Following a story similar to the unit circle, \(\sinh (\alpha )\) and \(\cosh (\alpha )\) are defined as the coordinates of the points on \(x^2 - y^2 = 1\), where \(A = \frac {\alpha }{2}\).

From the geometry, we can get formulas for \(\sinh (\alpha )\) and \(\cosh (\alpha )\) that relate back to our Elementary Functions.

Let’s zoom in on the diagram above. For formula purposes, let’s use \(b = \cosh (\alpha )\). Therefore, the corner of our right triangle is \((cosh(\alpha ), 0) = (b, 0)\).

Taking a little jump of faith, Calculus tells us that green area equals

The pink area is the area of the right triangle minus the green area.

This gives us

Now, solve for \(b\).

0.1 Graph

From \(\cosh (\alpha ) = \frac { e^{\alpha } + e^{-\alpha }}{2}\), we can see the function is an even function. That is, \(\cosh (\alpha ) = \cosh (-\alpha )\), making the graph symmetric.

We can also see that \(\lim \limits _{\alpha \to \infty } \cosh (\alpha ) = \infty \) \(\lim \limits _{\alpha \to -\infty } \cosh (\alpha ) = \infty \).

Finally, \(\cosh (0) = 1\)

Graph of \(y = \cosh (\alpha )\).

The graph isn’t a parabola. Compare the graphs of \(\cosh (x)\) and \(x^2\).

Instead, the graph is a catenary.

A catenary is what you get when you hang a telephone line from one pole to the next. The resulting curve has the unique property that the tension forces are all parallel to the curve. This minimizes its movements.

Now flip this curve upside down. It is an arch with no sheer stress. All of the forces are along the arch. This makes it very stable.

\(\blacktriangleright \) End-Behavior

Since \(\cosh (x)\) is exponential, it is going to dominate over power functions, like \(x^2\).