2\,\sinh u &\;=\; e^{u} -\, e^{-u} \\[4pt]

2\,\cosh u &\;=\; e^{u} + e^{-u}

\end{align}$$

$$\begin{align}

|\overline{OX}|\cdot|\overline{XY}| \;\equiv\; 1 &\quad\to\quad u = \int_{1}^{|\overline{OX}|} \frac{1}{t}dt \;=\; \ln |\overline{OX}| \\[6pt]

&\quad\to\quad |\overline{OX}| = e^{u} \quad\text{and}\quad |\overline{XY}| = e^{-u}

\end{align}$$

**Some background.** Geometrically, we define \(\sinh u\) and \(\cosh u\) by direct analogy with \(\sin\theta\) and \(\cos\theta\): as certain perpendicular segments associated with an arc of the “unit hyperbola”, \(x^2 – y^2 = 1\).

While \(\theta\) is usually interpreted as the length of a circular arc, we note that it is also *twice* the area of the corresponding circular sector. The hyperbolic parameter \(u\) is interpreted via area, as well; today’s trigonograph shows why:

Conveniently scaling lengths in the unit hyperbola figure by \(\sqrt{2}\) —and, thus, scaling areas by \(2\)—

we see that \(\sinh u\) and \(\cosh u\) are *directly computable from \(u\) via the exponential function!*

(In fact, we can say the same of \(\sin\theta\) and \(\cos\theta\), but we need *complex* exponentials for that.)

*Taken from this answer of mine on the Mathematics Stack Exchange.*