Hyperbolic Functions

Learning Outcomes

Identify the hyperbolic functions, their graphs, and basic identities

The hyperbolic functions are defined in terms of certain combinations of $e^x$ and $e^{−x}$ . These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary. If we introduce a coordinate system so that the low point of the chain lies along the $y$ -axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

A photograph of a spider web collecting dew drops.

Figure 6. The shape of a strand of silk in a spider’s web can be described in terms of a hyperbolic function. The same shape applies to a chain or cable hanging from two supports with only its own weight. (credit: “Mtpaley”, Wikimedia Commons)

Definition

Hyperbolic cosine

cosh x = \frac{e^{x} + e^{- x}}{2}

$\cosh x=\large \frac{e^x+e^{−x}}{2}$

Hyperbolic sine

sinh x = \frac{e^{x} - e^{- x}}{2}

$\sinh x=\large \frac{e^x-e^{−x}}{2}$

Hyperbolic tangent

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

$\tanh x=\large \frac{\sinh x}{\cosh x} \normalsize = \large \frac{e^x-e^{−x}}{e^x+e^{−x}}$

Hyperbolic cosecant

csch x = \frac{1}{sinh x} = \frac{2}{e^{x} - e^{- x}}

$\text{csch} \, x=\large \frac{1}{\sinh x} \normalsize = \large \frac{2}{e^x-e^{−x}}$

Hyperbolic secant

sech x = \frac{1}{cosh x} = \frac{2}{e^{x} + e^{- x}}

$\text{sech} \, x=\large \frac{1}{\cosh x} \normalsize = \large \frac{2}{e^x+e^{−x}}$

Hyperbolic cotangent

coth x = \frac{cosh x}{sinh x} = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}

$\coth x=\large \frac{\cosh x}{\sinh x} \normalsize = \large \frac{e^x+e^{−x}}{e^x-e^{−x}}$

The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.

Using the definition of $\cosh(x)$ and principles of physics, it can be shown that the height of a hanging chain, such as the one in Figure 6, can be described by the function $h(x)=a \cosh(x/a)+c$ for certain constants $a$ and $c$ .

But why are these functions called hyperbolic functions? To answer this question, consider the quantity $\cosh^2 t-\sinh^2 t$ . Using the definition of $\cosh$ and $\sinh$ , we see that

{cosh}^{2} t - {sinh}^{2} t = \frac{e^{2 t} + 2 + e^{- 2 t}}{4} - \frac{e^{2 t} - 2 + e^{- 2 t}}{4} = 1

$\cosh^2 t-\sinh^2 t=\large \frac{e^{2t}+2+e^{-2t}}{4}-\frac{e^{2t}-2+e^{-2t}}{4} \normalsize =1$

This identity is the analog of the trigonometric identity $\cos^2 t+\sin^2 t=1$ . Here, given a value $t$ , the point $(x,y)=(\cosh t,\sinh t)$ lies on the unit hyperbola $x^2-y^2=1$ (Figure 7).

An image of a graph. The x axis runs from -1 to 3 and the y axis runs from -3 to 3. The graph is of the relation “(x squared) - (y squared) -1”. The left most point of the relation is at the x intercept, which is at the point (1, 0). From this point the relation both increases and decreases in curves as x increases. This relation is known as a hyperbola and it resembles a sideways “U” shape. There is a point plotted on the graph of the relation labeled “(cosh(1), sinh(1))”, which is at the approximate point (1.5, 1.2).

Figure 7. The unit hyperbola $\cosh^2 t-\sinh^2 t=1$ .

Graphs of Hyperbolic Functions

To graph $\cosh x$ and $\sinh x$ , we make use of the fact that both functions approach $\left(\frac{1}{2}\right)e^x$ as $x \to \infty$ , since $e^{−x} \to 0$ as $x \to \infty$ . As $x \to −\infty, \, \cosh x$ approaches $\frac{1}{2}e^{−x}$ , whereas $\sinh x$ approaches $-\frac{1}{2}e^{−x}$ . Therefore, using the graphs of $\frac{1}{2}e^x, \, \frac{1}{2}e^{−x}$ , and $−\frac{1}{2}e^{−x}$ as guides, we graph $\cosh x$ and $\sinh x$ . To graph $\tanh x$ , we use the fact that $\tanh(0)=0, \, -1<\tanh(x)<1$ for all $x, \, \tanh x \to 1$ as $x \to \infty$ , and $\tanh x \to −1$ as $x \to −\infty$ . The graphs of the other three hyperbolic functions can be sketched using the graphs of $\cosh x, \, \sinh x$ , and $\tanh x$ (Figure 8).

An image of six graphs. Each graph has an x axis that runs from -3 to 3 and a y axis that runs from -4 to 4. The first graph is of the function “y = cosh(x)”, which is a hyperbola. The function decreases until it hits the point (0, 1), where it begins to increase. There are also two functions that serve as a boundary for this function. The first of these functions is “y = (1/2)(e to power of -x)”, a decreasing curved function and the second of these functions is “y = (1/2)(e to power of x)”, an increasing curved function. The function “y = cosh(x)” is always above these two functions without ever touching them. The second graph is of the function “y = sinh(x)”, which is an increasing curved function. There are also two functions that serve as a boundary for this function. The first of these functions is “y = (1/2)(e to power of x)”, an increasing curved function and the second of these functions is “y = -(1/2)(e to power of -x)”, an increasing curved function that approaches the x axis without touching it. The function “y = sinh(x)” is always between these two functions without ever touching them. The third graph is of the function “y = sech(x)”, which increases until the point (0, 1), where it begins to decrease. The graph of the function has a hump. The fourth graph is of the function “y = csch(x)”. On the left side of the y axis, the function starts slightly below the x axis and decreases until it approaches the y axis, which it never touches. On the right side of the y axis, the function starts slightly to the right of the y axis and decreases until it approaches the x axis, which it never touches. The fifth graph is of the function “y = tanh(x)”, an increasing curved function. There are also two functions that serve as a boundary for this function. The first of these functions is “y = 1”, a horizontal line function and the second of these functions is “y = -1”, another horizontal line function. The function “y = tanh(x)” is always between these two functions without ever touching them. The sixth graph is of the function “y = coth(x)”. On the left side of the y axis, the function starts slightly below the boundary line “y = 1” and decreases until it approaches the y axis, which it never touches. On the right side of the y axis, the function starts slightly to the right of the y axis and decreases until it approaches the boundary line “y = -1”, which it never touches.

Figure 8. The hyperbolic functions involve combinations of $e^x$ and $e^{−x}$ .

Identities Involving Hyperbolic Functions

The identity $\cosh^2 t-\sinh^2 t$ , shown in Figure 7, is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.

Identities Involving Hyperbolic Functions

$\cosh(−x)=\cosh x$
$\sinh(−x)=−\sinh x$
$\cosh x+\sinh x=e^x$
$\cosh x-\sinh x=e^{−x}$
$\cosh^2 x-\sinh^2 x=1$
$1-\tanh^2 x=\text{sech}^2 x$
$\coth^2 x-1=\text{csch}^2 x$
$\sinh(x \pm y)=\sinh x \cosh y \pm \cosh x \sinh y$
$\cosh (x \pm y)=\cosh x \cosh y \pm \sinh x \sinh y$

Example: Evaluating Hyperbolic Functions

Simplify $\sinh(5 \ln x)$ .
If $\sinh x=\frac{3}{4}$ , find the values of the remaining five hyperbolic functions.

Show Solution

Using the definition of the $\sinh$ function, we write
$\sinh(5 \ln x)=\large \frac{e^{5 \ln x}-e^{-5 \ln x}}{2} \normalsize = \large \frac{e^{\ln(x^5)}-e^{\ln(x^{-5})}}{2} \normalsize =\large \frac{x^5-x^{-5}}{2}$ .
Using the identity $\cosh^2 x-\sinh^2 x=1$ , we see that
$\cosh^2 x=1+\big(\frac{3}{4}\big)^2=\frac{25}{16}$ .

Since $\cosh x \ge 1$ for all $x$ , we must have $\cosh x=5/4$ . Then, using the definitions for the other hyperbolic functions, we conclude that $\tanh x=3/5, \, \text{csch} \, x=4/3, \, \text{sech} \, x=4/5$ , and $\coth x=5/3$ .

Watch the following video to see the worked solution to Example: Evaluating Hyperbolic Functions

Closed Captioning and Transcript Information for Video

Try It

Simplify $\cosh(2 \ln x)$ .

Hint

Show Solution

$\frac{(x^2+x^{-2})}{2}$

Inverse Hyperbolic Functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except $\cosh x$ and $\text{sech} \, x$ . If we restrict the domains of these two functions to the interval $[0,\infty)$ , then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

Definition

Inverse Hyperbolic Functions:

\begin{matrix} {sinh}^{- 1} x = arcsinh x = ln (x + \sqrt{x^{2} + 1}) & {cosh}^{- 1} x = arccosh x = ln (x + \sqrt{x^{2} - 1}) {tanh}^{- 1} x = arctanh x = \frac{1}{2} ln (\frac{1 + x}{1 - x}) & {coth}^{- 1} x = arccot x = \frac{1}{2} ln (\frac{x + 1}{x - 1}) {sech}^{- 1} x = arcsech x = ln (\frac{1 + \sqrt{1 - x^{2}}}{x}) & {csch}^{- 1} x = arccsch x = ln (\frac{1}{x} + \frac{\sqrt{1 + x^{2}}}{| x |}) \end{matrix}

$\begin{array}{cccc}\sinh^{-1} x=\text{arcsinh } x=\ln(x+\sqrt{x^2+1})\hfill & & & \cosh^{-1} x=\text{arccosh } x=\ln(x+\sqrt{x^2-1})\hfill \\ \tanh^{-1} x=\text{arctanh } x=\frac{1}{2}\ln\big(\frac{1+x}{1-x}\big)\hfill & & & \coth^{-1} x=\text{arccot } x=\frac{1}{2}\ln\big(\frac{x+1}{x-1}\big)\hfill \\ \text{sech}^{-1} x=\text{arcsech } x=\ln\big(\frac{1+\sqrt{1-x^2}}{x}\big)\hfill & & & \text{csch}^{-1} x=\text{arccsch } x=\ln\big(\frac{1}{x}+\frac{\sqrt{1+x^2}}{|x|}\big)\hfill \end{array}$

Let’s look at how to derive the first equation. The others follow similarly. Suppose $y=\sinh^{-1} x$ . Then, $x=\sinh y$ and, by the definition of the hyperbolic sine function, $x=\frac{e^y-e^{−y}}{2}$ . Therefore,

e^{y} - 2 x - e^{- y} = 0

$e^y-2x-e^{−y}=0$

Multiplying this equation by $e^y$ , we obtain

e^{2 y} - 2 x e^{y} - 1 = 0

$e^{2y}-2xe^y-1=0$

This can be solved like a quadratic equation, with the solution

e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} = x \pm \sqrt{x^{2} + 1}

$e^y=\large \frac{2x \pm \sqrt{4x^2+4}}{2} \normalsize =x \pm \sqrt{x^2+1}$

Since $e^y>0$ , the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

y = ln (x + \sqrt{x^{2} + 1})

$y=\ln(x+\sqrt{x^2+1})$

Example: Evaluating Inverse Hyperbolic Functions

Evaluate each of the following expressions

$\sinh^{-1}(2)$
$\tanh^{-1}\left(\frac{1}{4}\right)$

Show Solution

$\sinh^{-1}(2)=\ln(2+\sqrt{2^2+1})=\ln(2+\sqrt{5}) \approx 1.4436$

$\tanh^{-1}(\frac{1}{4})=\frac{1}{2}\ln(\frac{1+1/4}{1-1/4})=\frac{1}{2}\ln(\frac{5/4}{3/4})=\frac{1}{2}\ln(\frac{5}{3}) \approx 0.2554$

Try It

Evaluate $\tanh^{-1}\left(\frac{1}{2}\right)$ .

Hint

Use the definition of $\tanh^{-1} x$ and simplify.

Show Solution

$\frac{1}{2}\ln(3) \approx 0.5493$ .

Module 1: Functions and Graphs