I have seen this equality in some youtube video on hyperbolic geometry, but I want to understand the proof of it.
dist$((r_1,\theta_1),(r_2,\theta_2)) = \text{arcosh}(\text{cosh}r_1\text{cosh}r_2 - \text{sinh}r_1 \text{sinh}r_2\cos(\theta_1 -\theta_2))$
I understand, that it is some kind of a consequence of the law of cosines, but in hyperbolic form, but how it could be proved?
If i understand correctly, then
by hyperbolic law of cosines $\text{cosh}(r_1 r_2) = \text{cosh}r_1\text{cosh}r_2 - \text{sinh}r_1\text{sinh}r_2 cos(\theta_1 - \theta_2)$
then
$r_1 r_2 = \text{arcosh}(\text{cosh}r_1\text{cosh}r_2 - \text{sinh}r_1\text{sinh}r_2 cos(\theta_1 - \theta_2))$
where $r_1 r_2 = d((r_1, \theta_1) , (r_2, \theta_2))$. Hope I am right