Hyperbolic Functions: Definition & Examples

Jump to a key chapter

This article discusses the basic hyperbolic functions and their properties, identities, derivatives, integrals, inverses, and examples in detail.

Hyperbolic function definition
Hyperbolic functions: formulas
Hyperbolic functions: graphs
Hyperbolic functions: properties and identities
Derivatives of hyperbolic functions
Integrals of hyperbolic functions
Inverse hyperbolic functions
Hyperbolic functions: examples and applications

Hyperbolic Function Definition

What are the hyperbolic functions?

The hyperbolic functions are essentially the trigonometric functions of the hyperbola. They extend the notion of the parametric equations for the unit circle, where $(x = \cos (θ), y = \sin (θ))$ , to the parametric equations for the unit hyperbola, and are defined in terms of the natural exponential function (where $e$ is Euler's number), giving us the following two fundamental hyperbolic formulas:

$(x = \cos h (c) = \frac{e^{c} + e^{- c}}{2}, y = \sin h (c) = \frac{e^{c} - e^{- c}}{2})$

Based on these two definitions: hyperbolic cosine and hyperbolic sine, the rest of the six main hyperbolic functions can be derived as shown in the table below.

Hyperbolic Functions: Formulas

The formulas for the hyperbolic functions are listed below:

The 6 main hyperbolic formulas
Hyperbolic sine:	$\sin h (x) = \frac{e^{x} - e^{- x}}{2}$	Hyperbolic cosecant:	$c s c h (x) = \frac{1}{\sin h (x)} = \frac{2}{e^{x} - e^{- x}}$
Hyperbolic cosine:	$\cos h (x) = \frac{e^{x} + e^{- x}}{2}$	Hyperbolic secant:	$s e c h (x) = \frac{1}{\cos h (x)} = \frac{2}{e^{x} + e^{- x}}$
Hyperbolic tangent:	$\tan h (x) = \frac{\sin h (x)}{\cos h (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$	Hyperbolic cotangent:	$c o t h (x) = \frac{\cos h (x)}{\sin h (x)} = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$

Where $\sin h$ is pronounced "cinch", $\cos h$ is pronounced "cosh", $\tan h$ is pronounced "tanch", $\csc h$ is pronounced "coseech", ⁣ $\sec h$ is pronounced "seech", and $\cot h$ is pronounced "cotanch".

Deriving the Exponential Forms

A key characteristic of the hyperbolic trigonometric functions is their similarity to the trigonometric functions, which can be seen from Euler's formula:

$e^{\pm i θ} = \cos (θ) \pm i \sin (θ)$

Solving this for both cosine and sine gives us:

$\cos (θ) = \frac{e^{i θ} + e^{- i θ}}{2}, \sin (θ) = \frac{e^{i θ} - e^{- i θ}}{2 i}$

Which is strikingly similar to the hyperbolic cosine and sine functions:

$\cos h (c) = \frac{e^{c} + e^{- c}}{2}, \sin h (c) = \frac{e^{c} - e^{- c}}{2}$

But notice that the hyperbolic functions do not have the imaginary part that Euler's formula does.

Why are the imaginary numbers missing from the hyperbolic functions?

Because when we solve Euler's formula for the hyperbolic functions, the imaginary component does not exist within the solution to the hyperbolic functions.

Hyperbolic Functions: Graphs

The graphs of the two fundamental hyperbolic functions: hyperbolic sine and hyperbolic cosine, can be sketched using graphical addition as shown below.

The graph of $y = \sin h (x) = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} = \frac{e^{x} - e^{- x}}{2}$	The graph of $y = \cos h (x) = \frac{1}{2} e^{x} + \frac{1}{2} e^{- x} = \frac{e^{x} + e^{- x}}{2}$
The graph of hyperbolic sine using graphical addition - StudySmarter Originals	The graph of hyperbolic cosine using graphical addition - StudySmarter Originals

The graphs of the rest of the six main hyperbolic functions are shown below.

The graphs of hyperbolic cosecant, secant, tangent, and cotangent
The graph of $y = c s c h (x) = \frac{1}{\sin h (x)} = \frac{2}{e^{x} - e^{- x}}$	The graph of $y = s e c h (x) = \frac{1}{\cos h (x)} = \frac{2}{e^{x} + e^{- x}}$
The graph of hyperbolic cosecant - StudySmarter Originals	The graph of hyperbolic secant - StudySmarter Originals
The graph of $y = \tan h (x) = \frac{\sin h (x)}{\cos h (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$	The graph of $y = c o t h (x) = \frac{\cos h (x)}{\sin h (x)} = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
The graph of hyperbolic tangent - StudySmarter Originals	The graph of hyperbolic cotangent - StudySmarter Originals

Notice that these hyperbolic functions all have horizontal (green) and/or vertical (pink) asymptotes. The graph of hyperbolic secant has a global maximum at the point $(0, 1)$ .

Domain and Range of Hyperbolic Functions

While we are looking at the graphs of the hyperbolic functions, let's take note of their domains and ranges!

Function	Domain	Range
$y = \sin h (x)$	$(- \infty, \infty)$	$(- \infty, \infty)$
$y = \cos h (x)$	$(- \infty, \infty)$	$[1, \infty)$
$y = \tan h (x)$	$(- \infty, \infty)$	$(- 1, 1)$
$y = c s c h (x)$	$(- \infty, 0) \cup (0, \infty)$	$(- \infty, 0) \cup (0, \infty)$
$y = s e c h (x)$	$(- \infty, \infty)$	$(0, 1]$
$y = c o t h (x)$	$(- \infty, 0) \cup (0, \infty)$	$(- \infty, - 1) \cup (1, \infty)$

Hyperbolic Functions: Properties and Identities

The hyperbolic function properties and identities are also quite similar to those of their trigonometric counterparts:

Properties:
- $\sin h (- x) = - \sin h (x)$
- $\cos h (- x) = \cos h (x)$
- $\tan h (- x) = - \tan h (x)$
- $\sec h (- x) = \sec h (x)$
- $c s c h (- x) = - \csc (x)$
- $\cot h (- x) = - \cot h (x)$
- $\cos h (2 x) = \cos h^{2} (x) + \sin h^{2} (x) = 1 + 2 \sin h^{2} (x) = 2 \cos h^{2} (x) - 1$
- $\sin h (2 x) = 2 \sin h (x) \cos h (x)$
- They can be deduced from trigonometric functions that have complex arguments:
  - $\sin h (x) = - i \sin (i x)$
  - $\cos h (x) = \cos (i x)$
  - $\tan h (x) = - i \tan (i x)$
  - $\csc h (x) = i \csc (i x)$
  - $\sec h (x) = \sec (i x)$
  - $\cot h (x) = i \cot (i x)$
Identities:
- $\cos h^{2} (x) - \sin h^{2} (x) = 1$
- $\tan h^{2} (x) + s e c h^{2} (x) = 1$
- $c o t h^{2} (x) - c s c h^{2} (x) = 1$
- $\cos h (x) + \sin h (x) = e^{x}$
- $\cos h (x) - \sin h (x) = e^{- x}$
- $\sin h (x) + \sin h (y) = 2 \sin h (\frac{x + y}{2}) \cos h (\frac{x - y}{2})$
- $\sin h (x) - \sin h (y) = 2 \cos h (\frac{x + y}{2}) \sin h (\frac{x - y}{2})$
- $\cos h (x) + \cos h (y) = 2 \cos h (\frac{x + y}{2}) \cos h (\frac{x - y}{2})$
- $\cos h (x) - \cos h (y) = 2 \sin h (\frac{x + y}{2}) \sin h (\frac{x - y}{2})$
- $2 \sin h (x) \cos h (y) = \sin h (x + y) + \sin h (x - y)$
- $2 \cos h (x) \sin h (y) = \sin h (x + y) - \sin h (x - y)$
- $2 \sin h (x) \sin h (y) = \cos h (x + y) - \cos h (x - y)$
- $2 \cos h (x) \cos h (y) = \cos h (x + y) + \cos h (x - y)$
- $\sin h (x \pm y) = \sin h (x) \cos h (y) \pm \cos h (x) \sin h (y)$
- $\cos h (x \pm y) = \cos h (x) \cos h (y) \pm \sin h (x) \sin h (y)$
- $\sin h^{2} (x) = \frac{- 1 + \cos h (2 x)}{2}$
- $\cos h^{2} (x) = \frac{1 + \cos h (2 x)}{2}$

Let's test our understanding of these identities!

Prove that (a) $\cos h^{2} (x) - \sin h^{2} (x) = 1$ and (b) $1 - \tan h^{2} (x) = s e c h^{2} (x)$ .

Solution:

(a) We start with the definitions of hyperbolic cosine and hyperbolic sine, then simplify:

$\cos h^{2} (x) - \sin h^{2} (x) = {(\frac{e^{x} + e^{- x}}{2})}^{2} - {(\frac{e^{x} - e^{- x}}{2})}^{2} = \frac{e^{2 x} + e^{0} + e^{0} + e^{- 2 x}}{4} - \frac{e^{2 x} - e^{0} - e^{0} + e^{- 2 x}}{4} = \frac{e^{2 x} + 2 + e^{- 2 x}}{4} - \frac{e^{2 x} - 2 + e^{- 2 x}}{4} = \frac{4}{4} = 1$

(b) We start with the proof from part (a):

$\cos h^{2} (x) - \sin h^{2} (x) = 1$

Now, if we divide both sides by $\cos h^{2} (x)$ , we get:

$\frac{\cos h^{2} (x)}{\cos h^{2} (x)} - \frac{\sin h^{2} (x)}{\cos h^{2} (x)} = \frac{1}{\cos h^{2} (x)}$

This simplifies to:

$\frac{\cos h^{2} (x)}{\cos h^{2} (x)} - \frac{\sin h^{2} (x)}{\cos h^{2} (x)} = \frac{1}{\cos h^{2} (x)} \Rightarrow 1 - t a n h^{2} (x) = s e c h^{2} (x)$

This example gave us some insight into why these are called hyperbolic functions. Let's dive a bit deeper into it!

The Relationship Between Trigonometric and Hyperbolic Functions

♦ Say we have a real number, $t$ . Then the point $P (\cos (t), \sin (t))$ is on the unit circle:

$x^{2} + y^{2} = 1$

because

$\cos^{2} (t) + \sin^{2} (t) = 1$

In fact, the real number $t$ can also be interpreted as the radian measure of $∠ P O Q$ in the image below. This is the reason the trigonometric functions are sometimes called circular functions.

More importantly, the real number $t$ represents twice the area of the shaded section of the circle.

Hyperbolic functions graph of unit circle StudySmarter The graph of a unit circle with angle POQ shaded - StudySmarter Originals

♦ Likewise, if $t$ is any real number, then the point $P (\cos (t), \sin (t))$ is on the right half of the unit hyperbola:

$x^{2} - y^{2} = 1$

because

$\cos h^{2} (t) - \sin h^{2} (t) = 1$

where $\cos h (t) \geq 1$ .

In this case, however, $t$ does not represent the measure of an angle, but instead, it represents twice the area of the shaded hyperbolic section in the image below.

Hyperbolic functions graph of unit hyperbola StudySmarter The graph of a unit hyperbola with section P shaded - StudySmarter Originals

Derivatives of Hyperbolic Functions

The derivatives of hyperbolic functions are also analogous to the trigonometric functions. We list these derivatives in the table below.

Derivatives of the 6 main hyperbolic functions
$\frac{d}{d x} (\sin h (x)) = \cos h (x)$	$\frac{d}{d x} (c s c h (x)) = - c s c h (x) c o t h (x)$
$\frac{d}{d x} (\cos h (x)) = \sin h (x)$	$\frac{d}{d x} (s e c h (x)) = - s e c h (x) \tan h (x)$
$\frac{d}{d x} (\tan h (x)) = s e c h^{2} (x)$	$\frac{d}{d x} (c o t h (x)) = - c s c h^{2} (x)$

Beware! While the values of the derivatives are the same as with the trigonometric functions, the signs for the derivatives of hyperbolic cosine and hyperbolic secant are opposite to their trigonometric counterparts.

It is also important to note that any of these differentiation rules can be combined using The Chain Rule. For instance,

$\frac{d}{d x} (\cos h (\sqrt{x})) = \sin h (\sqrt{x}) \times \frac{d}{d x} (\sqrt{x}) = \frac{\sin h (\sqrt{x})}{2 \sqrt{x}}$

The derivatives of hyperbolic functions are simpler to calculate because of their use of $e^{x}$ and the simplicity of its derivation.

For instance,

$\frac{d}{d x} (\sin h (x)) = \frac{d}{d x} (\frac{e^{x} - e^{- x}}{2}) = \frac{e^{x} + e^{- x}}{2} = \cos h (x)$

Integrals of Hyperbolic Functions

Just as the derivatives of hyperbolic functions are analogous to their trigonometric counterparts, so are the integrals of hyperbolic functions. We list these integrals in the table below.

Integrals of the 6 main hyperbolic functions
$\int \sin h (x) d x = \cos h (x) + C$	$\int \csc h (x) d x = \ln \|\tan h (\frac{x}{2})\| + C$
$\int \cos h (x) d x = \sin h (x) + C$	$\int \sec h (x) d x = \tan^{- 1} \|\sin h (x)\| + C$
$\int \tan h (x) d x = \ln (\cos h (x)) + C$	$\int \cot h (x) d x = \ln \|\sin h (x)\| + C$

Other useful integrals of hyperbolic functions are listed below.

More integrals of hyperbolic functions
$\int \sec h^{2} (x) d x = \tan h (x) + C$	$\int \sec h (x) \tan h (x) d x = - \sec h (x) + C$
$\int \csc h^{2} (x) d x = - \cot h (x) + C$	$\int \csc h (x) \cot h (x) d x = - \csc h (x) + C$

Inverse Hyperbolic Functions

Based on the graphs of the hyperbolic functions, we can see that $\sin h$ (and its reciprocal, $\csc h$ ) and $\tan h$ (and its reciprocal, $\cot h$ ) are one-to-one functions, but $\cos h$ (and its reciprocal, $\sec h$ ) are not.

This is because cosine and secant are even functions, while sine, cosecant, tangent, and cotangent are odd functions.

Since cosine and secant are even functions, and are therefore not one-to-one, we have to restrict their domain to find their inverses.

So, with cosine's and secant's domains restricted to the interval $[0, \infty)$ , all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions as:

$y = \sin h^{- 1} (x) \Leftrightarrow \sin h (y) = x y = \cos h^{- 1} (x) \Leftrightarrow \cos h (y) = x a n d y \geq 0 y = \tan h^{- 1} (x) \Leftrightarrow \tan h (y) = x y = \csc h^{- 1} (x) \Leftrightarrow c sc h (y) = x y = \sec h^{- 1} (x) \Leftrightarrow \sec h (y) = x a n d y \geq 0 y = \cot h^{- 1} (x) \Leftrightarrow \cot h (y) = x$

Their formulas are:

The 6 main inverse hyperbolic functions
Inverse hyperbolic sine:	$\sin h^{- 1} (x) = a r c \sin h (x) = \ln (x + \sqrt{x^{2} + 1})$	Inverse hyperbolic cosecant:	$\csc h^{- 1} (x) = a r c \csc h (x) = \ln (\frac{1}{x} + \frac{\sqrt{x^{2} + 1}}{\|x\|})$
Inverse hyperbolic cosine:	$\cos h^{- 1} (x) = a r c \cos h (x) = \ln (x + \sqrt{x^{2} - 1})$	Inverse hyperbolic secant:	$\sec h^{- 1} (x) = a r c \sec h (x) = \ln (\frac{1 + \sqrt{1 - x^{2}}}{x})$
Inverse hyperbolic tangent:	$\tan h^{- 1} (x) = a r c \tan h (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$	Inverse hyperbolic cotangent:	$\cot h^{- 1} (x) = a r c \cot h (x) = \frac{1}{2} \ln (\frac{x + 1}{x - 1})$

Notice that the inverse hyperbolic functions all involve logarithmic functions. This is because the hyperbolic functions involve exponential functions, and exponential and logarithmic functions are inverses of each other!

Let's look at how the inverse of sinh (also called arc sinh) is derived. The derivation for the others follows a similar pattern.

Suppose that:

$y = \sin h^{- 1} (x)$

This means that:

$x = \sin h (y)$

By the definition of hyperbolic sine:

$x = \frac{e^{y} - e^{- y}}{2}$

Rearranging this, we get:

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

Then, multiplying both sides by $e^{y}$ , we have:

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

Now we solve it like we would a quadratic function, thinking of $e^{y}$ as $x$ , and get the solution:

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

Since $e^{y} > 0$ , the only possible solution is the positive one:

$e^{y} = x + \sqrt{x^{2} + 1}$

Finally, taking the natural logarithm of both sides, we get:

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

Graphs of Inverse Hyperbolic Functions

The graphs of the inverse hyperbolic functions are shown below.

The graphs of the inverse hyperbolic functions
The graph of $\sin h^{- 1} (x) = a r c \sin h (x) = \ln (x + \sqrt{x^{2} + 1})$	The graph of $\csc h^{- 1} (x) = a r c \csc h (x) = \ln (\frac{1}{x} + \frac{\sqrt{x^{2} + 1}}{\|x\|})$
The graph of inverse hyperbolic sine - StudySmarter Originals	The graph of inverse hyperbolic cosecant - StudySmarter Originals
The graph of $\cos h^{- 1} (x) = a r c \cos h (x) = \ln (x + \sqrt{x^{2} - 1})$	The graph of $\sec h^{- 1} (x) = a r c \sec h (x) = \ln (\frac{1 + \sqrt{1 - x^{2}}}{x})$
The graph of inverse hyperbolic cosine - StudySmarter Originals	The graph of inverse hyperbolic secant - StudySmarter Originals
The graph of $\tan h^{- 1} (x) = a r c \tan h (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$	The graph of $\cot h^{- 1} (x) = a r c \cot h (x) = \frac{1}{2} \ln (\frac{x + 1}{x - 1})$
The graph of inverse hyperbolic tangent - StudySmarter Originals	The graph of inverse hyperbolic cotangent - StudySmarter Originals

Notice that inverse hyperbolic cosecant, secant, tangent, and cotangent have horizontal (green) and/or vertical (pink) asymptotes. The graphs of inverse hyperbolic cosine and inverse hyperbolic secant have a definite beginning point at $(1, 0)$ .

Domain and Range of Inverse Hyperbolic Functions

While we are looking at the graphs of the inverse hyperbolic functions, let's take note of their domains and ranges!

Function	Domain	Range
$y = \sin h^{- 1} (x)$	$(- \infty, \infty)$	$(- \infty, \infty)$
$y = \cos h^{- 1} (x)$	$[1, \infty)$	$[0, \infty)$
$y = \tan h^{- 1} (x)$	$(- 1, 1)$	$(- \infty, \infty)$
$y = \csc h^{- 1} (x)$	$(- \infty, 0) \cup (0, \infty)$	$(- \infty, 0) \cup (0, \infty)$
$y = \sec h^{- 1} (x)$	$(0, 1]$	$[0, \infty)$
$y = \cot h^{- 1} (x)$	$(- \infty, - 1) \cup (1, \infty)$	$(- \infty, 0) \cup (0, \infty)$

Derivatives of Inverse Hyperbolic Functions

All the inverse hyperbolic functions are differentiable because all the hyperbolic functions are differentiable. The derivatives of the inverse hyperbolic functions are listed below.

The derivatives of the inverse hyperbolic functions
$\frac{d}{d x} (\sin h^{- 1} (x)) = \frac{1}{\sqrt{1 + x^{2}}}$	$\frac{d}{d x} (\csc h^{- 1} (x)) = - \frac{1}{\|x\| \sqrt{x^{2} + 1}}$
$\frac{d}{d x} (\cos h^{- 1} (x)) = \frac{1}{\sqrt{x^{2} - 1}}$	$\frac{d}{d x} (\sec h^{- 1} (x)) = - \frac{1}{x \sqrt{1 - x^{2}}}$
$\frac{d}{d x} (\tan h^{- 1} (x)) = \frac{1}{1 - x^{2}}$	$\frac{d}{d x} (\cot h^{- 1} (x)) = \frac{1}{1 - x^{2}}$

Let's prove that $\frac{d}{d x} (\sin h^{- 1} (x)) = \frac{1}{\sqrt{1 + x^{2}}}$ .

$\frac{d}{d x} (\sin h^{- 1} (x)) = \frac{d}{d x} (\ln (x + \sqrt{x^{2} + 1})) = \frac{1}{x + \sqrt{x^{2} + 1}} \times \frac{d}{d x} (x + \sqrt{x^{2} + 1}) (c h a i n r u l e) = \frac{1}{x + \sqrt{x^{2} + 1}} (1 + \frac{x}{\sqrt{x^{2} + 1}}) = \frac{\sqrt{x^{2} + 1} + x}{(x + \sqrt{x^{2} + 1}) \sqrt{x^{2} + 1}} = \frac{1}{\sqrt{x^{2} + 1}}$

Hyperbolic Functions: Examples and Applications

Find the value of $x$ if $3 \sin h (x) - 2 \cos h (x) - 2 = 0$ .

Solution:

Substitute the values: sinhx=ex-e-x2, coshx=ex+e-x2into the equation.
- $3 (\frac{e^{x} - e^{- x}}{2}) - 2 (\frac{e^{x} + e^{- x}}{2}) - 2 = 0$
Simplify:
- $3 (\frac{e^{x} - e^{- x}}{2}) - 2 (\frac{e^{x} + e^{- x}}{2}) - 2 = 0 \Rightarrow \frac{3 (e^{x} - e^{- x})}{2} - \frac{2 (e^{x} + e^{- x})}{2} - 2 = 0 \Rightarrow 3 (e^{x} - e^{- x}) - 2 (e^{x} + e^{- x}) - 4 = 0 (m u l t i p l y b o t h s i d e s b y 2) \Rightarrow 3 e^{x} - 3 e^{- x} - 2 e^{x} - 2 e^{- x} - 4 = 0 (e x p a n d) \Rightarrow e^{x} - 5 e^{- x} - 4 = 0 (c o l l e c t l i k e t e r m s) \Rightarrow e^{x} (e^{x} - 5 e^{- x} - 4) = e^{x} \cdot 0 (m u l t i p l y b o t h s i d e s b y e^{x}) \Rightarrow {(e^{x})}^{2} - 5 - 4 e^{x} = 0 (e x p a n d) \Rightarrow {(e^{x})}^{2} - 5 - 5 e^{x} + e^{x} = 0 (r e w r i t e - 4 e^{x}) \Rightarrow e^{x} (e^{x} + 1) - 5 (e^{x} + 1) = 0 (f a c t o r b y g r o u p i n g) \Rightarrow (e^{x} - 5) (e^{x} + 1) = 0 (c o m m o n f a c t o r e^{x} + 1)$
Since ex≠-1, the only solution is:
- $e^{x} = 5 \Rightarrow x = \ln (5)$

Express $e^{x}$ and $e^{- x}$ as a function of $\sin h (x)$ and $\cos h (x)$ .

Solution:

Add the two equations for coshx and sinhx.
- $\cos h (x) + \sin h (x) = \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2} = \frac{e^{x} + e^{- x} + e^{x} - e^{- x}}{2} = \frac{2 e^{x}}{2} = e^{x}$
- Therefore, $\cos h (x) + \sin h (x) = e^{x} (1)$
Subtract the two equations for coshx and sinhx.
- $\cos h (x) - \sin h (x) = \frac{e^{x} + e^{- x}}{2} - \frac{e^{x} - e^{- x}}{2} = \frac{e^{x} + e^{- x} - e^{x} + e^{- x}}{2} = \frac{2 e^{- x}}{2} = e^{- x}$
- Therefore, $\cos h (x) - \sin h (x) = e^{- x} (2)$
If we combine equations (1) and (2), we get:
- $e^{\pm x} = c o s h (x) \pm s i n h (x)$
- This is Euler's formula for the hyperbolic function.

There are several real-world applications for hyperbolic functions, such as:

describing the decay of light, velocity, electricity, or radioactivity
modeling the velocity of a wave as it moves across a body of water
the use of hyperbolic cosine to describe the shape of a hanging wire (called a catenary).

Perhaps the most famous of these is the description of the hanging wire.

Hyperbolic Functions - Key takeaways

The hyperbolic functions are essentially the trigonometric functions of the hyperbola.
- This is why they are sometimes called hyperbolic trig functions.
There are 6 hyperbolic functions:
- hyperbolic sine - $\sin h (x) = \frac{e^{x} - e^{- x}}{2}$
- hyperbolic cosine - $\cos h (x) = \frac{e^{x} + e^{- x}}{2}$
- hyperbolic tangent - $\tan h (x) = \frac{\sin h (x)}{\cos h (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$
- hyperbolic cosecant - $c s c h (x) = \frac{1}{\sin h (x)}$
- hyperbolic secant - $s e c h (x) = \frac{1}{\cos h (x)}$
- hyperbolic cotangent - $c o t h (x) = \frac{\cos h (x)}{\sin h (x)} = \frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}$
Their properties and identities are analogous to those of trigonometric functions.

Flashcards in Hyperbolic Functions 1

Start learning

Why do we have to restrict the domain of hyperbolic cosine and hyperbolic secant in order to find their inverses?

Because cosine and secant are even functions, and are therefore not one-to-one.

Already have an account? Log in

Frequently Asked Questions about Hyperbolic Functions

1. What are hyperbolic functions?

Hyperbolic functions are the trigonometric functions defined using a hyperbola instead of a circle. While the points (cos x, sin x) form a circle with a unit radius, the points (cosh x, sinh x) form the right half of a unit hyperbola. These functions are defined in terms of the exponential functions e^x and e^-x.

2. How do I evaluate hyperbolic functions?

Hyperbolic functions can be evaluated by plugging the value at which you wish to evaluate the function, and simplifying.

3. What are hyperbolic functions used for?

Hyperbolic functions are used in engineering and physics applications like the study of waves or vibrations of elastic membranes. They are also used to represent a hanging cable, or catenary, and to design arches to help stabilize structures.

4. How do you write a hyperbolic function?

Hyperbolic functions are written similarly to the trigonometric functions for a circle. The six main hyperbolic functions are written as:
sinh = (e^x - e^-x)/2
cosh = (e^x + e^-x)/2
tanh = (sinh x)/(cosh x)
csch = 1/(sinh x)
sech = 1/(cosh x)
coth = (cosh x)/(sinh x)

Save Article

Discover learning materials with the free StudySmarter app

About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more

StudySmarter Editorial Team

Team Math Teachers

12 minutes reading time
Checked by StudySmarter Editorial Team

Save Explanation

Hyperbolic Functions

StudySmarter Editorial Team

Hyperbolic Function Definition