Further Mathematics · AQA 7367 · Topic H

Hyperbolic functions

sinh, cosh and tanh: functions built from eˣ and e⁻ˣ that behave like the trigonometric functions. You graph them, use their identities, inverses and logarithmic forms, and differentiate and integrate them.

7 sub-topics
4 in Year 12, 3 in Year 13
78 practice questions
7 animated lessons

Practice questions (78) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

H1Hyperbolic functions and their graphsY12
H2Differentiate and integrate hyperbolic functionsY13
H3Inverse hyperbolic functions: definitions, domains and rangesY12
H4Derive and use logarithmic forms of the inverse hyperbolic functionsY12
H5Integrals leading to inverse hyperbolic functionsY13
H6Hyperbolic identities, including tanh x = sinh x / cosh xY12
H7Construct proofs involving hyperbolic functions and identitiesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 78 questions in this topic, free, no account needed.

Hyperbolic functions

Not needed for Year 12Easy

Solving a hyperbolic equation · Year 13 · 4 marks

Solve $\cosh x = 2$ , giving your answers as natural logarithms.
(4)

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Replace $\cosh x$ with its definition in exponentials. $\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$ ${}= 2$
Multiply through by $2\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 4\mathrm{e}^x$ ${}+ 1$ ${}= 0$
Solve the quadratic with the formula. $\mathrm{e}^x = 2 \pm \sqrt{3}$
Both values of $\mathrm{e}^x$ are positive, so both give solutions. $x = \ln(2 + \sqrt{3})$ or $x$ ${}= \ln(2 - \sqrt{3})$ ${}= -\ln(2 + \sqrt{3})$

Total4

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Hyperbolic functions

Not needed for Year 12Medium

A hyperbolic equation · Year 13 · 5 marks

Solve $3\cosh x - \sinh x = 3$ .
(5)

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Write both in exponentials. $\tfrac{3}{2}(\mathrm{e}^x + \mathrm{e}^{-x})$ ${}- \tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$ ${}= 3$
Collect like terms and halve. $\mathrm{e}^x$ ${}+ 2\mathrm{e}^{-x}$ ${}= 3$
Multiply by $\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 3\mathrm{e}^x$ ${}+ 2$ ${}= 0$
$(\mathrm{e}^x - 1)(\mathrm{e}^x - 2)$ ${}= 0$
$x = 0$ or $x = \ln 2$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Hyperbolic functions

Not needed for Year 12Hard

An equation with cosh 2x · Year 13 · 5 marks

Solve $\cosh 2x - 3\sinh x = 0$ , giving answers as logarithms.
(5)

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Use $\cosh 2x = 1 + 2\sinh^2 x$ to get one function. $1 + 2\sinh^2 x - 3\sinh x = 0$
Factorise the quadratic in $\sinh x$ . $(2\sinh x - 1)(\sinh x - 1) = 0$
$x$ ${}= \operatorname{arsinh} 1$ ${}= \ln(1 + \sqrt{2})$
$\tfrac{1}{2}$ ${}+ \sqrt{\tfrac{1}{4} + 1}$ ${}= \tfrac{1 + \sqrt{5}}{2}$ . $x$ ${}= \operatorname{arsinh}\tfrac{1}{2}$ ${}= \ln\dfrac{1 + \sqrt{5}}{2}$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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