Sorry guys, I don't know how to be more specific in the question without writing a way too long question...
Anyway my problem:
I have this formula to calculate the distance between two points on the earth. $$d=\operatorname{arcos}\left(\sin\theta_1\ \sin\theta_2+\cos\theta_1\ \cos\theta_2\ \cos(\phi_2-\phi_1)\right)$$
Now I need to calculate $\phi_2$.
I tried to change the formula myself but I failed to do it right. Now I am asking you guys to help me. It would be nice if someone could show me how he resolves it.
Thanks in advance
This basically amounts to some algebraic manipulation.
You have a formula of the form $$d=\operatorname{arcos}\left(\sin\theta_1\ \sin\theta_2+\cos\theta_1\ \cos\theta_2\ \cos(\phi_2-\phi_1)\right)$$ Now recall that $\operatorname{arcos}(x)$ is the inverse function of $\cos(\theta)$; therefore applying $\cos$ to both sides of the equation gives you the formula from the question I linked $$\cos d=\sin\theta_1\ \sin\theta_2+\cos\theta_1\ \cos\theta_2\ \cos(\phi_2-\phi_1)$$ But you actually want to know $\phi_2$ so we start by taking $\cos(\phi_2-\phi_1)$ to one side of the equality and leaving the rest on the other side $$\cos(\phi_2-\phi_1)=\frac{\cos d-\sin\theta_1\ \sin\theta_2}{\cos\theta_1\ \cos\theta_2}$$ Finally, we apply $\operatorname{arcos}$ to both sides of the equation and add $\phi_1$ $$ \begin{align} \phi_2-\phi_1&=\operatorname{arcos}\left(\frac{\cos d-\sin\theta_1\ \sin\theta_2}{\cos\theta_1\ \cos\theta_2}\right)\\ \phi_2&=\operatorname{arcos}\left(\frac{\cos d-\sin\theta_1\ \sin\theta_2}{\cos\theta_1\ \cos\theta_2}\right)+\phi_1\\ \end{align} $$