AS & A Level · Practice questions · Inverse hyperbolic functions

Turn it round, then take logs.

Six original Cambridge-style questions on the inverse hyperbolic functions: domains, the logarithmic forms and their derivations, exact values, equations with two solutions and compositions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Turn it round, solve for \(e^{y}\), take logs.

Domain first: \(\sinh^{-1}\) all \(x\); \(\cosh^{-1}\) needs \(x \ge 1\); \(\tanh^{-1}\) needs \(|x| \lt 1\).
Derive: let \(y = \) the inverse, write \(x = \sinh y\) (or \(\cosh y\), \(\tanh y\)) in exponentials and solve for \(e^{y}\).
Choose the root with a reason: \(e^{y} \gt 0\) for \(\sinh^{-1}\); the principal value \(y \ge 0\) for \(\cosh^{-1}\).
Count solutions: \(\cosh x = k\) (\(k \gt 1\)) gives \(x = \pm\ln\!\big(k + \sqrt{k^{2} - 1}\big)\).
Compositions: let \(y = \) the inverse, then use \(\cosh^{2} y - \sinh^{2} y = 1\) with the right sign.

Recall

[3 marks]

Express each of the following exactly as a single logarithm or a multiple of one: (a) \(\sinh^{-1} 2\) (b) \(\cosh^{-1} 3\) (c) \(\tanh^{-1}\tfrac{1}{2}\).

(a) \(\sinh^{-1} 2 = \ln\!\big(2 + \sqrt{4 + 1}\big) = \ln\!\big(2 + \sqrt{5}\big)\). ✓
(b) \(\cosh^{-1} 3 = \ln\!\big(3 + \sqrt{9 - 1}\big) = \ln\!\big(3 + 2\sqrt{2}\big)\). ✓
(c) \(\tanh^{-1}\tfrac{1}{2} = \tfrac{1}{2}\ln\!\left(\dfrac{3/2}{1/2}\right) = \tfrac{1}{2}\ln 3\). ✓

Calculation

[5 marks]

(a) Starting from the definition of \(\tanh y\) in terms of exponentials, show that \(\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right)\) for \(-1 \lt x \lt 1\).

(b) Hence find the exact value of \(x\) for which \(\tanh^{-1} x = \ln 2\).

(a) Let \(y = \tanh^{-1} x\). Then \(x = \tanh y = \dfrac{e^{y} - e^{-y}}{e^{y} + e^{-y}} = \dfrac{e^{2y} - 1}{e^{2y} + 1}\). ✓
\(x e^{2y} + x = e^{2y} - 1\), so \(e^{2y}(1 - x) = 1 + x\) and \(e^{2y} = \dfrac{1 + x}{1 - x}\) (positive, as \(|x| \lt 1\)). ✓
Taking logs: \(2y = \ln\!\left(\dfrac{1 + x}{1 - x}\right)\), so \(\tanh^{-1} x = \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right)\). ✓
(b) \(\tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right) = \ln 2\) gives \(\dfrac{1 + x}{1 - x} = e^{2\ln 2} = 4\). ✓
\(1 + x = 4 - 4x\), so \(5x = 3\) and \(x = \tfrac{3}{5}\), which satisfies \(|x| \lt 1\). ✓

Calculation

[4 marks]

(a) Given that \(\cosh^{-1} a = \ln 4\), find the exact value of \(a\).

(b) Given that \(\sinh^{-1} b = 2\ln 3\), find the exact value of \(b\).

(a) \(a = \cosh(\ln 4) = \dfrac{e^{\ln 4} + e^{-\ln 4}}{2} = \dfrac{4 + \frac{1}{4}}{2}\). ✓
\(a = \dfrac{17}{8}\). (Check: \(\ln 4 \ge 0\), so it is a valid principal value.) ✓
(b) \(2\ln 3 = \ln 9\), so \(b = \sinh(\ln 9) = \dfrac{9 - \frac{1}{9}}{2}\). ✓
\(b = \dfrac{80/9}{2} = \dfrac{40}{9}\). ✓

Analysis

[5 marks]

(a) Show that \(\cosh(\tanh^{-1} x) = \dfrac{1}{\sqrt{1 - x^{2}}}\) for \(-1 \lt x \lt 1\).

(b) Hence find the exact value of \(\cosh\!\left(\tanh^{-1}\tfrac{1}{2}\right)\), and check your answer using the logarithmic form of \(\tanh^{-1}\tfrac{1}{2}\).

(a) Let \(y = \tanh^{-1} x\), so \(\tanh y = x\). Divide \(\cosh^{2} y - \sinh^{2} y = 1\) by \(\cosh^{2} y\): \(1 - \tanh^{2} y = \operatorname{sech}^{2} y\). ✓
So \(\operatorname{sech}^{2} y = 1 - x^{2}\) and \(\cosh^{2} y = \dfrac{1}{1 - x^{2}}\). Since \(\cosh y \gt 0\), \(\cosh y = \dfrac{1}{\sqrt{1 - x^{2}}}\). ✓
(b) \(\cosh\!\left(\tanh^{-1}\tfrac{1}{2}\right) = \dfrac{1}{\sqrt{1 - \frac{1}{4}}} = \dfrac{1}{\sqrt{3}/2} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\). ✓
Check: \(\tanh^{-1}\tfrac{1}{2} = \tfrac{1}{2}\ln 3 = \ln\sqrt{3}\), so \(\cosh(\ln\sqrt{3}) = \dfrac{\sqrt{3} + \frac{1}{\sqrt{3}}}{2} = \dfrac{4/\sqrt{3}}{2} = \dfrac{2}{\sqrt{3}}\). The answers agree. ✓

Analysis

[5 marks]

Find the exact solutions of the equation \(\cosh 2x - 5\cosh x + 4 = 0\), giving non-zero answers in the form \(\pm\ln\!\left(\dfrac{a + \sqrt{b}}{c}\right)\).

Use \(\cosh 2x = 2\cosh^{2} x - 1\): \(2\cosh^{2} x - 1 - 5\cosh x + 4 = 0\), so \(2\cosh^{2} x - 5\cosh x + 3 = 0\). ✓
Factorise: \((2\cosh x - 3)(\cosh x - 1) = 0\), so \(\cosh x = \tfrac{3}{2}\) or \(\cosh x = 1\). ✓
\(\cosh x = 1\) gives the single solution \(x = 0\) (the minimum of \(\cosh\)). ✓
\(\cosh x = \tfrac{3}{2}\) gives \(x = \pm\cosh^{-1}\tfrac{3}{2} = \pm\ln\!\left(\tfrac{3}{2} + \sqrt{\tfrac{9}{4} - 1}\right) = \pm\ln\!\left(\dfrac{3 + \sqrt{5}}{2}\right)\). ✓
Solutions: \(x = 0\) and \(x = \pm\ln\!\left(\dfrac{3 + \sqrt{5}}{2}\right)\). ✓

Proof

[6 marks]

(a) Using the logarithmic forms, prove that \(\sinh^{-1}\!\left(\dfrac{x}{\sqrt{1 - x^{2}}}\right) = \tanh^{-1} x\) for \(-1 \lt x \lt 1\).

(b) Hence find the exact value of \(x\) for which \(\sinh^{-1}\!\left(\dfrac{x}{\sqrt{1 - x^{2}}}\right) = \ln 3\).

(a) Let \(u = \dfrac{x}{\sqrt{1 - x^{2}}}\). Then \(u^{2} + 1 = \dfrac{x^{2} + 1 - x^{2}}{1 - x^{2}} = \dfrac{1}{1 - x^{2}}\), so \(\sqrt{u^{2} + 1} = \dfrac{1}{\sqrt{1 - x^{2}}}\). ✓
\[\sinh^{-1} u = \ln\!\left(\frac{x}{\sqrt{1 - x^{2}}} + \frac{1}{\sqrt{1 - x^{2}}}\right) = \ln\!\left(\frac{1 + x}{\sqrt{(1 - x)(1 + x)}}\right) = \ln\sqrt{\frac{1 + x}{1 - x}}.\] ✓
\(= \tfrac{1}{2}\ln\!\left(\dfrac{1 + x}{1 - x}\right) = \tanh^{-1} x\), as required. (The step \(\dfrac{1 + x}{\sqrt{1 + x}} = \sqrt{1 + x}\) is valid because \(1 + x \gt 0\).) ✓
(b) By (a), \(\tanh^{-1} x = \ln 3\), so \(\dfrac{1 + x}{1 - x} = e^{2\ln 3} = 9\). ✓
\(1 + x = 9 - 9x\), so \(x = \tfrac{4}{5}\). ✓
Check: \(\dfrac{0.8}{\sqrt{1 - 0.64}} = \dfrac{0.8}{0.6} = \dfrac{4}{3}\), and \(\sinh^{-1}\tfrac{4}{3} = \ln\!\left(\tfrac{4}{3} + \tfrac{5}{3}\right) = \ln 3\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.