Hyperbolic Functions

CIE A-Level Further Mathematics· 6 min read

1. Definitions and Graphs of Hyperbolic Functions★★☆☆☆⏱ 15 min

📘 Definition

Core Hyperbolic Functions

$sinh x$ , $cosh x$ , $tanh x$ , $coth x$ , $sech x$ , $csch x$

All hyperbolic functions are defined in terms of exponentials: $sinh x = \frac{e ^{x} - e ^{- x}}{2}$ , $cosh x = \frac{e ^{x} + e ^{- x}}{2}$ , $tanh x = \frac{s i n h x}{c o s h x}$ , $coth x = \frac{1}{t a n h x}$ , $sech x = \frac{1}{c o s h x}$ , $csch x = \frac{1}{s i n h x}$

Example:

$sinh 0 = 0$ , $cosh 0 = 1$ , $tanh 0 = 0$

Hyperbolic functions share many similarities with trigonometric functions, but have different parity and range properties:

$cosh x$ is even ( $cosh (- x) = cosh x$ ), range $[1, \infty)$
$sinh x$ is odd ( $sinh (- x) = - sinh x$ ), range $(- \infty, \infty)$
$tanh x$ is odd, range $(- 1, 1)$

📐 Worked Example

Find the exact value of $cosh (ln 3)$

1
Start with the definition of $cosh x$ :
$\cosh x = \frac{e^x + e^{-x}}{2}$
2
Substitute $x = ln 3$ , using the property $e^{k l n a} = a^{k}$ :
$e^{\ln 3} = 3, \quad e^{-\ln 3} = \frac{1}{3}$
3
Calculate and simplify:
$\cosh(\ln 3) = \frac{3 + \frac{1}{3}}{2} = \frac{10}{6} = \frac{5}{3}$

2. Hyperbolic Identities★★☆☆☆⏱ 20 min

📘 Definition

Fundamental Pythagorean Hyperbolic Identity

The core identity for all hyperbolic functions is: $cosh^{2} x - sinh^{2} x = 1$

$sinh (A + B) = sinh A cosh B + cosh A sinh B$
$cosh (A + B) = cosh A cosh B + sinh A sinh B$
$cosh 2 x = cosh^{2} x + sinh^{2} x = 2 cosh^{2} x - 1 = 2 sinh^{2} x + 1$
$sinh 2 x = 2 sinh x cosh x$

📐 Worked Example

Prove that $cosh^{2} x - sinh^{2} x = 1$ using exponential definitions

1
Substitute the definitions of $cosh x$ and $sinh x$ into the left-hand side:
$\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2$
2
Expand both squares:
$= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}$
3
Simplify the numerator:
$= \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1$
4
This matches the right-hand side, completing the proof.

3. Inverse Hyperbolic Functions★★★☆☆⏱ 20 min

Restricted domains make core hyperbolic functions one-to-one, so we can define inverse functions that can be written exactly in terms of natural logarithms.

📘 Definition

Logarithmic Forms of Inverse Hyperbolic Functions

$arsinh x$ , $arcosh x$ , $artanh x$

$arsinh x = ln (x + x^{2} + 1)$ for all $x \in R$ ; $arcosh x = ln (x + x^{2} - 1)$ for $x \geq 1$ ; $artanh x = \frac{1}{2} ln (\frac{1 + x}{1 - x})$ for $∣ x ∣ < 1$

📐 Worked Example

Express $arsinh 3$ as an exact natural logarithm

1
Use the standard logarithmic form for $arsinh x$ :
$\text{arsinh}\,x = \ln\left(x + \sqrt{x^2 + 1}\right)$
2
Substitute $x = 3$ :
$x^2 + 1 = 9 + 1 = 10$
3
Final result:
$\text{arsinh}\,3 = \ln\left(3 + \sqrt{10}\right)$

✓ Quick check

Test your understanding of domain restrictions

What is the domain of $arcosh x$ ?
- All real numbers
- $x \geq 1$
- $∣ x ∣ < 1$
- $x > 0$
Reveal answer
1 —
Correct: $cosh x$ has range $[1, \infty)$ , so the inverse function $arcosh x$ has domain $[1, \infty)$

4. Calculus of Hyperbolic Functions★★★☆☆⏱ 25 min

🚫 No Calculator

Differentiating and integrating hyperbolic and inverse hyperbolic functions gives standard results that are frequently used for integrals involving quadratics under square roots.

Function	Derivative
$sinh x$	$cosh x$
$cosh x$	$sinh x$
$tanh x$	$sech^{2} x$
$arsinh x$	$\frac{1}{x ^{2} + 1}$
$arcosh x$	$\frac{1}{x ^{2} - 1}$
$artanh x$	$\frac{1}{1 - x ^{2}}$

📐 Worked Example

Find $\int \frac{1}{9 x ^{2} - 16} d x$ for $x > \frac{4}{3}$

1
Factor out constants to match the standard integral form for inverse hyperbolic functions:
$\int \frac{1}{\sqrt{9x^2 - 16}} dx = \frac{1}{3} \int \frac{1}{\sqrt{(3x)^2 - 4^2}} d(3x)$
2
Use the standard result $\int \frac{1}{u ^{2} - a ^{2}} d u = arcosh (\frac{u}{a}) + C$ :
$= \frac{1}{3} \text{arcosh}\left(\frac{3x}{4}\right) + C$
3
This can also be rewritten in logarithmic form if required:
$= \frac{1}{3} \ln\left(3x + \sqrt{9x^2 - 16}\right) + C'$

5. Common Pitfalls

Wrong move:

Sign error in hyperbolic identities when using Osborn's rule

Why:

Students often forget to flip the sign for terms containing a product of two sines

Correct move:

Apply Osborn's rule systematically: flip the sign of any term that has a product of two hyperbolic sines

Wrong move:

Using $arcosh x$ for $x < 1$

Why:

The function is undefined for $x < 1$ , so any result with this domain is invalid

Correct move:

Always check the argument of $arcosh x$ is at least 1 before using the function

Wrong move:

Writing derivative of $tanh x$ as $- sech^{2} x$

Why:

Confusion with derivative of $tan x$ leads to incorrect negative sign

Correct move:

Recall $\frac{d}{d x} tanh x = sech^{2} x$ , there is no negative sign for this derivative

Wrong move:

Missing the constant factor for linear substitutions in integrals

Why:

When substituting $u = k x$ , $d u = k d x$ so $d x = \frac{d u}{k}$ , this factor is often missed

Correct move:

Always adjust the integral for the derivative of the substitution to get the correct constant factor in the final result

6. Quick Reference Cheatsheet

Function	Definition	Derivative	Inverse Log Form
$sinh x$	$\frac{e ^{x} - e ^{- x}}{2}$	$cosh x$	$ln (x + x^{2} + 1)$
$cosh x$	$\frac{e ^{x} + e ^{- x}}{2}$	$sinh x$	$ln (x + x^{2} - 1), x \geq 1$
$tanh x$	$\frac{s i n h x}{c o s h x}$	$sech^{2} x$	$\frac{1}{2} ln (\frac{1 + x}{1 - x}), ∣ x ∣ < 1$
Key Identity			$cosh^{2} x - sinh^{2} x = 1$
$\int \frac{1}{x ^{2} + a ^{2}} d x$			$arsinh (\frac{x}{a}) + C$
$\int \frac{1}{x ^{2} - a ^{2}} d x$			$arcosh (\frac{x}{a}) + C$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Prove identity, solve hyperbolic equation
2023 · 2
Differentiate inverse hyperbolic function
2021 · 1
Integral using inverse hyperbolic result

Going deeper

syllabusCIE 9231 Further Mathematics Syllabus

What's Next

Hyperbolic functions are a foundational topic for Further Pure 2, and they appear regularly in integration problems, arc length calculations, coordinate geometry, and solving differential equations. Mastery of their identities and calculus results simplifies more advanced topics like reduction formulae and improper integrals. Many integration results that use inverse trigonometric functions can also be expressed in terms of inverse hyperbolic functions, so this topic helps you connect different areas of calculus and prepare for mixed questions in the exam.