How do I show $\mathbb{R}^2$ with $dr^2 + \sinh^2r d \theta^2$ is isometric to the Poincare disc with $g = \frac{4(dx^2 + dy^2)}{(1-x^2-y^2)^2}$?
I tried converting to polar coordinates, so that $x = r \cos{\theta}$ and $y = r\sin{\theta}$. Then $$g = \frac{4}{(1-r^2)^2} (dr^2 + r^2 d{\theta}^2). $$
I don't really know where to go from here. I saw somewhere that $$\sinh^2{r} = \frac{4r^2}{(1-r^2)^2},$$ but I don't know how to get to that. Even if I did, that would only take care of the second term, but $dr^2$ would still have that additional factor in front.
Define a map $\mathbb{R}^2 \to D$ by $x=\tanh{(r/2)}\cos{\theta}$, $y=\tanh{(r/2)}\sin{\theta}$. To compute the new metric, we use $$ g(x)_{ij} = g(y)_{kl} \frac{\partial y^k}{\partial x^i} \frac{\partial y^l}{\partial x^j} ,$$ the transformation law for a covariant tensor (in particular, we have to start with the metric on $D$ and pull back to the metric on $\mathbb{R}^2$). In this case, $$ \frac{\partial x}{\partial r} = \frac{1}{2}\sech^2{(r/2)}\cos{\theta}, \quad \frac{\partial y}{\partial r} = \frac{1}{2}\sech^2{(r/2)}\sin{\theta} \\ \frac{\partial x}{\partial \theta} = -\tanh{(r/2)}\sin{\theta} \quad \frac{\partial y}{\partial \theta} = \tanh{(r/2)}\cos{\theta}, $$ and the metric on $D$, written in the new coordinates, is $$ \frac{4}{\sech^4{(r/2)}}\delta_{kl} $$ so in matrix notation, $$ g_{rr} = \frac{4}{\sech^4{(r/2)}}\begin{pmatrix} \frac{1}{2}\sech^2{(r/2)}\cos{\theta} \\ \frac{1}{2}\sech^2{(r/2)}\sin{\theta} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} \sech^2{(r/2)}\cos{\theta} \\ \frac{1}{2}\sech^2{(r/2)}\sin{\theta} \end{pmatrix} = \dots = 1, $$ and by the same process, $$ g_{r\theta}=g_{\theta r}=0, \quad g_{\theta\theta} = \sinh^2{r}, $$ as required.