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\noindent
Note: $g_{\text{hyp}} = \frac{4(\dd u^2 + \dd
v^2)}{(1 - u^2 - v^2)^2}$, so $E = G$, $F = 0$ so
this is \emph{conformal} to the flat metric.
Angles computed with respect to $g_{\text{hyp}}$
agree with Euclidean angles.

\subsubsection*{Hyperbolic cosine formula}

\begin{flashcard}[hyperbolic-cosine-formula]
\prompt{Hyperbolic cosine formula? \\}
\cloze{$\cosh C = \cosh A \cosh B - \sinh A \sinh B \cos
\gamma$.}
\end{flashcard}

\begin{proof}
    To simplify, by an isometry, put vertex with
    angle $\gamma$ at $0 \in D$ and put vertex
    with angle $\beta$ on $\RR_+ \cap D$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/b787d2a6bf3411ed.png}
    \end{center}
    Then
    \[ A = d_{\text{hyp}}(0, a) = 2\tanh^{-1}(a) \]
    \[ a = \tanh \frac{A}{2} \]
    and
    \[ b = e^{i\gamma} \tanh(B / 2) \]
    \[ \left| \frac{b - a}{1 - \ol{a}b} \right| =
    \tanh \left( \frac{C}{2} \right) \]
    If $t = \tanh(\lambda / 2)$, ``recall''
    \begin{align*}
        \cosh(\lambda)
        &= \frac{1 + t^2}{1 - t^2} \\
        \sinh(\lambda)
        &= \frac{2t}{1 - t^2}
    \end{align*}
    so
    \begin{align*}
        \cosh(A)
        &= \frac{1 + |a|^2}{1 - |a|^2} \\
        \cosh(B)
        &= \frac{1 + |b|^2}{1 - |b|^2} \\
        \cosh C
        &= \frac{|1 - \ol{a}b|^2 + |b - a|^2}{|1 -
        \ol{a}b|^2 - |b - a|^2} \\
        &= \frac{(1 + |a|^2)(1 + |b|^2) -
        2(\ol{a}b + a\ol{b})}{(1 - |a|^2)(1 -
        |b|^2)}
    \end{align*}
    But $a \in \RR$ and $b + \ol{b} = 2\Re(b) = 2b
    \cos \gamma$. Using the above one can check
    that
    \[ \cosh C = \cosh A \cosh B - \sinh A \sinh B
    \cos \gamma \]
    as desired.
\end{proof}

\begin{hiddenflashcard}[hyperbolic-sine-formula]
Hyperbolic sine rule?
\[ \cloze{\frac{\sin(A)}{\sinh(\alpha)} =
\frac{\sin(B)}{\sinh(\beta)} =
\frac{\sin(C)}{\sinh(\gamma)}} \]
\end{hiddenflashcard}

\begin{hiddenflashcard}[hyperbolic-trig-half-angle]
If $t = \tanh(\lambda / 2)$, then
\cloze{
\begin{align*}
    \cosh(\lambda)
    &= \frac{1 + t^2}{1 - t^2} \\
    \sinh(\lambda)
    &= \frac{2t}{1 - t^2}
\end{align*}
}
\end{hiddenflashcard}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item If $A, B$ and $C$ small, then $\sinh
        \approx A$, $\cosh \approx 1 +
        \frac{A^2}{2}$ then the formula reduces to
        \[ C^2 = A^2 + B^2 - 2AB \cos \gamma \]
        (up to higher order terms), which is just
        the usual Euclidean cosine formula.
        Recall: dilating a surface in $\RR^3$
        rescaled its curvature. Similarly, ``zooming in'' to
        any point on any abstract surface with a
        Riemannian metric, the surface looks
        closer and closer to being flat.
    \item $\cos\gamma \ge -1$, so formula gives
        \[ \cosh C \le \cosh A \cosh B + \sinh A
        \sinh B = \cosh (A + B) \]
        $\cosh$ is increasing, so this implies $C
        \le A + B$, which is the triangle
        inequality! So hyperbolic cosine formula
        is a refinement of the triangle
        inequality.
\end{enumerate}

\subsubsection*{Areas of triangles}

Let $T \subset \HH$ be a hyperbolic triangle, with
internal angles $\alpha, \beta, \gamma$.
\begin{flashcard}[hyperbolic-triangle-area-formula]
\prompt{Area of a hyperbolic triangle? \\}
\[ \boxed{\Area_{\text{hyp}}(T) = \cloze{\pi - \alpha -
\beta - \gamma}} \]
\end{flashcard}
\begin{hiddenflashcard}[spherical-triangle-area-formula]
\prompt{Area of a spherical triangle? \\}
\[ \boxed{\Area_{\text{spherical}}(T) = \cloze{
\alpha + \beta + \gamma - \pi}} \]
\end{hiddenflashcard}
This is a version of the \emph{Gauss-Bonnet}
theorem. We allow $T$ to have \emph{ideal
vertices}.

\begin{hiddenflashcard}[hyperbolic-triangle-area-formula-proof]
\prompt{Proof of hyperbolic triangle area formula?
\\}
\cloze{
Work in the half plane model, first doing the case
where one of the vertices is on $\partial H$. Put one
of the vertices at infinity, and the
other two vertices on the unit circle, and compute
the area.

\myskip
For the general case, use an isometry to take one
of the edges of the triangle to a vertical line,
and then write the area as a difference of areas
of triangles with a vertex at infinity, and
compute the area.
}
\end{hiddenflashcard}

\begin{proof}
    We can work in any model. We take $\HH =
    \mathcal{H}$. First we compute the area of the
    traignel assuming $\gamma = 0$ (so one vertex
    is at $\infty$). Using an isometry we can
    assume that two vertices are  on the unit
    circle as in the figure:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/acb85630bf3811ed.png}
    \end{center}
    Then
    \begin{align*}
        \Area(T)
        &= \int_{\cos(\pi - \alpha)}^{\cos\beta}
        \dd x \int_{\sqrt{1 - x^2}}^\infty
        \frac{\dd y}{y^2} \\
        &= \int_{\cos(\pi - \alpha}^{\cos\beta}
        \frac{\dd x}{\sqrt{1 - x^2}} \\
        &= \int_{\pi - \alpha}^\beta -\dd \theta
        \\
        &= \pi - \alpha - \beta
    \end{align*}
    For the general case, we can always arrange
    for one of the sides of a triangle to be a
    vertical line:
    \begin{center}
        \includegraphics[width=0.4\linewidth]
        {images/f12f6d08bf3811ed.png}
    \end{center}
    Then
    \begin{align*}
        \Area(T)
        &= \pi - \alpha - (\beta + \delta) - (\pi
        - \delta - \pi + \gamma) \\
        &= \pi - \alpha - \beta - \gamma \qedhere
    \end{align*}
\end{proof}

\begin{corollary}
    The area of hyperbolic $n$-gon (with sides
    being arcs of hyperbolic lines) is given by
    the formula
    \[ (n - 2)\pi - (\alpha_1 + \cdots + \alpha_n) \]
    where $\alpha_1, \ldots, \alpha_n$ are the
    internal angles.
\end{corollary}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/450512c0bf3911ed.png}
\end{center}

\begin{proof}
    Divide into $n$ triangles
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/66fb7234bf3911ed.png}
    \end{center}
    The $n$ triangles have interior angles adding
    up to $2\pi + \sum_{i = 1}^n \alpha_i$.
    \begin{align*}
        \implies \Area(\text{$n$-gon})
        &= n\pi - \left( 2\pi + \sum_{i = 1}^n
        \alpha_i \right) \\
        &= (n - 2)\pi - \sum_{i = 1}^n \alpha_i
        \qedhere
    \end{align*}
\end{proof}

\begin{lemma}
    If $g \ge 2$, there is a regular $4g$-gon in
    $\HH$ with internal angle $\frac{\pi}{2g}$.
\end{lemma}

\begin{proof}
    Take an \emph{ideal} $4g$-gon in $D$ with all
    vertices at $\partial D$ being the $4g$-roots
    of unity
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/8342ae70bf3a11ed.png}
    \end{center}
    ``slide'' all vertices radially inwards:
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/969ffd7ebf3a11ed.png}
    \end{center}
    Then $A(r) = (4g - 2)\pi - 4g\alpha(r)$. $A(r)
    \to 0$ as $r \to 0$, and $A(1) = (4g - 2)
    \pi$. Hence, by the intermediate value theorem
    there exists $r_0$ for which $4\pi g - 4\pi =
    A(r_0) = (4g - 2)\pi - 4g\alpha(r_0)$, which
    gives $\alpha(r_0) = \frac{\pi}{2g}$.
\end{proof}

\myskip
We'll use this lemma to construct a Riemannian
metric with $\kappa = -1$ a surface of genus $\ge
2$.
