Inverse of $\phi + A/B \arcsin(B \sin(\phi))$ I'm looking for the inverse of the function (family)
$$ f(\phi)=\phi + \frac{A}{B}\cdot\arcsin\left(B\cdot\sin(\phi)\right),$$
where $A=\sin\theta,$ and $B=\cos\theta$, where $\theta$ can be thought of as a parameter (i.e. constant).
My hunch is that this has no solution that can be written down using elementary functions. Now, $f$ is well-behaved, continuous and monotonically increasing, so there should be no problem making an algorithm to find the inverse. 


*

*Is this a function that is known to not have an inverse that can be written down using elementary functions?

*What is the "best" way to invert this function using an algorithm?
 A: This is not a complete answer at all. 
I should not expect a closed form for the inverse.
I think that is would be better to not consider $A,B$ as parameters but to use their definition. So
$$f(\phi)=\phi+\tan (\theta ) \sin ^{-1}(\cos (\theta ) \sin (\phi ))$$ If you are concerned by small values of $x$, you could make a Taylor expansion and then use series reversion. Around $x=0$, this would give
$$\phi=\frac{1}{\sin (\theta )+1}f(\phi)+\frac{ \sin ^3(\theta )}{6 (\sin (\theta
   )+1)^4}f(\phi)^3+O\left(f(\phi)^5\right)$$
I wonder if it could or not be better to consider instead
$$g(\phi)=f(\phi)\cos (\theta )=\phi  \cos (\theta )+\sin (\theta ) \sin ^{-1}(\cos (\theta ) \sin (\phi ))$$ for which the same process would give
$$\phi=\frac{\sec (\theta )+(\sin (\theta ) (\sin (\theta )+3)+3) \tan (\theta )}{(\sin
   (\theta )+1)^4}g(\phi)+\frac{ \tan ^3(\theta )}{6 (\sin (\theta )+1)^4}g^3(\phi)+O\left(g(\phi)^5\right)$$
For sure, the same could be done building the series around $\phi=a$ but the formulae are really messy to be reported here.
May I suggest you play with that ? And let me know.
If it works more or less, we could improve using instead $[1,n]$ Padé approximant.
