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Hyperbolic functions — chapter review — 9231 FP2

Use exponential definitions and identities consistently. Use logarithmic forms with their real domains.

Notes and key methods

Definitions are \(\sinh x=(e^x-e^{-x})/2\), \(\cosh x=(e^x+e^{-x})/2\), \(\tanh x=\sinh x/\cosh x\), and \(\operatorname{sech}x=1/\cosh x\). They imply cosh²x−sinh²x=1.

Unlike cos x, cosh x is at least one for real x. Sinh is odd and strictly increasing; cosh is even. Work from exponentials when proving a hyperbolic identity rather than borrowing an unchecked circular-trigonometric sign.

For real x, \(\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})\). On x≥1, \(\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1})\) is nonnegative. For |x|<1, \(\operatorname{artanh}x=\frac12\ln[(1+x)/(1-x)]\).

These are inverse functions, not reciprocals. Solving an equation involving cosh requires choosing a branch unless its argument is explicitly restricted to be nonnegative.

Check the conditions

Three worked examples

Example 1 · Hyperbolic functions

Let \(u=\ln 4\). Use exact exponential definitions.

Diagram for this worked example
Use the labelled scales alongside the numerical method.
  1. Find sinh u.
  2. Find cosh u.
  3. Find tanh u.
  4. Find sech u.
Show the reasoning

(a)

\((4-1/4)/2=\frac{15}{8}\)

Answer: \(\frac{15}{8}\).

(b)

\((4+1/4)/2=\frac{17}{8}\)

Answer: \(\frac{17}{8}\).

(c)

\(\sinh u/\cosh u=\frac{15}{17}\)

Answer: \(\frac{15}{17}\).

(d)

\(1/\cosh u=\frac{8}{17}\)

Answer: \(\frac{8}{17}\).

Example 2 · Inverse hyperbolic functions

Use the real inverse functions, with \(a=5,\ p=\frac{2}{3}\).

  1. Find arsinh(a).
  2. Find the nonnegative value arcosh(a).
  3. Find artanh(p).
  4. Find exp(2 artanh(p)).
Show the reasoning

(a)

\(\ln(a+\sqrt{a^2+1})=2.31243834127275\)

Answer: \(2.31243834127275\).

(b)

\(\ln(a+\sqrt{a^2-1})=2.29243166956118\)

Answer: \(2.29243166956118\).

(c)

\(\frac12\ln[(1+p)/(1-p)]=0.80471895621705\)

Answer: \(0.80471895621705\).

(d)

\((1+p)/(1-p)=5\)

Answer: \(5\).

Example 3 · Hyperbolic functions

Let \(u=\ln 6\). Use exact exponential definitions.

Diagram for this worked example
Use the labelled scales alongside the numerical method.
  1. Find sinh u.
  2. Find cosh u.
  3. Find tanh u.
  4. Find sech u.
Show the reasoning

(a)

\((6-1/6)/2=\frac{35}{12}\)

Answer: \(\frac{35}{12}\).

(b)

\((6+1/6)/2=\frac{37}{12}\)

Answer: \(\frac{37}{12}\).

(c)

\(\sinh u/\cosh u=\frac{35}{37}\)

Answer: \(\frac{35}{37}\).

(d)

\(1/\cosh u=\frac{12}{37}\)

Answer: \(\frac{12}{37}\).

Common mistakes

Practise with answer checking

Original exam-style problems

Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.

Examination practice · 2020–2026

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