TITLE: Geometric construction of hyperbolic trigonometric functions
QUESTION [15 upvotes]: If we have a circle we can geometrically construct the trigonometric functions as shown. The functions all derive from sin and cos. If we say that the circle is a conic section and imagine it on the cone we can draw a hyperbola perpendicular to it. I believe that the hyperbolic trigonometric functions can be plotted geometrically as well but I cannot find and representation of it. What do the hyperbolic trigonometric functions actually tell you about a hyperbola?

REPLY [2 votes]: $$
\begin{align}
&\int_1^{\cosh(u)}\sqrt{x^2-1}\,\mathrm{d}x\\
&=\int_0^u\sinh(t)\,\mathrm{d}\cosh(t)\tag1\\
&=\sinh(u)\cosh(u)-\int_0^u\cosh(t)\,\mathrm{d}\sinh(t)\tag2\\
&=\frac12\sinh(u)\cosh(u)-\frac12\int_0^u\cosh^2(t)\,\mathrm{d}t+\frac12\int_0^u\sinh^2(t)\,\mathrm{d}t\tag3\\[3pt]
&=\frac12\sinh(u)\cosh(u)-\frac u2\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $x=\cosh(t)$
$(2)$: integrate by parts
$(3)$: average $(1)$ and $(2)$
$(4)$: $\cosh^2(t)-\sinh^2(t)=1$
Consider the diagram

Subtracting the area of the violet region, which by $(4)$ is $\frac12\sinh(u)\cosh(u)-\frac u2$, from the area of the violet and green right triangle, which is $\frac12\sinh(u)\cosh(u)$, we get that the area of the green region is  $u/2$.
This is reminiscent of the situation with circular functions when the angle is measured by the area of its sector: