Trigonometric function for cosh and sinh How do I approach this correlation to find $q^1$ and $q^2$ as functions of $x_1,x_2$?
\begin{align}x_1 &=a\cosh q^1\cos q^2, \\
x_2 &=a\sinh q^1\sin q^2.
\end{align}
Which trigonometric function should I use?
Considering $q^1\geq 0$ and $0 \leq q^2 \leq 2 \pi$.
 A: Edit: This answer is a completely re-written version (I have erased the old version which was a dead end).
The context I should have taken into account is that it is a change of coordinates between

*

*cartesian coordinates (rectangular axes) and


*elliptic coordinates (curvilinear axes), somewhat cousins of polar coordinates when one sees them "from far away":

Fig. 1: Extracted from https://encyclopediaofmath.org/wiki/Elliptic_coordinates. See Remark 1 below.
Indeed, from
$$\begin{cases} x_1 &= a \cosh q_1 \cos q_2 \\ x_2 &= a \sinh q_1 \sin q_2, \end{cases}\tag{1}$$
one gets either the equation of an ellipse characterized by constant $q_1$:
$$\dfrac{x_1^2}{\cosh^2 q_1} + \dfrac{x_2^2}{\sinh^2 q_1} = a^2\tag{2}$$
or that of a hyperbola characterized by constant $q_2$:
$$\dfrac{x_1^2}{\cos^2 q_2} - \dfrac{x_2^2}{\sin^2 q_2} = a^2\tag{3}$$
Let us consider (3), written under the form:
$$x_1^2\sin^2 q_2 - x_2^2 \cos^2 q_2 = a^2 \sin^2 q_2 \cos^2 q_2\tag{4}$$
Let us concentrate on (4) ; let us apply the method mentionned by @A rural reader. Replace $\cos^2 q_2$ by $1-\sin^2 q_2$ into (4):
$$x_1^2\sin^2 q_2 - x_2^2 (1-\sin^2 q_2) = a^2 \sin^2 q_1 (1-\sin^2 q_2) \tag{5}$$
Setting
$$p=\sin^2 q_2, \tag{6}$$
(5) becomes a quadratic equation:
$$x_1^2 p - x_2^2 (1-p) = a^2 p(1-p) \tag{7}$$
with solutions :
$$\left. \begin{matrix}p'\\p''\end{matrix} \right\}= \dfrac{1}{2a^2}\left(- a^2 + x_1^2 + x_2^2 \pm \sqrt{(a^2 + x_1^2 + x_2^2)^2-4a^2x_1^2}\right)\tag{8}$$
Taking into account (6), this will give rise to four solutions, one in each quadrant :
$$q_2 \ = \ -\sin^{-1} \sqrt{p'}, \ \ +\sin^{-1}\sqrt{p'}, \ \  -\sin^{-1}\sqrt{p''}, \ \ +\sin^{-1}\sqrt{p''}$$
Indeed this "1 to 4" correspondence could be expected because each hyperbola intersects each ellipse 4 times.
Same type of computation for $q_1$.
I stop there because, making some researches, I found this article:
Che Sun. "Explicit Equations to Transform from Cartesian to Elliptic Coordinates", by Che Sun, Mathematical Modelling and Applications.   Vol. 2, No. 4, 2017, pp. 43-46 (that you will find on ResearchGate) where all this is well explained.
Remarks:

*

*Fig. 1 uses Lamé coefficients giving a so-called bipolar representation which is yet another coordinate system...


*An interesting generalization can be found here.
A: The given components are real and imaginary parts complex cosine type  $\cos Q = \cos (q_1+i q_2)$ with complex argument.
Using subscript instead of superscript, and taking a=1,
$$ \cos ( q_2-i q_1)= \cos q_2 \cos i q_1 +
\sin q2 \;\sin i q1= \cos q_2 \cosh q_1 +i
\sin q2 \;\sinh q1 $$
$$ = x_1 + i x_2$$
