Building a different trigonometry on an inside-out circle
The world of trigonometry is based upon the unit circle, a circle of radius $r = 1$ centered at $(0, 0)$. Coordinate pairs anywhere on the unit circle are given by $(\text{cos}(\theta), \text{sin}(\theta))$. For a circle of radius $r \ne 1$, we simply multiply the radius by each unit-circle coordinates: $(x,y)=(r \, \text{cos}(\theta), r \, \text{sin}(\theta))$.
Unit-circle trig. construction
Note that unit-circle trigonometry is built upon the figure with equation
$$x^2 + y^2 = 1.$$
Now one way to look at the hyperbola,
$$x^2 - y^2 = 1$$
(note the minus sign instead of the plus) is that the hyperbola is an inside-out circle. We base hyperbolic trigonometry on that hyperbola in a way very similar to the development of unit-circle trig. Here's the construction:
Hyperbolic trig. construction
The hyperbolic trig. functions (usually just "hyperbolic functions") are defined in a similar way, as a function of the angle between the pointing vector (corresponds to the radius of the unit circle) and the $x$-axis. Any coordinate on the hyperbola has coordinates $(\text{cosh}(\theta), \text{sinh}(\theta)).$ The angle is related to the area between the pointing vector, the hyperbola and the x-axis: $a = \theta/2$ (see box below).
Any study of hyperbolic functions must rely heavily on Euler's equation, $e^{ix} = \text{cos}(x) + i\, \text{sin}(x).$ Knowledge of how to use Euler's equation is essential for understanding this section and many other concepts in math and physics.
Relationship between angle and area
The angle between our vector from the origin to a location on the parabola and the area bouned by the x-axis, that vector and the hyperbola (area a) are related. To discover that relationship, consider the close-up of the figure above.
To find the green area, we just need to find the area of the triangle $(\frac{1}{2} bh = \frac{1}{2} \text{cosh}(\theta) \cdot \text{sinh}(\theta)),$ and subtract the area of the crescent-shaped region.
That small area is just the integral from 1 to \text{cosh}(θ) of the right part of the hyperbola. When we solve for the equation of that part, it's
$$y = \sqrt{x^2 - 1}.$$
So our area equation is:
$$a = \frac{1}{2} \text{sinh}(\theta) \cdot \text{cosh}(\theta) - \int_1^{\text{cosh}(\theta)} \sqrt{x^2 - 1} \, dx.$$
The way to do that integral is by a substitution:
$$ \begin{matrix} \text{Let } \; \theta = \text{cosh}^{-1}(x) && \\[5pt] \text{then } \; x = \text{cosh}(\theta) && \text{and } \; dx = \text{sinh}(\theta) \, d\theta \end{matrix}$$
So our area is
$$a = \frac{1}{2} \text{sinh}(\theta) \cdot \text{cosh}(\theta) - \int_1^x \sqrt{\text{cosh}^2(\theta) - 1} \, d\theta$$
Now using $\text{cosh}^2(x) - \text{sinh}^2(x) = 1,$ we have
$$a = \frac{1}{2} \text{sinh}(\theta) \cdot \text{cosh}(\theta) - \int_1^x \text{sinh}^2(\theta) \, d\theta$$
One of the convenient things about hyperbolic trig functions is that we can go back and forth between the standard and exponential notation. An exponential change will make this integral easy:
$$ \begin{align} a &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \int_1^x \left( \frac{e^x - e^{-x}}{2}\right)^2 \, d\theta \\[5pt] &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \int_1^x \frac{e^{2\theta} - 2 + e^{-2\theta}}{4} \, d\theta \\[5pt] &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \frac{1}{4}\left[ \frac{e^{2\theta}}{2} - 2\theta - \frac{e^{-2\theta}}{2} \right]_0^{\theta} \\[5pt] &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \frac{1}{4}\left[ \frac{e^{2\theta} - e^{-2\theta}}{2} - 2\theta \right]_0^{\theta} \\[5pt] &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \frac{1}{4}\left[ \text{sinh}(2\theta) - 2\theta \right]_0^{\theta} \\[5pt] &= \frac{1}{2} \text{sinh}(\theta) \text{cosh}(\theta) - \frac{1}{4}\left[ 2 \text{sinh}(\theta)\text{cosh}(\theta) - 2\theta \right]_0^{\theta} \\[5pt] &= \frac{1}{2}\text{sinh}(\theta)\text{cosh}(\theta) - \frac{1}{2}\text{sinh}(\theta)\text{cosh}(\theta) + \frac{\theta}{2} \\[5pt] &= \frac{\theta}{2} \end{align}$$
So the area is half of the angle.
The hyperbolic functions
Recall that, through Euler's equation, $e^{±ix} = cos(x) ± i\, sin(x),$ we can express the sine and cosine functions using exponentials:
$$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
$$sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
Likewise, we can define the hyperbolic functions using exponentials. They are listed below.
The six hyperbolic functions
| $$\text{sinh}(x) = \frac{e^x - e^{-x}}{2}$$ |
| $$\text{cosh}(x) = \frac{e^x + e^{-x}}{2}$$ |
| $$\text{tanh}(x) = \frac{\text{sinh}(x)}{\text{cosh}(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$ |
| $$\text{coth}(x) = \frac{\text{cosh}(x)}{\text{sinh}(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$$ |
| $$\text{sech}(x) = \frac{1}{\text{cosh}(x)} = \frac{1}{e^x + e^{-x}}$$ |
| $$\text{csch}(x) = \frac{1}{\text{sech}(x)} = \frac{1}{e^x - e^{-x}}$$ |
Pro tip:
The hyperbolic trig. functions are prounounced by putting a ch (as in "ranch") behind them:
| $\text{sinh}(x)$ | "sinch (as in pinch) of x" |
| $\text{cosh}(x)$ | "kosh (as in posh) of x" |
| $\text{tanh}(x)$ | "tanch (as in ranch) of x" |
| $\text{coth}(x)$ | "koth (as in moth) of x" |
| $\text{sech}(x)$ | "seech (as in peach) of x" |
| $\text{csch}(x)$ | "coseech of x" |
Hyperbolic function plots
The graphs of $f(x) = \text{cosh}(x)$ and $f(x) = \text{sinh}(x)$ are plotted below. The sinh function is purely odd, $\text{sinh}(-x) = -\text{sinh}(x)$, while the cosh function is even; that is, $\text{cosh}(-x) = \text{cosh}(x)$.
As $x$ gets large ($x \rightarrow \infty$), the difference between $\text{sinh}(x)$ and $\text{cosh}(x)$ goes to zero, so they overlap (each is the asymptote of the other). We can show that using the exponential forms.
Click for details of the limit
$$ \begin{align} \lim_{x\to\infty} &\; \text{sinh}(x) - \text{cosh}(x) \\[5pt] &= \lim_{x\to\infty} \frac{e^x - e^{-x}}{2} - \frac{e^x + e^{-x}}{2} \\[5pt] &= \frac{1}{2} \lim_{x\to\infty} \, \left( e^x - \frac{1}{e^x} - e^x - \frac{1}{e^x} \right) \\[5pt] &= -\lim_{x\to\infty} \frac{1}{e^x} = 0 \end{align}$$
As $x \rightarrow \infty$, the difference increases.
Click for details of the limit
$$ \begin{align} &= \frac{1}{2} \lim_{x -\to\infty} \, \left( e^x - \frac{1}{e^x} - e^x - \frac{1}{e^x} \right) \\[5pt] &= -\lim_{x\to -\infty} \frac{1}{e^x} \\[5pt] &= -\lim_{x\to \infty} e^x \rightarrow \infty \end{align}$$
The next plot shows the graphs of $\text{tanh}(x)$ and $\text{coth}(x)$. Because the graph of \text{tanh}(x) has a zero, and $\text{coth}(x) = \frac{1}{\text{tanh}(x)},$ the $\text{coth}(x)$ function has a vertical asyptote at $x = 0$. The functions converge to ±1 as $x \rightarrow \infty$.
The graphs of $\text{sech}(x)$ and $\text{csch}(x)$ are plotted below. One is even, the other odd, and both converge to zero as $x \rightarrow \infty$.
Relationships between hyperbolic functions
Because of the construction of hyperbolic functions, many of the relationships between them is the same as or similar to their unit-circle trig. functions. Let's work through some of those.
Even/odd
From the graphs of the hyperbolic functions above, we can see that \text{cosh}(-x) = \text{cosh}(x), so \text{cosh}(x) is an even function. Likewise, \text{sinh}(-x) = - \text{sinh}(x), so \text{sinh}(x) is odd. It's then easy to find the even/odd-ness of the other functions. They are:
| Even | Odd |
|---|---|
| $\text{cosh}(x)$ | $\text{sinh}(x)$ |
| $\text{sech}(x)$ | $\text{csch}(x)$ |
| $\text{tanh}(x)$ | |
| $\text{coth}(x)$ |
Pythagorean
If we work through the exponential representations, we can find the analog of the Pythagorean identity, $\text{sin}^2(x) + \text{cos}^2(x) = 1$ except (with prior knowledge that this how it will work out), we'll subtract the hyperbolic functions:
$$ \begin{align} &\text{sinh}^2(x) - \text{cosh}^2(x) \\[5pt] &= \left( \frac{e^x + e^{-x}}{2} \right)^2 - \left( \frac{e^x - e^{-x}}{2} \right)^2 \\[5pt] &= \frac{1}{4}\left( e^{2x} + 1 + 1 + e^{-2x} - e^{2x} + 1 + 1 - e^{-2x} \right) \\[5pt] &= \frac{1}{4}\cdot 4 = 1 \end{align}$$
In a similar way we can develop all three Pythagorean identities:
Pythagorean identities
$$\text{cosh}^2(x) - \text{sinh}^2(x) = 1$$
$$\text{sech}^2(x) + \text{tanh}^2(x) = 1$$
$$\text{coth}^2(x) - \text{csch}^2(x) = 1$$
We can also develop analogs of the power-reduction formulae, which are very helpful when integrating trig. functions:
Power-reduction formulas
$$\text{sinh}^2(x) = \frac{\text{cosh}(2x) - 1}{2}$$
$$\text{cosh}^2(x) = \frac{\text{cosh}(2x) + 1}{2}$$
$$\text{tanh}^2(x) = \frac{2 \, \text{tanh}(x)}{\text{tanh}(2x)} - 1$$
The sum and difference formulas are easily derived using the eponential representation. Here is the derivation for $\text{cosh}(a + b)$; the sum and difference formulae for sinh, $\text{cosh}$ and tanh are summarized in the box.
Click to view derivation
First express \text{cosh}(a + b) in exponential form. Use the properties of the exponential to write the terms with sum exponents as products of exponentials. Then translate back to sinh and cosh and continue:
$$ \begin{align} &\text{cosh}(a + b) = \frac{e^{a + b} + e^{-(a + b)}}{2} \\[5pt] &= \frac{1}{2}[e^a e^b + e^{-a} e^{-b}] \\[5pt] &= \frac{1}{2}[(\text{cosh}(a) + i \text{sinh}(a))(\text{cosh}(b) + i \text{sinh}(b)) + \\[5pt] &\phantom{0000} = (\text{cosh}(a) - i \text{sinh}(a))(\text{cosh}(b) - i \text{sinh}(b))] \\[5pt] &= \frac{1}{2} [\text{cosh}(a)\text{cosh}(b) + i \text{cosh}(a)\text{sinh}(b) \\[5pt] &\phantom{0000} + i \text{sinh}(a)\text{cosh}(b) - \text{sinh}(a)\text{sinh}(b) \\[5pt] &\phantom{0000}+ \text{cosh}(a)\text{cosh}(b) - i \text{cosh}(a)\text{sinh}(b) \\[5pt] &\phantom{0000}- i \text{sinh}(a)\text{cosh}(b) - \text{sinh}(a)\text{sinh}(b)] \\[5pt] &= \frac{1}{2}[2 \text{cosh}(a)\text{cosh}(b) - 2 \text{sinh}(a)\text{sinh}(b)] \\[5pt] &= \text{cosh}(a)\text{cosh}(b) - \text{sinh}(a)\text{sinh}(b) \end{align}$$
Sum & difference formulas
$$ \begin{align} \text{sinh}(a ± b) &= \text{sinh}(a) \, \text{cosh}(b) \\[5pt] &\phantom{0000} ± \text{cosh}(a) \, \text{sinh}(b) \\[5pt] \text{cosh}(a ± b) &= \text{cosh}(a) \, \text{cosh}(b) \\[5pt] &\phantom{0000} ± \text{sinh}(a) \, \text{sinh}(b) \\[5pt] \text{tanh}(a ± b) &= \frac{\text{tanh}(a) ± \text{tanh}(b)}{1 ± \text{tanh}(a) \, \text{tanh}(b)} \end{align}$$
You can see that the relationships between the hyperbolic functions are very similar (but not identical) to those between the unit-circle trig. functions. You can download a complete table of identities using the button below.
In other sections we'll develop derivatives, integrals and Taylor-series representations of the hyperbolic functions.
The catenary
The
Consider the force of gravity on a strand of something (wire, rope, chain) that has a uniform mass distribution along it length. To think about how the force of gravity works on that strand, consider a small part of it, designated $\Delta s$ in the figure above. The force of gravity on this section is $F_g$
$$F_g = \rho g \Delta s,$$
where $\rho$ is the density of the chain (mass per unit length), $g$ is the acceleration of gravity on Earth (9.81 m/s2), and $\Delta s$ is the length of the strand between $x$ and $x+\Delta x$.
The $T$ values in the figure are tensions in the strand, and they are not necessarily the same. If we have equilibrium (the strand isn't accelerating), then the up-down and right-left forces must be in balance. The left-right balance is
$$-T(x) \text{cos}(\phi_x) + T(x + \Delta x) \text{cos}(\phi_{x + \Delta x}) = 0$$
Now the only way for
$$T(x) \text{cos}(\phi_x) = T(x + \Delta x) \text{cos}(\phi_{x + \Delta x})$$
to be true is if the horizontal force is constant throughout the system, or $T(x) = T(x + \Delta x),$ for all $x.$ From here on we'll call that constant value $T_o.$
The up down balance, which includes $F_g$, is
$$-T(x) \text{sin}(\phi_x) + T(x + \Delta x) \text{sin}(\phi_{x + \Delta x}) - F_g = 0$$
We can rewrite that equation using $T_o,$ as
$$-T_o \text{sin}(\phi_x) + T_o \text{sin}(\phi_{x + \Delta x}) - F_g = 0$$
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