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# Inverse Hyperbolic Functions

$$\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\csch}{cosech}\DeclareMathOperator{\coth}{coth}$$

Get a short length of string and put it in a straight line on a flat surface. Now, put your finger onto one end of the string, and drag it along the surface perpendicular to the original direction of the string. What curve is mapped out by the position of the other end of the string, as you drag it along the table?

It turns out that this curve is called a tractrix, which, if the original piece of string was facing in the direction of the $$y-$$ axis, is mapped out by the equation,

$x = \sech^{-1}{\frac{y}{a}} - \sqrt{a^2 - y^2},$

where $$a$$ is the length of the string.

You will notice the presence of the inverse hyperbolic secant function in this formula. Thus, to understand this curve, you must first understand the inverse hyperbolic functions, and this is what we will explore here in this article.

If you track the end of a straight piece of string pulled from from the other end, perpendicular to the original line, it will create a tractrix.

## Graphs of Inverse Hyperbolic Functions

Remember that if we have a function $$f$$ such that $$f(x) = y$$, then the inverse of $$f$$ is the function $$f^{-1}$$ such that $$f^{-1}(y) = x$$. This is exactly the same way that the inverse hyperbolic functions work too.

The standard inverse hyperbolic functions are,

• Inverse hyperbolic sine: $$\sinh^{-1}{x}$$,

• Inverse hyperbolic cosine: $$\cosh^{-1}{x}$$,

• Inverse hyperbolic tangent: $$\tanh^{-1}{x}$$.

The inverse reciprocal hyperbolic functions are,

• Inverse hyperbolic secant: $$\sech^{-1}{x}$$,

• Inverse hyperbolic cosecant: $$\csch^{-1}{x}$$,

• Inverse hyperbolic cotangent: $$\coth^{-1}{x}$$.

Remember that you can only find an inverse function if that function is one-to-one. This means that every value in the domain of the function maps to exactly one unique value in the range of the function. Here are the domains and ranges of our hyperbolic functions.

 Function Domain Range $$y = \sinh{x}$$ $$\left( -\infty, \infty \right)$$ $$\left( -\infty, \infty \right)$$ $$y=\cosh{x}$$ $$\left( -\infty, \infty \right)$$ $$\left[ 1, \infty \right)$$ $$y=\tanh{x}$$ $$\left( -\infty, \infty \right)$$ $$\left( -1, 1 \right)$$ $$y=\csch{x}$$ $$\left( -\infty, 0 \right) \cup \left( 0, \infty \right)$$ $$\left( -\infty, 0 \right) \cup \left( 0, \infty \right)$$ $$y=\sech{x}$$ $$\left( -\infty, \infty \right)$$ $$\left(0, 1\right]$$ $$y=\coth{x}$$ $$\left( -\infty, 0 \right) \cup \left( 0, \infty \right)$$ $$\left( -\infty, -1 \right) \cup \left( 1, \infty \right)$$

Remember that the domain of a function is the set of valid inputs into the function, and the range is the set of all possible outputs of the function.

Inverse hyperbolic sine, tangent, cotangent, and cosecant are all one-to-one functions, and hence their inverses can be found without any need to modify them.

Hyperbolic cosine and secant, however, are not one-to-one. For this reason, to find their inverses, you must restrict the domain of these functions to only include positive values. This is because these are even functions, meaning that $$\cosh{(-x)} = \cosh{x}$$, and $$\sech{(-x)} = \sech{(x)}$$ for any value of $$x$$. Hence, if you only allow the functions to take inputs that are positive, each input has its own unique output.

As with any inverse functions, the graphs of the $$\sinh^{-1}{x}$$, $$\cosh^{-1}{x}$$ and $$\tanh^{-1}{x}$$ are the same as the graphs of $$\sinh{x}$$, $$\cosh{x}$$ and $$\tanh{x}$$, but reflected in the line $$y = x$$.

The inverses of the hyperbolic sine, cosine, and tangent functions are the original lines, reflected in the line $$y = x$$.

You may notice that inverse cosine is only shown on the positive quadrant of the graph. This is the result of the domain being restricted, as mentioned before.

Similarly, you will notice that $$\tanh^{-1}{x}$$ is only defined for values of $$x$$ between -1 and 1. This is because for any $$x$$, $$\tanh{x}$$ is always between -1 and 1. Hence, its inverse can only take input values in this range.

The domains and ranges of the standard inverse hyperbolic functions are:

 Function Domain Range $$y = \sinh^{-1}{x}$$ $$\left( -\infty, \infty \right)$$ $$\left( -\infty, \infty \right)$$ $$y=\cosh^{-1}{x}$$ $$\left[ 1, \infty \right)$$ $$\left[ 0, \infty \right)$$ $$y=\tanh^{-1}{x}$$ $$\left( -1, 1 \right)$$ $$\left( -\infty, \infty \right)$$

Don't confuse $$\sinh^{-1}{x}$$ with $$\frac{1}{\sinh{x}}$$, these are two separate functions. $$\frac{1}{\sinh{x}}$$ is the reciprocal hyperbolic function of $$\sinh{x}$$, known as $$\csch{x}$$, whereas $$\sinh^{-1}{x}$$ is the inverse hyperbolic function of $$\sinh(x)$$.

### Graphs of Reciprocal Inverse Hyperbolic Functions

Below you can see the graphs of the inverse reciprocal hyperbolic functions, $$\csch^{-1}{x}$$, $$\sech^{-1}{x}$$ and $$\coth^{-1}{x}$$.

The inverses of the reciprocal hyperbolic functions, hyperbolic secant, cosecant and cotangent, are the original graphs reflected in the line $$y = x$$.

Again, the domain must be restricted when working with the inverse of $$\sech{x}$$, as this is a many-to-one function, just like its non-reciprocal counterpart $$\cosh{x}$$. The domains and ranges of the inverse reciprocal hyperbolic functions are:

 Function Domain Range $$y = \csch^{-1}{x}$$ $$\left( -\infty, 0 \right) \cup \left( 0, \infty \right)$$ $$\left( -\infty, 0 \right) \cup \left( 0, \infty \right)$$ $$y=\sech^{-1}{x}$$ $$\left( 0, 1 \right]$$ $$\left[ 0, \infty \right)$$ $$y=\coth^{-1}{x}$$ $$\left( -\infty, -1 \right)$$ $$\left( 1, \infty \right)$$

## Inverse Hyperbolic Functions Formulas

Just as the standard hyperbolic functions have exponential forms, the inverse hyperbolic functions have logarithmic forms. This makes sense, given that taking the natural logarithm of a number is the inverse of raising that number to the exponential constant $$e$$.

The logarithmic forms of the inverse hyperbolic functions, $$\sinh^{-1}{x}$$, $$\cosh^{-1}{x}$$ and $$\tanh^{-1}{x}$$ are,

\begin{align} \sinh^{-1}{x} & = \ln{\left(x + \sqrt{x^2 + 1} \right)}, \\ \cosh^{-1}{x} & = \ln{\left(x + \sqrt{x^2 - 1} \right)}, \\ \tanh^{-1}{x} & = \frac{1}{2} \ln{\left( \frac{1+x}{1-x} \right) }. \end{align}

### Inverse Reciprocal Hyperbolic Functions Formulas

There are also logarithmic forms of the inverse reciprocal hyperbolic functions $$\sech^{-1}{x}$$, $$\csch^{-1}{x}$$ and $$\coth^{-1}{x}$$. These are,

\begin{align} \sech^{-1}{x} & = \ln{\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} - 1} \right)}, \\ \csch^{-1}{x} & = \ln{\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \right)}, \\ \coth^{-1}{x} & = \frac{1}{2} \ln{\left(\frac{x+1}{x-1}\right)}. \end{align}

## Examples of Inverse Hyperbolic Functions

A common question is to prove one of the logarithmic forms shown above. To do this, it is important to make use of the exponential form of the standard hyperbolic functions.

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

Solution

First, write $y = \sinh^{-1}{x}.$ Now, take the hyperbolic sine of both sides to get, $\sinh{y} = x.$ Write $$\sinh{y}$$ in exponential form, $x = \sinh{y} = \frac{e^y - e^{-y}}{2}.$ From here, you can solve for $$y.$$ Start by multiplying both sides by 2, and then by $$e^y$$,

\begin{align} 2 x & = e^y - e^{-y} \\ 2 x e^y & = e^{2y} - 1 \\ 0 & = e^{2y} - 2x e^{y} - 1. \end{align}

This is a quadratic in $$e^y$$. This can be solved this using the quadratic formula:

\begin{align} e^y & = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \\ & = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \\ & = \frac{2x \pm 2 \sqrt{x^2 + 1}}{2} \\ & = x \pm \sqrt{x^2 + 1}. \end{align}

There is a choice of plus or minus here. The natural logarithm is undefined for numbers under 0, so if you take the minus, that is $$x-\sqrt{x^2+1}$$, we always have $$x < \sqrt{x^2}+1$$, and thus $$e^y=x-\sqrt{x^2+1}$$ is undefined. Thus, the plus sign is the correct choice,

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

Finally, take the natural logarithm,

$y = \ln{\left(x + \sqrt{x^2 + 1}\right)},$ and the proof is complete.

It is also important to be able to manipulate hyperbolic functions in such a way to make a question easier, as well as to practice substituting numbers into the logarithmic formulas.

Solve $$\cosh^{3}{x} - 3 \cosh{x} = 0$$, giving your answer in logarithmic form.

Solution

First, take out the common factor of $$\cosh{x}$$,

\begin{align} \cosh^{3}{x} - 3 \cosh{x} & = 0 \\ \cosh{x} \left( \cosh^2{x} - 3 \right) & = 0. \end{align}

For this to be true, it must be that either $$\cosh{x} = 0$$, or $$\cosh^2{x} = 3 \implies \cosh{x} = \pm \sqrt{3}.$$ You can see from the graphs above that $$\cosh{x}$$ never goes below 1. Thus, it cannot be that $$\cosh{x} = 0$$, or that $$\cosh{x} = -\sqrt{3}$$. Thus, it must be the case that,

$\cosh{x} = \sqrt{3}.$

Take the inverse hyperbolic cosine of this, to get,

$x = \cosh^{-1}{\sqrt{3}},$

and finally write this using the logarithmic formula for inverse hyperbolic cosine,

\begin{align} x & = \ln{\left(3 + \sqrt{\sqrt{3}^2 - 1}\right)} \\ & = \ln{\left(3 + \sqrt{2} \right)}. \end{align}

## Derivatives of Inverse Hyperbolic trig Functions

The derivatives of the inverse hyperbolic functions $$\sinh^{-1}{x}$$, $$\cosh^{-1}{x}$$ and $$\tanh^{-1}{x}$$ are,

\begin{align} \frac{d}{dx} \sinh^{-1}{x} & = \frac{1}{\sqrt{1+x^2}}, \\ \frac{d}{dx} \cosh^{-1}{x} & = \frac{1}{\sqrt{x^2 - 1}}, \\ \frac{d}{dx} \tanh^{-1}{x} & = \frac{1}{1-x^2}. \end{align}

Here you will notice a resemblance to the derivatives of the inverse trigonometric functions.

Knowing all the inverse hyperbolic and trigonometric derivatives will make solving many complicated integrals much easier, as you can use integration by substitution with a hyperbolic function as the substitution. See Integration of Hyperbolic Functions for more information on this.

### Derivatives of Inverse Reciprocal Hyperbolic Functions

The derivatives of the inverse reciprocal hyperbolic functions $$\sech^{-1}{x}$$, $$\csch^{-1}{x}$$ and $$\coth^{-1}{x}$$ are,

\begin{align} \frac{d}{dx} \sech^{-1}{x} & = \frac{-1}{x \sqrt{1 - x^2}}, \\ \frac{d}{dx} \csch^{-1}{x} & = \frac{-1}{|x|\sqrt{1+x^2}}, \\ \frac{d}{dx} \coth^{-1}{x} & = \frac{1}{1-x^2}. \end{align}

You may notice that the derivatives of hyperbolic tangent and hyperbolic cotangent appear to be the same. This is normal, as they are defined on different domains. Hyperbolic tangent and its derivative are defined on $$|x| < 1$$, whereas hyperbolic cotangent and its derivative are defined on $$|x| > 1$$.

## Inverse Hyperbolic Functions of Complex Numbers

Hyperbolic functions of complex numbers are not something that you will have to consider in the Further Mathematics course, but it can be interesting to study them nonetheless. For a recap on complex numbers, see Core Complex Numbers.

If you have studied roots of unity, you will know that a number can have multiple roots in the complex plane, for example, $$16^{\frac{1}{4}}$$ could be $$2, 2i , -2$$ or $$-2i.$$ See Roots of Unity for more information.

Similarly, in the first example under the subheading "Examples of Inverse Hyperbolic Functions", you solve for $$e^y$$. However, in the complex plane, $$e^y$$ will have multiple solutions when you solve for $$y.$$ Thus, the inverse hyperbolic functions will be multi-valued functions in the complex plane.

It is common to define a principal value of these functions, as a way to make them single valued. The usual principle values for the standard inverse hyperbolic functions are,

\begin{align} \sinh^{-1}{z} & = \ln{\left(z + \sqrt{z^2 + 1} \right)}, \\ \cosh^{-1}{z} & = \ln{\left(z + \sqrt{z^2 - 1} \right)}, \\ \tanh^{-1}{z} & = \frac{1}{2} \ln{\left( \frac{1+z}{1-z} \right) }. \end{align}

You may notice that these are the same formulas as the logarithmic forms of the inverse hyperbolic functions on the real numbers, but with $$x$$ replaced with $$z.$$

## Inverse Hyperbolic Functions - Key takeaways

• Due to the range of the hyperbolic functions, the inverse hyperbolic functions are not all defined on the whole real line. In particular:
• $$\sinh^{-1}{x}$$ has domain $$\mathbb{R}$$ and range $$\mathbb{R}.$$
• $$\cosh^{-1}{x}$$ has domain $$\{ x: x \geq 1 \}$$ and range $$\{ f(x): f(x) \geq 0 \} .$$
• $$\tanh^{-1}{x}$$ has domain $$\{ x: -1 < x < 1 \}$$ and range $$\mathbb{R} .$$
• Just as the standard hyperbolic functions can be written in exponential form, the inverse hyperbolic functions can be written in logarithmic forms. These are, \begin{align} \sinh^{-1}{x} & = \ln{\left(x + \sqrt{x^2 + 1} \right)}, \\ \cosh^{-1}{x} & = \ln{\left(x + \sqrt{x^2 - 1} \right)}, \\ \tanh^{-1}{x} & = \frac{1}{2} \ln{\left( \frac{1+x}{1-x} \right) }. \end{align}
• To prove the logarithmic forms of the inverse hyperbolic functions, you must invert the function so it is in terms of a standard hyperbolic function, then solve for $$y$$.
• The derivatives of the inverse hyperbolic functions can be very useful for solving tricky integrals. These derivatives are, \begin{align} \frac{d}{dx} \sinh^{-1}{x} & = \frac{1}{\sqrt{1+x^2}}, \\ \frac{d}{dx} \cosh^{-1}{x} & = \frac{1}{\sqrt{x^2 - 1}}, \\ \frac{d}{dx} \tanh^{-1}{x} & = \frac{1}{1-x^2}. \end{align}

The inverse hyperbolic sine function, sinh-1(x), has the following formula:

sinh-1(x) = ln(x + (x2 + 1)1/2).

No, hyperbolic sine and inverse sine are different functions. Hyperbolic sine (sinh(x)) maps out the unit hyperbola in the same way as the usual sine maps out the unit circle, while inverse sine (sin-1(x) or arsin(x)) is the inverse function of sine. This means that if y = sin(x), x = sin-1(y).

To find the inverse of a hyperbolic function, first write the hyperbolic function in terms of the exponential form, and then solve for ey. From here, take the natural logarithm of both sides to get a function in terms of x. For example,

sinh-1(x) = ln(x + (x2 + 1)1/2).

Yes, though for cosh(x) you must restrict the domain. Inverses only exist on one-to-one functions, and cosh(x) is a one-to-many function. Thus, to find the inverse of cosh(x), we must restrict its domain to x > 0.

sinh(x) and tanh(x) are invertible on the real line, but cosh(x) must be restricted to the domain of x > 0 to find its inverse. This is because cosh(x) is symmetrical, so cosh(x) = cosh(-x) for all x, and thus is not a one-to-one function.

## Final Inverse Hyperbolic Functions Quiz

Question

What is the domain of the inverse hyperbolic sine function?

The set of real numbers.

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Question

What is the range of the inverse hyperbolic sine function?

The set of real numbers.

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Question

What is the range of the inverse hyperbolic tangent function?

The set of real numbers.

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Question

What is the domain of inverse hyperbolic secant?

(0, 1].

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