Approximating Distances in the Hyperbolic Space I am currently trying to understand the paper by Krioukov et. al. on hyperbolic networks, but since I do not have a background in hyperbolic geometry (or, in that sense, in geometry at all) I struggle to understand some points they are making.
For instance, they aim to compute the distance $x$ between two points $(r, \theta)$ and $(r', \theta')$ in the hyperbolic plane by using the hyperbolic law of cosines:
\begin{equation}
  \cosh(kx) = \cosh(kr) \cosh(kr') - \sinh(kr)\sinh(kr')\cos(\Delta\theta)
\end{equation}
Here, $K=-k^2<0$ refers to the curvature, and $\Delta \theta = \pi - |\pi - |\theta - \theta'||$. I can intuitively follow this equation since it is similar to the regular law of cosines in the Euclidean plane. However, in the next step they approximate it by assuming that $kr, kr'$ is large, and $\Delta \theta > 2 \sqrt{e^{-2kr}+e^{-2kr'}}$ as follows:
\begin{equation}
x = r + r' + \frac{2}{k} \ln \sin \frac{\Delta\theta}{2} \approx r + r' + \frac{2}{k} \ln \frac{\Delta\theta}{2}.
\end{equation}
I am completely lost there, especially their assumption on $\Delta \theta$. Why does it make sense and how does it help simplify the formula?
 A: The derivation is a little long, and I admit to being mystified by their assumption on $\Delta \theta$ - it merely needs to be that $\Delta \theta$ is small compared to something $O(1)$.  But anyway...write
$$\cosh{k x} = \cosh{k r} \cosh{k r'} (1-\tanh{k r} \tanh{k r'} \cos{\Delta \theta})$$
and note that $\lim_{t \rightarrow \infty} \tanh{t} = 1$.  We can then write
$$\cosh{k x} \approx 2 \cosh{k r} \cosh{k r'} \sin^2{\left ( \frac{\Delta \theta}{2} \right )}$$
Use the definition of cosh and that $k r$ and $k r'$ are both large to derive the following equation:
$$\exp{(2 k x)} - \exp{(k (r + r'))} \sin^2{\left ( \frac{\Delta \theta}{2} \right )} \exp{(k x)} + 1 \approx 0 $$
This gives
$$  \exp{(k x)} \approx \exp{(k (r + r'))} \sin^2{\left ( \frac{\Delta \theta}{2} \right )} \frac{1}{2} \left [ 1 \pm \sqrt{1- 4 \exp{(-k (r + r'))} \csc^2{\left ( \frac{\Delta \theta}{2} \right )}} \right ] $$
The sqrt term on the right is very small.  Assume the +ve solution, so the term in brackets is about 2.  Take logs of both sides and get
$$x \approx r + r' + \frac{2}{k} \log{\sin{\left ( \frac{\Delta \theta}{2} \right )}} $$ 
I hope this helps.
