# Geometric construction of hyperbolic trigonometric functions

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?

• Thee is a similar question here: math.stackexchange.com/questions/321343/… , but no answer. Commented Jul 24, 2013 at 14:14
• It's worth noting that the line tangent to the point with coordinates $(\cosh x, \sinh x)$ on the unit hyperbola has slope $\cosh x/\sinh x = \coth x$, echoing the fact that the line tangent to the unit circle at $(\cos\theta, \sin\theta)$ has slope $-\cot\theta$.
– Blue
Commented Jul 24, 2013 at 14:20
• (Whoops, timed-out.) So, the line through the origin, parallel to that tangent line, meets the vertical line through the appropriate hyperbola vertex, at a point that happens to be $\coth x$ units away from the hyperbola's axis. There's a construction of $\coth$.
– Blue
Commented Jul 24, 2013 at 14:28

If you consider this alternative rendering of the circular trig diagram ...

... then there's this hyperbolic analogue:

In each case, the point where the inclined ray meets the curve determines sine and cosine, and the points where it meets the vertical and horizontal tangent lines determine tangent and cotangent. (In the hyperbolic case, the "horizontal tangent line" is actually tangent to the (invisible) conjugate hyperbola.)

What about the secant and cosecant segments? In the circular case, these are the portions of the inclined ray that form hypotenuses of the $1$-and-$\tan$ and $1$-and-$\cot$ right triangles. In the hyperbolic case ... well ... there don't seem to be direct analogues to the circular arrangement.

Edit (7 April, 2018). But, there is this:

Here, the (now-visible) conjugate hyperbola fittingly hosts the cosine as well as cotangent ... though not cosecant, which, like secant, is introduced via an interesting ---if non-obvious--- reciprocal construction. (In an arbitrary hyperbola, the construction relates two segments whose geometric mean is the conjugate radius.) Still, there's a nice balance across the two hyperbolas.

The hyperbolic figure isn't as identity-rich as its circular counterpart, but we see that it covers the essentials. For instance, we have

$$\sinh \cdot \operatorname{csch} = 1 \qquad \cosh \cdot \operatorname{sech} = 1 \qquad \tanh \cdot \coth = 1$$

the last of which is implied by similar triangles; similarity also yields

$$\frac{\tanh}{1} = \frac{\sinh}{\cosh} \qquad\qquad \frac{\coth}{1} = \frac{\cosh}{\sinh}$$

Moreover, since the hyperbola has equation $x^2-y^2=1$, we have these hyperbolic Pythagorean relations $$\cosh^2 - \sinh^2 = 1 \qquad\qquad \coth^2 - \operatorname{csch}^2 = 1$$ while a guest appearance by the unit circle $x^2+y^2=1$ shows $$\operatorname{sech}^2 + \tanh^2 = 1$$

• In the last paragraph, you'd said "and doesn't provide much guidance about the corresponding placement of sech"; on the bright side, at least {sec(x) = 1/cos(x)} sees analytic continuation to {sech(z) = 1/cosh(z)}. mathworld.wolfram.com/HyperbolicSecant.html Commented Jul 4, 2016 at 5:34
• I've been looking into this lately, and I'm wondering if the best way to bring sech and csch into your second diagram is just to measure distances along the arm of the angle using a Minkowski metric instead of a Euclidean metric. Then all the same line segments can be used between the circle diagram and the hyperbola diagram, and you can even say the distance from the origin to the circle/hyperbola is always 1 in both cases. Commented Sep 25, 2017 at 18:32
• +1 for the new diagram added on April 7th, 2018, almost five years after the initial post. Commented Apr 20, 2018 at 7:00
• @Blue ... FYI: I have put a version of the diagram onto a t-shirt (via Teespring). (I'll delete this comment if there are objections.)
– Blue
Commented Mar 27, 2021 at 10:05

\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:

• If you rotate everything by $45^\circ$, the integration becomes significantly easier, since the hyperbola becomes (a multiple of) $y=1/x$. See, for instance, this answer of mine, which scales lengths by $\sqrt{2}$ to provide a diagrammatic derivation of the exponential definitions of $\sinh$ and $\cosh$. (The answer also observes that measuring angles by twice the sector area is the key to bridging the circular and hyperbolic functions.)
– Blue
Commented Jun 16, 2021 at 17:52