Inverse of an ordered pair?

Question

Let $f: A \to B$ be a bijective function where $A = [0, 2\pi)$ and $B$ is the unit circle. Find the inverse of $f(\theta) = (\cos\theta, \sin\theta)$.

I don't understand what it means to take the inverse of an ordered pair. I see that the function is mapping points from the interval $[0, 2\pi)$ to coordinates on the unit circle in $\mathbb{R^2}$ plane, so we have to take those coordinates back into the interval $[0, 2\pi)$. I would guess this involves some two-variable function $f(x,y)$ with inverse functions $\cos^{-1}x$ and $\sin^{-1}x$ since we want the value of the inverse function to be a real number in the interval $[0, 2\pi)$.

Answer 1

Use the inverse functions of $\sin$, and $\cos$ $$\arcsin : [-1, 1] \rightarrow [\frac{-\pi}{2}, \frac{\pi}{2}]$$ $$\arccos : [-1, 1] \rightarrow [0, \pi]$$

$f^{-1}(x, y) = \arcsin x$, if $\arcsin x = \arccos y$

$f^{-1}(x, y) = 2\pi -\arccos x $, if $\arcsin x < 0$ and $\arccos y > \frac{\pi}{2}$

$f^{-1}(x, y) = \pi -\arcsin x$, if $\arcsin x \geq 0$ and $\arccos y > \frac{\pi}{2}$

$f^{-1}(x, y) = \arcsin x + 2\pi$, if $\arcsin x < 0$ and $\arccos y \leq \frac{\pi}{2}$

Answer 2

Hint $g(\cdot)$ is an inverse of $f(\cdot)$ if and only if $g(f(x)) = f(g(x)) = x$ for all valid $x$.

Answer 3

The following $f^{-1}$ is not the inverse function, because you need to add $2\pi$ to negative outputs. Notice that the image has to be $[0,2\pi)$, not $(-\pi,\pi]$.

$$f^{-1}(c,s) = \begin{cases} \arctan(\frac{s}{c}),& c > 0 \\ \frac{\pi}{2}s, & c = 0\\ \pi + \arctan(\frac{s}{c}),& c < 0 \end{cases}$$