I am interested in the function
$$ y(x) := \left( x +\frac{3\pi}{2} \right) \sin(x) + \cos(x). $$
Over the range $ x \in \left[ -\frac{\pi}{2} ,\frac{\pi}{2} \right]$, this function grows monotonically from $-\pi$ to $2\pi$, as illustrated in this WolframAlpha plot:
Hence, clearly its inverse exists and is well-behaved. I am wondering, however, whether its inverse can be written in terms of known transcendental functions, i.e. whether there is some 'nice' expression $x(y)$.
Let me first define a rescaled function $\tilde y(x) \equiv \frac{2}{3\pi} y(x) - \frac{1}{3}$, just so that the domain $[-\pi/2,\pi/2]$ maps to the range $[-1,1]$. For completeness:
$$ \boxed{ \tilde y(x) = \left( \frac{2x}{3\pi} + 1 \right) \sin(x) + \frac{2}{3\pi} \cos(x) - \frac{1}{3} }. $$
It seems that the inverse is well-approximated by the following function:
$$ \boxed{ x_\textrm{guess}(\tilde y) = \frac{\pi}{2} + \left( \alpha + \beta \tilde y \right) \arccos(\tilde y) + \frac{\gamma}{\pi} \left(\arccos(\tilde y) \right)^2 }. $$
More precisely, if I try to fit this function, I obtain
$$ \begin{array}{ccc} \alpha_\textrm{fit} &= & -0.817 \pm 0.002 \\ \beta_\textrm{fit} &= & -0.032 \pm 0.002 \\ \gamma_\textrm{fit} &= & -0.215 \pm 0.004 \end{array}$$
The fact that this is close to the true inverse is demonstrated in the following plot:
The black curve is $x(\tilde y)$ (this is a numerical result and can be taken to be exact for our practical purposes), and the red dashed curve is $x_\textrm{guess}(\tilde y)$ with the above fitted values of $\alpha$, $\beta$ and $\gamma$.
Do note that the above guess is not the true/complete inverse, since there are still a finite difference between the two curves. The following plot shows $x(\tilde y) - x_\textrm{guess}(\tilde y)$:
The above suggests that it is not crazy to think that there might exist a closed form for $x(\tilde y)$ in terms of known transcendental functions. In case anyone has an idea of what extra term would be sensible to add to my Ansatz inverse $x_\textrm{guess}(\tilde y)$, do let me know. It would be very entertaining if we could make the second plot (i.e. the error plot) zero (within machine precision).
Nice expression, I do not know.
However, the function is quite well represented using Taylor expansion built at $x=0$. This write $$y=\sum_{n=0}^\infty \frac{3 \pi \sin \left(\frac{\pi n}{2}\right)-2 (n-1) \cos \left(\frac{\pi n}{2}\right)}{2\,n!}\, x^n$$
So, we can use series reversion to get $$x=\sum_{n=1}^p a_n t^n +O(t^{n+1})\qquad \text{where} \qquad t=\frac{2 (y-1)}{3 \pi }$$ where the first coefficients are $$a_1=1 \qquad a_2=-\frac{1}{3 \pi }\qquad a_3=\frac{1}{6}+\frac{2}{9 \pi ^2}\qquad a_4=-\frac{5}{27 \pi ^3}-\frac{7}{36 \pi }$$ $$a_5=\frac{3}{40}+\frac{14}{81 \pi ^4}+\frac{2}{9 \pi ^2}\qquad a_6=-\frac{14}{81 \pi ^5}-\frac{7}{27 \pi ^3}-\frac{53}{360 \pi }$$ $$a_7=\frac{5}{112}+\frac{44}{243 \pi ^6}+\frac{25}{81 \pi ^4}+\frac{371}{1620 \pi ^2}$$ $$a_8=-\frac{143}{729 \pi ^7}-\frac{121}{324 \pi ^5}-\frac{143}{432 \pi ^3}-\frac{823}{6720 \pi }$$ The higher coefficients become more and more messy and they will not be reported here.
To check how good or bad is this approximation, give $x$ a values, compute the correspond $y$ and recompute $x$ using the last formula. This would give the following table $$\left( \begin{array}{ccc} x_{given} & y_{calc} & x_{calc} \\ -1.50 & -3.13360 & -1.24997 \\ -1.25 & -2.97043 & -1.16175 \\ -1.00 & -2.58357 & -0.97992 \\ -0.75 & -1.96923 & -0.74766 \\ -0.50 & -1.14194 & -0.49992 \\ -0.25 & -0.13510 & -0.25000 \\ +0.00 & +1.00000 & +0.00000 \\ +0.25 & +2.19663 & +0.25000 \\ +0.50 & +3.37653 & +0.49990 \\ +0.75 & +4.45506 & +0.74733 \\ +1.00 & +5.34711 & +0.97727 \\ +1.25 & +5.97354 & +1.15378 \\ +1.50 & +6.26756 & +1.23905 \end{array} \right)$$ which not bad at least over the range $-1 \leq x \leq 1$.
For sure, we could improve these results building the series expansions centered at $x=-\frac \pi 2$, $x=0$ and $x=\frac \pi 2$ and proceed the same way using the reversed series for $-\frac \pi 2 \leq x \leq -1$, $-1 \leq x \leq 1$ and $1 \leq x \leq \frac \pi 2$.
