Inverse hyperbolic functionsEdexcel International A Level Further Maths: Revision notes

Section 1

Inverse hyperbolic functions and their graphs

The inverse hyperbolic functions undo $\sinh$ , $\cosh$ and $\tanh$ . They are written $\operatorname{arsinh}x$ (or $\sinh^{-1}x$ ), $\operatorname{arcosh}x$ and $\operatorname{artanh}x$ . If $y=\operatorname{arsinh}x$ then $x=\sinh y$ , and similarly for the others. Each graph is the reflection in the line $y=x$ of the graph of the original function. $\cosh x$ is not one-to-one, so $\operatorname{arcosh}$ is the inverse of the branch $x\ge0$ .

Key termsinverse hyperbolic functionreflection in $y=x$

Exam tip

The domain of an inverse is the range of the original, and the range of the inverse is the domain of the original.

Section 2

Domains, ranges and properties

$\operatorname{arsinh}x$ : domain all real $x$ , range all real numbers; odd and increasing, with gradient $1$ at the origin.
$\operatorname{arcosh}x$ : domain $x\ge1$ , range $y\ge0$ ; not defined for $x<1$ , and the graph starts at $(1,0)$ with a vertical tangent.
$\operatorname{artanh}x$ : domain $-1<x<1$ , range all real numbers; odd and increasing, with vertical asymptotes $x=1$ and $x=-1$ .

Key termsdomainrange

Common mistake

Giving $\operatorname{arcosh}x$ a domain that includes $x<1$ , or a negative range.

Section 3

Logarithmic equivalents

The inverse functions have exact forms in terms of the natural logarithm: $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right),\quad\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)\ (x\ge1),\quad\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}\ (|x|<1).$ Examples: $\operatorname{arsinh}\frac34=\ln\left(\frac34+\frac54\right)=\ln2$ ; $\operatorname{arcosh}5=\ln\left(5+2\sqrt6\right)$ ; $\operatorname{artanh}\frac12=\frac12\ln3$ .

Key termsnatural logarithm

Common mistake

Forgetting the $\frac12$ in $\operatorname{artanh}$ , or putting $x^2-1$ under the root for $\operatorname{arsinh}$ .

Section 4

Proving the logarithmic forms

To prove $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$ , let $y=\operatorname{arsinh}x$ , so $x=\sinh y=\frac{e^y-e^{-y}}{2}$ . Multiply by $2e^y$ : $e^{2y}-2xe^y-1=0$ , a quadratic in $e^y$ : $e^y=x\pm\sqrt{x^2+1}.$ Since $\sqrt{x^2+1}>|x|$ , the root $x-\sqrt{x^2+1}$ is negative, which is impossible as $e^y>0$ . So $y=\ln\left(x+\sqrt{x^2+1}\right)$ . For $\operatorname{artanh}$ : $x=\frac{e^{2y}-1}{e^{2y}+1}$ gives $e^{2y}=\frac{1+x}{1-x}$ , which is positive only if $|x|<1$ , so $y=\frac12\ln\frac{1+x}{1-x}$ . For $\operatorname{arcosh}$ the same method gives $e^y=x\pm\sqrt{x^2-1}$ and the range $y\ge0$ selects the $+$ sign.

Key termsquadratic in $e^y$

Exam tip

Reject any root where $e^y\le0$ , and say why.

Section 5

Using the logarithmic forms

To solve an equation involving inverse hyperbolic functions, convert to logarithms, use the log laws and solve. Example: $\operatorname{artanh}x=\operatorname{arsinh}\frac43-\ln2$ . Since $\operatorname{arsinh}\frac43=\ln3$ , we get $\frac12\ln\frac{1+x}{1-x}=\ln\frac32$ , so $\frac{1+x}{1-x}=\frac94$ and $x=\frac{5}{13}$ . Properties can also be shown from the logarithmic forms. For example $\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)=1$ , so $\operatorname{arsinh}(-x)=-\operatorname{arsinh}x$ .

Exam tip

Check any solution of an $\operatorname{artanh}$ equation satisfies $|x|<1$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Inverse hyperbolic functions

The inverse hyperbolic functions have the logarithmic forms $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$ for all real $x$ , $\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)$ for $x\ge1$ , and $\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}$ for $|x|<1$ .
Find the exact value of $\operatorname{arcosh}5$ , giving your answer in the form $\ln\left(a+b\sqrt6\right)$ .2 marks
The function $f$ is defined by $f(x)=\operatorname{arcosh}x$ .
Explain how the graph of $y=\operatorname{arcosh}x$ is related to the graph of $y=\cosh x$ .2 marks
For real $x$ , $y=\operatorname{arsinh}x$ means that $x=\sinh y$ .
Prove that $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).