\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Let $T$ be a triangle in the hyperbolic plane with angles
$\alpha$, $0$, and $\frac{\pi}{2}$.  Prove that
$\cosh(b)\sin(\alpha)= 1$, where $b$ is the hyperbolic length of
the side of $T$ opposite the vertex with angle $0$. [Note that
since the triangle is not compact, the hyperbolic trigonometric
rules as we have stated and proved them in class do not apply.]
\medskip

{\bf Answer} %note ref to previous problem%

Using the transitivity properties of M\"{o}b$^+({\bf{H}})$, assume
that the vertex with angle 0 is at $\infty$, the vertex with angle
$\ds\frac{\pi}{2}$ is at $i$, and the side joining the vertices
with angles $\ds\frac{\pi}{2}$ and $\alpha$ lies on the unit
circle $S_1$, as pictured below:

\begin{center}
$ \begin{array}{c}
\epsfig{file=407-10-1.eps, width=50mm}
\end{array}
\ \ \
\begin{array}{c}
v = e^{i\alpha} =\cos(\alpha)+i\sin(\alpha)
\end{array} $
\end{center}

\begin{eqnarray*} 1+\ds\frac{|v-i|^2}{2 \cdot 1 \cdot
\sin(\alpha)} & = &
1+\ds\frac{\cos^2(\alpha)+(\sin(\alpha)-1)}{2\sin(\alpha)}\\ & = &
\ds\frac{2\sin(\alpha)+2+-2\sin(\alpha)}{2\sin(\alpha)}\\ & = &
\ds\frac{1}{\sin(\alpha)} = \csc(\alpha) \end{eqnarray*}

So, by previous problem, $\cosh(b)=\csc(\alpha)$ and so
$\cosh(b)\sin(\alpha)=1$ as desired.

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