Hyperbolic Functions

Q: What notation does Edexcel use for inverse hyperbolic functions?

Edexcel uses the ar- prefix for inverse hyperbolic functions: , , . You must not use or similar notation for logarithmic inverse forms in exam answers.

Q: Do I need to memorize the logarithmic forms of inverse hyperbolic functions?

The logarithmic forms are provided in the official formula book, but you may be asked to derive them from first principles, so you must understand the full proof process.

Q: Can I use Osborn's rule to remember hyperbolic identities?

Yes, Osborn's rule is a valid shortcut, but you will only be examined on the core identities listed in the FP3 specification. For proof questions, always derive identities from exponential definitions for full marks.

Edexcel International A-Level Further Mathematics· FP3 1.1-1.2· 45 min read

1. Definitions and Graphs of Hyperbolic Functions★★☆☆☆⏱ 10 min

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📘 Definition

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:

$tanh x = \frac{s i n h x}{c o s h x}$ , $sech x = \frac{1}{c o s h x}$ , $cosech x = \frac{1}{s i n h x}$ , $coth x = \frac{1}{t a n h x}$

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
$cosh x$	$R$	$[1, \infty)$
$sinh x$	$R$	$R$
$tanh x$	$R$	$(- 1, 1)$
$sech x$	$R$	$(0, 1]$
$cosech x$	$R ∖ {0}$	$R ∖ {0}$
$coth x$	$R ∖ {0}$	$R ∖ [- 1, 1]$

📐 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: $e^{l n k} = k$ and $e^{- l n k} = 1/ k$ :
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.

2. Core Hyperbolic Identities★★★☆☆⏱ 12 min

📘 Definition

Core hyperbolic identity

$cosh^{2} x$ means $(cosh x)^{2}$ , matching standard trigonometric notation

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.

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 $cosh 2 x \equiv cosh^{2} x + sinh^{2} x$ using exponential definitions.

1
Start with the exponential definition of $cosh 2 x$ :
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 $cosh 2 x$ , completing the proof:
9
$= \frac{1}{2}(e^{2x} + e^{-2x}) = \cosh 2x$

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.

3. Solving Linear Hyperbolic Equations★★★★☆⏱ 12 min

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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 $e^{x} = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ 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.

4. Inverse Hyperbolic Functions★★★★☆⏱ 11 min

📘 Definition

Inverse hyperbolic functions

The inverses of the primary hyperbolic functions, denoted $arsinh x$ , $arcosh x$ , and $artanh x$ 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 $arcosh x$ 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 $arsinh x = ln (x + x^{2} + 1)$ for all real $x$ .

1
Let $y = arsinh x$ . 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 $2 e^{y}$ 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: $x - x^{2} + 1 < 0$ 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
$arsinh x$	$R$	$ln (x + x^{2} + 1)$
$arcosh x$	$x \geq 1$	$ln (x + x^{2} - 1)$ (principal value $\geq 0$ )
$artanh x$	$∣ x ∣ < 1$	$\frac{1}{2} ln (\frac{1 + x}{1 - x})$

Exam tip:

Always state the domain restriction for inverse hyperbolic functions when using them, especially for $arcosh x$ (must have $x \geq 1$ ) and $artanh x$ (must have $∣ x ∣ < 1$ ).

5. Common Pitfalls

Wrong move:

Using $sinh^{- 1} x$ notation for inverse hyperbolic functions

Why:

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 move:

Always write $arsinh x$ , $arcosh x$ , or $artanh x$ when referring to inverse hyperbolic functions.

Wrong move:

Forgetting to reject negative roots for $e^{x}$ when solving hyperbolic equations

Why:

$e^{x}$ is always positive for all real values of $x$ , so any negative solution for $e^{x}$ is not valid for real $x$ .

Correct move:

After solving the quadratic in $e^{x}$ , check that all roots are positive, and discard any negative roots before taking logarithms.

Wrong move:

Using the trigonometric Pythagorean identity sign for hyperbolic functions

Why:

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 move:

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 move:

Quoting the logarithmic form of an inverse hyperbolic function instead of deriving it when asked to prove it

Why:

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 move:

When asked to prove a logarithmic inverse form, start with $y = [inverse function]$ , 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 move:

Using $arcosh x$ for values of $x$ less than 1, or $artanh x$ for $∣ x ∣ \geq 1$

Why:

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 move:

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.

6. Quick Reference Cheatsheet

Concept	Key Details
Exponential definitions	$sinh x = \frac{1}{2} (e^{x} - e^{- x})$ , $cosh x = \frac{1}{2} (e^{x} + e^{- x})$ , $tanh x = \frac{s i n h x}{c o s h x}$
Core identities	$cosh^{2} x - sinh^{2} x \equiv 1$ , $cosh 2 x = cosh^{2} x + sinh^{2} x$ , $sinh 2 x = 2 sinh x cosh x$
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 $x = ln (x + x^{2} + 1)$ (all $x$ ), arcosh $x = ln (x + x^{2} - 1)$ ( $x \geq 1$ ), artanh $x = \frac{1}{2} ln (\frac{1 + x}{1 - x})$ ( $∣ x ∣ < 1$ )

7. Frequently Asked

What notation does Edexcel use for inverse hyperbolic functions?

Edexcel uses the ar- prefix for inverse hyperbolic functions: $arsinh x$ , $arcosh x$ , $artanh x$ . You must not use $sinh^{- 1} x$ or similar $^{- 1}$ notation for logarithmic inverse forms in exam answers.

Do I need to memorize the logarithmic forms of inverse hyperbolic functions?

The logarithmic forms are provided in the official formula book, but you may be asked to derive them from first principles, so you must understand the full proof process.

Can I use Osborn's rule to remember hyperbolic identities?

Yes, Osborn's rule is a valid shortcut, but you will only be examined on the core identities listed in the FP3 specification. For proof questions, always derive identities from exponential definitions for full marks.

Going deeper

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.