We have defined the values of sine and cosine as coordinates of points on the unit circle. These have been tied to values of the rotating angle about the origin. Finding a domain relationship for desired function values is called parameterizing. In this case, we have the function values of sine and cosine and then we establish the angle as the parameter.
We have the same situation when defining the hyperbolic trigonometric functions. The unit hyperbola gives the function values. However, obtaining the domain parameter is not so straightforward. Rotating angles will not work, because the unit hyperbola does not go around the origin.
Instead, area is the domain parameter.
Let’s parameterize sine and cosine by area and then duplicate this process for the unit hyperbola.
0.1 Area Parameterization
We say that \(\sin (\theta )\) and \(\cos (\theta )\) are parameterized by angle \(\theta \). Given an angle \(\theta \), draw a line from the origin at that angle. The line intersects the unit circle. \(\sin (\theta )\) and \(\cos (\theta )\) are defined to be the coordinates of that point.
We could parameterize \(\sin (\theta )\) and \(\cos (\theta )\) with something besides the angle.
0.2 Arc Length
If you begin at \((1,0)\) and move counterclockwise on the unit circle, then the arc length you have travelled identifies points just as well as an angle does.
If you travel an arc distance of \(\ell \) on the unit circle, then you arrive at a point and its coordinates depend on \(\ell \) insead of \(\theta \): \(\sin (\ell )\) and \(\cos (\ell )\).
Of course, on the unit circle \(\theta \) and \(\ell \) have the exact same measurement. The unit circle is \(2\pi \) radians around. The circumference of the unit circle is \(2\pi \). Numerically, the angle and the arc length mark the same points on the unit circle.
0.3 Sector Area
There are other choices for parameter in addition to angle and arc length.
If you begin at \((1,0)\) and move counterclockwise on the unit circle, then you sweep out a sector area - only one area.
Or, thinking the other way, if you sweep out a particular sector area, then you arrive at a point and its coordinates depend on \(A\) insead of \(\theta \): \(\sin (A)\) and \(\cos (A)\).
Actually, this isn’t much different that the other two parameterizations.
If the sector sweeps out an angle \(\theta \), then the area of the sector is \(A = \frac {\theta }{2\pi } \pi 1^2 = \frac {\theta }{2}\)
The area is just half the angle.
Therefore, we can connect this back up to the two other parameterizations, by \(\theta = 2 A\).
All of our trigonometric functions are based off of these coordinates of the unit circle.
All of our trigonometric functions are based off of this curve defined by \(x^2 + y^2 = 1\).
But, a circle is just one type of conic section.
We could do the same thing with the curve defined by \(x^2 - y^2 = 1\) - a unit hyperbola.
1 Hyperbolic Functions
Let’s repeat the whole process with \(x^2 - y^2 = 1\).
The graph is a unit hyperbola, rather than a circle.
The points on the parabola have two coordinates and we will call them hyperbolic-sine and hyperbolic-cosine.
We’ll keep \(\theta \) for circular trig functions and adopt \(\alpha \) as the hyperbolic angle.
We need a way to walk along the hyperbola, like we moved around the unit circle.
Since the unit hyperbola doesn’t wrap around the origin, it is easier for the hyperbolic trig functions to be parameterized by area.
We can still think that there is an angle \(\alpha \) and \(area = A = \frac {\alpha }{2}\). However, due to the hyperbola configuration, the angle is not the geometric focus. Instead, the diagrams focus on area.
Geometrically, we see the area that defines the point. But, we are used to parameterizing with an angle. So, we’ll use \(\alpha \) for domain purposes. It the back or our heads, we know that \(A\) comes from the drawing and \(\alpha = 2 A\). That way our trigonometric and hyperbolic trigonometric relations will look alike.
From this definition we get
Compare this to the relationship for the circular trig functions: