Study guides › 9231 FP2 › Hyperbolic functions Maths4U Original · Learn, check, explain
Inverse hyperbolic functions — 9231 FP2
Use logarithmic forms with their real domains.
1. Understand 2. See examples 3. Check answers 4. Explain your working Notes and key methods For real x, \(\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})\). On \(\displaystyle x\ge 1\), \(\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})\) is nonnegative. For |x|<1, \(\displaystyle \operatorname{artanh}x= \frac{1}{2}\ln [\frac{1+ x}{1- x}]\).
These are inverse functions, not reciprocals. Solving an equation involving cosh requires choosing a branch unless its argument is explicitly restricted to be nonnegative.
Check the conditions Arcosh has real domain [1,∞), with range [0,∞). Artanh has real domain (−1,1). Three worked examples Example 1 · Inverse hyperbolic functions Use the real inverse functions, with \(\displaystyle a=6,\ p=\frac{5}{7}\).
Find arsinh(a). Find the nonnegative value arcosh(a). Find artanh(p). Find exp(2 artanh(p)). Show the reasoning (a) \(\ln(a+\sqrt{a^2+1})=2.49177985264491\)
Answer: \(2.49177985264491\).
(b) \(\ln(a+\sqrt{a^2-1})=2.47788873028848\)
Answer: \(2.47788873028848\).
(c) \(\displaystyle \frac{1}{2}\ln [\frac{1+ p}{1- p}]= 0.895879734614027\)
Answer: \(0.895879734614027\).
(d) \(\displaystyle \frac{1+ p}{1- p}= 6\)
Answer: \(6\).
Example 2 · Inverse hyperbolic functions Use the real inverse functions, with \(\displaystyle a=7,\ p=\frac{3}{4}\).
Find arsinh(a). Find the nonnegative value arcosh(a). Find artanh(p). Find exp(2 artanh(p)). Show the reasoning (a) \(\ln(a+\sqrt{a^2+1})=2.64412076105863\)
Answer: \(2.64412076105863\).
(b) \(\ln(a+\sqrt{a^2-1})=2.63391579384963\)
Answer: \(2.63391579384963\).
(c) \(\displaystyle \frac{1}{2}\ln [\frac{1+ p}{1- p}]= 0.972955074527657\)
Answer: \(0.972955074527657\).
(d) \(\displaystyle \frac{1+ p}{1- p}= 7\)
Answer: \(7\).
Example 3 · Inverse hyperbolic functions Use the real inverse functions, with \(\displaystyle a=8,\ p=\frac{7}{9}\).
Find arsinh(a). Find the nonnegative value arcosh(a). Find artanh(p). Find exp(2 artanh(p)). Show the reasoning (a) \(\ln(a+\sqrt{a^2+1})=2.77647228072372\)
Answer: \(2.77647228072372\).
(b) \(\ln(a+\sqrt{a^2-1})=2.76865938331357\)
Answer: \(2.76865938331357\).
(c) \(\displaystyle \frac{1}{2}\ln [\frac{1+ p}{1- p}]= 1.03972077083992\)
Answer: \(1.03972077083992\).
(d) \(\displaystyle \frac{1+ p}{1- p}= 8\)
Answer: \(8\).
Common mistakes Sinh⁻¹x is not \(\displaystyle \frac{1}{\sinh x}\) when inverse notation is intended. Do not drop the negative solution of \(\displaystyle \cosh u=c\) without a branch condition. Practise with answer checking Original exam-style problems Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.
Next skill: Logarithmic form for inverse hyperbolic functions →
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