AS & A Level · Practice questions · Hyperbolic functions and their graphs

Write it in e^x, then solve.

Six original Cambridge-style questions on the definitions and graphs of the hyperbolic functions and on solving equations through a quadratic in \(e^{x}\), each with a full model solution.

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

From definitions to answers.

Definitions: \(\sinh x = \tfrac{1}{2}(e^{x} - e^{-x})\), \(\cosh x = \tfrac{1}{2}(e^{x} + e^{-x})\), \(\tanh x = \dfrac{\sinh x}{\cosh x}\); the reciprocals are sech, cosech and coth.
Exact values: \(e^{\ln a} = a\) and \(e^{-\ln a} = \tfrac{1}{a}\).
Graphs: \(\cosh x \ge 1\), \(-1 \lt \tanh x \lt 1\); cosech and coth have the asymptote \(x = 0\); tanh and coth share \(y = \pm 1\).
Equations: write in exponentials, multiply by \(e^{x}\), solve the quadratic in \(u = e^{x}\), reject \(u \le 0\), then \(x = \ln u\).
Check the value is in the range first, and look for two answers when the function is even.

Recall

[3 marks]

Find the exact values of \(\sinh(\ln 4)\), \(\cosh(\ln 4)\), \(\tanh(\ln 4)\) and \(\operatorname{sech}(\ln 4)\).

\(e^{\ln 4} = 4\) and \(e^{-\ln 4} = \tfrac{1}{4}\). ✓
\(\sinh(\ln 4) = \tfrac{1}{2}\left(4 - \tfrac{1}{4}\right) = \tfrac{15}{8}\) and \(\cosh(\ln 4) = \tfrac{1}{2}\left(4 + \tfrac{1}{4}\right) = \tfrac{17}{8}\). ✓
\(\tanh(\ln 4) = \dfrac{15/8}{17/8} = \dfrac{15}{17}\) and \(\operatorname{sech}(\ln 4) = \dfrac{1}{\cosh(\ln 4)} = \dfrac{8}{17}\). ✓

Proof

[4 marks]

(a) Show that \(\cosh x + \sinh x = e^{x}\).

(b) Hence show that \((\cosh x + \sinh x)^{n} = \cosh nx + \sinh nx\) for any positive integer \(n\), and write \((\cosh x + \sinh x)^{3}\) in terms of \(\cosh 3x\) and \(\sinh 3x\).

(a) \(\cosh x + \sinh x = \dfrac{e^{x} + e^{-x}}{2} + \dfrac{e^{x} - e^{-x}}{2} = \dfrac{2e^{x}}{2} = e^{x}\). ✓
(b) \((\cosh x + \sinh x)^{n} = (e^{x})^{n} = e^{nx}\). ✓
Applying part (a) with \(nx\) in place of \(x\): \(e^{nx} = \cosh nx + \sinh nx\). ✓
So \((\cosh x + \sinh x)^{n} = \cosh nx + \sinh nx\); in particular \((\cosh x + \sinh x)^{3} = \cosh 3x + \sinh 3x\). ✓

Analysis

[6 marks]

(a) Show that \(\coth x = \dfrac{e^{2x} + 1}{e^{2x} - 1}\) for \(x \ne 0\).

(b) Sketch the graph of \(y = \coth x\), stating the equations of its asymptotes and its range.

(a) \(\coth x = \dfrac{\cosh x}{\sinh x} = \dfrac{e^{x} + e^{-x}}{e^{x} - e^{-x}}\); multiply top and bottom by \(e^{x}\) to get \(\dfrac{e^{2x} + 1}{e^{2x} - 1}\). ✓
(b) Two branches, as in Fig. 2: for \(x \gt 0\) the curve comes down from \(+\infty\) near the \(y\)-axis towards \(y = 1\) from above; for \(x \lt 0\) it is the rotation of this branch about the origin (coth is odd), rising from \(-\infty\) to \(y = -1\) from below. ✓
Asymptotes: \(x = 0\), \(y = 1\) and \(y = -1\). Range: \(y \lt -1\) or \(y \gt 1\). ✓
(c) \(\dfrac{e^{2x} + 1}{e^{2x} - 1} = 3\), so \(e^{2x} + 1 = 3e^{2x} - 3\) and \(2e^{2x} = 4\). ✓
\(e^{2x} = 2\), so \(x = \tfrac{1}{2}\ln 2\). (Check: \(\tanh(\tfrac{1}{2}\ln 2) = \tfrac{2 - 1}{2 + 1} = \tfrac{1}{3}\), so \(\coth = 3\).) ✓

Calculation

[5 marks]

Solve \(4\cosh x + \sinh x = 4\), giving your answers in exact form.

\(4 \times \dfrac{e^{x} + e^{-x}}{2} + \dfrac{e^{x} - e^{-x}}{2} = 4\). Multiply by \(2\): \(4e^{x} + 4e^{-x} + e^{x} - e^{-x} = 8\), so \(5e^{x} + 3e^{-x} = 8\). ✓
Multiply by \(u = e^{x}\): \(5u^{2} - 8u + 3 = 0\), so \((5u - 3)(u - 1) = 0\). ✓
\(u = 1\) or \(u = \tfrac{3}{5}\); both are positive, so both are valid. ✓
\(x = \ln 1 = 0\) or \(x = \ln\tfrac{3}{5}\) (which is negative, about \(-0.511\)). ✓

Calculation

[5 marks]

Solve \(2\sinh x + \cosh x = 2\), giving your answer as an exact logarithm.

\(2 \times \dfrac{e^{x} - e^{-x}}{2} + \dfrac{e^{x} + e^{-x}}{2} = 2\). Multiply by \(2\): \(2e^{x} - 2e^{-x} + e^{x} + e^{-x} = 4\), so \(3e^{x} - e^{-x} = 4\). ✓
Multiply by \(u = e^{x}\): \(3u^{2} - 4u - 1 = 0\). ✓
\(u = \dfrac{4 \pm \sqrt{16 + 12}}{6} = \dfrac{4 \pm 2\sqrt{7}}{6} = \dfrac{2 \pm \sqrt{7}}{3}\). ✓
\(\dfrac{2 - \sqrt{7}}{3} \lt 0\) because \(\sqrt{7} \gt 2\), so reject it. ✓
\(x = \ln\left(\dfrac{2 + \sqrt{7}}{3}\right)\), about \(0.437\). ✓

Proof

[5 marks]

A point \(P\) has coordinates \((\cosh t, \sinh t)\), where \(t\) is a real parameter.

(a) Show that \(P\) lies on the curve \(x^{2} - y^{2} = 1\).

(b) Explain why \(P\) can only lie on the branch of this curve with \(x \ge 1\).

(a) \(\cosh^{2} t = \dfrac{e^{2t} + 2 + e^{-2t}}{4}\) and \(\sinh^{2} t = \dfrac{e^{2t} - 2 + e^{-2t}}{4}\). ✓
So \(x^{2} - y^{2} = \cosh^{2} t - \sinh^{2} t = \dfrac{2 + 2}{4} = 1\), and \(P\) lies on \(x^{2} - y^{2} = 1\). ✓
(b) \(x = \cosh t = 1 + \dfrac{\left(e^{t/2} - e^{-t/2}\right)^{2}}{2} \ge 1\) for every \(t\), so \(P\) is never on the branch with \(x \le -1\). ✓
(c) Adding the coordinates: \(\cosh t + \sinh t = e^{t}\), so \(e^{t} = \tfrac{5}{4} + \tfrac{3}{4} = 2\) and \(t = \ln 2\). ✓
Check: \(\cosh(\ln 2) = \tfrac{5}{4}\) and \(\sinh(\ln 2) = \tfrac{3}{4}\), as required. ✓

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.