what is the inverse of f(x) = sin(2x)cos(2x) ?

Answer 1

Well, strictly speaking, it doesn't have one, since f is not injective. However, we can find an inverse of f restricted to some interval on which it is injective, say, [-π/8, π/8]. Then we have:

f(x) = sin (2x) cos (2x) = 1/2 sin (4x)

So since f(f⁻¹(x)) = x, we have:

x = 1/2 sin (4f⁻¹(x))
2x = sin (4f⁻¹(x))
4f⁻¹(x) = arcsin (2x)
f⁻¹(x) = 1/4 arcsin (2x)

Note that the range of this function is indeed [-π/8, π/8], which was the interval to which we restricted the original function.