# Hyperbolic Functions and Similarities with Trigonometric Functions

There are some great similarities between the hyperbolic functions and trigonometric functions. Let’s first define the hyperbolic function as they truly are. The hyperbolic functions are exponential functions and are unlike the trigonometric counterparts, which are periodic functions.

Below are graphs comparing the functions.

The hyperbolic cosine looks like a parabola but is unlike the parabola since the hyperbolic cosine grows exponentially. Both functions exhibit line symmetry about the y-axis.

The hyperbolic sine resembles the third degree polynomial. Both functions are odd functions and exhibit point symmetry about the origin. Unlike polynomial functions, there is no point on the hyperbolic sine where the slope is 0.

When the hyperbolic sine and cosine are graphed together, we can see that the two are asymptotic on the positive x-axis. Therefore, we expect the hyperbolic tangent to have a horizontal asymptote. The hyperbolic cosine is a little more than $$\frac{e^x}{2}$$ and the hyperbolic sine is a little less than $$\frac{e^x}{2}$$. We expect $$\frac{e^x}{2}$$ to fit right in between the two curves. Also, if we add the two, then the sum equal $$e^x$$. So, $$\cosh x + \sinh x = e^x$$.

The hyperbolic tangent function has a limit of 1 on the positive axis and –1 on the negative axis. The curve gets close to the asymptotes rather quickly. The hyperbolic cotangent is also asymptotic with the same lines. However, it has no values between –1 and 1.

Below, the functions are given in their exponential form.

The exponential equivalents of the hyperbolic functions:

$$\sinh x = \frac{e^x - e^{-x}}{2}$$, $$\cosh x = \frac{e^x + e^{-x}}{2}$$, $$\tanh x = \frac{\sinh x}{\cosh x}$$

Selected Hyperbolic Identities

$$\cosh^2 x - \sinh^2 x = 1$$, $$\cosh x + \sinh x = e^x$$

$$\sinh(x\pm y) =$$ $$\sinh x \cosh y \pm \cosh x \sinh y$$

$$\cosh(x\pm y) =$$ $$\cosh x \cosh y \pm \sinh x \sinh y$$

$$\sinh(2x) = 2\sinh x \cosh x$$, $$\cosh(2x) = \cosh^2 x + \sinh^2 x$$

Derivatives of Hyperbolic Functions

$$\frac{d}{dx}\sinh x = \cosh x$$

$$\frac{d}{dx}\cosh x = \sinh x$$

$$\frac{d}{dx}\tanh x = 1 - \tanh^{2}x = \text{sech}^{2}x$$

$$\frac{d}{dx}\coth x = 1 - \coth^{2}x = -\text{csch}^{2}x$$

### Inverse Hyperbolic Functions

Since the hyperbolic functions are exponential functions, we expect the inverse hyperbolic functions to be logarithmic. And that is exactly so when switching x and y and solving for y.

Inverse Hyperbolic Functions

$$\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)$$

$$\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right)$$ (x ≥ 1)

$$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$ (|x| < 1)

$$\coth^{-1} x = \tanh^{-1}\left(\frac{1}{x}\right)$$

Derivatives of Inverse Hyperbolic Functions

$$\frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{x^2+1}}$$

$$\frac{d}{dx}\cosh^{-1} x = \frac{1}{\sqrt{x^2-1}}$$

$$\frac{d}{dx}\tanh^{-1} x = \frac{1}{1-x^2}$$ (–1 < x < 1)

$$\frac{d}{dx}\coth^{-1} x = \frac{1}{1-x^2}$$ (x < –1 or x > 1)

The derivative of both the tanh–1x and coth–1x is the same. This is a great time to study their graphs.

The blue curve is the hyperbolic arctangent. Because the hyperbolic tangent has horizontal asymptotes, its inverse has vertical asymptotes. Therefore, the domain of tanh–1x is –1 < x < 1.

The red curve shows coth–1x, which is really tanh–1(1/x). Therefore, its domain is x < –1 or x > 1.

The derivative is the same for both functions, but it applies only within their respective domains.

### Miscellaneous Calculations

We can also evaluate the hyperbolic functions of their inverses. For example, $$\cosh (\sinh^{-1} x) = \sqrt{x^2+1}$$. This results from evaluating the exponential form of the cosh function. Similarly, $$\sinh (\cosh^{-1} x) = \sqrt{x^2-1}$$, $$\cosh (\tanh^{-1} x) = \frac{1}{\sqrt{1-x^2}}$$, $$\sinh (\tanh^{-1} x) = \frac{x}{\sqrt{1-x^2}}$$, $$\tanh (\sinh^{-1} x) = \frac{x}{\sqrt{x^2+1}}$$, and $$\tanh (\cosh^{-1} x) = \frac{\sqrt{x^2-1}}{x}$$

## Taylor Series of Hyperbolic Functions

Both cosine and hyperbolic cosine functions are even functions. Sine and hyperbolic sine are odd functions. This can be seen by the Taylor Series representation.

$$\sin x = \frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} ...$$

$$\sinh x = \frac{x^1}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \frac{x^9}{9!}...$$

$$\cos x = \frac{x^0}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^8}{8!} + \frac{x^{10}}{10!} ...$$

$$\cosh x = \frac{x^0}{0!} + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^8}{8!} + \frac{x^{10}}{10!}...$$

## Relationship with the Conics

### Similarity in the Parametric Equations

The trigonometric functions are related to the circle. Similarly, the hyperbolic functions are related to the hyperbola. We know the famous identity $$\cos^2 x + \sin^2 x = 1$$ as it relates to the circle. As stated above, $$\cosh^2 x - \sinh^2 x = 1$$, and this is related to the hyperbola.

Recall that the equation of the unit circle centered at the origin is $$x^2 + y^2 = 1$$. In the same way, a “unit” hyperbola centered at the origin has the equation $$x^2 - y^2 = 1$$. The parallel in the signs is noteworthy.

We can parameterize the equation of the circle in terms of sine and cosine. The parametric equation of the circle is $$(\cos t, \sin t)$$. In the same way, the parametric equation of the hyperbola is $$(\cosh t, \sinh t)$$. For the circle, t represents the angle, created by the origin and a point on the circle with respect to the x-axis, as we move counterclockwise around the circle. As t increases and passes 2π, we get the periodic nature of the trigonometric functions. For the hyperbolic functions, as t increases, the distance of the point on the hyperbola from the origin simply increases exponentially, and it does not represent an angle created by the origin and a point on the hyperbola.

The image above shows the two conics and the points on the circle and the hyperbola for a given t. For the circle, you can sum the squares of x and y at point C and you will get 1. For the hyperbola, you can subtract the squares of the x and y at point H and you will get 1.

Note that the parameter t, although it may have the same value, will not give points on the circle and the hyperbola that are collinear with the origin. Therefore, a t of 1, which is the radian measure for a circle or the angle AOC, gives point C on the circle; whereas a t of 1 (not a radian measure and is not the measure of angle AOH) for the hyperbola gives point H, where points C and H are not in line with the origin.

### Similar Area Formula

An interesting parallel is the area swept by the parameter t. For a circle, the area created by a sector is equal to $$\frac{1}{2}t$$. Remember that t represents the angle measure or the arc length of the circle since it is a unit circle. Therefore, there is a hidden π in this formula.

For a hyperbola, the area enclosed by the hyperbola, the segment from the origin to the point on the hyperbola, and the x-axis is $$\frac{1}{2}t$$.

The image below shows the sector of a circle given by the angle t. The area is easy to calculate. The ratio of the arc created by angle t to the circumference of the circle is equal to the ratio of the area of the sector to the area of the whole circle: $$\frac{t}{2\pi} = \frac{\text{area of sector}}{\text{area of whole circle}}$$ or $$\text{area of sector} = \frac{t\pi}{2\pi} = \frac{1}{2}t$$.

Now, we focus on the hyperbolic sector.

The region shaded in blue above represent the the area enclosed by the segment OH, the hyperbola, and the x-axis. The easiest method is to find the area, AH, under the curve by integrating the parametric equation, then subtracting that from the area, AT, of triangle OHC. This eliminates the messy calculus, which we will do anyway later to prove that this area grows at a logarithmic rate with respect to x.

The area under the parametric curve is $$\int_{a}^{b} y \text{ } dx$$. For our equation, $$y = \sinh t$$ and $$\frac{dx}{dt}(\cosh t) = \sinh t$$ or $$dx = \sinh t \text{ } dt$$. The area under the hyperbola from t = 0, to t = t1 is given by:

(i) $$A_H = \int_{0}^{t_1} \sinh^2 t \text{ } dt$$.

Using the “double-angle” (not really an angle) identity, we can restate (i) as:

(ii) $$A_H = \int_{0}^{t_1} \frac{\cosh(2t) - 1}{2} \text{ } dt =$$ $$\left[ \frac{\sinh(2t)}{4} - \frac{1}{2}t \right ]_{0}^{t_1}=$$ $$A_H = \frac{\sinh(2t_1)}{4}-\frac{1}{2}t_1$$.

We can simply replace t1 with t and say that the area under the hyperbola from t = 0 to t = t is $$\frac{\sinh(2t)}{4}-\frac{1}{2}t$$.

The area of the triangle is $$A_T = \frac{1}{2}(\sinh t)(\cosh t)$$. Using the “double-angle” identity, this is equal to $$A_T = \frac{1}{4}\sinh(2t)$$. So, the area of the shaded region is:

(iii) $$A = \frac{1}{4}\sinh(2t) - \left(\frac{\sinh(2t)}{4}-\frac{1}{2}t\right) =$$ $$\frac{1}{2}t$$.

### The Growth Rate of the Hyperbolic Sector

Before we continue, let’s make a note that the formulas for the circular sector and hyperbolic sector are the same, giving the same area for any given t. For the circle, when t reaches 2π, the same area will simply repeat and the region is retraced. So, the area increases because of retracing, not because the region is getting larger, albeit at a very slow rate. For the hyperbola, this never happens. As t increases toward infinity, the area region actually gets larger. Therefore, the areas are the same with regards to an increasing t. That also leads us to the next question: how fast is the area growing in the hyperbolic sector?

If we look at the rate of change based on the parameter t, it simply grows at a constant rate of 1/2 since the area is t/2. But let’s find the rate based on the x coordinate. We need to integrate using the rectangular equation, as we alluded to the messy calculations earlier. The area, AH, under the hyperbola from x = 0 to x1 is given by:

(iv) $$A_H = \int_{1}^{x_1}\sqrt{x^2-1}\text{ }dx =$$ $$\left[ \frac{1}{2}x\sqrt{x^2-1} - \frac{1}{2}\ln\left(\sqrt{x^2-1} +x \right) \right]_{1}^{x_1}$$

(v) $$A_H = \frac{1}{2}x_1\sqrt{x_1^2-1} -$$ $$\frac{1}{2}\ln\left(\sqrt{x_1^2-1} +x_1 \right)$$

The area of the triangle created by the point on the hyperbola at x1, which is at $$\left(x_1, \sqrt{x_1^2-1}\right)$$, is: $$A_T = \frac{1}{2}x_1\sqrt{x_1^2-1}$$. When we subtract the two, our area, A, of the sector between the line segment and the hyperbola, in terms of x, is:

(vi) $$A = \frac{1}{2}x_1\sqrt{x_1^2-1}$$ $$- \left[\frac{1}{2}x_1\sqrt{x_1^2-1} - \frac{1}{2}\ln\left(\sqrt{x_1^2-1} +x_1 \right)\right]$$ $$=\frac{1}{2}\ln\left(\sqrt{x_1^2-1} +x_1 \right)$$

So, our area is $$\frac{1}{2}\ln\left(\sqrt{x^2-1} +x \right)$$ in terms of x. (This is actually half the inverse hyperbolic cosine function: cosh–1x. In the proofs section, we have shown that they are related.) We can approximate this area by removing the 1: $$A \approx \frac{1}{2}\ln\left(\sqrt{x^2} + x \right) \approx \frac{1}{2}\ln (2x)$$. Therefore, the area grows at a logarithmic rate. This observation agrees with the area under the curve of the function $$f(x) = \frac{1}{x}$$, which is the unit hyperbola rotated by 45°. The area under this curve is given by $$\ln x$$.

## Applications

### The Rolling Parabola

There are some cool things that can happen with the hyperbolic functions. For example, the focus of a parabola rolling on the x-axis traces the catenary, or the hyperbolic cosine curve. Below is a Geogebra activity that you can play with to see this in action.

I have the complete proof that this parabola focus traces a catenary here: Proof the Focus of a Parabola Rolling Traces a Catenary.

### The Rolling Square

Another cool property of the catenary is that if a square rolls on the catenary, the center of the square follows a linear path. This can be seen in the following Geogebra activity.

You can change the shape of the catenary by increasing or decreasing the coefficient a. The length of the side of the square will be 2a. Notice that at some point, the square no longer touches the catenary; however, its center remains on the x-axis. This point is when the bottom right corner of the square coincides with point P. This occurs when the square has rolled half its side, so the arc lenth of the catenary is equal to a. The arc length of the catenary is given by: $$a\sinh\frac{x}{a}$$. We can find the x value of this point to be:

(i) $$a\sinh\frac{x}{a} = a$$

(ii) $$\sinh\frac{x}{a} = 1$$

(iii) $$\sinh^{-1}(1) = \frac{x}{a}$$

(iv) $$x = a\sinh^{-1}(1)$$

The y value can be calculated by substituting the x value we found:

(i) $$y = a\cosh\left(\frac{a\sinh^{-1}(1)}{a}\right) =$$ $$a\cosh(\sinh^{-1}(1)) =$$ $$a\sqrt{2}$$   (See Miscellaneous Calculations above.)

Therefore, the square no longer touches the caternary after this point: $$\left(a\sinh^{-1}(1), a\sqrt{2} \right)$$. This is approximately (0.88×a, 1.41×a).

### Proof of Construction

The center of the square always falls on the x-axis. This can be proved easily. We will use the image below to prove this.

First, we note that point H is the midpoint of the side of the square that will roll on the catenary. The point H was initially on the vertex of the catenary before it started to roll. Therefore, as the square rolls, the length HP is equal to the arc length of the catenary from the vertex of the catenary to point P. The arc length is given by the integral $$\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\text{ }dx$$. For our catenary, the equation is $$y = -a\cosh\frac{x}{a}$$. The derivative is $$\frac{dy}{dx} = -\sinh\frac{x}{a}$$. We will find the arc length, L, from x = 0 to x = t, where t is the x value that will give us the x position of point P.

(i) $$L = \int_{0}^{t}\sqrt{1+\sinh^{2}\frac{x}{a}}\text{ }dx =$$ $$\int_{0}^{t}\sqrt{\cosh^{2}\frac{x}{a}}\text{ }dx =$$ $$\int_{0}^{t}\cosh\frac{x}{a}\text{ }dx$$

(ii) $$L = \int_{0}^{t}\cosh\frac{x}{a}\text{ }dx =$$ $$a\sinh\frac{t}{a}$$

That’s a nice surprise. Hence, the arc length from the vertex to point P (after changing our dummy t to an x) is asinh(x/a). This is equal to HP since the square is rolling on the catenary. We already know that OP is equal to –acosh(x/a). We need to find OH. If OH is equal to a, then we know that it has to lie on the x-axis. Since OHP is a right triangle, we can use the Pythagorean theorem to find OH.

(i) $$\text{OH}^2 = \left(a\cosh\frac{x}{a}\right)^2-\left(a\sinh\frac{x}{a}\right)^2 =$$ $$a^2\left[\left(\cosh\frac{x}{a}\right)^2-\left(\sinh\frac{x}{a}\right)^2\right]$$

(ii) $$\text{OH}^2 = a^2$$

Therefore, point OH and OP both have a common point that falls on the x-axis. This allows us to find the midpoint of the side of the square rolling on the catenary. If one wanted to find point H, here are the steps:

1. Pick point P first on the catenary.
2. Draw a tangent at point P.
3. Draw a perpendicular to the x-axis at P.
4. Draw a perpendicular to the tangent which crosses point O.
5. Draw a circle at H with a radius of HO. The diameter of this circle that falls on the tangent will be the length of the side of the square. Constructing the whole square after that is elementary.

## Proofs

### Proof of the Squares Identity

The easiest proof is the difference of squares for hyperbolic functions.

(i) $$\cosh^{2}x - \sinh^{2}x = 1$$

(ii) $$\left(\frac{e^x+e^{-x}}{2}\right)^2 -$$ $$\left(\frac{e^x-e^{-x}}{2}\right)^2 = 1$$

(iii) $$\frac{e^{2x}+2e^{x}e^{-x}+e^{-2x}}{4}$$ $$-\frac{e^{2x}-2e^{x}e^{-x}+e^{-2x}}{4} = 1$$

(iv) $$\frac{4e^{x}e^{-x}}{4} = 1$$    ($$e^xe^{-x} = e^0 = 1$$)

Our last statement is correct.

### Deriving the Inverse Functions

It’s a classic exercise to derive the inverse functions. We will show the inverse of hyperbolic cosine by switching the x and y and solving for y.

(i) $$x = \frac{e^y + e^{-y}}{2}$$

(ii) $$2x = e^y + e^{-y}$$

(iii) $$2x\cdot e^y = e^{2y} + 1$$    (Multiply both sides by $$e^y$$)

(iv) $$(e^{y})^2 - 2x\cdot e^y + 1 = 0$$ (This is a nice quadratic equation.)

(v) $$e^y = \frac{2x \pm\sqrt{4x^2-4}}{2}$$

(vi) $$e^y = x \pm\sqrt{x^2-1}$$

(vii) $$y = \ln\left(x \pm\sqrt{x^2-1}\right)$$

Since the range of coshx is x ≥ 1, the domain of (vi) must by x ≥ 1. The equation in (vii) gives both values of y for which coshx is equal to y. The equation in (vii) can also be written as $$y = \pm\ln\left(x + \sqrt{x^2-1}\right)$$ because both are equivalent. I will let the reader prove this. (Hint: take the inverse argument of the logarithm and change the sign in the front to a negative. Then, rationalize the denominator.)

The method of deriving the other inverse functions is similar.

### Proof of Hyperbolic Function from Hyperbolic Sector

We found the area of the hyperbolic sector to be $$\frac{1}{2}\ln\left(\sqrt{x^2-1} +x \right)$$ in terms of x. The area is also equal to t/2 in terms of t. This is the same area, and so both are equivalent. We can solve for x and find that it gives us the hyperbolic function.

(i) $$\frac{1}{2}t = \frac{1}{2}\ln\left(\sqrt{x^2-1} +x \right)$$

(ii) $$t = \ln\left(\sqrt{x^2-1} +x \right)$$

(iii) $$e^t = \sqrt{x^2-1} +x$$ (Take the exponent of both sides.)

(iii) $$e^t - x = \sqrt{x^2-1}$$

(iv) $$(e^t)^2 - 2e^{t}x + x^2 = x^2-1$$

(v) $$e^{2t} + 1 = 2e^{t}x$$

(v) $$x = \frac{e^{-t}(e^{2t} + 1)}{e^{-t}2e^t}$$

(v) $$x = \frac{e^{t} + e^{-t}}{2}$$

This proves the equivalence of the hyperbolic function cosht is equal to x.