I'm trying to figure out how the hyperbolic functions are derived using a unit hyperbola.

According to this walkthrough, argument $u$ in $(\cosh(u), \sinh(u))$ should be equal to $2A$, where $A$ is the area of an intercepted hyperbolic sector from $(0,0)$ to $(\cosh(u), \sinh(u))$.

Confused as to why this is defined as such, I found on Wikipedia:

The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.

However, I couldn't find a explanation for this anywhere. Can anyone help show me why this is?

  • $\begingroup$ See the last paragraph of this answer. $\endgroup$
    – Blue
    Commented Jan 12, 2019 at 5:11
  • $\begingroup$ The middle portion of your answer is a very difficult argument to follow. I've managed to figure out the "e" definition by means of integration, but I still don't understand from your answer what necessarily implies u = 2A. I'm guessing this is somehow in the geometry of your included picture, but I would really prefer a more algebraic answer because I'm more easily convinced that way. Could you elaborate on all of this more specifically? I appreciate your work by the way, in fact your answer was one of the first I read when I started trying to figure this out and it helped get me started. $\endgroup$ Commented Jan 12, 2019 at 5:52

1 Answer 1


The notion of "hyperbolic angle" is really a contrived concept that is only used to better define an analogy between the hyperbolic functions as a parameterization of the hyperbola and the trigonometric functions as a parameterization of the circle.

The hyperbolic functions are not defined based on the traditional notion of an angle. They are, in fact, defined based on the area of an intercepted arc on the hyperbola.

To find the area of this intercepted arc using calculus and geometry, subtract the area under the curve from 1 to some x in quadrant I from the triangle created from the line that connects the origin $(0,0)$ with this point $(x,\sqrt{x^2 -1})$:

$\hskip2in$Area under curve

\begin{align} A &= \frac{1}{2}x\sqrt{x^2 -1} - \int _1 ^x \sqrt{x^2 -1}dx \\ &= \frac{1}{2}x\sqrt{x^2 -1} - \frac{1}{2}x\sqrt{x^2 -1} + \frac{1}{2}\ln(x + \sqrt{x^2-1}) \\ 2A &= \ln(x + \sqrt{x^2-1})\\ e^{2A}&= x + \sqrt{x^2-1}\\ \end{align}

At this point, we can substitute $\sqrt{x^2-1}$ with $y$ for simplicity.

\begin{align} e^{2A}&= x + y\\ e^{2A}(x-y)&= (x+y)(x-y)\\ e^{2A}(x-y)&= x^2-y^2\\ \end{align}

Because these equations describe a unit hyperbola $x^2 - y^2 = 1$, we can make this crucial substitution and derive the $e^x$ definition.

\begin{align} e^{2A}(x-y)&= 1\\ e^{2A}&= \frac{1}{x-y}\\ e^{-2A}&= x-y\\ \end{align}

If we add $x+y$ (which, remember, is equal to $e^{2A}$) to this equation, the $y$'s will cancel. \begin{align} 2x &= e^{-2A} + x + y \\ 2x &= e^{-2A} + e^{2A} \\ x &= \frac{e^{-2A} + e^{2A}}{2} \\ \end{align}

We can find $y$ if we calculate $(x+y) - (x-y)$ \begin{align} (x+y) - (x-y) &= e^{2A} - e^{-2A} \\ 2y &= e^{2A} - e^{-2A} \\ y &= \frac{e^{-2A} - e^{2A}}{2} \end{align}

If we treat these equations as functions themselves, we can now see where this $u = 2A$ definition comes from more evidently. Every $(x,y)$ on the graph $x^2 - y^2 = 1$ can be described in a relationship involving area $A$ times $2$ as proven above. We can substitute $2A$ with $u$ and call our argument $u$ whatever we want henceforth.

Here's an interactive Desmos graph I made to help illustrate the connection. It isn't a "formal proof", but I think its pretty cool.


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