What is the simplified version of Sqrt((-2sin2t)^2 + (2cos2t)^2)

Sqrt((-2sin2t)^2 + (2cos2t)^2) = ?

It simplifies to 2.

First step: simplify the squares by squaring the terms in ()

4 sin^2(2t) + 4 cos^2(2t)

= 4 (sin^2(2t) + cos^2(2t))

Next step: apply a trigonometric identity, sin^2 u + cos^2 u = 1 for all values of u, substituting u = 2t.
This is an application of the Pythagorean Theorem: you have the Unit Circle, and C = (0,0) , X = (cos u, 0), Y = (0, sin u). The hypotenuse is 1, the horizontal leg is cos u, the vertical leg is sin u.

= 4 (1) = 4

Next step: put the Sqrt() back on it

Sqrt(4) = 2

The End

Sqrt((-2sin2t)^2 + (2cos2t)^2) = Sqrt(4sin^2(2t) + 4cos^2(2t))
=Sqrt[4{sin^2(2t) + cos^2(2t)}]
=Sqrt[4]
=2