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Inverse Hyperbolic Functions

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\(\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\csch}{cosech}\DeclareMathOperator{\coth}{coth}\)

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.

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** \( y = x \).

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 \(\sinh(x)\).

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

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 |

\( y = \csch^{-1}{x} \) | \(\left( -\infty, 0 \right) \cup \left( 0, \infty \right) \) | \(\left( -\infty, 0 \right) \cup \left( 0, \infty \right) \) |

\(y=\sech^{-1}{x}\) | \(\left( 0, 1 \right] \) | \(\left[ 0, \infty \right) \) |

\(y=\coth^{-1}{x}\) | \(\left( -\infty, -1 \right) \) | \(\left( 1, \infty \right) \) |

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,

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

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

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

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,

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

This is the final answer.

**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.

**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 \( |x| > 1 \).

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 \( z. \)

- 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 + (x^{2} + 1)^{1/2}).

^{-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 e^{y}. From here, take the natural logarithm of both sides to get a function in terms of x. For example,

sinh^{-1}(x) = ln(x + (x^{2} + 1)^{1/2}).

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