Prove that the expression (sin2x-cos2x)^2+(cos2x+sin2x)^2 does not depend on x.

We'll raise to square both binomials, using the special
products:

`(a+b)^2 = a^2 + 2ab +
b^2`

`(a-b)^2 = a^2 - 2ab +
b^2`

`(sin2x-cos2x)^2 = sin^2 (2x) - 2sin 2x*cos 2x + cos^2
(2x) ` (1)

`(cos2x+sin2x)^2 = cos^2 (2x) + 2sin 2x*cos 2x +
sin^2 (2x)`  (2)

We'll add (1) +
(2):

`(sin2x-cos2x)^2 + (cos2x+sin2x)^2 = sin^2 (2x) - 2sin
2x*cos 2x + cos^2 (2x) + cos^2 (2x) + 2sin 2x*cos 2x + sin^2
(2x)`

We'll eliminate like
terms:

`(sin2x-cos2x)^2 + (cos2x+sin2x)^2 = sin^2 (2x) +
cos^2 (2x) + cos^2 (2x) + sin^2 (2x)`

We'll use Pythagorean
identity:

`sin^2 (2x) + cos^2 (2x) =
1`

`(sin2x-cos2x)^2 + (cos2x+sin2x)^2 = 1 + 1 =
2`

We notice that the result of the sum of
squares is a constant
value:

`(sin2x-cos2x)^2 +
(cos2x+sin2x)^2 = 2`