Home / A-Level Further Maths 9231 / Hyperbolic functions / Inverse hyperbolic functions
Cambridge International AS & A Level · Topic 2.1
A-Level Further Maths 9231 / Topic 2.1 / AS & A Level

Undo the hyperbolic, and a logarithm appears.

Because \(\sinh x\), \(\cosh x\) and \(\tanh x\) are built from \(e^{x}\), their inverses are built from \(\ln\). Learn the domains and ranges of \(\sinh^{-1}\), \(\cosh^{-1}\) and \(\tanh^{-1}\), derive their logarithmic forms from a quadratic in \(e^{y}\), and use them to give exact answers.

Topic 2.1All hyperbolic functions lessons →
The key idea

\(\sinh^{-1} x = \ln\!\big(x + \sqrt{x^{2} + 1}\big)\) for all real \(x\); \(\cosh^{-1} x = \ln\!\big(x + \sqrt{x^{2} - 1}\big)\) for \(x \ge 1\), taking the principal value \(\cosh^{-1} x \ge 0\); \(\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right)\) for \(|x| \lt 1\). Each comes from writing \(x\) as a hyperbolic function of \(y\) and solving for \(e^{y}\). Because \(\cosh\) is even, \(\cosh x = k\) with \(k \gt 1\) has TWO solutions, \(x = \pm\cosh^{-1} k\).

x y −3 −2 −1 2 3 −2 −1 1 2 3 y = sinh⁻¹ x y = cosh⁻¹ x y = x (1, 0) x ≥ 1, y ≥ 0 all real x x y −2 −1 1 2 −2 −1 1 2 3 y = tanh⁻¹ x x = 1 x = −1 y = x −1 < x < 1
Fig. 1: left, y = sinh⁻¹ x (defined for every x) and y = cosh⁻¹ x (x ≥ 1, principal value y ≥ 0, starting at (1, 0)); right, y = tanh⁻¹ x, defined only for -1 < x < 1, with vertical asymptotes x = ±1. Each graph is the reflection in the dashed line y = x of the hyperbolic function (for cosh, of its right-hand half)
Section 01

Inverses on the right domain.

A function has an inverse only if it is one-one. \(\sinh x\) is increasing for every \(x\), so \(\sinh^{-1}\) is defined for all real \(x\). \(\tanh x\) is increasing too, but its outputs lie strictly between \(-1\) and \(1\), so \(\tanh^{-1} x\) needs \(|x| \lt 1\). \(\cosh x\) is not one-one (\(\cosh(-2) = \cosh 2\)), so we restrict it to \(x \ge 0\); its inverse \(\cosh^{-1} x\) then takes the principal value, the non-negative one. Each graph is the reflection of the original in the line \(y = x\) (Fig. 1).

FunctionDomainRangeGraph
\(\sinh^{-1} x\)all real \(x\)all real \(y\)odd, through the origin, gradient 1 there
\(\cosh^{-1} x\)\(x \ge 1\)\(y \ge 0\)starts at \((1, 0)\) with a vertical tangent
\(\tanh^{-1} x\)\(-1 \lt x \lt 1\)all real \(y\)odd, asymptotes \(x = \pm 1\)

Notation: \(\sinh^{-1} x\) means the inverse function, NOT \(\dfrac{1}{\sinh x}\) (that is \(\operatorname{cosech} x\)). You may also see \(\operatorname{arsinh} x\), \(\operatorname{arcosh} x\), \(\operatorname{artanh} x\); Cambridge papers use \(\sinh^{-1}\), \(\cosh^{-1}\), \(\tanh^{-1}\).

Worked example Where is it defined?
Find the set of values of \(x\) for which each expression is defined: (a) \(\cosh^{-1}(2x + 5)\) (b) \(\tanh^{-1}\!\left(\dfrac{x}{3}\right)\) (c) \(\sinh^{-1}(x^{2} - 7)\).
  1. (a) \(\cosh^{-1}\) needs its input at least 1: \(2x + 5 \ge 1\), so \(x \ge -2\).
  2. (b) \(\tanh^{-1}\) needs its input strictly between \(-1\) and \(1\): \(-1 \lt \dfrac{x}{3} \lt 1\), so \(-3 \lt x \lt 3\).
  3. (c) \(\sinh^{-1}\) accepts every real input, so there is no restriction.
Answer: (a) \(x \ge -2\)   (b) \(-3 \lt x \lt 3\)   (c) all real \(x\)
Worked example Principal values
Find the exact value of (a) \(\sinh(\sinh^{-1} 4)\) (b) \(\sinh^{-1}(\sinh(-2))\) (c) \(\cosh^{-1}(\cosh(-2))\).
  1. (a) \(\sinh\) undoes \(\sinh^{-1}\) for every input: the answer is \(4\).
  2. (b) \(\sinh\) is one-one on all of \(\mathbb{R}\), so \(\sinh^{-1}\) undoes it too: \(-2\).
  3. (c) \(\cosh(-2) = \cosh 2\), and \(\cosh^{-1}\) returns the non-negative value whose \(\cosh\) is \(\cosh 2\).
  4. So \(\cosh^{-1}(\cosh(-2)) = 2\), not \(-2\). In general \(\cosh^{-1}(\cosh x) = |x|\).
Answer: (a) \(4\)   (b) \(-2\)   (c) \(2\)
Your turn
State the set of values of \(x\) for which \(\tanh^{-1}(1 - 2x)\) is defined.
Examiner trap

\(\cosh^{-1} x\) is never negative, and \(\cosh^{-1}(\cosh x) = |x|\), not \(x\). Also, \(\tanh^{-1} x\) needs STRICT inequalities: \(\tanh^{-1} 1\) does not exist, because \(\tanh x\) never reaches 1.

Section 02

Deriving the logarithmic forms.

Write \(y\) for the inverse function, turn it round into a hyperbolic equation, replace the hyperbolic function by exponentials and solve for \(e^{y}\). For \(\sinh\) and \(\cosh\) this gives a quadratic in \(e^{y}\); for \(\tanh\) it is linear. Then choose the root using a REASON, and take logs. These forms come back in sections 2.3 and 2.4, where \(\sinh^{-1}\) and \(\cosh^{-1}\) appear as answers to integrals and the log forms turn them into exact numbers.

The three results (all in the formula list, but you must be able to derive them):

\[\sinh^{-1} x = \ln\!\big(x + \sqrt{x^{2} + 1}\big), \qquad \cosh^{-1} x = \ln\!\big(x + \sqrt{x^{2} - 1}\big)\ (x \ge 1), \qquad \tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\frac{1 + x}{1 - x}\right)\ (|x| \lt 1).\]
Worked example Derive the form for sinh⁻¹
Show that \(\sinh^{-1} x = \ln\!\big(x + \sqrt{x^{2} + 1}\big)\).
  1. Let \(y = \sinh^{-1} x\), so \(x = \sinh y = \dfrac{e^{y} - e^{-y}}{2}\).
  2. Multiply by \(2e^{y}\): \(2xe^{y} = e^{2y} - 1\), so \((e^{y})^{2} - 2x\,e^{y} - 1 = 0\), a quadratic in \(e^{y}\).
  3. \[e^{y} = \frac{2x \pm \sqrt{4x^{2} + 4}}{2} = x \pm \sqrt{x^{2} + 1}.\]
  4. Since \(\sqrt{x^{2} + 1} \gt \sqrt{x^{2}} = |x| \ge x\), the root \(x - \sqrt{x^{2} + 1}\) is negative, but \(e^{y} \gt 0\) for every real \(y\). Reject it.
  5. So \(e^{y} = x + \sqrt{x^{2} + 1}\), and taking logs gives \(y = \ln\!\big(x + \sqrt{x^{2} + 1}\big)\).
Answer: \(\sinh^{-1} x = \ln\!\big(x + \sqrt{x^{2} + 1}\big)\) for all real \(x\)
Worked example Derive the form for cosh⁻¹
Show that, for \(x \ge 1\), \(\cosh^{-1} x = \ln\!\big(x + \sqrt{x^{2} - 1}\big)\).
  1. Let \(y = \cosh^{-1} x\), so \(x = \cosh y = \dfrac{e^{y} + e^{-y}}{2}\) with \(y \ge 0\).
  2. Multiply by \(2e^{y}\): \((e^{y})^{2} - 2x\,e^{y} + 1 = 0\), so \(e^{y} = x \pm \sqrt{x^{2} - 1}\).
  3. This time both roots are positive, and their product is \(\big(x + \sqrt{x^{2} - 1}\big)\big(x - \sqrt{x^{2} - 1}\big) = x^{2} - (x^{2} - 1) = 1\). So their logs are \(\pm\ln\!\big(x + \sqrt{x^{2} - 1}\big)\): the two values of \(y\) with \(\cosh y = x\).
  4. The principal value needs \(y \ge 0\), so \(e^{y} \ge 1\), which is the larger root \(x + \sqrt{x^{2} - 1}\).
Answer: \(\cosh^{-1} x = \ln\!\big(x + \sqrt{x^{2} - 1}\big)\), the non-negative value; the other root gives \(-\cosh^{-1} x\)
Your turn
Show that \(\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right)\) for \(|x| \lt 1\).
Examiner trap

'Show that' questions want the reason for the root you keep. For \(\sinh^{-1}\) the minus root is rejected because it is NEGATIVE and \(e^{y} \gt 0\). For \(\cosh^{-1}\) that reason is false (both roots are positive); the minus root is rejected because it gives \(y \le 0\), and \(\cosh^{-1}\) takes the principal value \(y \ge 0\).

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What is the domain of \(\cosh^{-1} x\)?

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Which is the logarithmic form of \(\sinh^{-1} x\)?

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Deriving \(\cosh^{-1} x\), you reach \(e^{y} = x \pm \sqrt{x^{2} - 1}\). Why is the \(+\) sign taken?

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What is the value of \(\cosh^{-1}(\cosh(-3))\)?

Section 03

Exact values and equations.

The log forms give exact answers. To solve an equation, rearrange to \(\sinh x = k\), \(\cosh x = k\) or \(\tanh x = k\) and count the solutions from the graph: \(\sinh x = k\) has one solution for every \(k\); \(\tanh x = k\) has one if \(|k| \lt 1\) and none otherwise; \(\cosh x = k\) has two if \(k \gt 1\), one (\(x = 0\)) if \(k = 1\) and none if \(k \lt 1\).

Worked example Three ways to ln 2
Show that \(\sinh^{-1}\tfrac{3}{4}\), \(\cosh^{-1}\tfrac{5}{4}\) and \(\tanh^{-1}\tfrac{3}{5}\) are all equal to \(\ln 2\).
  1. \(\sinh^{-1}\tfrac{3}{4} = \ln\!\left(\tfrac{3}{4} + \sqrt{\tfrac{9}{16} + 1}\right) = \ln\!\left(\tfrac{3}{4} + \tfrac{5}{4}\right) = \ln 2\).
  2. \(\cosh^{-1}\tfrac{5}{4} = \ln\!\left(\tfrac{5}{4} + \sqrt{\tfrac{25}{16} - 1}\right) = \ln\!\left(\tfrac{5}{4} + \tfrac{3}{4}\right) = \ln 2\).
  3. \(\tanh^{-1}\tfrac{3}{5} = \tfrac{1}{2}\ln\!\left(\dfrac{1 + \frac{3}{5}}{1 - \frac{3}{5}}\right) = \tfrac{1}{2}\ln 4 = \ln 2\).
  4. Check: \(\sinh(\ln 2) = \dfrac{2 - \frac{1}{2}}{2} = \dfrac{3}{4}\) and \(\cosh(\ln 2) = \dfrac{2 + \frac{1}{2}}{2} = \dfrac{5}{4}\), and \(\tanh(\ln 2) = \dfrac{3/4}{5/4} = \dfrac{3}{5}\).
Answer: Each equals \(\ln 2\)
Worked example One solution or two?
Find the exact solutions of (a) \(\cosh x = 3\) (b) \(\sinh x = 3\), giving answers as logarithms.
  1. (a) \(3 \gt 1\), so there are two solutions, \(x = \pm\cosh^{-1} 3\).
  2. \(\cosh^{-1} 3 = \ln\!\big(3 + \sqrt{9 - 1}\big) = \ln\!\big(3 + 2\sqrt{2}\big)\).
  3. The negative solution can also be written \(-\ln(3 + 2\sqrt{2}) = \ln\!\big(3 - 2\sqrt{2}\big)\), since \((3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 1\).
  4. (b) \(\sinh\) is one-one, so there is exactly one solution: \(x = \sinh^{-1} 3 = \ln\!\big(3 + \sqrt{10}\big)\).
Answer: (a) \(x = \pm\ln\!\big(3 + 2\sqrt{2}\big)\)   (b) \(x = \ln\!\big(3 + \sqrt{10}\big)\)
Your turn
Show that \(\tanh(\sinh^{-1} x) = \dfrac{x}{\sqrt{1 + x^{2}}}\).
Examiner trap

\(\cosh^{-1} k\) gives only the non-negative solution of \(\cosh x = k\). Writing just \(x = \ln(3 + 2\sqrt{2})\) loses a mark: the evenness of \(\cosh\) gives the second solution \(-\ln(3 + 2\sqrt{2})\). And leave answers as exact logs when the question says 'exact'; \(1.763\) scores nothing there.

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Find the exact value of \(\sinh^{-1}\tfrac{3}{4}\).

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Find the exact solutions of \(3\cosh x = 5\).

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Express \(\tanh(\sinh^{-1} x)\) in terms of \(x\).

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Given that \(\tanh^{-1} x = \tfrac{1}{2}\ln 5\), find \(x\).

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Inverse hyperbolic functions turned into exact logs.

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Six original Cambridge-style questions
Exact log forms, the derivation of \(\tanh^{-1}\) with an equation, recovering \(x\) from a log, simplifying \(\cosh(\tanh^{-1} x)\), a \(\cosh 2x\) equation with three roots, and an identity linking \(\tanh^{-1}\) and \(\sinh^{-1}\).