cos(3π/2-x) = cos(3π/2)cos(x) + sin(3π/2)sin(x)
cos(3π/2) = 0 (cosine of 3π/2 is 0)
sin(3π/2) = -1 (sine of 3π/2 is -1)
Therefore, cos(3π/2-x) = 0cos(x) + (-1)sin(x) = -sin(x)

Now, let's square both sides:
cos^2(3π/2-x) = (-sin(x))^2
cos^2(3π/2-x) = sin^2(x) (since the square of a negative number is positive)

So, in conclusion, cos(3π/2-x) = -sin(x) and cos^2(3π/2-x) = sin^2(x)