# Hyperbolic Functions

> Edexcel International A-Level Further Mathematics · FP3 (2018 Spec)
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-hyperbolic-functions/

This guide covers Edexcel IAL FP3 hyperbolic function content: exponential definitions, graphs, core identities, solving equations of the form $a \cosh x + b \sinh x = c$, and inverse hyperbolic functions with their logarithmic forms.

**Prerequisites:** [Exponential and logarithmic function manipulation](https://www.owlsprep.com/study/edexcel-ial-maths-pure-3-exponentials-logarithms/); [Quadratic equation solving, including root rejection](https://www.owlsprep.com/study/edexcel-ial-maths-pure-1-quadratics/)

## Learning objectives

- Define all six hyperbolic functions in terms of exponentials and recall their graphs, domains and ranges
- Derive and apply core hyperbolic identities including $\cosh^2x - \sinh^2x \equiv 1$ and double angle formulas
- Solve equations of the form $a \cosh x + b \sinh x = c$ using exponential substitution and quadratic solving
- Derive the logarithmic forms of inverse hyperbolic functions (arsinh, arcosh, artanh) and apply their properties

## Definitions and Graphs of Hyperbolic Functions

**Hyperbolic functions** — Six functions defined in terms of $e^x$ and $e^{-x}$, analogous to trigonometric functions. The primary definitions are $\sinh x = \frac{1}{2}(e^x - e^{-x})$ and $\cosh x = \frac{1}{2}(e^x + e^{-x})$, with the remaining four derived as reciprocals or ratios of these two.

*Example:* $\tanh x = \frac{\sinh x}{\cosh x}$, $\text{sech } x = \frac{1}{\cosh x}$, $\text{cosech } x = \frac{1}{\sinh x}$, $\coth x = \frac{1}{\tanh x}$

Each hyperbolic function has a defined domain and range you must recall for the exam. $\cosh x$ is always greater than or equal to 1, $\sinh x$ can take any real value, and $\tanh x$ is bounded between -1 and 1. Reciprocal functions have domains excluding points where the original function equals zero.

| Function | Domain | Range |
| --- | --- | --- |
| $\cosh x$ | $\mathbb{R}$ | $[1, \infty)$ |
| $\sinh x$ | $\mathbb{R}$ | $\mathbb{R}$ |
| $\tanh x$ | $\mathbb{R}$ | $(-1, 1)$ |
| $\text{sech } x$ | $\mathbb{R}$ | $(0, 1]$ |
| $\text{cosech } x$ | $\mathbb{R} \setminus \{0\}$ | $\mathbb{R} \setminus \{0\}$ |
| $\coth x$ | $\mathbb{R} \setminus \{0\}$ | $\mathbb{R} \setminus [-1, 1]$ |

**Worked example:** Calculate the exact value of $\cosh(\ln 3)$ using its exponential definition.

1. Substitute $x = \ln 3$ into the definition of $\cosh x$:
2. $$\cosh(\ln 3) = \frac{1}{2}(e^{\ln 3} + e^{-\ln 3})$$
3. Simplify using logarithm rules: $e^{\ln k} = k$ and $e^{-\ln k} = 1/k$:
4. $$= \frac{1}{2}\left(3 + \frac{1}{3}\right) = \frac{1}{2}\left(\frac{10}{3}\right) = \frac{5}{3}$$

> **Exam tip:** Always simplify exponential expressions using logarithm rules when asked for exact values of hyperbolic functions with logarithmic inputs, as this avoids calculator rounding errors.

*Calculator:* allowed

## Core Hyperbolic Identities

**Core hyperbolic identity** — The identity $\cosh^2 x - \sinh^2 x \equiv 1$, derived directly from the exponential definitions of $\cosh x$ and $\sinh x$. This is analogous to the Pythagorean trigonometric identity, with a sign change for the squared $\sinh$ term.

*Notation:* $\cosh^2 x$ means $(\cosh x)^2$, matching standard trigonometric notation

Other core identities, including double angle formulas, are derived from this base identity and exponential definitions. While these identities are provided in the formula book, you may be asked to prove them to demonstrate your understanding of their origin.

**Worked example:** Prove that $\cosh 2x \equiv \cosh^2 x + \sinh^2 x$ using exponential definitions.

1. Start with the exponential definition of $\cosh 2x$:
2. $$\cosh 2x = \frac{1}{2}(e^{2x} + e^{-2x})$$
3. Expand $\cosh^2 x$ and $\sinh^2 x$ separately using their definitions:
4. $$\cosh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}$$
5. $$\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}$$
6. Add the two expanded expressions together:
7. $$\cosh^2 x + \sinh^2 x = \frac{e^{2x} + 2 + e^{-2x} + e^{2x} - 2 + e^{-2x}}{4} = \frac{2(e^{2x} + e^{-2x})}{4}$$
8. Simplify to match the definition of $\cosh 2x$, completing the proof:
9. $$= \frac{1}{2}(e^{2x} + e^{-2x}) = \cosh 2x$$

> **mnemonic**
>
> Osborn's Rule: Convert trigonometric identities to hyperbolic by replacing $\sin$ with $\sinh$, $\cos$ with $\cosh$, and flipping the sign of any term containing a product of two $\sinh$ terms. E.g. $\cos^2 x + \sin^2 x = 1$ becomes $\cosh^2 x - \sinh^2 x = 1$.

> **Exam tip:** If asked to prove an identity, always start from the exponential definitions unless explicitly told otherwise, as this guarantees full marks even if you forget Osborn's rule.

## Solving Linear Hyperbolic Equations

Equations of the form $a \cosh x + b \sinh x = c$ are solved by substituting the exponential definitions of $\cosh x$ and $\sinh x$, rearranging to form a quadratic in $e^x$, solving the quadratic, and rejecting any non-positive roots (since $e^x > 0$ for all real $x$).

**Worked example:** Solve the equation $3 \cosh x + 2 \sinh x = 5$, giving your answers as exact logarithms.

1. Substitute the exponential definitions of $\cosh x$ and $\sinh x$ into the equation:
2. $$3\left(\frac{e^x + e^{-x}}{2}\right) + 2\left(\frac{e^x - e^{-x}}{2}\right) = 5$$
3. Multiply through by 2 to eliminate denominators:
4. $$3(e^x + e^{-x}) + 2(e^x - e^{-x}) = 10$$
5. Expand and simplify like terms:
6. $$5e^x + e^{-x} = 10$$
7. Multiply through by $e^x$ to eliminate the negative exponent, forming a quadratic in $e^x$:
8. $$5(e^x)^2 - 10e^x + 1 = 0$$
9. Solve using the quadratic formula $e^x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=5$, $b=-10$, $c=1$:
10. $$e^x = \frac{10 \pm \sqrt{80}}{10} = \frac{5 \pm 2\sqrt{5}}{5}$$
11. Check both roots are positive (both are ~1.89 and ~0.11, so valid), then take natural logs of both sides:
12. $$x = \ln\left(\frac{5 + 2\sqrt{5}}{5}\right) \text{ or } x = \ln\left(\frac{5 - 2\sqrt{5}}{5}\right)$$

> **Exam tip:** Always verify that your roots for $e^x$ are positive, as $e^x$ can never be zero or negative for real $x$. You will lose marks if you include invalid negative solutions.

*Calculator:* allowed

## Inverse Hyperbolic Functions

**Inverse hyperbolic functions** — The inverses of the primary hyperbolic functions, denoted $\text{arsinh } x$, $\text{arcosh } x$, and $\text{artanh } x$ per Edexcel notation. Each has a restricted domain to ensure a one-to-one mapping, and can be written as an equivalent logarithmic function.

*Example:* The principal value of $\text{arcosh } x$ is the non-negative value for $x \geq 1$, as required by the syllabus.

You may be asked to derive the logarithmic form of any of the three inverse hyperbolic functions from first principles, even though the forms are given in the formula book.

**Worked example:** Prove that $\text{arsinh } x = \ln\left(x + \sqrt{x^2 + 1}\right)$ for all real $x$.

1. Let $y = \text{arsinh } x$. By definition of inverse functions, this means $x = \sinh y$.
2. Substitute the exponential definition of $\sinh y$:
3. $$x = \frac{e^y - e^{-y}}{2}$$
4. Multiply through by $2e^y$ to eliminate the negative exponent, forming a quadratic in $e^y$:
5. $$e^{2y} - 2x e^y - 1 = 0$$
6. Solve using the quadratic formula, treating $e^y$ as the variable:
7. $$e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1}$$
8. Reject the negative root: $x - \sqrt{x^2 + 1} < 0$ for all $x$, and $e^y > 0$ for all real $y$.
9. Take the natural logarithm of both sides to solve for $y$:
10. $$y = \ln\left(x + \sqrt{x^2 + 1}\right) = \text{arsinh } x, \text{ as required}$$

| Inverse Function | Domain | Logarithmic Form |
| --- | --- | --- |
| $\text{arsinh } x$ | $\mathbb{R}$ | $\ln\left(x + \sqrt{x^2 + 1}\right)$ |
| $\text{arcosh } x$ | $x \geq 1$ | $\ln\left(x + \sqrt{x^2 - 1}\right)$ (principal value $\geq 0$) |
| $\text{artanh } x$ | $\|x\| < 1$ | $\frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$ |

> **Exam tip:** Always state the domain restriction for inverse hyperbolic functions when using them, especially for $\text{arcosh } x$ (must have $x \geq 1$) and $\text{artanh } x$ (must have $|x| < 1$).

## Common pitfalls

- **Wrong:** Using $\sinh^{-1} x$ notation for inverse hyperbolic functions
  - Why it fails: Edexcel specifies the use of the `ar-` prefix (arsinh, arcosh, artanh) for logarithmic inverse forms; $^{-1}$ notation is reserved for reciprocal functions in this context and will lose marks.
  - Correct: Always write $\text{arsinh } x$, $\text{arcosh } x$, or $\text{artanh } x$ when referring to inverse hyperbolic functions.
- **Wrong:** Forgetting to reject negative roots for $e^x$ when solving hyperbolic equations
  - Why it fails: $e^x$ is always positive for all real values of $x$, so any negative solution for $e^x$ is not valid for real $x$.
  - Correct: After solving the quadratic in $e^x$, check that all roots are positive, and discard any negative roots before taking logarithms.
- **Wrong:** Using the trigonometric Pythagorean identity sign for hyperbolic functions
  - Why it fails: The core hyperbolic identity is $\cosh^2 x - \sinh^2 x \equiv 1$, not $\cosh^2 x + \sinh^2 x \equiv 1$. The sign flips because of the negative term in the definition of $\sinh x$.
  - Correct: Use Osborn's rule to remember the sign change for terms containing products of two $\sinh$ functions, or derive identities from exponential definitions to avoid sign errors.
- **Wrong:** Quoting the logarithmic form of an inverse hyperbolic function instead of deriving it when asked to prove it
  - Why it fails: Exam questions explicitly asking for a proof require you to show full working from first principles, not just state the given formula from the formula book.
  - Correct: When asked to prove a logarithmic inverse form, start with $y = \text{[inverse function]}$, rearrange to get the hyperbolic function of $y$ equal to $x$, substitute the exponential definition, solve the quadratic in $e^y$, reject invalid roots, and take logs to get the final form.
- **Wrong:** Using $\text{arcosh } x$ for values of $x$ less than 1, or $\text{artanh } x$ for $|x| \geq 1$
  - Why it fails: These inverse functions are only defined for their specified domains, so using them outside these ranges gives undefined or non-real results, which are invalid for real number exam questions.
  - Correct: Always check that the input to an inverse hyperbolic function falls within its valid domain before using it, and state the domain restriction in your working where relevant.

## Cheatsheet

| Concept | Key Details |
| --- | --- |
| Exponential definitions | $\sinh x = \frac{1}{2}(e^x - e^{-x})$, $\cosh x = \frac{1}{2}(e^x + e^{-x})$, $\tanh x = \frac{\sinh x}{\cosh x}$ |
| Core identities | $\cosh^2 x - \sinh^2 x \equiv 1$, $\cosh 2x = \cosh^2 x + \sinh^2 x$, $\sinh 2x = 2\sinh x \cosh x$ |
| Solving $a \cosh x + b \sinh x = c$ | Substitute exponentials, form quadratic in $e^x$, reject negative roots, take natural log |
| Inverse function log forms | arsinh $x = \ln(x + \sqrt{x^2 + 1})$ (all $x$), arcosh $x = \ln(x + \sqrt{x^2 -1})$ ($x \geq 1$), artanh $x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ ($\|x\|<1$) |

## What's next

Now that you have mastered the core definitions, identities, equations and inverse forms of hyperbolic functions, you are ready to move on to applying these concepts to differentiation and integration problems in FP3. Hyperbolic functions are used extensively in further calculus topics, including integration by substitution and solving differential equations, so having a solid grasp of their basic properties is critical for success in the rest of the FP3 unit. You may also want to practice past paper questions focused on this topic to familiarize yourself with Edexcel's exam phrasing and mark scheme expectations.

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