\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Inverse Hyperbolic Functions

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Inverse Hyperbolic Functions

Before you study this page, make sure you have gone through Hyperbolic Functions and understand those first.

As you probably expect, we can define inverses of hyperbolic functions. Some of the hyperbolic functions require a restricted domain since only one-to-one functions can have inverses. (For more detail, see the inverse functions page.)

Let's think about what the equations might look like. Since the hyperbolic functions are defined in terms of \(e^x\), we would except the inverse hyperbolic functions to be some form involving \(\ln(x)\). This is indeed the case. Here is a table listing the inverse hyperbolic functions.

\(\arcsinh(x) = \ln\left[ x + \sqrt{x^2+1} \right]\)

Domain: all real numbers

\(\arccosh(x) = \ln\left[ x + \sqrt{x^2-1} \right]\)

Domain: \( [1, +\infty) \)

\(\displaystyle{ \arctanh(x) = \frac{1}{2} \ln\left[ \frac{1+x}{1-x} \right] }\)

Domain: \( (-1,1) \)

\(\displaystyle{ \arccoth(x) = \frac{1}{2}\ln\left[ \frac{x+1}{x-1} \right] }\)

Domain: \( (-\infty, -1) \cup ( 1, \infty ) \)

\(\displaystyle{ \arcsech(x) = \ln \left[ \frac{1}{x} + \sqrt{\frac{1}{x^2}-1} \right] }\)

Domain: \((0,1]\)

\(\displaystyle{ \arccsch(x) = \ln \left[ \frac{1}{x} + \sqrt{\frac{1}{x^2}+1} \right] }\)

Domain: all real numbers except for zero

This GeoGebra page shows the graphs of the inverse hyperbolic functions.

This Wikipedia page contains more information about inverse hyperbolic functions.

To get used to working with inverse hyperbolic functions, let's work these practice problems.

Practice

Unless otherwise instructed, use the definitions of the related hyperbolic functions in terms of \(e^x\) to get the equations for the given functions involving the natural logarithm.

\(\text{arcsinh}(x)\)

Problem Statement

Use the definition of the hyperbolic function \(\sinh(x)\) in terms of \(e^x\) to get the equation for \(\text{arcsinh}(x)\) involving the natural logarithm.

Hint

\(\sinh(x) = \) \(\displaystyle{ \frac{e^x - e^{-x}}{2} }\)

Problem Statement

Use the definition of the hyperbolic function \(\sinh(x)\) in terms of \(e^x\) to get the equation for \(\text{arcsinh}(x)\) involving the natural logarithm.

Hint

\(\sinh(x) = \) \(\displaystyle{ \frac{e^x - e^{-x}}{2} }\)

Solution

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\(\text{arccosh}(x)\)

Problem Statement

Use the definition of the hyperbolic function \(\cosh(x)\) in terms of \(e^x\) to get the equation for \(\text{arccosh}(x)\) involving the natural logarithm.

Hint

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

Problem Statement

Use the definition of the hyperbolic function \(\cosh(x)\) in terms of \(e^x\) to get the equation for \(\text{arccosh}(x)\) involving the natural logarithm.

Hint

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

Solution

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\(\text{arctanh}(x)\)

Problem Statement

Use the definition of the hyperbolic function \(\tanh(x)\) in terms of \(e^x\) to get the equation for \(\text{arctanh}(x)\) involving the natural logarithm.

Hint

\(\tanh(x) = \) \(\displaystyle{ \frac{\sinh(x)}{\cosh(x)} = }\) \(\displaystyle{ \frac{e^x - e^{-x}}{e^x + e^{-x}} = }\) \(\displaystyle{ \frac{e^{2x} - 1}{e^{2x} + 1} }\)

Problem Statement

Use the definition of the hyperbolic function \(\tanh(x)\) in terms of \(e^x\) to get the equation for \(\text{arctanh}(x)\) involving the natural logarithm.

Hint

\(\tanh(x) = \) \(\displaystyle{ \frac{\sinh(x)}{\cosh(x)} = }\) \(\displaystyle{ \frac{e^x - e^{-x}}{e^x + e^{-x}} = }\) \(\displaystyle{ \frac{e^{2x} - 1}{e^{2x} + 1} }\)

Solution

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Really UNDERSTAND Precalculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, use the definitions of the related hyperbolic functions in terms of \(e^x\) to get the equations for the given functions involving the natural logarithm.

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