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\documentclass{amsart}
\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath, fullpage}
\title{The addition formulas for the hyperbolic sine and cosine functions via linear algebra}
\author{David Radcliffe}
\begin{document}
\begin{abstract}
We present a geometric proof of the addition formulas for the hyperbolic sine and cosine functions,
using elementary properties of linear transformations.
\end{abstract}
\maketitle
\thispagestyle{empty}
\section{Introduction}
By analogy with the unit circle, the \emph{unit hyperbola} is the set of points in the plane satisfying
the equation $x^2 - y^2 = 1$. The hyperbola is not connected -- it has two branches. The right branch ($x > 0$)
is parameterized by $x = \cosh t$ and $y = \sinh t$ for $t \in \mathbb{R}$.
A \emph{hyperbolic sector} is the curvilinear triangular region bounded by an arc of the hyperbola and by two
line segments from the origin to the endpoints of the arc. If $t > 0$ then the area of the hyperbolic
sector bounded by the arc from $(1, 0)$ to $(\cosh t, \sinh t)$ is $t/2$.
This fact about hyperbolic sectors provides a $\emph{geometric}$ definition of the hyperbolic sine and
cosine functions.
The hyperbolic sine and cosine functions satisfy addition rules that are strikingly similar to
the analogous formulas for sine and cosine.
\begin{align*}
\cosh (s+t) &= \cosh s \cosh t + \sinh s \sinh t \\
\sinh (s+t) &= \sinh s \cosh t + \cosh s \sinh t
\end{align*}
We will prove these formulas under the assumption that $s$ and $t$ are positive, although they are in fact
valid for all real values of $s$ and $t$.
\section{Proof}
Let $s$ and $t$ be positive real numbers. The linear transformation
$$T(x, y) = (x \cosh t + y \sinh t, x \sinh t + y \cosh t)$$
preserves the right branch of the unit hyperbola $$x^2-y^2=1$$
and it preserves areas since $\det T = 1$.
Let $A$ be the hyperbolic sector bounded by the arc from $(1, 0)$ to $(\cosh s, \sinh s)$, and
let $B$ be the hyperbolic sector bounded by the arc from $(1, 0)$ to $(\cosh t, \sinh t)$.
Note that $A$ has area $s/2$, and $B$ has area $t/2$.
The image $A' := T(A)$ is a hyperbolic sector since $T$ preserves the right branch of the unit hyperbola;
and it has area $s/2$ since $T$ preserves areas. $A'$ is bounded by the arc from
$T(1,0) = (\cosh t, \sinh t)$
to
$$T(\cosh s, \sinh s) = (\cosh s \cosh t + \sinh s \sinh t,\ \sinh s \cosh t + \cosh s \sinh t).$$
Now, $A' \cup B$ is a hyperbolic sector, bounded by the arc from $(1, 0)$ to
$$(\cosh s \cosh t + \sinh s \sinh t,\ \sinh s \cosh t + \cosh s \sinh t).$$
Since the area of $A'\cup B$ is $(s+t)/2$, the upper endpoint can be expressed as
$$(\cosh (s+t),\ \sinh (s+t)).$$
Therefore,
$$\cosh (s+t) = \cosh s \cosh t + \sinh s \sinh t$$
and
$$\sinh (s+t) = \sinh s \cosh t + \cosh s \sinh t.$$
\end{document}
```