AS & A Level · Practice questions · Hyperbolic identities

Prove it, then use it.

Six original Cambridge-style questions on hyperbolic identities: proofs from the exponential definitions with model solutions, quadratic equations in \(\sinh x\) and \(\cosh x\) with exact logarithmic answers, and an addition formula put to work.

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

Exponentials to prove, identities to solve.

Prove from \(\cosh x = \tfrac{1}{2}(e^{x} + e^{-x})\), \(\sinh x = \tfrac{1}{2}(e^{x} - e^{-x})\): work on one side, use \(e^{x}e^{-x} = 1\), recognise the result.
Predict with Osborn's rule (change the sign of each product of two sines), but never offer it as a proof.
Solve: pick the form of \(\cosh 2x\) that leaves ONE function, then solve the quadratic.
Convert: \(\sinh x = k \Rightarrow x = \ln\big(k + \sqrt{k^{2} + 1}\big)\); \(\cosh x = k \ge 1 \Rightarrow x = \pm\ln\big(k + \sqrt{k^{2} - 1}\big)\).
Check: reject \(\cosh x \lt 1\) and \(|\tanh x| \ge 1\); give both signs for \(\cosh\).

Recall

[3 marks]

Prove from the definitions of \(\cosh x\) and \(\sinh x\) that \(\cosh 2x = 2\cosh^{2}x - 1\).

\(2\cosh^{2}x - 1 = 2 \times \tfrac{1}{4}(e^{x} + e^{-x})^{2} - 1 = \tfrac{1}{2}(e^{2x} + 2 + e^{-2x}) - 1\). ✓
\(= \tfrac{1}{2}e^{2x} + 1 + \tfrac{1}{2}e^{-2x} - 1 = \tfrac{1}{2}(e^{2x} + e^{-2x})\). ✓
This is the definition of \(\cosh 2x\), so \(\cosh 2x = 2\cosh^{2}x - 1\). ✓

Proof

[4 marks]

(a) Prove from the definitions that \(\cosh^{2}x - \sinh^{2}x = 1\).

(b) Hence show that \(\coth^{2}x - 1 = \operatorname{cosech}^{2}x\), stating the values of \(x\) for which this holds.

(a) \(\cosh^{2}x = \tfrac{1}{4}(e^{2x} + 2 + e^{-2x})\) and \(\sinh^{2}x = \tfrac{1}{4}(e^{2x} - 2 + e^{-2x})\). ✓
Subtracting, \(\cosh^{2}x - \sinh^{2}x = \tfrac{1}{4}(2 + 2) = 1\). ✓
(b) For \(x \ne 0\), \(\sinh x \ne 0\), so divide (a) by \(\sinh^{2}x\): \(\dfrac{\cosh^{2}x}{\sinh^{2}x} - 1 = \dfrac{1}{\sinh^{2}x}\). ✓
That is \(\coth^{2}x - 1 = \operatorname{cosech}^{2}x\), valid for all real \(x \ne 0\) (where \(\coth x\) and \(\operatorname{cosech}x\) are undefined). ✓

Proof

[5 marks]

(a) Use Osborn's rule on \(\cos 3x = 4\cos^{3}x - 3\cos x\) to write down an identity for \(\cosh 3x\).

(b) Prove your identity from the definitions.

(a) No term contains a sine, so nothing changes sign: \(\cosh 3x = 4\cosh^{3}x - 3\cosh x\). ✓
(b) \((e^{x} + e^{-x})^{3} = e^{3x} + 3e^{x} + 3e^{-x} + e^{-3x}\), using \(e^{2x}e^{-x} = e^{x}\) and \(e^{x}e^{-2x} = e^{-x}\). ✓
\(4\cosh^{3}x = \tfrac{4}{8}(e^{x} + e^{-x})^{3} = \tfrac{1}{2}(e^{3x} + e^{-3x}) + \tfrac{3}{2}(e^{x} + e^{-x}) = \cosh 3x + 3\cosh x\). ✓
So \(4\cosh^{3}x - 3\cosh x = \cosh 3x\), as required. ✓

Calculation

[5 marks]

Solve the equation \(3\cosh 2x + 2\sinh x = 11\), giving your answers in exact logarithmic form.

Use \(\cosh 2x = 1 + 2\sinh^{2}x\): \(3 + 6\sinh^{2}x + 2\sinh x = 11\), so \(6\sinh^{2}x + 2\sinh x - 8 = 0\). ✓
Divide by 2: \(3\sinh^{2}x + \sinh x - 4 = 0\), so \((3\sinh x + 4)(\sinh x - 1) = 0\). ✓
\(\sinh x = 1\): \(x = \ln\big(1 + \sqrt{2}\big)\). ✓
\(\sinh x = -\tfrac{4}{3}\): \(x = \ln\Big(-\tfrac{4}{3} + \sqrt{\tfrac{16}{9} + 1}\Big) = \ln\big(-\tfrac{4}{3} + \tfrac{5}{3}\big) = \ln\tfrac{1}{3} = -\ln 3\). ✓
Both are valid, since \(\sinh x\) takes every real value: \(x = \ln\big(1 + \sqrt{2}\big)\) or \(x = -\ln 3\). ✓

Analysis

[5 marks]

Solve the equation \(\cosh 2x - 3\cosh x = 1\), giving your answers in exact logarithmic form. Explain why one root of your quadratic gives no solution.

Use \(\cosh 2x = 2\cosh^{2}x - 1\): \(2\cosh^{2}x - 1 - 3\cosh x = 1\), so \(2\cosh^{2}x - 3\cosh x - 2 = 0\). ✓
\((2\cosh x + 1)(\cosh x - 2) = 0\), so \(\cosh x = 2\) or \(\cosh x = -\tfrac{1}{2}\). ✓
\(\cosh x \ge 1\) for every real \(x\), so \(\cosh x = -\tfrac{1}{2}\) has no solution. ✓
\(\cosh x = 2\): \(e^{x} + e^{-x} = 4 \Rightarrow e^{2x} - 4e^{x} + 1 = 0 \Rightarrow e^{x} = 2 \pm \sqrt{3}\). ✓
Both values are positive and \((2 + \sqrt{3})(2 - \sqrt{3}) = 1\), so \(x = \pm\ln\big(2 + \sqrt{3}\big)\). ✓

Proof

[7 marks]

(a) Using \(\sinh(A + B) = \sinh A\cosh B + \cosh A\sinh B\) and \(\cosh(A + B) = \cosh A\cosh B + \sinh A\sinh B\), prove that \[\tanh(A + B) = \dfrac{\tanh A + \tanh B}{1 + \tanh A\tanh B}.\]

(b) Show that if \(\tanh x = k\), where \(-1 \lt k \lt 1\), then \(x = \tfrac{1}{2}\ln\dfrac{1 + k}{1 - k}\).

(c) Given that \(\tanh A = \tfrac{1}{2}\) and \(\tanh B = \tfrac{1}{3}\), find \(\tanh(A + B)\) and hence find the exact value of \(A + B\).

(a) \(\tanh(A + B) = \dfrac{\sinh A\cosh B + \cosh A\sinh B}{\cosh A\cosh B + \sinh A\sinh B}\). ✓
Divide numerator and denominator by \(\cosh A\cosh B\) (never zero): \(\dfrac{\tanh A + \tanh B}{1 + \tanh A\tanh B}\), as required. ✓
(b) \(\tanh x = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = \dfrac{e^{2x} - 1}{e^{2x} + 1} = k\), so \(e^{2x} - 1 = ke^{2x} + k\) and \(e^{2x}(1 - k) = 1 + k\). ✓
\(e^{2x} = \dfrac{1 + k}{1 - k}\), which is positive for \(-1 \lt k \lt 1\), so \(x = \tfrac{1}{2}\ln\dfrac{1 + k}{1 - k}\). ✓
(c) \(\tanh(A + B) = \dfrac{\frac{1}{2} + \frac{1}{3}}{1 + \frac{1}{6}} = \dfrac{\frac{5}{6}}{\frac{7}{6}} = \dfrac{5}{7}\). ✓
By (b), \(A + B = \tfrac{1}{2}\ln\dfrac{1 + \frac{5}{7}}{1 - \frac{5}{7}} = \tfrac{1}{2}\ln\dfrac{12}{2} = \tfrac{1}{2}\ln 6\). ✓
Check: \(A = \tfrac{1}{2}\ln 3\) and \(B = \tfrac{1}{2}\ln 2\), whose sum is \(\tfrac{1}{2}\ln 6\). ✓

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.