Hyperbolic functions

Q: What is Osborn's Rule and how is it applied?

Replace trigonometric functions with hyperbolic ones, but change the sign in front of any term involving the product of two sine terms (explicit or implicit).

Q: How do you solve a linear equation mixing \(\sinh x\) and \(\cosh x\)?

Substitute the exponential definitions \(\frac{e^x - e^{-x}}{2}\) and \(\frac{e^x + e^{-x}}{2}\), multiply by \(e^x\), and solve the resulting quadratic in \(e^x\).

Q: What are the logarithmic forms of \(\operatorname{arsinh} x\), \(\operatorname{arcosh} x\), and \(\operatorname{artanh} x\)?

\(\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right)\), \(\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right)\) (for \(x \ge 1\)), and \(\operatorname{artanh} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) (for \(|x|

Q: What are the derivatives of \(\sinh x\), \(\cosh x\), and \(\tanh x\)?

\(\frac{d}{dx}(\sinh x) = \cosh x\), \(\frac{d}{dx}(\cosh x) = \sinh x\), and \(\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x\).

Q: What are the standard derivatives of the inverse hyperbolic functions \(\operatorname{arsinh} x\), \(\operatorname{arcosh} x\), and \(\operatorname{artanh} x\)?

\(\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}}\), \(\frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}}\) (for \(x > 1\)), and \(\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2}\) (for \(|x|

Further Mathematics 2330 · CCEA A Level · 9 min read · Updated: 31 Aug 2026

Introduction to Hyperbolic Functions

Welcome to one of the most elegant and practical topics in A2 Further Mathematics: Hyperbolic Functions! If you have ever wondered what shape a power cable makes when hanging freely between two pylons, or how the gateway arch in St. Louis was designed, you are looking at hyperbolic geometry in action. This shape is called a catenary, and it is modeled by a hyperbolic function.

Don't worry if this seems intimidating at first. If you are already comfortable with standard trigonometry (\(\sin\), \(\cos\), \(\tan\)) and working with exponential functions (\(e^x\)), you already have all the tools you need. Hyperbolic functions are simply combinations of exponential curves that share remarkable parallels with ordinary trigonometric functions.

1. Definitions of Hyperbolic Functions

The Core Exponential Definitions

Unlike circular trigonometric functions which relate to the unit circle \(x^2 + y^2 = 1\), hyperbolic functions relate to the standard hyperbola \(x^2 - y^2 = 1\). They are defined using the exponential function \(e^x\):

• Hyperbolic Sine (pronounced "shine" or "sinch"):
\(\sinh x = \frac{e^x - e^{-x}}{2}\)

• Hyperbolic Cosine (pronounced "cosh"):
\(\cosh x = \frac{e^x + e^{-x}}{2}\)

• Hyperbolic Tangent (pronounced "than" or "tansh"):
\(\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1}\)

Reciprocal Hyperbolic Functions

Just like standard trigonometry, each primary function has a reciprocal:

• Hyperbolic Secant: \(\operatorname{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}\)
• Hyperbolic Cosecant: \(\operatorname{cosech} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}\) (where \(x \neq 0\))
• Hyperbolic Cotangent: \(\operatorname{coth} x = \frac{1}{\tanh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}\) (where \(x \neq 0\))

Quick Review:
What happens at \(x = 0\)?
\(\sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0\)
\(\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1\)
\(\tanh(0) = \frac{\sinh(0)}{\cosh(0)} = \frac{0}{1} = 0\)

Key Takeaway: \(\sinh x\) and \(\cosh x\) are just the odd and even exponential combinations divided by \(2\). Always remember that \(\cosh(0) = 1\) and \(\sinh(0) = 0\).

2. Graphs, Symmetries, Domains, and Ranges

Symmetry Properties

• \(\cosh x\) is an even function: \(\cosh(-x) = \frac{e^{-x} + e^{-(-x)}}{2} = \frac{e^{-x} + e^x}{2} = \cosh x\). Its graph is symmetrical about the \(y\)-axis.
• \(\sinh x\) is an odd function: \(\sinh(-x) = \frac{e^{-x} - e^x}{2} = -\left(\frac{e^x - e^{-x}}{2}\right) = -\sinh x\). Its graph has rotational symmetry of order \(2\) about the origin.
• \(\tanh x\) is an odd function: \(\tanh(-x) = -\tanh x\).

Graph Characteristics

• The graph of \(y = \cosh x\):
Domain: \(x \in \mathbb{R}\)
Range: \(y \ge 1\)
Shape: A smooth U-shaped curve (the catenary) with a global minimum at \((0, 1)\). As \(x \to \pm\infty\), \(\cosh x \to +\infty\).

• The graph of \(y = \sinh x\):
Domain: \(x \in \mathbb{R}\)
Range: \(y \in \mathbb{R}\)
Shape: Strictly increasing curve passing through \((0, 0)\). As \(x \to +\infty\), \(\sinh x \to +\infty\); as \(x \to -\infty\), \(\sinh x \to -\infty\).

• The graph of \(y = \tanh x\):
Domain: \(x \in \mathbb{R}\)
Range: \(-1 < y < 1\)
Shape: An S-shaped curve passing through \((0, 0)\) with two horizontal asymptotes: \(y = 1\) (as \(x \to +\infty\)) and \(y = -1\) (as \(x \to -\infty\)).

Key Takeaway: Notice that while circular trigonometric functions oscillate between \(-1\) and \(1\), hyperbolic functions are not periodic and grow exponentially!

3. Hyperbolic Identities and Osborn's Rule

The Fundamental Identity

In trigonometry, we have \(\cos^2 x + \sin^2 x = 1\). For hyperbolic functions, the fundamental relationship has a minus sign:

\(\cosh^2 x - \sinh^2 x = 1\)

Quick Proof:
\(\cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2\)
\(= \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}\)
\(= \frac{4}{4} = 1\)

Dividing the Fundamental Identity

• Divide by \(\cosh^2 x\): \(1 - \tanh^2 x = \operatorname{sech}^2 x\)
• Divide by \(\sinh^2 x\): \(\operatorname{coth}^2 x - 1 = \operatorname{cosech}^2 x\)

Osborn's Rule

You do not need to memorize entirely new formula lists for hyperbolic functions! Osborn's Rule allows you to convert any standard trigonometric identity into its hyperbolic counterpart:

1. Replace every trigonometric function with its corresponding hyperbolic function (\(\sin \to \sinh\), \(\cos \to \cosh\), etc.).
2. Change the sign in front of any term that involves the product of two sine terms (explicitly or implicitly, such as \(\sin^2\), \(\tan^2\), \(\sin A \sin B\), or \(\tanh^2\)).

Examples of Osborn's Rule:
• \(\cos(2x) = \cos^2 x - \sin^2 x \implies \cosh(2x) = \cosh^2 x + \sinh^2 x\)
• \(\cos(2x) = 1 - 2\sin^2 x \implies \cosh(2x) = 1 + 2\sinh^2 x\)
• \(\cos(2x) = 2\cos^2 x - 1 \implies \cosh(2x) = 2\cosh^2 x - 1\)
• \(\sin(2x) = 2\sin x \cos x \implies \sinh(2x) = 2\sinh x \cosh x\) (no sign change, as only one sine term is present)
• \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \implies \tanh(A + B) = \frac{\tanh A + \tanh B}{1 + \tanh A \tanh B}\) (sign changes in denominator because \(\tan A \tan B\) involves \(\sin A \sin B\))

Common Mistake to Avoid: Be careful with \(\tan^2 x\) and \(\tanh^2 x\)! Since \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\), it contains a product of two sines, so \(\sec^2 x = 1 + \tan^2 x\) becomes \(\operatorname{sech}^2 x = 1 - \tanh^2 x\).

4. Solving Hyperbolic Equations

There are two primary methods for solving equations containing hyperbolic functions:

Method 1: Expressing in terms of Exponentials (\(e^x\))

This is the most dependable method when solving equations mixing \(\sinh x\), \(\cosh x\), and constants.

Worked Example: Solve \(6\sinh x - 2\cosh x = 7\).

Step 1: Replace \(\sinh x\) and \(\cosh x\) with their exponential definitions:
\(6\left(\frac{e^x - e^{-x}}{2}\right) - 2\left(\frac{e^x + e^{-x}}{2}\right) = 7\)

Step 2: Simplify the coefficients:
\(3(e^x - e^{-x}) - (e^x + e^{-x}) = 7\)
\(3e^x - 3e^{-x} - e^x - e^{-x} = 7\)
\(2e^x - 4e^{-x} = 7\)

Step 3: Multiply every term by \(e^x\) to remove the negative power, and let \(u = e^x\):
\(2(e^x)^2 - 7e^x - 4 = 0\)
\(2u^2 - 7u - 4 = 0\)

Step 4: Factorise and solve for \(u\):
\((2u + 1)(u - 4) = 0 \implies u = -\frac{1}{2} \text{ or } u = 4\)

Step 5: Substitute back \(e^x = u\):
Since \(e^x > 0\) for all real \(x\), \(e^x = -\frac{1}{2}\) gives no real solutions.
For \(e^x = 4 \implies x = \ln 4\).

Method 2: Using Identities

If an equation involves quadratic powers of hyperbolic terms, use identities like \(\cosh^2 x - \sinh^2 x = 1\).

Worked Example: Solve \(2\cosh^2 x + \sinh x = 5\).

Step 1: Use \(\cosh^2 x = 1 + \sinh^2 x\):
\(2(1 + \sinh^2 x) + \sinh x = 5\)
\(2\sinh^2 x + \sinh x - 3 = 0\)

Step 2: Factorise the quadratic in \(\sinh x\):
\((2\sinh x + 3)(\sinh x - 1) = 0\)
\(\sinh x = -\frac{3}{2}\) or \(\sinh x = 1\)

Since the range of \(\sinh x\) is all real numbers, both values yield valid solutions for \(x\)!

5. Inverse Hyperbolic Functions

Definitions and Domains

• \(\operatorname{arsinh} x\): The inverse of \(\sinh x\). Defined for all \(x \in \mathbb{R}\).
• \(\operatorname{arcosh} x\): To make \(\cosh x\) a one-to-one function, its domain is restricted to \(x \ge 0\). Therefore, \(\operatorname{arcosh} x\) is defined only for \(x \ge 1\), with range \(y \ge 0\).
• \(\operatorname{artanh} x\): Defined for \(-1 < x < 1\).

Logarithmic Forms of Inverse Hyperbolic Functions

Because hyperbolic functions are defined using \(e^x\), their inverses can be expressed as natural logarithms:

• \(\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right)\) for all \(x \in \mathbb{R}\)
• \(\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right)\) for \(x \ge 1\)
• \(\operatorname{artanh} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\) for \(-1 < x < 1\)

Deriving the Logarithmic Form (Step-by-Step)

Exam questions frequently ask you to prove these formulas. Let's derive \(\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right)\):

Step 1: Let \(y = \operatorname{arsinh} x \implies x = \sinh y\).
Step 2: Write \(\sinh y\) using exponentials:
\(x = \frac{e^y - e^{-y}}{2} \implies 2x = e^y - e^{-y}\)
Step 3: Multiply through by \(e^y\):
\((e^y)^2 - 2x(e^y) - 1 = 0\)
Step 4: Use the quadratic formula to solve for \(e^y\):
\(e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2} = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1}\)
Step 5: Since \(\sqrt{x^2 + 1} > x\), choosing the minus sign would make \(e^y < 0\), which is impossible for real values. Thus:
\(e^y = x + \sqrt{x^2 + 1}\)
Step 6: Take the natural logarithm of both sides:
\(y = \ln\left(x + \sqrt{x^2 + 1}\right)\)

Key Takeaway: When finding exact values like \(\operatorname{arcosh}(3)\), you can quickly use the logarithmic form: \(\operatorname{arcosh}(3) = \ln(3 + \sqrt{3^2 - 1}) = \ln(3 + \sqrt{8}) = \ln(3 + 2\sqrt{2})\).

6. Differentiation and Integration

Derivatives of Hyperbolic Functions

Watch out: the derivative of \(\cosh x\) does not have a minus sign!

• \(\frac{d}{dx}(\sinh x) = \cosh x\)
• \(\frac{d}{dx}(\cosh x) = \sinh x\)
• \(\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x\)
• \(\frac{d}{dx}(\operatorname{sech} x) = -\operatorname{sech} x \tanh x\)
• \(\frac{d}{dx}(\operatorname{cosech} x) = -\operatorname{cosech} x \coth x\)
• \(\frac{d}{dx}(\operatorname{coth} x) = -\operatorname{cosech}^2 x\)

Standard Integrals

• \(\int \sinh x \, dx = \cosh x + c\)
• \(\int \cosh x \, dx = \sinh x + c\)
• \(\int \operatorname{sech}^2 x \, dx = \tanh x + c\)

Derivatives of Inverse Hyperbolic Functions

• \(\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}}\)
• \(\frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}}\) (for \(x > 1\))
• \(\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2}\) (for \(|x| < 1\))

Standard Integration Forms Involving Inverses

By using substitution or standard formula sheet results, we get:

• \(\int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \operatorname{arsinh}\left(\frac{x}{a}\right) + c = \ln\left(x + \sqrt{x^2 + a^2}\right) + c'\)
• \(\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \operatorname{arcosh}\left(\frac{x}{a}\right) + c = \ln\left(x + \sqrt{x^2 - a^2}\right) + c'\) (where \(x > a\))
• \(\int \frac{1}{a^2 - x^2} \, dx = \frac{1}{a}\operatorname{artanh}\left(\frac{x}{a}\right) + c = \frac{1}{2a}\ln\left|\frac{a + x}{a - x}\right| + c\)

Worked Example: Find \(\int \frac{1}{\sqrt{4x^2 + 9}} \, dx\).

Step 1: Factor out \(4\) from inside the square root:
\(\sqrt{4x^2 + 9} = \sqrt{4\left(x^2 + \frac{9}{4}\right)} = 2\sqrt{x^2 + \left(\frac{3}{2}\right)^2}\)

Step 2: Pull the constant out of the integral:
\(\frac{1}{2}\int \frac{1}{\sqrt{x^2 + \left(\frac{3}{2}\right)^2}} \, dx\)

Step 3: Apply the standard form with \(a = \frac{3}{2}\):
\(= \frac{1}{2}\operatorname{arsinh}\left(\frac{x}{3/2}\right) + c = \frac{1}{2}\operatorname{arsinh}\left(\frac{2x}{3}\right) + c\)

Summary Checklist & Exam Reminders

• No minus sign for \(\cosh\): Remember \(\frac{d}{dx}(\cos x) = -\sin x\), but \(\frac{d}{dx}(\cosh x) = +\sinh x\).
• Look out for Osborn's Rule: Sign change occurs only when two \(\sinh\) terms are multiplied together.
• Domain of \(\operatorname{arcosh} x\): You cannot evaluate \(\operatorname{arcosh}(x)\) for values less than \(1\). Always discard non-valid solutions.
• Express in log form: When an exam asks for an "exact value" of an inverse hyperbolic expression, provide it in logarithmic form using \(\ln\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the exponential definitions of \(\sinh x\), \(\cosh x\), and \(\tanh x\)?

\(\sinh x = \frac{e^x - e^{-x}}{2}\), \(\cosh x = \frac{e^x + e^{-x}}{2}\), and \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).

What are the definitions of the reciprocal hyperbolic functions \(\operatorname{sech} x\), \(\operatorname{cosech} x\), and \(\operatorname{coth} x\)?

\(\operatorname{sech} x = \frac{1}{\cosh x}\), \(\operatorname{cosech} x = \frac{1}{\sinh x}\) (for \(x \neq 0\)), and \(\operatorname{coth} x = \frac{1}{\tanh x}\) (for \(x \neq 0\)).

Which hyperbolic functions are even and which are odd?

\(\cosh x\) is an even function, whereas \(\sinh x\) and \(\tanh x\) are odd functions.

What are the domains and ranges of \(y = \cosh x\), \(y = \sinh x\), and \(y = \tanh x\)?

\(\cosh x\): Domain \(x \in \mathbb{R}\), Range \(y \ge 1\); \(\sinh x\): Domain \(x \in \mathbb{R}\), Range \(y \in \mathbb{R}\); \(\tanh x\): Domain \(x \in \mathbb{R}\), Range \(-1 < y < 1\).

What is the fundamental hyperbolic identity relating \(\cosh x\) and \(\sinh x\)?

\(\cosh^2 x - \sinh^2 x = 1\)

What is Osborn's Rule and how is it applied?

Replace trigonometric functions with hyperbolic ones, but change the sign in front of any term involving the product of two sine terms (explicit or implicit).

How do you solve a linear equation mixing \(\sinh x\) and \(\cosh x\)?

Substitute the exponential definitions \(\frac{e^x - e^{-x}}{2}\) and \(\frac{e^x + e^{-x}}{2}\), multiply by \(e^x\), and solve the resulting quadratic in \(e^x\).

What are the logarithmic forms of \(\operatorname{arsinh} x\), \(\operatorname{arcosh} x\), and \(\operatorname{artanh} x\)?

\(\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right)\), \(\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right)\) (for \(x \ge 1\)), and \(\operatorname{artanh} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\) (for \(|x| < 1\)).

What are the derivatives of \(\sinh x\), \(\cosh x\), and \(\tanh x\)?

\(\frac{d}{dx}(\sinh x) = \cosh x\), \(\frac{d}{dx}(\cosh x) = \sinh x\), and \(\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x\).

What are the standard derivatives of the inverse hyperbolic functions \(\operatorname{arsinh} x\), \(\operatorname{arcosh} x\), and \(\operatorname{artanh} x\)?

\(\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}}\), \(\frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}}\) (for \(x > 1\)), and \(\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2}\) (for \(|x| < 1\)).

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