Eliminating Variables in Trigonometric Equations for Research Purposes

[highflyyer]

Consider the following set of equations:

$r = \cosh\rho \cos\tau + \sinh\rho \cos\varphi$

$rt = \cosh\rho \sin\tau$

$rl\phi = \sinh\rho \sin\varphi$

Is there some way to combine the equations to get rid of $\varphi$ and $\tau$ and express $\rho$ in terms of $r, t, \phi$?

I tried the following:

$r^{2} = (\cosh\rho \cos\tau + \sinh\rho \cos\varphi)^{2}$

$r^{2}(t-l\phi)^{2} = (\cosh\rho \sin\tau - \sinh\rho \sin\varphi)^{2}$

so that we have

$r^{2} + r^{2}(t-l\phi)^{2} = \cosh^{2}\rho + \sinh^{2}\rho + 2\cos(\tau+\varphi)\sinh\rho\cosh\rho.$

The above line is not exactly what I want, because of the factor $\cos(\tau+\varphi)$!

Is there some neat way to get rid of $\varphi$ and $\tau$ and express $\rho$ in terms of $r, t, \phi$?

 

[jedishrfu]

Is this a homework problem? What course are you taking?

We need some context here. Where did this problem come from and what do you need it for?

 

[Svein]

There are connections between the hyperbolic and trigonometric functions:. For example:
\sinh(x)=-i \sin(ix)\cosh(x)=\cos(ix)
See https://en.wikipedia.org/wiki/Hyperbolic_function

 

[mfb]

You can express $\varphi$ and $\tau$ as function of the other variables using the second and third equation. Not nice, but possible.

 

[highflyyer]

Is this a homework problem? What course are you taking?

We need some context here. Where did this problem come from and what do you need it for?

This is not a homework problem. This is part of my research work.

The equations are a modified form of (1.17) on page 13 of https://esc.fnwi.uva.nl/thesis/centraal/files/f37733672.pdf.

I need it to make progress in my work.