Logarithmic form of \operatorname{arcosh}x, and its domain?

\ln\left(x+\sqrt{x^2-1}\right), for x\ge1

Logarithmic form of \operatorname{artanh}x, and its domain?

\frac12\ln\frac{1+x}{1-x}, for |x|<1

Why is \operatorname{arcosh} the inverse of \cosh only for x\ge0?

\cosh is not one-to-one, so only the branch x\ge0 has an inverse.

First step in proving \operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)?

Let y=\operatorname{arsinh}x, so x=\sinh y, then form a quadratic in e^y.

Is \operatorname{arsinh} odd or even?

Odd: \operatorname{arsinh}(-x)=-\operatorname{arsinh}x.

Inverse hyperbolic functionsEdexcel International A Level Further Maths: Flashcards

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What does $y=\operatorname{arsinh}x$ mean?

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What does $y=\operatorname{arsinh}x$ mean?: $x=\sinh y$
Logarithmic form of $\operatorname{arsinh}x$ ?: $\ln\left(x+\sqrt{x^2+1}\right)$
Logarithmic form of $\operatorname{arcosh}x$ , and its domain?: $\ln\left(x+\sqrt{x^2-1}\right)$ , for $x\ge1$
Logarithmic form of $\operatorname{artanh}x$ , and its domain?: $\frac12\ln\frac{1+x}{1-x}$ , for $|x|<1$
Domain and range of $\operatorname{arsinh}x$ ?: All real $x$ ; all real $y$ .
Domain and range of $\operatorname{arcosh}x$ ?: $x\ge1$ ; $y\ge0$ .
Domain and range of $\operatorname{artanh}x$ ?: $-1<x<1$ ; all real $y$ .
How is the graph of $y=\operatorname{arsinh}x$ obtained from $y=\sinh x$ ?: Reflect in the line $y=x$ .
Why is $\operatorname{arcosh}$ the inverse of $\cosh$ only for $x\ge0$ ?: $\cosh$ is not one-to-one, so only the branch $x\ge0$ has an inverse.
First step in proving $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$ ?: Let $y=\operatorname{arsinh}x$ , so $x=\sinh y$ , then form a quadratic in $e^y$ .
Why reject $e^y=x-\sqrt{x^2+1}$ ?: It is negative, but $e^y>0$ .
$\operatorname{arsinh}\frac34$ ?: $\ln2$
Vertical asymptotes of $y=\operatorname{artanh}x$ ?: $x=1$ and $x=-1$ .
Is $\operatorname{arsinh}$ odd or even?: Odd: $\operatorname{arsinh}(-x)=-\operatorname{arsinh}x$ .

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).