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:

b^2

b^2

(2x)

sin^2 (2x)

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

2x*cos 2x + cos^2 (2x) + cos^2 (2x) + 2sin 2x*cos 2x + sin^2
(2x)

We'll eliminate like
terms:

cos^2 (2x) + cos^2 (2x) + sin^2 (2x)

We'll use Pythagorean
identity:

1

2

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

(cos2x+sin2x)^2 = 2