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

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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` **