Why is "Swapping the signs in the definitions ($\sinh x = \frac{e^x + e^{-x}}{2}$)" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: The two definitions differ only by one sign, so they are easy to confuse under pressure. How to avoid it: Mnemonic: 'sinh has a minus, cosh is positive (cuddly)'. Or check: 0 = 0 needs the minus; 0 = 1 needs the plus. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Forgetting the sign change in $\cosh^2 x - \sinh^2 x = 1$" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: Copying the trig identity ^2 + ^2 = 1 directly without applying Osborn's rule. How to avoid it: Osborn: ^2 x is a product of two sinhs, so its sign flips — giving a minus. Verify once from the definitions. Source: 9231 Paper 2 mark schemes

Why is "Dividing by $e^x$ instead of multiplying when forming a quadratic" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: Trying to cancel the e^-x term directly rather than clearing it. How to avoid it: To turn ae^x + be^-x = c into a quadratic, MULTIPLY every term by e^x, giving ae^2x + b = ce^x. Source: 9231 Paper 2 Examiner Reports

Why is "Accepting a non-positive value of $e^x$ as a solution" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: Solving the quadratic in u = e^x and forgetting that e^x > 0 for all real x. How to avoid it: After solving, reject any root with u ≤ 0 and state the rejection clearly — it is a marked step. Source: 9231 Paper 2 Examiner Reports

Why is "Giving only one solution to a $\cosh x = k$ equation" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: Forgetting that is even, so x = k (with k > 1) has two solutions x = c. How to avoid it: Whenever appears, expect a pair. The quadratic in e^x naturally produces both reciprocal roots. Source: 9231 Paper 2 mark schemes

Why is "Quoting an identity instead of proving it on a 'prove' question" a common mistake in Hyperbolic functions in terms of exponential functions?

Why it happens: Treating a 'prove'/'show that' as a 'state'; the result is well known so students skip the working. How to avoid it: For 'prove from the definitions', substitute e^x e^-x2 and show every algebraic step — the stated result earns nothing on its own. Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Hyperbolic functions in terms of exponential functions

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Detailed Study Notes

Detailed notes on Hyperbolic functions for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Hyperbolic Functions in Terms of Exponentials — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (2026-2027 syllabus)

Hyperbolic functions are built directly from $e^x$ and $e^{-x}$ . Learn the three core definitions, the identities they generate, and how to prove identities and solve equations from first principles.

At a glance

$\sinh x = \frac{e^x - e^{-x}}{2}$ , $\cosh x = \frac{e^x + e^{-x}}{2}$ — these exponential definitions are NOT in MF19, so memorise them.
$\tanh x = \frac{\sinh x}{\cosh x}$ , with reciprocals $\text{sech}, \text{cosech}, \coth$ .
Fundamental identity $\cosh^2 x - \sinh^2 x = 1$ (note the minus) — this and the double-angle results ARE in MF19.
Osborn's rule converts a trig identity to its hyperbolic form: flip the sign of any product of two sinhs.
Solving equations: rewrite in exponentials and form a quadratic in $e^x$ , OR use $\cosh^2 - \sinh^2 = 1$ to get a quadratic in one function.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.1a — Understand the definitions of $\sinh x$ , $\cosh x$ and $\tanh x$ in terms of the exponential function $e^x$ .
2.1b — Prove and use the standard hyperbolic identities (e.g. $\cosh^2 x - \sinh^2 x = 1$ , $\sinh 2x = 2\sinh x\cosh x$ ).
2.1c — Solve equations involving hyperbolic functions, giving exact answers in logarithmic form where appropriate.

The three definitions — built from eˣ

sinh and cosh are the odd and even parts of eˣ; tanh is their ratio.

Where do hyperbolic functions come from? Take the exponential function $e^x$ and split it into an even part and an odd part. The even part is $\cosh$ , the odd part is $\sinh$ :

$\cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}$

Adding these gives back $\cosh x + \sinh x = e^x$ — a neat check.

The tangent and reciprocals. Just like in trigonometry:

$\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}$
$\text{sech}\,x = \dfrac{1}{\cosh x}$ , $\quad \text{cosech}\,x = \dfrac{1}{\sinh x}$ , $\quad \coth x = \dfrac{1}{\tanh x}$

Critical exam fact. The exponential definitions of $\sinh$ , $\cosh$ and $\tanh$ are NOT printed in MF19. You must have them memorised, because every proof and most equation-solving starts by substituting them.

cosh (blue) is even with minimum 1; sinh (red) is odd and passes through the origin. Both grow like ½eˣ for large positive x.

Cambridge tip. Write the definitions at the very top of your answer before you do anything else — it both jogs your memory and earns the first method mark on a 'prove from definitions' question.

$\cosh x = \frac{e^x + e^{-x}}{2}$ (even), $\sinh x = \frac{e^x - e^{-x}}{2}$ (odd).
$\cosh x + \sinh x = e^x$ — a quick self-check.
These definitions are NOT in MF19; memorise them.

See the full worked example for hyperbolic functions in terms of exponential functions →

The identities and Osborn's rule

Hyperbolic identities mirror trig ones — flip the sign of any product of two sinhs.

Hyperbolic functions obey a whole family of identities, closely paralleling trigonometry. The most important:

Fundamental identity (the hyperbolic Pythagoras): $\cosh^2 x - \sinh^2 x = 1$

Double-angle identities: $\sinh 2x = 2\sinh x\cosh x, \qquad \cosh 2x = \cosh^2 x + \sinh^2 x$

These three ARE printed in MF19, but you must still be able to prove them from the exponential definitions when a question says 'show that' or 'prove'.

Osborn's rule is the bridge between trig and hyperbolic identities:

Take any trigonometric identity. Replace $\sin \to \sinh$ and $\cos \to \cosh$ . Then change the sign of every term containing a product of two sinhs (an explicit $\sinh \times \sinh$ , or a hidden one like $\sin^2$ , or a $\tan^2 = \frac{\sin^2}{\cos^2}$ ).

Examples of Osborn in action:

$\cos^2 x + \sin^2 x = 1$ → $\cosh^2 x \,{-}\, \sinh^2 x = 1$ (the $\sin^2$ term flips sign).
$\cos 2x = \cos^2 x - \sin^2 x$ → $\cosh 2x = \cosh^2 x \,{+}\, \sinh^2 x$ (the $-\sin^2$ flips to $+\sinh^2$ ).
$\sin 2x = 2\sin x\cos x$ → $\sinh 2x = 2\sinh x\cosh x$ (no product of two sinhs, so no sign change).

Osborn's rule: sin→sinh, cos→cosh, and flip the sign of any product of two sinhs (here sin² becomes −sinh²).

Cambridge tip. Use Osborn's rule to recall or check an identity quickly in the exam — but if the question says 'prove', you must still substitute the exponential definitions and show every line.

$\cosh^2 x - \sinh^2 x = 1$ (note the minus).
Osborn: flip the sign of any product of two sinhs (including $\sin^2$ ).
MF19 lists the identities, but 'prove' means prove from definitions.

See the full worked example for hyperbolic functions in terms of exponential functions →

Solving hyperbolic equations

Two routes: a quadratic in eˣ, or reduce to one function using the identity.

Hyperbolic equations come in two flavours, and each has a standard method.

Route 1 — quadratic in $e^x$ . When the equation mixes $\sinh$ , $\cosh$ and constants, substitute the definitions, then multiply through by $e^x$ to clear the $e^{-x}$ terms. You get a quadratic in $u = e^x$ :

Replace each function with $\frac{e^x \pm e^{-x}}{2}$ .
Multiply every term by $e^x$ to obtain $au^2 + bu + c = 0$ where $u = e^x$ .
Solve the quadratic.
Reject any root with $u \leq 0$ (since $e^x > 0$ always), then take logs: $x = \ln u$ .

Route 2 — reduce to one function. When a double angle appears (e.g. $\cosh 2x$ ), use an identity to express everything in terms of a single function:

$\cosh 2x = 1 + 2\sinh^2 x$ (to get a quadratic in $\sinh x$ ), or
$\cosh 2x = 2\cosh^2 x - 1$ (to get a quadratic in $\cosh x$ ).

Then solve the quadratic in $\sinh x$ or $\cosh x$ and convert back using the inverse log-forms.

Two important warnings:

$\sinh x = k$ has exactly one real solution (sinh is one-to-one), found from $\text{arsinh}\,k = \ln(k + \sqrt{k^2+1})$ .
$\cosh x = k$ (with $k > 1$ ) has two solutions $x = \pm c$ , because cosh is even. Expect a $\pm$ pair.

Cambridge tip. A 'give your answer in exact logarithmic form' instruction means no decimals. Leave surds inside the log, e.g. $x = \ln\!\left(\frac{3 + \sqrt{13}}{2}\right)$ , and never round.

Mixed equation → quadratic in $e^x$ (multiply by $e^x$ , reject $u \leq 0$ ).
Double angle → reduce to one function via $\cosh 2x = 1 + 2\sinh^2 x$ .
$\cosh x = k$ gives two ( $\pm$ ) solutions; $\sinh x = k$ gives one.

See the full worked example for hyperbolic functions in terms of exponential functions →

Quick recap

$\cosh x = \frac{e^x + e^{-x}}{2}$ , $\sinh x = \frac{e^x - e^{-x}}{2}$ , $\tanh x = \frac{\sinh x}{\cosh x}$ .
These exponential definitions are NOT in MF19 — memorise them.
$\cosh^2 x - \sinh^2 x = 1$ ; $\sinh 2x = 2\sinh x\cosh x$ ; $\cosh 2x = \cosh^2 x + \sinh^2 x$ (all in MF19).
Osborn's rule: flip the sign of any product of two sinhs.
Solve equations by a quadratic in $e^x$ (reject $u \leq 0$ ) or by reducing to one function.
$\cosh x = k$ gives a $\pm$ pair; give exact logarithmic answers.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$ (NOT in MF19).
$\cosh^2 x - \sinh^2 x = 1$ (note the minus sign).
$\sinh 2x = 2\sinh x\cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$ .
Osborn's rule — flip the sign of any product of two sinhs.
$e^x > 0$ always — reject non-positive roots when solving in $e^x$ .

How it’s examined

Topic 2.1 underpins much of Paper 2 (Further Pure 2). Expect a 'prove from the exponential definitions' part (3-5 marks) and an equation-solving part (4-7 marks) asking for exact logarithmic answers. Marks are lost through (i) sign errors in the definitions or the identity $\cosh^2 - \sinh^2 = 1$ , (ii) failing to reject non-positive roots of the $e^x$ quadratic, and (iii) quoting an identity instead of proving it. Because the exponential definitions are not in MF19, secure recall is essential.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2); 9231/21 & 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Hyperbolic functions in terms of exponential functions

Step-by-step solutions to past-paper-style questions on hyperbolic functions in terms of exponential functions, written exactly the way a tutor would explain them at the board.

Question type:

1Evaluating sinh and cosh from the definition
Getting startedDirect calculation• definition, evaluate
Question
Using the exponential definitions, find the exact value of $\cosh(\ln 3)$ .
Step-by-step solution
1. Step 1
  Quote the definition of $\cosh$ — this is NOT in MF19, so you must know it.
  $\cosh x = \frac{e^x + e^{-x}}{2}$
2. Step 2
  Substitute $x = \ln 3$ . Recall $e^{\ln 3} = 3$ and $e^{-\ln 3} = \frac{1}{3}$ .
  $\cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2}$
3. Step 3
  Simplify the fraction.
  $\cosh(\ln 3) = \frac{\frac{10}{3}}{2} = \frac{10}{6} = \frac{5}{3}$
Answer
$\cosh(\ln 3) = \dfrac{5}{3}$ .
Examiner tip
The key skill is $e^{\ln a} = a$ and $e^{-\ln a} = \frac{1}{a}$ . Students who reach for a calculator lose the 'exact value' marks.
2Writing tanh in exponential form
Getting startedDirect calculation• definition, tanh
Question
Show that $\tanh x = \dfrac{e^{2x} - 1}{e^{2x} + 1}$ .
Step-by-step solution
1. Step 1
  Start from $\tanh x = \dfrac{\sinh x}{\cosh x}$ and substitute the exponential definitions.
  $\tanh x = \frac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$
2. Step 2
  Multiply top and bottom by $e^x$ to clear the negative powers.
  $\tanh x = \frac{e^x(e^x - e^{-x})}{e^x(e^x + e^{-x})} = \frac{e^{2x} - 1}{e^{2x} + 1}$
Answer
$\tanh x = \dfrac{e^{2x} - 1}{e^{2x} + 1}$ , as required.
Examiner tip
Multiplying by $e^x$ is the standard trick to convert from $e^{-x}$ form. Note $e^x \cdot e^x = e^{2x}$ and $e^x \cdot e^{-x} = 1$ .
3Proving cosh²x − sinh²x = 1
Building confidenceShow that / prove• identity, proof
Question
Prove, from the exponential definitions, that $\cosh^2 x - \sinh^2 x = 1$ .
Step-by-step solution
1. Step 1
  Square each definition.
  $\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}$
2. Step 2
  Square the sinh definition. Note the middle term is $-2$ here.
  $\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}$
3. Step 3
  Subtract. The $e^{2x}$ and $e^{-2x}$ terms cancel.
  $\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1$
Answer
$\cosh^2 x - \sinh^2 x = 1$ , as required.
Examiner tip
The cross term is $+2$ for cosh and $-2$ for sinh — getting this sign right is the whole proof. Use $(e^x)(e^{-x}) = e^0 = 1$ .
4Proving the double-angle identity for sinh
Building confidenceShow that / prove• identity, double-angle
Question
Prove from the definitions that $\sinh 2x = 2\sinh x \cosh x$ .
Step-by-step solution
1. Step 1
  Expand the right-hand side by substituting both definitions.
  $2\sinh x \cosh x = 2 \cdot \frac{e^x - e^{-x}}{2} \cdot \frac{e^x + e^{-x}}{2}$
2. Step 2
  Multiply out — the two brackets form a difference of squares $a^2 - b^2$ with $a = e^x$ , $b = e^{-x}$ .
  $2\sinh x \cosh x = \frac{(e^x - e^{-x})(e^x + e^{-x})}{2} = \frac{e^{2x} - e^{-2x}}{2}$
3. Step 3
  Recognise the left-hand side. By definition $\sinh 2x = \dfrac{e^{2x} - e^{-2x}}{2}$ .
  $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2} = 2\sinh x \cosh x$
Answer
$\sinh 2x = 2\sinh x \cosh x$ , as required.
Examiner tip
Spotting the difference of two squares $(e^x)^2 - (e^{-x})^2 = e^{2x} - e^{-2x}$ saves a line of working. Compare with the trig identity $\sin 2x = 2\sin x\cos x$ (Osborn: no sign change, no product of two sinhs).
5Solving a hyperbolic equation via a quadratic in eˣ
StretchMulti-step problem• solve, equation
Question
Solve $2\cosh x - \sinh x = 2$ , giving your answer in exact logarithmic form.
Step-by-step solution
1. Step 1
  Substitute the definitions and multiply through by $2$ .
  $2 \cdot \frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2} = 2 \;\Rightarrow\; 2(e^x + e^{-x}) - (e^x - e^{-x}) = 4$
2. Step 2
  Collect like terms. $2e^x - e^x = e^x$ and $2e^{-x} + e^{-x} = 3e^{-x}$ .
  $e^x + 3e^{-x} = 4$
3. Step 3
  Multiply by $e^x$ to form a quadratic, and let $u = e^x$ .
  $e^{2x} - 4e^x + 3 = 0 \;\Rightarrow\; u^2 - 4u + 3 = 0$
4. Step 4
  Factorise and solve for $u = e^x$ .
  $(u - 1)(u - 3) = 0 \;\Rightarrow\; e^x = 1 \text{ or } e^x = 3$
5. Step 5
  Take logarithms. Both roots are positive, so both are valid.
  $x = \ln 1 = 0 \quad \text{or} \quad x = \ln 3$
Answer
$x = 0$ or $x = \ln 3$ .
Examiner tip
Always check $e^x > 0$ : reject any non-positive root. Here both $1$ and $3$ are positive, so both survive. A common slip is dividing by $e^x$ instead of multiplying — multiply to keep all solutions.
6Solving an equation using cosh² − sinh² = 1
StretchMulti-step problem• solve, identity
Question
Solve $\cosh 2x - 3\sinh x = 1$ for $x$ , giving answers in exact form.
Step-by-step solution
1. Step 1
  Use the double-angle identity $\cosh 2x = 1 + 2\sinh^2 x$ (derived from $\cosh 2x = \cosh^2 x + \sinh^2 x$ and $\cosh^2 x = 1 + \sinh^2 x$ ).
  $1 + 2\sinh^2 x - 3\sinh x = 1$
2. Step 2
  Form a quadratic in $\sinh x$ . Let $s = \sinh x$ .
  $2s^2 - 3s = 0 \;\Rightarrow\; s(2s - 3) = 0$
3. Step 3
  Solve for $\sinh x$ .
  $\sinh x = 0 \quad \text{or} \quad \sinh x = \frac{3}{2}$
4. Step 4
  Convert to $x$ using the logarithmic inverse $\text{arsinh}\,x = \ln(x + \sqrt{x^2 + 1})$ . For $\sinh x = 0$ , $x = 0$ .
  $x = 0 \quad \text{or} \quad x = \ln\left(\frac{3}{2} + \sqrt{\tfrac{9}{4} + 1}\right) = \ln\left(\frac{3 + \sqrt{13}}{2}\right)$
Answer
$x = 0$ or $x = \ln\!\left(\dfrac{3 + \sqrt{13}}{2}\right)$ .
Examiner tip
$\sinh x$ is one-to-one, so each value of $\sinh x$ gives exactly one real $x$ (unlike $\cosh x$ ). Using $\cosh 2x = 1 + 2\sinh^2 x$ produces a quadratic in $\sinh x$ alone — the cleanest route.

Model Answers — Hyperbolic functions in terms of exponential functions

High-scoring sample answers for hyperbolic functions in terms of exponential functions on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Using the exponential definition, find the exact value of $\sinh(\ln 2)$ .
Model answer
Quote the definition (not in MF19): $\sinh x = \frac{e^x - e^{-x}}{2}$

Substitute $x = \ln 2$ , using $e^{\ln 2} = 2$ and $e^{-\ln 2} = \frac{1}{2}$ : $\sinh(\ln 2) = \frac{2 - \frac{1}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$
Why this scores
M1 for quoting the definition, M1 for $e^{-\ln 2} = \frac{1}{2}$ , A1 for $\frac{3}{4}$ . The negative-exponent step is where most marks are lost.
Question 2
4 marks
Prove from the exponential definitions that $\cosh 2x = \cosh^2 x + \sinh^2 x$ .
Model answer
Right-hand side, squaring each definition: $\cosh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4}, \qquad \sinh^2 x = \frac{e^{2x} - 2 + e^{-2x}}{4}$

Add them — the $+2$ and $-2$ cancel: $\cosh^2 x + \sinh^2 x = \frac{2e^{2x} + 2e^{-2x}}{4} = \frac{e^{2x} + e^{-2x}}{2}$

Left-hand side, by definition with $2x$ in place of $x$ : $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$

The two sides are equal, so $\cosh 2x = \cosh^2 x + \sinh^2 x$ as required.
Why this scores
M1 for both squared expansions, M1 for adding and simplifying, M1 for stating $\cosh 2x$ from the definition, A1 for the conclusion. As a 'prove', every line earns credit — never quote the trig analogue as proof.
Question 3
5 marks
Solve $3\cosh x = 5$ , giving your answers in exact logarithmic form.
Model answer
Substitute the definition of $\cosh x$ : $3 \cdot \frac{e^x + e^{-x}}{2} = 5 \;\Rightarrow\; 3(e^x + e^{-x}) = 10$

Multiply by $e^x$ and let $u = e^x$ : $3e^{2x} + 3 = 10e^x \;\Rightarrow\; 3u^2 - 10u + 3 = 0$

Factorise: $(3u - 1)(u - 3) = 0 \;\Rightarrow\; u = \frac{1}{3} \text{ or } u = 3$

Take logarithms (both roots positive, so both valid): $x = \ln 3 \quad \text{or} \quad x = \ln\frac{1}{3} = -\ln 3$

The two answers are $\pm\ln 3$ , reflecting that $\cosh$ is an even function.
Why this scores
M1 for substitution, M1 for forming the quadratic, A1 for the roots $\frac{1}{3}, 3$ , A1 A1 for $x = \pm\ln 3$ . Note $\cosh$ gives two symmetric solutions — losing the negative one is the standard error.
Question 4
6 marks
The trigonometric identity $\sin(A+B) = \sin A\cos B + \cos A\sin B$ has a hyperbolic analogue. (a) State the hyperbolic version $\sinh(A+B)$ . (b) Prove it from the exponential definitions.
Model answer
(a) By Osborn's rule, the expansion contains no product of two sinhs, so the sign is unchanged: $\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B$

(b) Expand the right-hand side using the definitions: $\sinh A\cosh B = \frac{(e^A - e^{-A})(e^B + e^{-B})}{4} = \frac{e^{A+B} + e^{A-B} - e^{-A+B} - e^{-A-B}}{4}$ $\cosh A\sinh B = \frac{(e^A + e^{-A})(e^B - e^{-B})}{4} = \frac{e^{A+B} - e^{A-B} + e^{-A+B} - e^{-A-B}}{4}$

Add them — the $e^{A-B}$ and $e^{-A+B}$ terms cancel: $\sinh A\cosh B + \cosh A\sinh B = \frac{2e^{A+B} - 2e^{-A-B}}{4} = \frac{e^{A+B} - e^{-(A+B)}}{2} = \sinh(A+B)$
Why this scores
(a) B1 for the correct identity. (b) M1 M1 for the two expansions, M1 for adding and cancelling, A1 for recognising $\sinh(A+B)$ . Osborn's rule predicts the form; the proof must come from the definitions.
Question 5
7 marks
Solve the equation $\cosh x + 2\sinh x = 2$ , giving your answers in exact logarithmic form. State clearly any rejected solutions.
Model answer
Substitute the definitions: $\frac{e^x + e^{-x}}{2} + 2 \cdot \frac{e^x - e^{-x}}{2} = 2$

Multiply through by $2$ and collect: $(e^x + e^{-x}) + 2(e^x - e^{-x}) = 4 \;\Rightarrow\; 3e^x - e^{-x} = 4$

Multiply by $e^x$ and let $u = e^x$ : $3e^{2x} - 4e^x - 1 = 0 \;\Rightarrow\; 3u^2 - 4u - 1 = 0$

Quadratic formula: $u = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{4 \pm \sqrt{28}}{6} = \frac{2 \pm \sqrt{7}}{3}$

Reject the negative root: $\dfrac{2 - \sqrt{7}}{3} < 0$ (since $\sqrt{7} \approx 2.646 > 2$ ), and $e^x$ must be positive. So: $e^x = \frac{2 + \sqrt{7}}{3} \;\Rightarrow\; x = \ln\!\left(\frac{2 + \sqrt{7}}{3}\right)$
Why this scores
M1 substitution, M1 forming the quadratic, A1 correct quadratic, M1 quadratic formula, A1 both roots, B1 rejecting the negative root with reason, A1 final answer. The explicit rejection (with $e^x > 0$ justification) is a marked step — omitting it loses the B1.
Question 6
8 marks
Prove from the exponential definitions that $\tanh 2x = \dfrac{2\tanh x}{1 + \tanh^2 x}$ .
Model answer
Start from the definition of $\tanh 2x$ in terms of $\sinh$ and $\cosh$ , then use the double-angle results: $\tanh 2x = \frac{\sinh 2x}{\cosh 2x} = \frac{2\sinh x\cosh x}{\cosh^2 x + \sinh^2 x}$

(using $\sinh 2x = 2\sinh x\cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$ , both proved earlier).

Divide numerator and denominator by $\cosh^2 x$ to introduce $\tanh x$ : $\tanh 2x = \frac{\dfrac{2\sinh x\cosh x}{\cosh^2 x}}{\dfrac{\cosh^2 x + \sinh^2 x}{\cosh^2 x}} = \frac{2\dfrac{\sinh x}{\cosh x}}{1 + \dfrac{\sinh^2 x}{\cosh^2 x}}$

Recognise $\dfrac{\sinh x}{\cosh x} = \tanh x$ : $\tanh 2x = \frac{2\tanh x}{1 + \tanh^2 x}$

as required.
Why this scores
M1 for writing $\tanh 2x$ as $\frac{\sinh 2x}{\cosh 2x}$ , M1 M1 for the two double-angle substitutions, M1 for dividing through by $\cosh^2 x$ , A1 each for the numerator and denominator simplifications, A1 for the conclusion. Compare with the trig version $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ — Osborn flips the sign in the denominator (a product of two tanhs).

Key Formulae — Hyperbolic functions in terms of exponential functions

The formulae you need to memorise for hyperbolic functions in terms of exponential functions on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Definition of sinh
$\sinh x = \frac{e^x - e^{-x}}{2}$
$x$
any real number
$e$
the base of natural logarithms, $\approx 2.718$
When to use
Whenever you must prove an identity or solve an equation from first principles. NOT in MF19 — memorise it.
Definition of cosh
$\cosh x = \frac{e^x + e^{-x}}{2}$
$x$
any real number
When to use
The even partner of sinh; note the $+$ sign. NOT in MF19 — memorise it.
Definition of tanh and reciprocals
$\tanh x = \frac{\sinh x}{\cosh x}, \quad \text{sech}\,x = \frac{1}{\cosh x}, \quad \text{cosech}\,x = \frac{1}{\sinh x}, \quad \coth x = \frac{1}{\tanh x}$
$x$
any real number (with $x \neq 0$ for cosech and coth)
When to use
Convert any of the six functions into sinh/cosh, then into exponentials, to manipulate.
Fundamental hyperbolic identity
$\cosh^2 x - \sinh^2 x = 1$
$x$
any real number
When to use
The hyperbolic analogue of $\cos^2 + \sin^2 = 1$ (note the minus). IN MF19. Use to eliminate one function.
Double-angle identities
$\sinh 2x = 2\sinh x\cosh x, \qquad \cosh 2x = \cosh^2 x + \sinh^2 x$
$x$
any real number
When to use
Reduce a $2x$ expression to functions of $x$ (or vice versa) when solving equations. IN MF19.
Osborn's rule
$\text{trig identity} \to \text{hyperbolic identity: replace } \sin\to\sinh,\; \cos\to\cosh,\; \text{flip sign of any product of two sinhs}$
$\sin^2$
counts as a product of two sinhs, so its sign flips
When to use
To recall or check a hyperbolic identity by analogy with the trig one. A memory aid — Cambridge still wants a proof from definitions.

Key Definitions and Keywords — Hyperbolic functions in terms of exponential functions

Definitions to memorise and the exact keywords mark schemes credit for hyperbolic functions in terms of exponential functions answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Hyperbolic sine ( $\sinh$ )
Examiner keyword
The function $\sinh x = \frac{e^x - e^{-x}}{2}$ . It is odd: $\sinh(-x) = -\sinh x$ . Read aloud as 'shine' or 'sinch'.
Hyperbolic cosine ( $\cosh$ )
Examiner keyword
The function $\cosh x = \frac{e^x + e^{-x}}{2}$ . It is even: $\cosh(-x) = \cosh x$ , with minimum value $1$ . Read aloud as 'cosh'.
Hyperbolic tangent ( $\tanh$ )
Examiner keyword
The function $\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^{2x}-1}{e^{2x}+1}$ . It is odd and bounded between $-1$ and $1$ .
sech, cosech, coth
Examiner keyword
The reciprocal hyperbolic functions: $\text{sech}\,x = \frac{1}{\cosh x}$ , $\text{cosech}\,x = \frac{1}{\sinh x}$ , $\coth x = \frac{1}{\tanh x}$ .
Osborn's rule
Examiner keyword
A mnemonic: to convert a trig identity to its hyperbolic form, replace $\sin\to\sinh$ , $\cos\to\cosh$ , and change the sign of any term containing a product of two sinhs (including $\sin^2$ ).
Fundamental identity
Examiner keyword
$\cosh^2 x - \sinh^2 x = 1$ — the hyperbolic counterpart of the Pythagorean identity, with a minus sign (predicted by Osborn's rule). Given in MF19.

Common Mistakes and Misconceptions — Hyperbolic functions in terms of exponential functions

The traps other students keep falling into on hyperbolic functions in terms of exponential functions questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Swapping the signs in the definitions ( $\sinh x = \frac{e^x + e^{-x}}{2}$ )
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
The two definitions differ only by one sign, so they are easy to confuse under pressure.
How to avoid it
Mnemonic: 'sinh has a minus, cosh is positive (cuddly)'. Or check: $\sinh 0 = 0$ needs the minus; $\cosh 0 = 1$ needs the plus.
✕Forgetting the sign change in $\cosh^2 x - \sinh^2 x = 1$
9231 Paper 2 mark schemes
Why it happens
Copying the trig identity $\cos^2 + \sin^2 = 1$ directly without applying Osborn's rule.
How to avoid it
Osborn: $\sin^2 x$ is a product of two sinhs, so its sign flips — giving a minus. Verify once from the definitions.
✕Dividing by $e^x$ instead of multiplying when forming a quadratic
9231 Paper 2 Examiner Reports
Why it happens
Trying to cancel the $e^{-x}$ term directly rather than clearing it.
How to avoid it
To turn $ae^x + be^{-x} = c$ into a quadratic, MULTIPLY every term by $e^x$ , giving $ae^{2x} + b = ce^x$ .
✕Accepting a non-positive value of $e^x$ as a solution
9231 Paper 2 Examiner Reports
Why it happens
Solving the quadratic in $u = e^x$ and forgetting that $e^x > 0$ for all real $x$ .
How to avoid it
After solving, reject any root with $u \leq 0$ and state the rejection clearly — it is a marked step.
✕Giving only one solution to a $\cosh x = k$ equation
9231 Paper 2 mark schemes
Why it happens
Forgetting that $\cosh$ is even, so $\cosh x = k$ (with $k > 1$ ) has two solutions $x = \pm c$ .
How to avoid it
Whenever $\cosh$ appears, expect a $\pm$ pair. The quadratic in $e^x$ naturally produces both reciprocal roots.
✕Quoting an identity instead of proving it on a 'prove' question
9231 Paper 2 Examiner Reports
Why it happens
Treating a 'prove'/'show that' as a 'state'; the result is well known so students skip the working.
How to avoid it
For 'prove from the definitions', substitute $\frac{e^x \pm e^{-x}}{2}$ and show every algebraic step — the stated result earns nothing on its own.

Hyperbolic functions in terms of exponential functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Evaluating sinh and cosh from the definition

2Writing tanh in exponential form

3Proving cosh²x − sinh²x = 1

4Proving the double-angle identity for sinh

5Solving a hyperbolic equation via a quadratic in eˣ

6Solving an equation using cosh² − sinh² = 1

Definition of sinh

Definition of cosh

Definition of tanh and reciprocals

Fundamental hyperbolic identity

Double-angle identities

Osborn's rule

Hyperbolic sine (sinh⁡\sinhsinh)

Hyperbolic cosine (cosh⁡\coshcosh)

Hyperbolic tangent (tanh⁡\tanhtanh)

sech, cosech, coth

Osborn's rule

Fundamental identity

✕Swapping the signs in the definitions (sinh⁡x=ex+e−x2\sinh x = \frac{e^x + e^{-x}}{2}sinhx=2ex+e−x​)

✕Forgetting the sign change in cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1cosh2x−sinh2x=1

✕Dividing by exe^xex instead of multiplying when forming a quadratic

✕Accepting a non-positive value of exe^xex as a solution

✕Giving only one solution to a cosh⁡x=k\cosh x = kcoshx=k equation

✕Quoting an identity instead of proving it on a 'prove' question

Hyperbolic sine ( $\sinh$ )

Hyperbolic cosine ( $\cosh$ )

Hyperbolic tangent ( $\tanh$ )

✕Swapping the signs in the definitions ( $\sinh x = \frac{e^x + e^{-x}}{2}$ )

✕Forgetting the sign change in $\cosh^2 x - \sinh^2 x = 1$

✕Dividing by $e^x$ instead of multiplying when forming a quadratic

✕Accepting a non-positive value of $e^x$ as a solution

✕Giving only one solution to a $\cosh x = k$ equation