The goal is to transform the following coordinates: $$x(t)= R(\Phi-\sin\Phi)$$ and $$z(t)=R(2 +\cos\Phi)$$
with the substitution: $u=\cos\left(\Phi/2\right)$
in order to get: $$x(t)=2R(\arccos(u)-u\sqrt{1-u^2})$$ and $$z(t)=R(1+2u^2)$$
How do I go about solving this problem? I already tried using the following addition theorems: $\cos(\Phi/2)=\sqrt{(1+\cos\Phi)/2)}$ and $\sin\Phi=\sqrt{1-\cos^2\Phi}$ and identities for $\arccos$, but somehow I failed to transform the coordinates.
On
As I always have problems to memorize trigonometric identities, I prefer to start by a basic identity:
$$e^{i\phi}=\left(e^{i\phi/2}\right)^2$$
an use Euler's formula for real and imaginary parts:
$$\cos(\phi)=\left(\cos(\phi/2)\right)^2-\left(\sin(\phi/2)\right)^2$$ $$\sin(\phi)=2\cos(\phi/2)\sin(\phi/2)$$
From these two trigonometric identities and introducing the definition of $u$: $$\cos(\phi)=u^2-(\sqrt{1-u^2})^2=...$$ $$\sin(\phi)=...$$
you will finish in the target equations.
This is quite straightforward, I'm not sure where you're having a problem. I won't give the full answer since this is a classic "homework-style" problem, I'll just show you how to do one of them, and the other can be done in almost exactly the same way. Take the first equation, you want to go from:
$$x(t) = R(\Phi - \sin\Phi) \quad \quad \text{to} \quad \quad x(t) = R ( 2 \cos^{-1}u - 2u\sqrt{1- u^2}).$$
The most sensible substitution to start with is $$\Phi = 2 \cos^{-1}u \quad \quad \text{or}\quad \quad u = \cos\left({\frac{\Phi}{2}}\right).$$
Draw out a little triangle (or use trigonometric identities), and you should be able to show that this means that $$\sin{\left(\frac{\Phi}{2}\right)} = \sqrt{1-u^2}.$$
Then, using the simple result $$\sin\Phi = 2\sin{\left(\frac{\Phi}{2}\right)}\cos{\left(\frac{\Phi}{2}\right)},$$ you should be done. You should now be able to do the next one, after you find a relation between $\cos\Phi$ and $\cos(\Phi/2)$ and $\sin(\Phi/2)$.