Optimization Quick Problem

It is a common trick to use trigonometric identities to get ride of square roots in computations. For example if we let $x=\rho\cos \theta$, then 
\[\sqrt{\rho^2-x^2}=\sqrt{\rho^2-\rho^2\cos^2 \theta}=\sqrt{\rho^2(1-\cos^2 \theta)}=\sqrt{\rho^2\sin^2 \theta}=\rho\sin \theta. \]
In many problems computationally working with $\rho\sin \theta$ could be easier that working with $\sqrt{\rho^2-x^2}$. 

In your problem we have $\sqrt{h^2+(r-R)^2}$. So we would like to make changes of variables that helps us get rid of the quare root.  If we let $h=\rho\cos \theta$, it would be great if 
\[(1)        (r-R)=\rho \sin \theta,\]
so that 
\[\sqrt{h^2+(r-R)^2}=\sqrt{\rho^2\cos^2 \theta+\rho^2\sin^2 \theta}=\sqrt{\rho^2(\sin^2 \theta +\cos^2 \theta)}\]
\[=\sqrt{\rho^2}=\rho.\]

Indeed if we let 
$$   r = \frac{s+\rho \sin \theta}{2}, \\ R = \frac{s-\rho \sin \theta}{2},  $$
then (1) holds and this change of variable will replace the term $\sqrt{h^2+(r-R)^2}$ by just  $\rho$, and computations will be simplified a lot. Notice that the most computationally challenging term in your equations is $\sqrt{h^2+(r-R)^2}$ and replacing it with only $\rho$ would significantly simply your equation. 

The following system of 4 equations just follows from replacing $h, r$ and $R$ in terms of $\rho, \theta, s$ from the above formulas. 

I hope this makes it clear why they chose this change of variable. Let me know if you have any further questions, and I will be happy to add more details to this answer.