🚀 Welcome to Hyperbolic Functions: Your FP3 Study Guide!

Hello future Further Mathematician! Welcome to the exciting world of Hyperbolic Functions. Don
't worry if the name sounds intense; these functions are simply partners to the trigonometric (circular) functions you already know, but instead of relating to a circle, they relate to a hyperbola.

In this chapter, we will learn their definitions, discover their unique identities, and master their differentiation and integration. This is a foundational chapter for advanced calculus and geometry, so let's dive in!

Section 1: The Building Blocks – Definitions and Graphs

Hyperbolic functions are defined using the exponential function, \(\text{e}^x\). This makes them incredibly useful in engineering, physics (especially relativity), and architecture.

1.1 The Primary Definitions

The two fundamental hyperbolic functions are sinh (pronounced "shine") and cosh (pronounced "kosh").

  • Hyperbolic Sine (\(\sinh x\)):
    \[ \sinh x = \frac{\text{e}^x - \text{e}^{-x}}{2} \]

    Analogy: Since the minus sign is present, \(\sinh x\) is an odd function (like \(\sin x\))—it passes through the origin.

  • Hyperbolic Cosine (\(\cosh x\)):
    \[ \cosh x = \frac{\text{e}^x + \text{e}^{-x}}{2} \]

    Analogy: Since the plus sign is present, \(\cosh x\) is an even function (like \(\cos x\))—it is symmetric about the y-axis and has a minimum value at \(x=0\).

💡 Memory Aid: The 'C' in \(\cosh x\) can remind you of the word Catenary, which is the shape a hanging chain or cable makes. This curve looks like a U-shape, which confirms \(\cosh x\) is always positive.

1.2 The Secondary Definitions

Just like in trigonometry, we define tangent, secant, cosecant, and cotangent based on the primary functions:

  • Hyperbolic Tangent (\(\tanh x\)): \[ \tanh x = \frac{\sinh x}{\cosh x} = \frac{\text{e}^x - \text{e}^{-x}}{\text{e}^x + \text{e}^{-x}} \]
  • Hyperbolic Secant (\(\text{sech } x\)): \[ \text{sech } x = \frac{1}{\cosh x} \]
  • Hyperbolic Cosecant (\(\text{cosech } x\)): \[ \text{cosech } x = \frac{1}{\sinh x} \]
  • Hyperbolic Cotangent (\(\coth x\)): \[ \coth x = \frac{1}{\tanh x} \]
1.3 Key Takeaways on Definitions


The exponential definitions are fundamental. If you ever forget an identity or derivative, you can always revert to these definitions to prove the identity or perform the differentiation!

Section 2: Fundamental Identities – The Rules of Hyperbola

This section is critical! Hyperbolic identities look very similar to trigonometric identities, but pay extremely close attention to the signs—they are often flipped.

2.1 The Master Identity

The most important identity connects \(\cosh x\) and \(\sinh x\).

THE HYPERBOLIC PYTHAGOREAN IDENTITY: \[ \cosh^2 x - \sinh^2 x = 1 \]

🔥 Crucial Note: In trigonometry, it's \(\cos^2 \theta + \sin^2 \theta = 1\). In hyperbolic functions, the sign is subtractive! This minus sign is what defines the hyperbola.

Step-by-Step Proof (Quick Review):

  1. Start with \(\cosh^2 x - \sinh^2 x\).
  2. Substitute definitions: \( \left(\frac{\text{e}^x + \text{e}^{-x}}{2}\right)^2 - \left(\frac{\text{e}^x - \text{e}^{-x}}{2}\right)^2 \)
  3. Expand the numerators: \( \frac{(\text{e}^{2x} + 2 + \text{e}^{-2x}) - (\text{e}^{2x} - 2 + \text{e}^{-2x})}{4} \)
  4. Simplify: \( \frac{4}{4} = 1 \). Q.E.D.

2.2 Derived Identities

We derive the other two primary identities by dividing the master identity by \(\cosh^2 x\) or \(\sinh^2 x\):

  • Divide by \(\cosh^2 x\): \[ 1 - \frac{\sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} \] \[ 1 - \tanh^2 x = \text{sech}^2 x \quad \implies \quad \mathbf{1 - \tanh^2 x = \text{sech}^2 x} \]

    (Compare to trig: \(1 + \tan^2 \theta = \sec^2 \theta\). Again, the sign is flipped!)

  • Divide by \(\sinh^2 x\): \[ \frac{\cosh^2 x}{\sinh^2 x} - 1 = \frac{1}{\sinh^2 x} \] \[ \coth^2 x - 1 = \text{cosech}^2 x \quad \implies \quad \mathbf{\coth^2 x - 1 = \text{cosech}^2 x} \]

    (Compare to trig: \(1 + \cot^2 \theta = \csc^2 \theta\). Sign is flipped!)

2.3 Addition Formulae

These are the analogues of the double angle and compound angle formulae. They are incredibly useful for solving equations or simplifying expressions.

Compound Angle Formulae:

  • \(\cosh(A+B) = \cosh A \cosh B + \sinh A \sinh B\)
  • \(\cosh(A-B) = \cosh A \cosh B - \sinh A \sinh B\)
  • \(\sinh(A+B) = \sinh A \cosh B + \cosh A \sinh B\)
  • \(\sinh(A-B) = \sinh A \cosh B - \cosh A \sinh B\)
  • \(\tanh(A+B) = \frac{\tanh A + \tanh B}{1 + \tanh A \tanh B}\)

💡 Observation: Notice that the addition formulae for \(\sinh\) keep the original sign (plus stays plus, minus stays minus), just like \(\sin\). However, the \(\cosh\) formula is different from \(\cos\): for \(\cosh(A+B)\) the sign is PLUS, whereas for \(\cos(\alpha+\beta)\) the sign is MINUS.

Double Angle Formulae (Let \(A=B=x\)):

  • \(\cosh(2x) = \cosh^2 x + \sinh^2 x\) (No simple trig analogue!)
  • \(\cosh(2x) = 2\cosh^2 x - 1\)
  • \(\cosh(2x) = 1 + 2\sinh^2 x\)
  • \(\sinh(2x) = 2\sinh x \cosh x\)

Key Takeaway: When working with hyperbolic identities, always assume the sign is opposite to the trigonometric version, unless you prove otherwise! The relationship \(\cosh^2 x - \sinh^2 x = 1\) is your anchor.

Section 3: Inverse Hyperbolic Functions (Logarithmic Form)

Just like finding \(\arcsin\) gives you an angle, finding \(\text{arsinh}\) gives you the value \(x\) that produced the function value.

We use the notation \(\text{arsinh } x\), \(\text{arcosh } x\), etc., although you may also see \(\sinh^{-1} x\).

A key requirement of the FP3 curriculum is to express these inverse functions in terms of natural logarithms (\(\ln\)).

3.1 Deriving the Logarithmic Forms

Let's see how \(\text{arsinh } x\) is derived. This process is a common exam technique.

Step 1: Set \(y = \text{arsinh } x\).
This means \(x = \sinh y\).

Step 2: Use the exponential definition.
\[ x = \frac{\text{e}^y - \text{e}^{-y}}{2} \]

Step 3: Clear the fraction and multiply by \(\text{e}^y\).
\( 2x = \text{e}^y - \text{e}^{-y} \)
\( 2x\text{e}^y = (\text{e}^y)^2 - 1 \)

Step 4: Rearrange into a quadratic equation in terms of \(\text{e}^y\).
\[ (\text{e}^y)^2 - 2x\text{e}^y - 1 = 0 \]

Step 5: Solve using the quadratic formula.
Let \(A=\text{e}^y\). \( A = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)} \)
\( A = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \frac{2x \pm 2\sqrt{x^2 + 1}}{2} \)
\( A = x \pm \sqrt{x^2 + 1} \)

Step 6: Choose the correct root and solve for \(y\).
Since \(A = \text{e}^y\) must be positive, and \(\sqrt{x^2+1}\) is always greater than \(x\), we must choose the positive root: \( \text{e}^y = x + \sqrt{x^2 + 1} \).
Therefore, \( y = \ln \left( x + \sqrt{x^2 + 1} \right) \).

3.2 The Standard Logarithmic Forms

You must know these three standard results:


1. Inverse Hyperbolic Sine (Domain: \(x \in \mathbb{R}\)) \[ \mathbf{\text{arsinh } x = \ln \left( x + \sqrt{x^2 + 1} \right)} \]

2. Inverse Hyperbolic Cosine (Domain: \(x \ge 1\)) \[ \mathbf{\text{arcosh } x = \ln \left( x + \sqrt{x^2 - 1} \right)} \]

(Note the domain restriction: since \(\cosh x \ge 1\), the input \(x\) must be at least 1.)

3. Inverse Hyperbolic Tangent (Domain: \(-1 < x < 1\)) \[ \mathbf{\text{artanh } x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)} \]

(Note the domain restriction: \(\tanh x\) is always between -1 and 1.)

Quick Review: The log forms are essential for solving equations and for certain integration techniques. Practice the derivation process, especially for \(\text{arsinh } x\).

Section 4: Calculus of Hyperbolic Functions

This is where hyperbolic functions truly shine in Further Maths. Their derivatives and integrals are elegant and have beautiful symmetry.

4.1 Differentiation of Standard Hyperbolic Functions

When differentiating hyperbolic functions, the rules are almost identical to trig functions, but the signs are much kinder!

The key difference: You rarely introduce a negative sign.

  • \[ \frac{\text{d}}{\text{d}x} (\sinh x) = \cosh x \]
  • \[ \frac{\text{d}}{\text{d}x} (\cosh x) = \sinh x \]
  • \[ \frac{\text{d}}{\text{d}x} (\tanh x) = \text{sech}^2 x \]
  • \[ \frac{\text{d}}{\text{d}x} (\text{sech } x) = -\text{sech } x \tanh x \]
  • \[ \frac{\text{d}}{\text{d}x} (\text{cosech } x) = -\text{cosech } x \coth x \]
  • \[ \frac{\text{d}}{\text{d}x} (\coth x) = -\text{cosech}^2 x \]

💡 Memory Aid: In trigonometry, derivatives starting with 'co' (\(\cos, \cot, \csc\)) usually result in a negative. In hyperbolic functions, the only ones that result in a negative are the reciprocals (\(\text{sech}, \text{cosech}, \coth\)).

Example: Chain Rule (Don't forget the inner derivative!)
If \(y = \cosh(3x^2)\), then \(\frac{\text{d}y}{\text{d}x} = \sinh(3x^2) \times (6x) = 6x \sinh(3x^2)\).

4.2 Differentiation of Inverse Hyperbolic Functions

These derivatives are often tested, as they are essential for standard integral forms. You can prove these by differentiating their logarithmic forms.

  • \[ \frac{\text{d}}{\text{d}x} (\text{arsinh } x) = \frac{1}{\sqrt{x^2 + 1}} \]
  • \[ \frac{\text{d}}{\text{d}x} (\text{arcosh } x) = \frac{1}{\sqrt{x^2 - 1}} \quad (x>1) \]
  • \[ \frac{\text{d}}{\text{d}x} (\text{artanh } x) = \frac{1}{1 - x^2} \quad (|x|<1) \]

Did you know? The derivative of \(\text{artanh } x\) is extremely important in complex analysis and engineering. Notice how similar these forms are to the inverse trig derivatives, but again, watch those signs!

4.3 Integration of Standard Hyperbolic Functions

Integration is simply the reverse of differentiation. Since the differentiation rules were mostly positive, the integration rules are very straightforward:

  • \[ \int \sinh x \text{d}x = \cosh x + C \]
  • \[ \int \cosh x \text{d}x = \sinh x + C \]
  • \[ \int \text{sech}^2 x \text{d}x = \tanh x + C \]
4.4 Integration Leading to Inverse Hyperbolic Functions

A major part of the FP3 curriculum involves identifying integrals that result in inverse hyperbolic functions, especially the forms derived from the differentiation section (4.2). These are typically handled using substitution or by recognizing the standard forms below:

Standard Forms to Memorize (where \(a\) is a constant):

  1. Integral leading to \(\text{arsinh}\): \[ \mathbf{\int \frac{1}{\sqrt{x^2 + a^2}} \text{d}x = \text{arsinh} \left(\frac{x}{a}\right) + C} \] (Or, expressed in log form: \( \ln \left( x + \sqrt{x^2 + a^2} \right) + C \))
  2. Integral leading to \(\text{arcosh}\): \[ \mathbf{\int \frac{1}{\sqrt{x^2 - a^2}} \text{d}x = \text{arcosh} \left(\frac{x}{a}\right) + C} \] (Or, expressed in log form: \( \ln \left( x + \sqrt{x^2 - a^2} \right) + C \). Must have \(|x| > a\).)
  3. Integral leading to \(\text{artanh}\): \[ \mathbf{\int \frac{1}{a^2 - x^2} \text{d}x = \frac{1}{2a} \ln \left( \frac{a+x}{a-x} \right) + C} \] (This is the result of \(\frac{1}{a} \text{artanh} \left(\frac{x}{a}\right) + C \). Must have \(|x| < a\).)

Common Mistake Alert: Students often confuse these hyperbolic integral forms with the inverse trigonometric forms (like \(\int \frac{1}{\sqrt{a^2 - x^2}} \text{d}x = \arcsin (\frac{x}{a}) \)). Note where the signs are placed!

Summary: Hyperbolic Calculus

Mastering hyperbolic calculus involves two skills: 1) knowing the standard derivatives/integrals, and 2) recognizing when a definite integral requires you to substitute the final answer into the logarithmic form for evaluation. Practice converting between \(\text{arsinh}\) and its \(\ln\) form frequently!