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

Six new functions, every one built from ex.

The hyperbolic functions look like the trigonometric functions on paper, but they are made from \(e^{x}\) and \(e^{-x}\). Learn the three definitions, the three reciprocals, the shape of each graph, and how to turn an equation such as \(3\sinh x - \cosh x = 1\) into a quadratic in \(e^{x}\).

Topic 2.1All hyperbolic functions lessons →
The key idea

\(\sinh x = \dfrac{e^{x} - e^{-x}}{2}\), \(\cosh x = \dfrac{e^{x} + e^{-x}}{2}\), \(\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\), and \(\operatorname{sech} x = \dfrac{1}{\cosh x}\), \(\operatorname{cosech} x = \dfrac{1}{\sinh x}\), \(\coth x = \dfrac{1}{\tanh x}\). \(\sinh\) and \(\tanh\) are odd, \(\cosh\) is even; \(\cosh x \ge 1\) and \(-1 \lt \tanh x \lt 1\). To solve an equation, replace each function by its exponential form, multiply by \(e^{x}\), and solve the quadratic in \(u = e^{x}\), keeping only \(u \gt 0\).

x y −3 −2 −1 1 2 3 −4 −3 −2 −1 1 2 3 4 y = sinh x y = cosh x y = tanh x y = 1 y = −1 (0, 1)
Fig. 1: y = sinh x, y = cosh x and y = tanh x. cosh has its minimum at (0, 1) and sinh always lies below it; sinh and tanh pass through the origin; tanh stays between the dashed asymptotes y = 1 and y = −1
Section 01

Built from exponentials.

Each hyperbolic function is a simple combination of \(e^{x}\) and \(e^{-x}\). \(\cosh x\) is the average of the two, \(\sinh x\) is half their difference, and \(\tanh x\) is the ratio \(\sinh x \div \cosh x\). The three reciprocals, \(\operatorname{sech}\), \(\operatorname{cosech}\) and \(\coth\), follow the same pattern as \(\sec\), \(\operatorname{cosec}\) and \(\cot\).

FunctionDefinitionValue at \(x = 0\)Odd or even
\(\sinh x\)\(\dfrac{e^{x} - e^{-x}}{2}\)\(0\)odd: \(\sinh(-x) = -\sinh x\)
\(\cosh x\)\(\dfrac{e^{x} + e^{-x}}{2}\)\(1\)even: \(\cosh(-x) = \cosh x\)
\(\tanh x\)\(\dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\)\(0\)odd
\(\operatorname{sech} x\)\(\dfrac{2}{e^{x} + e^{-x}}\)\(1\)even
\(\operatorname{cosech} x\)\(\dfrac{2}{e^{x} - e^{-x}}\), \(x \ne 0\)undefinedodd
\(\coth x\)\(\dfrac{e^{x} + e^{-x}}{e^{x} - e^{-x}}\), \(x \ne 0\)undefinedodd

Exact values come from \(e^{\ln a} = a\) and \(e^{-\ln a} = \dfrac{1}{a}\). So \(\cosh(\ln a) = \dfrac{1}{2}\left(a + \dfrac{1}{a}\right)\) and \(\sinh(\ln a) = \dfrac{1}{2}\left(a - \dfrac{1}{a}\right)\).

Why 'hyperbolic'? Squaring the definitions,

\[\cosh^{2} t - \sinh^{2} t = \frac{e^{2t} + 2 + e^{-2t}}{4} - \frac{e^{2t} - 2 + e^{-2t}}{4} = 1.\]

So the point \((\cosh t, \sinh t)\) lies on the hyperbola \(x^{2} - y^{2} = 1\), just as \((\cos t, \sin t)\) lies on the circle \(x^{2} + y^{2} = 1\). The next lesson builds a whole family of identities from this one.

Worked example Exact values at ln 3
Find the exact values of \(\sinh(\ln 3)\), \(\cosh(\ln 3)\) and \(\tanh(\ln 3)\).
  1. \(e^{\ln 3} = 3\) and \(e^{-\ln 3} = \dfrac{1}{3}\).
  2. \(\sinh(\ln 3) = \dfrac{1}{2}\left(3 - \dfrac{1}{3}\right) = \dfrac{1}{2} \times \dfrac{8}{3} = \dfrac{4}{3}\).
  3. \(\cosh(\ln 3) = \dfrac{1}{2}\left(3 + \dfrac{1}{3}\right) = \dfrac{1}{2} \times \dfrac{10}{3} = \dfrac{5}{3}\).
  4. \(\tanh(\ln 3) = \dfrac{\sinh(\ln 3)}{\cosh(\ln 3)} = \dfrac{4/3}{5/3} = \dfrac{4}{5}\).
Answer: \(\sinh(\ln 3) = \tfrac{4}{3}\), \(\cosh(\ln 3) = \tfrac{5}{3}\), \(\tanh(\ln 3) = \tfrac{4}{5}\)
Worked example Odd, even, and back to e^x
Use the definitions to show that \(\sinh x\) is an odd function and \(\cosh x\) is an even function. Show also that \(\cosh x + \sinh x = e^{x}\) and \(\cosh x - \sinh x = e^{-x}\).
  1. Replace \(x\) by \(-x\): \(\sinh(-x) = \dfrac{e^{-x} - e^{x}}{2} = -\dfrac{e^{x} - e^{-x}}{2} = -\sinh x\), so \(\sinh\) is odd.
  2. \(\cosh(-x) = \dfrac{e^{-x} + e^{x}}{2} = \cosh x\), so \(\cosh\) is even.
  3. Add the definitions: \(\cosh x + \sinh x = \dfrac{e^{x} + e^{-x} + e^{x} - e^{-x}}{2} = \dfrac{2e^{x}}{2} = e^{x}\).
  4. Subtract: \(\cosh x - \sinh x = \dfrac{e^{x} + e^{-x} - e^{x} + e^{-x}}{2} = e^{-x}\).
Answer: \(\sinh(-x) = -\sinh x\), \(\cosh(-x) = \cosh x\); \(\cosh x \pm \sinh x = e^{\pm x}\)
Your turn
Find the exact values of \(\sinh(\ln 2)\), \(\cosh(\ln 2)\) and \(\tanh(\ln 2)\), and hence \(\operatorname{sech}(\ln 2)\), \(\operatorname{cosech}(\ln 2)\) and \(\coth(\ln 2)\).
Examiner trap

\(e^{-\ln 2}\) is \(\tfrac{1}{2}\), not \(-2\): the minus sign in the power gives a reciprocal, never a negative number. And keep the \(2\) in the denominator: \(\cosh(\ln 2) = \tfrac{5}{4}\), not \(\tfrac{5}{2}\). Do not mix the signs either: \(\cosh\) has the plus, \(\sinh\) the minus (the \(s\) in sinh goes with subtract).

Section 02

Six graphs, and their asymptotes.

For large positive \(x\), \(e^{-x}\) is tiny, so \(\sinh x\) and \(\cosh x\) both behave like \(\tfrac{1}{2}e^{x}\) and their graphs merge. For large negative \(x\), \(e^{x}\) is tiny: \(\cosh x \approx \tfrac{1}{2}e^{-x}\) climbs again while \(\sinh x \approx -\tfrac{1}{2}e^{-x}\) plunges. Since \(\cosh x - \sinh x = e^{-x} \gt 0\), the sinh curve is always below the cosh curve and the two never meet (Fig. 1).

x y −3 −2 −1 1 2 3 −4 −3 −2 −1 1 2 3 4 y = sech x y = coth x y = coth x y = cosech x y = 1 y = −1
Fig. 2: the reciprocals. sech is a bell with its peak at (0, 1); cosech and coth split at the asymptote x = 0; coth stays outside the dashed lines y = 1 and y = −1, where tanh stays inside

Sketch a reciprocal graph from the original: where \(\cosh x \ge 1\), \(\operatorname{sech} x\) lies in \((0, 1]\); where \(\sinh x = 0\) (at the origin), \(\operatorname{cosech} x\) and \(\coth x\) have a vertical asymptote; where \(\tanh x \to \pm 1\), so does \(\coth x\).

Worked example Why cosh x is at least 1
Show that \(\cosh x \ge 1\) for all real \(x\), with equality only when \(x = 0\). Deduce the range of \(\operatorname{sech} x\).
  1. \(\cosh x - 1 = \dfrac{e^{x} - 2 + e^{-x}}{2} = \dfrac{\left(e^{x/2} - e^{-x/2}\right)^{2}}{2}\), since \(e^{x/2} \times e^{-x/2} = 1\).
  2. A square is never negative, so \(\cosh x - 1 \ge 0\), that is \(\cosh x \ge 1\).
  3. Equality needs \(e^{x/2} = e^{-x/2}\), so \(e^{x} = 1\) and \(x = 0\).
  4. \(\operatorname{sech} x = \dfrac{1}{\cosh x}\) with \(\cosh x \ge 1\), so \(0 \lt \operatorname{sech} x \le 1\), with the maximum \(1\) at \(x = 0\).
Answer: \(\cosh x \ge 1\) (equal to \(1\) only at \(x = 0\)); range of \(\operatorname{sech} x\) is \(0 \lt y \le 1\)
Worked example tanh between the asymptotes
Show that \(\tanh x = \dfrac{e^{2x} - 1}{e^{2x} + 1}\). Hence show that \(-1 \lt \tanh x \lt 1\), and state the asymptotes of \(y = \tanh x\) and of \(y = \coth x\).
  1. Multiply the top and bottom of \(\dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\) by \(e^{x}\): \(\tanh x = \dfrac{e^{2x} - 1}{e^{2x} + 1}\).
  2. Write it as \(\tanh x = 1 - \dfrac{2}{e^{2x} + 1}\). Since \(e^{2x} \gt 0\), the fraction \(\dfrac{2}{e^{2x} + 1}\) lies strictly between \(0\) and \(2\), so \(-1 \lt \tanh x \lt 1\).
  3. As \(x \to \infty\), \(e^{2x} \to \infty\) and \(\tanh x \to 1\); as \(x \to -\infty\), \(e^{2x} \to 0\) and \(\tanh x \to -1\). Asymptotes \(y = 1\) and \(y = -1\).
  4. \(\coth x = \dfrac{1}{\tanh x}\): the reciprocal of a number strictly between \(-1\) and \(1\) (and not \(0\)) has size greater than \(1\), so \(|\coth x| \gt 1\). It has the same horizontal asymptotes \(y = \pm 1\), plus \(x = 0\), where \(\tanh x = 0\).
Answer: \(-1 \lt \tanh x \lt 1\); \(\tanh\): asymptotes \(y = \pm 1\); \(\coth\): asymptotes \(y = \pm 1\) and \(x = 0\)
Your turn
State the range of \(\cosh x\), of \(\tanh x\) and of \(\coth x\), and say which of the six hyperbolic functions are not defined at \(x = 0\).
Examiner trap

\(\tanh x\) never reaches \(\pm 1\), so the curve must flatten towards the dashed lines without touching them; in the same way \(\coth x\) never reaches \(\pm 1\) from outside. \(\operatorname{sech} x\) has NO vertical asymptote (unlike \(\sec x\)): \(\cosh x\) is never zero. And never draw \(\sinh x\) crossing \(\cosh x\): their difference is \(e^{-x}\), which is always positive.

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Find the exact value of \(\cosh(\ln 2)\).

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Which statement is true for every real \(x\)?

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What is the range of \(y = \tanh x\)?

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Which of these graphs has the vertical asymptote \(x = 0\)?

Section 03

Equations as a quadratic in ex.

An equation that mixes \(\sinh x\) and \(\cosh x\) (and a constant) becomes a quadratic once you write everything in exponentials. Put \(u = e^{x}\), so \(e^{-x} = \dfrac{1}{u}\); multiply through by \(u\); solve; and remember that \(u = e^{x}\) must be positive.

For example, \(\tanh x = \tfrac{1}{3}\) gives \(\dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{1}{3}\), so \(3e^{2x} - 3 = e^{2x} + 1\), \(e^{2x} = 2\) and \(x = \tfrac{1}{2}\ln 2\). Before you start, check the value is in the range: \(\cosh x = \tfrac{1}{2}\) and \(\tanh x = 2\) have no solutions at all.

Worked example A mixed equation
Solve \(3\sinh x - \cosh x = 1\), giving the answer in exact form.
  1. \(3 \times \dfrac{e^{x} - e^{-x}}{2} - \dfrac{e^{x} + e^{-x}}{2} = 1\). Multiply by \(2\): \(3e^{x} - 3e^{-x} - e^{x} - e^{-x} = 2\), so \(2e^{x} - 4e^{-x} = 2\).
  2. Divide by \(2\): \(e^{x} - 2e^{-x} = 1\). Multiply by \(e^{x}\) and put \(u = e^{x}\): \(u^{2} - u - 2 = 0\).
  3. \((u - 2)(u + 1) = 0\), so \(u = 2\) or \(u = -1\). Reject \(u = -1\), because \(e^{x} \gt 0\).
  4. \(e^{x} = 2\), so \(x = \ln 2\). Check: \(3\sinh(\ln 2) - \cosh(\ln 2) = \tfrac{9}{4} - \tfrac{5}{4} = 1\).
Answer: \(x = \ln 2\)
Worked example cosh x = 2 has two answers
Solve \(\cosh x = 2\), giving the answers as logarithms.
  1. \(\dfrac{e^{x} + e^{-x}}{2} = 2\), so \(e^{x} + e^{-x} = 4\). Multiply by \(u = e^{x}\): \(u^{2} - 4u + 1 = 0\).
  2. \(u = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\). Both are positive (\(2 - \sqrt{3} \approx 0.268\)), so keep both.
  3. \(x = \ln(2 + \sqrt{3})\) or \(x = \ln(2 - \sqrt{3})\).
  4. Since \((2 + \sqrt{3})(2 - \sqrt{3}) = 1\), \(\ln(2 - \sqrt{3}) = -\ln(2 + \sqrt{3})\). The two answers are \(\pm\ln(2 + \sqrt{3})\), as the even graph of \(\cosh\) predicts.
Answer: \(x = \pm\ln(2 + \sqrt{3})\)
Your turn
Solve \(\sinh x = \tfrac{3}{4}\), giving your answer in exact form.
Examiner trap

Do not throw away a positive root just because it is less than \(1\): \(u = 2 - \sqrt{3}\) is a valid \(e^{x}\) and gives a negative \(x\). Reject only \(u \le 0\). And with \(\cosh x = k\) for \(k \gt 1\) there are ALWAYS two answers, \(\pm\) the same logarithm; giving only one loses a mark.

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Solve \(2\cosh x = 3\sinh x\).

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Find the exact value of \(\tanh(\ln 3)\).

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

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Six original Cambridge-style questions
Exact values at a logarithm, the e^x identities and powers, the graph and range of coth with an equation, two mixed sinh and cosh equations, and the hyperbola behind the name.