Zenithal perspective projections are generated from a point P and carried through the sphere to the plane of projection as illustrated in the figure below.
By a simple geometric relationship, we can show that the radius satisfies the following:
with . The inverse equation is given by:
where
I could reproduce the first equation but not the second one. I think the derivation is purely based on algebraic operation of the first one but I have no idea how. Does anyone have any idea how to derive the second equation?
I will work in radians allowing to neglect factor $\frac{180}{\pi}$.
From your drawing, I see that you can express the tangent of the angle in $P$ in two ways :
$$\underbrace{\frac{R_{\theta}}{(\mu+1)}}_{\rho}=\frac{\cos \theta}{\mu+\sin \theta}.$$
Otherwise said :
$$\rho (\mu+\sin \theta)=\cos \theta$$
which is equivalent to :
$$\cos \theta-\rho \sin \theta=\rho \mu$$
Let us divide both sides by $\sqrt{1+\rho^2}$ :
$$\frac{1}{\sqrt{1+\rho^2}}\cos \theta-\frac{\rho}{\sqrt{1+\rho^2}} \sin \theta=\frac{\rho}{\sqrt{1+\rho^2}} \mu \tag{1}$$
There is an angle $\alpha$ such that :
$$\begin{cases}\frac{1}{\sqrt{1+\rho^2}}&=&\sin\alpha\\\frac{\rho}{\sqrt{1+\rho^2}}&=&\cos \alpha\end{cases}\tag{1'}$$
A little pause here : let us observe that the ratio of the previous equations gives
$$\frac{1}{\rho}=\tan \alpha \tag{1''}$$
(which is a way to define $\alpha$)
Using (1'), (1) can be transformed into :
$$\sin\alpha \cos \theta-\cos\alpha \sin \theta=\frac{\rho}{\sqrt{1+\rho^2}} \mu $$
We recognize here an addition formula :
$$\sin(\alpha -\theta)=\frac{\rho}{\sqrt{1+\rho^2}} \mu$$
giving
$$\alpha - \theta = \sin^{-1}\frac{\rho \mu}{\sqrt{1+\rho^2}}$$
$$ \theta = \alpha - \sin^{-1}\frac{\rho \mu}{\sqrt{1+\rho^2}}$$
with $\alpha$ given indeed by relationship (1'').