Inverse of two variable function I do not have much experience in finding the inverse of a multivariable function, so any help is appreciated.
I have function $f(\theta_a,\theta_b) =\begin{pmatrix}\frac{\cos(\theta_a)+\cos(\theta_b)}{1+\cos(\theta_a)\cos(\theta_b)}\\ \frac{\sin(\theta_a)+\sin(\theta_b)}{1+\sin(\theta_a)\sin(\theta_b)}\end{pmatrix} $
The right hand side above is a vector, so the function takes two real numbers $\theta_a,\theta_b \in [0,2\pi)$ in a vector and outputs a vector. My strategy was to try and solve the system of equations:
$u = \frac{\cos(\theta_a)+\cos(\theta_b)}{1+\cos(\theta_a)\cos(\theta_b)}, \quad v = \frac{\sin(\theta_a)+\sin(\theta_b)}{1+\sin(\theta_a)\sin(\theta_b)} $.
by first isolating $\theta_a$ in the first equation, and inserting this expression into the second equation, then isolating $\theta_b$ and reinserting to get $\theta_a$ and $\theta_b$ as functions of $u$ and $v$. This quickly becomes some very complex expressions however, and I have not been able to complete the computations, leading me to wondering if there is a different way.
Thanks in advance!
 A: Basically the best you can do, if you can't calculate actual inverse of your function, is the process of "inverse image". Given $f: X\rightarrow Y$:
$$ f^{-1}(B) = \{ x \in X | f(x)\in B \}$$
Basically this filters out those values of $x$ from $X$ where $f(x) : Y$ doesn't belong to $B$.
This can be made more explicit by deciding a test function $g:: Y\rightarrow 2$, which chooses which values of $Y$ belong to the set $B$. $g$ returns true, if an element $y:Y$ belongs to the set $B$. Then the function composition $g(f(x))$ returns true, if $x$ transformed through $f$ belongs to the subset $B$ of $Y$. Thus it's enough to focus on elements of $x$ where $g(f(x))=true$.
That's the theory, now returning to the question at hand, obviously $X=(\theta_a,\theta_b)$, and $Y=(u,v)$, and you can choose $B$ to be a set $B \subset Y$, where B is the set where some arbitrary property is true. Inverse image allows transferring that property $B$ through $f$ backwards the arrows in such way that $g(f(x))$ decides true only where property $B$ is true in $X$, or more compactly:
$$ g(f(x))=true \Leftrightarrow (B\subset Y)\mapsto (f^{-1}(B)\subset X)$$
