other out in any way. But you could do a couple of things to it I suppose:

cos^2(t) sin^2(t) =
[ cos(t) sin(t) ]^2 =
[ sin(2t) / 2 ]^2 =
sin^2(2t) / 4

or:
cos^2(t) * [ 1 - cos^2(t) ] =
cos^2(t) - cos^4(t)
Or similarly get sin^2(t) - sin^4(t).

Answer 3

pascal... just if you noticed any number squared times another number squared = n^4 so this would equal
sin²(Θ) · cos²(Θ) = (sin(Θ) · cos(Θ))^4