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

Trig identities with a sign change.

Every trigonometric identity has a hyperbolic twin, and most differ by a single sign. Prove them from the exponential definitions, remember them with Osborn's rule, and use them to turn equations in \(\cosh 2x\) and \(\sinh x\) into quadratics with exact logarithmic answers.

Topic 2.1All hyperbolic functions lessons →
The key idea

From \(\cosh x = \tfrac{1}{2}(e^{x} + e^{-x})\) and \(\sinh x = \tfrac{1}{2}(e^{x} - e^{-x})\): \(\cosh^{2}x - \sinh^{2}x = 1\), \(\sinh 2x = 2\sinh x\cosh x\), \(\cosh 2x = \cosh^{2}x + \sinh^{2}x = 2\cosh^{2}x - 1 = 1 + 2\sinh^{2}x\). Osborn's rule turns a trig identity into its hyperbolic twin: replace each trig function by its hyperbolic one and change the sign of every product (or implied product) of two sines. It is a memory aid only: a proof must use the definitions.

x y P(cosh t, sinh t) cosh t sinh t 1 x² − y² = 1 cosh²t − sinh²t = 1 x² + y² = 1 cos²t + sin²t = 1 y = ±x asymptotes cosh t ≥ 1, so P is always on the right branch
Fig. 1: the point (cosh t, sinh t) lies on the hyperbola x² - y² = 1, just as (cos t, sin t) lies on the circle x² + y² = 1. One minus sign separates the two identities, and because cosh t is never less than 1 the point never reaches the left branch
Section 01

Proofs from the definitions.

Every hyperbolic identity can be proved by writing each function in exponentials, simplifying, and recognising the result. Square or multiply out the brackets, collect the powers of \(e\), and look for \(\tfrac{1}{2}(e^{u} + e^{-u})\) or \(\tfrac{1}{2}(e^{u} - e^{-u})\). The key fact is \(e^{x} \times e^{-x} = 1\).

IdentityHow it is proved
\(\cosh^{2}x - \sinh^{2}x = 1\)expand both squares in exponentials and subtract
\(\operatorname{sech}^{2}x = 1 - \tanh^{2}x\)divide \(\cosh^{2}x - \sinh^{2}x = 1\) by \(\cosh^{2}x\)
\(\coth^{2}x - 1 = \operatorname{cosech}^{2}x\)divide \(\cosh^{2}x - \sinh^{2}x = 1\) by \(\sinh^{2}x\) (\(x \ne 0\))
\(\sinh 2x = 2\sinh x\cosh x\)multiply the two definitions
\(\cosh 2x = \cosh^{2}x + \sinh^{2}x\)add the two squares
\(\cosh 2x = 2\cosh^{2}x - 1 = 1 + 2\sinh^{2}x\)replace \(\sinh^{2}x\) or \(\cosh^{2}x\) using \(\cosh^{2}x - \sinh^{2}x = 1\)
\(\sinh(A + B) = \sinh A\cosh B + \cosh A\sinh B\)expand the right side in exponentials
\(\cosh(A + B) = \cosh A\cosh B + \sinh A\sinh B\)expand the right side in exponentials

Work on one side only, usually the more complicated one, and finish with a sentence that names the identity. Replacing \(B\) by \(-B\) (with \(\cosh(-B) = \cosh B\) and \(\sinh(-B) = -\sinh B\)) gives the subtraction formulae.

Worked example The fundamental identity
Prove from the definitions that \(\cosh^{2}x - \sinh^{2}x = 1\), and deduce that \(\operatorname{sech}^{2}x = 1 - \tanh^{2}x\).
  1. \(\cosh^{2}x = \tfrac{1}{4}(e^{x} + e^{-x})^{2} = \tfrac{1}{4}(e^{2x} + 2 + e^{-2x})\), since \(e^{x}e^{-x} = 1\).
  2. \(\sinh^{2}x = \tfrac{1}{4}(e^{x} - e^{-x})^{2} = \tfrac{1}{4}(e^{2x} - 2 + e^{-2x})\).
  3. Subtract: \(\cosh^{2}x - \sinh^{2}x = \tfrac{1}{4}\big(2 - (-2)\big) = \tfrac{4}{4} = 1\).
  4. \(\cosh x \ge 1\), so \(\cosh x \ne 0\) and we may divide by \(\cosh^{2}x\): \(1 - \dfrac{\sinh^{2}x}{\cosh^{2}x} = \dfrac{1}{\cosh^{2}x}\), that is \(1 - \tanh^{2}x = \operatorname{sech}^{2}x\).
Answer: \(\cosh^{2}x - \sinh^{2}x = 1\) and \(\operatorname{sech}^{2}x = 1 - \tanh^{2}x\), as required
Worked example An addition formula
Prove from the definitions that \(\sinh(A + B) = \sinh A\cosh B + \cosh A\sinh B\).
  1. Start with the right side: \[\sinh A\cosh B + \cosh A\sinh B = \tfrac{1}{4}\Big[(e^{A} - e^{-A})(e^{B} + e^{-B}) + (e^{A} + e^{-A})(e^{B} - e^{-B})\Big].\]
  2. First product: \(e^{A+B} + e^{A-B} - e^{-A+B} - e^{-A-B}\).
  3. Second product: \(e^{A+B} - e^{A-B} + e^{-A+B} - e^{-A-B}\).
  4. Add: the middle terms cancel, leaving \(\tfrac{1}{4}\big(2e^{A+B} - 2e^{-(A+B)}\big) = \tfrac{1}{2}\big(e^{A+B} - e^{-(A+B)}\big)\).
  5. This is the definition of \(\sinh(A + B)\).
Answer: \(\sinh(A + B) = \sinh A\cosh B + \cosh A\sinh B\), as required
Your turn
Prove from the definitions that \(\sinh 2x = 2\sinh x\cosh x\).
Examiner trap

Do not start from the identity you are proving and 'show both sides are equal' by working on both at once: that assumes the result. Begin with one side, transform it, and arrive at the other. And write the square properly: \((e^{x} + e^{-x})^{2} = e^{2x} + 2 + e^{-2x}\), not \(e^{2x} + e^{-2x}\).

Section 02

Osborn's rule: a memory aid.

You know dozens of trig identities already. Osborn's rule converts any of them to the hyperbolic version: replace each trig function by the matching hyperbolic function, then change the sign of every term that contains a product of two sines. Remember the hidden sines: \(\tan^{2}x = \dfrac{\sin^{2}x}{\cos^{2}x}\), \(\cot^{2}x\), \(\operatorname{cosec}^{2}x\) and \(\tan A\tan B\) all contain a product of two sines.

Trigonometric identityHyperbolic identity
\(\cos^{2}x + \sin^{2}x = 1\)\(\cosh^{2}x - \sinh^{2}x = 1\)
\(1 + \tan^{2}x = \sec^{2}x\)\(1 - \tanh^{2}x = \operatorname{sech}^{2}x\)
\(\cos 2x = 1 - 2\sin^{2}x\)\(\cosh 2x = 1 + 2\sinh^{2}x\)
\(\sin 2x = 2\sin x\cos x\)\(\sinh 2x = 2\sinh x\cosh x\) (only one sine: no change)
\(\cos(A + B) = \cos A\cos B - \sin A\sin B\)\(\cosh(A + B) = \cosh A\cosh B + \sinh A\sinh B\)
\(\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\)\(\tanh(A + B) = \dfrac{\tanh A + \tanh B}{1 + \tanh A\tanh B}\)

Osborn's rule is NOT a proof. If a question says 'prove' or 'show', work from the exponential definitions (or from identities you have already proved that way) unless it tells you otherwise.

Worked example Predict, with the hidden sines
Use Osborn's rule to write down the hyperbolic versions of (a) \(\cos(A - B) = \cos A\cos B + \sin A\sin B\) (b) \(1 + \cot^{2}x = \operatorname{cosec}^{2}x\).
  1. (a) \(\sin A\sin B\) is a product of two sines, so its sign changes: \(\cosh(A - B) = \cosh A\cosh B - \sinh A\sinh B\).
  2. (b) \(\cot^{2}x = \dfrac{\cos^{2}x}{\sin^{2}x}\) and \(\operatorname{cosec}^{2}x = \dfrac{1}{\sin^{2}x}\) both contain \(\sin^{2}x\), so both signs change: \(1 - \coth^{2}x = -\operatorname{cosech}^{2}x\).
  3. Tidy (b) by multiplying by \(-1\).
Answer: (a) \(\cosh(A - B) = \cosh A\cosh B - \sinh A\sinh B\)   (b) \(\coth^{2}x - 1 = \operatorname{cosech}^{2}x\)
Worked example Predict, then prove
Use Osborn's rule on \(\sin 3x = 3\sin x - 4\sin^{3}x\) to suggest an identity for \(\sinh 3x\), then prove it from the definitions.
  1. \(\sin^{3}x = \sin^{2}x \times \sin x\) contains a product of two sines, so that term changes sign: \(\sinh 3x = 3\sinh x + 4\sinh^{3}x\).
  2. Proof: \((e^{x} - e^{-x})^{3} = e^{3x} - 3e^{x} + 3e^{-x} - e^{-3x}\) (binomial expansion, with \(e^{2x}e^{-x} = e^{x}\)).
  3. So \(4\sinh^{3}x = \tfrac{4}{8}(e^{3x} - e^{-3x}) - \tfrac{4}{8} \times 3(e^{x} - e^{-x}) = \sinh 3x - 3\sinh x\).
  4. Rearrange: \(\sinh 3x = 3\sinh x + 4\sinh^{3}x\).
Answer: \(\sinh 3x = 3\sinh x + 4\sinh^{3}x\), proved from the definitions
Your turn
Use Osborn's rule on \(\tan 2x = \dfrac{2\tan x}{1 - \tan^{2}x}\) to write down a formula for \(\tanh 2x\).
Examiner trap

The commonest slip is missing a hidden product of sines. \(\tan^{2}x\) and \(\tan A\tan B\) each contain two sines, so \(1 + \tan^{2}x = \sec^{2}x\) becomes \(1 - \tanh^{2}x = \operatorname{sech}^{2}x\), not \(1 + \tanh^{2}x = \operatorname{sech}^{2}x\). And an answer that only quotes Osborn's rule scores nothing on a 'prove' question.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Which of these identities is correct for all \(x\)?

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Which expression is equal to \(\cosh 2x\)?

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By Osborn's rule, \(1 + \tan^{2}x = \sec^{2}x\) becomes

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Solve \(\sinh x = 2\) exactly.

Section 03

Solving equations with exact answers.

An equation that mixes \(\cosh 2x\), \(\cosh^{2}x\) and \(\sinh x\) (or \(\cosh x\)) becomes a quadratic once every term is in ONE function. Choose the form of \(\cosh 2x\) that matches: \(1 + 2\sinh^{2}x\) next to \(\sinh x\), \(2\cosh^{2}x - 1\) next to \(\cosh x\). Solve the quadratic, then turn each value into \(x\) as an exact logarithm.

Solving \(\sinh x = k\): \(e^{x} - e^{-x} = 2k\), so \(e^{2x} - 2ke^{x} - 1 = 0\) and \(e^{x} = k \pm \sqrt{k^{2} + 1}\). The minus sign gives a negative value, impossible for \(e^{x}\). For example \(\sinh x = 2\) gives \(x = \ln(2 + \sqrt{5})\). The same method gives:

EquationSolutionsCheck
\(\sinh x = k\)\(x = \ln\big(k + \sqrt{k^{2} + 1}\big)\), exactly oneany real \(k\)
\(\cosh x = k\)\(x = \pm\ln\big(k + \sqrt{k^{2} - 1}\big)\)needs \(k \ge 1\); none if \(k \lt 1\)
\(\tanh x = k\)\(x = \tfrac{1}{2}\ln\dfrac{1 + k}{1 - k}\)needs \(-1 \lt k \lt 1\)

The next lesson names these as the inverse functions \(\operatorname{arsinh}\), \(\operatorname{arcosh}\) and \(\operatorname{artanh}\) and proves the logarithmic forms. For now, derive each one or use the table, and always check: \(\cosh x \ge 1\) for every \(x\), so a root such as \(\cosh x = -2\) must be rejected, and \(\cosh x = k \gt 1\) gives TWO values of \(x\), one positive and one negative.

Worked example A quadratic in sinh x
Solve \(2\cosh 2x - \sinh x = 5\), giving your answers as exact logarithms.
  1. \(\sinh x\) appears, so use \(\cosh 2x = 1 + 2\sinh^{2}x\): \(2 + 4\sinh^{2}x - \sinh x = 5\).
  2. \(4\sinh^{2}x - \sinh x - 3 = 0\), so \((4\sinh x + 3)(\sinh x - 1) = 0\).
  3. \(\sinh x = 1\): \(x = \ln\big(1 + \sqrt{2}\big)\).
  4. \(\sinh x = -\tfrac{3}{4}\): \(x = \ln\Big(-\tfrac{3}{4} + \sqrt{\tfrac{9}{16} + 1}\Big) = \ln\big(-\tfrac{3}{4} + \tfrac{5}{4}\big) = \ln\tfrac{1}{2} = -\ln 2\).
  5. \(\sinh x\) can take any real value, so both roots give a solution.
Answer: \(x = \ln\big(1 + \sqrt{2}\big)\) or \(x = -\ln 2\)
Worked example A quadratic in cosh x, with a root to reject
Solve \(\cosh 2x + \cosh x = 5\), giving your answers as exact logarithms.
  1. \(\cosh x\) appears, so use \(\cosh 2x = 2\cosh^{2}x - 1\): \(2\cosh^{2}x + \cosh x - 6 = 0\).
  2. \((2\cosh x - 3)(\cosh x + 2) = 0\), so \(\cosh x = \tfrac{3}{2}\) or \(\cosh x = -2\).
  3. \(\cosh x \ge 1\) for all \(x\), so reject \(\cosh x = -2\).
  4. \(\cosh x = \tfrac{3}{2}\): \(x = \pm\ln\Big(\tfrac{3}{2} + \sqrt{\tfrac{9}{4} - 1}\Big) = \pm\ln\dfrac{3 + \sqrt{5}}{2}\).
Answer: \(x = \pm\ln\dfrac{3 + \sqrt{5}}{2}\)
Your turn
Solve \(\cosh^{2}x - 3\sinh x = 3\), giving your answers to 3 significant figures.
Examiner trap

Using the trig form \(\cos 2x = 1 - 2\sin^{2}x\) by habit turns \(\cosh 2x\) into \(1 - 2\sinh^{2}x\) and wrecks the quadratic: in hyperbolic work it is \(1 + 2\sinh^{2}x\). Then check every root: reject \(\cosh x \lt 1\), give both \(\pm\) values for \(\cosh x \gt 1\), but only one value for each \(\sinh x\).

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Solve \(\cosh 2x - 3\sinh x = 3\), giving exact answers.

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Solve \(\cosh 2x - 3\cosh x + 2 = 0\).

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Given that \(\sinh x = \tfrac{3}{4}\), find the exact values of \(\cosh x\) and \(\tanh x\).

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Which of these is a correct hyperbolic identity?

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Go deeper · practice
Six original Cambridge-style questions
Model proofs of \(\cosh 2x\), \(\coth^{2}x - 1 = \operatorname{cosech}^{2}x\) and \(\cosh 3x\) from the definitions, two equations solved with exact logarithms, and the \(\tanh(A + B)\) formula used to add two artanh values.