Further Mathematics (9665) Study Notes: FP2.9 Hyperbolic Functions
Hello there, future expert! Welcome to the fascinating world of hyperbolic functions. Don't worry, even though they look similar to trigonometric functions (\(\sin, \cos, \tan\)), they are actually defined using the wonderful exponential function \(e^x\). This chapter is central to advanced calculus and geometry, so let's dive in!
What will you learn? You will learn the definitions, fundamental identities, graphs, derivatives, and integration techniques related to \(\sinh x\) and \(\cosh x\), and how to solve equations involving them.
Section 1: Defining Hyperbolic Functions (The Building Blocks)
Hyperbolic functions are essentially combinations of \(e^x\) and \(e^{-x}\). They are called "hyperbolic" because they relate to the unit hyperbola \(x^2 - y^2 = 1\) in the same way that trigonometric functions relate to the unit circle \(x^2 + y^2 = 1\).
1.1 The Core Definitions
The two fundamental hyperbolic functions are Hyperbolic Sine (\(\sinh x\)) and Hyperbolic Cosine (\(\cosh x\)).
- Hyperbolic Cosine (cosh): The even combination of exponentials.
$$\cosh x = \frac{e^x + e^{-x}}{2}$$ - Hyperbolic Sine (sinh): The odd combination of exponentials.
$$\sinh x = \frac{e^x - e^{-x}}{2}$$
Memory Aid: Think of the letter 'H' in \(\mathbf{C} \cosh x\) looking like a plus sign (+) laid on its side, helping you remember the definition involves addition.
1.2 Secondary Hyperbolic Functions
Just like in trigonometry, we have reciprocal and ratio functions:
- Hyperbolic Tangent (tanh): $$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
- Hyperbolic Secant (sech): $$\mathrm{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$$
- Hyperbolic Cosecant (cosech): $$\mathrm{cosech} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$$
- Hyperbolic Cotangent (coth): $$\coth x = \frac{1}{\tanh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$$
Key Takeaway: All hyperbolic functions are fundamentally defined using the exponential forms. When in doubt, convert back to \(e^x\) and \(e^{-x}\)!
Section 2: Fundamental Hyperbolic Identities
The relationships between hyperbolic functions are very similar to trigonometric ones, but watch out for the crucial difference in signs!
2.1 The Master Identity (Proof Required)
The equivalent of \(\sin^2 \theta + \cos^2 \theta = 1\) is the fundamental identity for hyperbolics, but with a minus sign:
Identity 1: $$\cosh^2 x - \sinh^2 x = 1$$
Proof: This proof is often requested, so let's walk through it step-by-step:
- Start with the left side (LHS): \(\cosh^2 x - \sinh^2 x\)
- Substitute the exponential definitions: $$LHS = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2$$
- Expand the squares (remember \((e^x)^2 = e^{2x}\) and \(e^x e^{-x} = e^0 = 1\)): $$LHS = \frac{1}{4} (e^{2x} + 2(1) + e^{-2x}) - \frac{1}{4} (e^{2x} - 2(1) + e^{-2x})$$
- Factor out \(1/4\): $$LHS = \frac{1}{4} [ (e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x}) ]$$
- Simplify the terms inside the bracket (notice \(e^{2x}\) and \(e^{-2x}\) cancel out): $$LHS = \frac{1}{4} [ 2 - (-2) ] = \frac{1}{4} [ 4 ] = 1$$ This proves \(\cosh^2 x - \sinh^2 x = 1\).
2.2 Secondary Identities
We derive these by dividing the Master Identity by \(\cosh^2 x\) or \(\sinh^2 x\):
Dividing by \(\cosh^2 x\): $$\frac{\cosh^2 x}{\cosh^2 x} - \frac{\sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x}$$ $$\mathbf{1 - \tanh^2 x = \mathrm{sech}^2 x}$$
Dividing by \(\sinh^2 x\): $$\frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\sinh^2 x} = \frac{1}{\sinh^2 x}$$ $$\mathbf{\coth^2 x - 1 = \mathrm{cosech}^2 x}$$
2.3 Addition Formulae (Use and Proof Required)
You must be familiar with proving identities like \(\sinh(x+y)\) using the exponential definitions.
- $$\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$$
- $$\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$$
(Notice: For \(\cosh(x+y)\), the plus sign stays, unlike in trigonometry where it changes to minus for \(\cos(A+B)\)).
Quick Review Box: Key Identities to Memorise
$$ \cosh^2 x - \sinh^2 x = 1 $$ $$ 1 - \tanh^2 x = \mathrm{sech}^2 x $$ $$ \coth^2 x - 1 = \mathrm{cosech}^2 x $$
Key Takeaway: The fundamental identity is \(\cosh^2 x - \sinh^2 x = 1\). Remember that minus sign!
Section 3: Graphs of Hyperbolic Functions
Understanding the shape of these graphs is crucial for domain, range, and finding inverse functions.
3.1 Graph of y = \(\sinh x\)
- Shape: Looks like a stretched cubic function. It passes through the origin (0, 0).
- Symmetry: It is an odd function: \(\sinh(-x) = -\sinh x\).
- Domain: \(x \in \mathbb{R}\) (all real numbers).
- Range: \(y \in \mathbb{R}\) (all real numbers).
3.2 Graph of y = \(\cosh x\)
- Shape: This curve is called a catenary. It is the shape adopted by a perfectly flexible chain or cable hanging freely from two points.
- Symmetry: It is an even function: \(\cosh(-x) = \cosh x\).
- Minimum Point: (0, 1).
- Domain: \(x \in \mathbb{R}\).
- Range: \(y \ge 1\). (This restriction is crucial when defining the inverse function!)
Did you know? The arches of the Gateway Arch in St. Louis, USA, are designed in the shape of an inverted catenary curve!
3.3 Graph of y = \(\tanh x\)
- Shape: Increases monotonically (always increasing).
- Asymptotes: As \(x \to \infty\), \(e^{-x} \to 0\), so \(\tanh x \to \frac{e^x}{e^x} = 1\). As \(x \to -\infty\), \(e^x \to 0\), so \(\tanh x \to \frac{-e^{-x}}{e^{-x}} = -1\).
- Range: \(-1 < y < 1\).
Key Takeaway: \(\cosh x\) has a restricted range (\(y \ge 1\)), which impacts its inverse function. \(\tanh x\) is always between -1 and 1.
Section 4: Inverse Hyperbolic Functions (Logarithmic Forms)
Because hyperbolic functions are constructed from exponentials, their inverses can always be expressed in terms of logarithms. These logarithmic forms are provided in the formula booklet, but you may be asked to prove them.
4.1 Inverse Hyperbolic Sine: \(\mathrm{arsinh} x\) or \(\sinh^{-1} x\)
Logarithmic Form: $$\sinh^{-1} x = \ln (x + \sqrt{x^2 + 1}), \quad x \in \mathbb{R}$$
Step-by-Step Proof for \(\sinh^{-1} x\):
- Start with \(y = \sinh^{-1} x\). This means \(x = \sinh y\).
- Substitute the exponential definition: $$x = \frac{e^y - e^{-y}}{2}$$
- Rearrange to clear the fraction and multiply by \(2e^y\) to eliminate \(e^{-y}\): $$2x e^y = e^{2y} - e^0$$ $$2x e^y = e^{2y} - 1$$
- Rearrange into a quadratic equation in terms of \(e^y\): $$(e^y)^2 - 2x(e^y) - 1 = 0$$
- Use the quadratic formula to solve for \(e^y\): $$e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)}$$ $$e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \frac{2x \pm 2\sqrt{x^2 + 1}}{2}$$ $$e^y = x \pm \sqrt{x^2 + 1}$$
- Since \(e^y\) must be positive, we must take the positive root (\(x + \sqrt{x^2+1}\) is always positive), so: $$e^y = x + \sqrt{x^2 + 1}$$
- Take the natural logarithm (ln) of both sides: $$y = \ln (x + \sqrt{x^2 + 1})$$ (Proof complete!)
4.2 Inverse Hyperbolic Cosine: \(\mathrm{arcosh} x\) or \(\cosh^{-1} x\)
Since \(\cosh x\) has a restricted range (\(y \ge 1\)), its inverse must have a restricted domain.
Logarithmic Form: $$\cosh^{-1} x = \ln (x + \sqrt{x^2 - 1}), \quad x \ge 1$$
Note: The proof follows the same quadratic method as \(\sinh^{-1} x\), leading to the term \(\sqrt{x^2 - 1}\). This is why the domain must be \(x \ge 1\).
4.3 Inverse Hyperbolic Tangent: \(\mathrm{artanh} x\) or \(\tanh^{-1} x\)
Logarithmic Form: $$\tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad -1 < x < 1$$
Key Takeaway: Proofs for inverse hyperbolic functions involve converting to exponential form and solving a quadratic equation in \(e^y\).
Section 5: Calculus of Hyperbolic Functions
One great feature of hyperbolic functions is that their derivatives are very tidy! We must know the proofs of the differentiation results and apply them to integration.
5.1 Differentiation (Proofs Required)
The proofs for differentiation rely directly on the exponential definitions.
- Derivative of \(\sinh x\): $$\frac{d}{dx}(\sinh x) = \frac{d}{dx} \left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x$$ $$\mathbf{\frac{d}{dx}(\sinh x) = \cosh x}$$
- Derivative of \(\cosh x\): $$\frac{d}{dx}(\cosh x) = \frac{d}{dx} \left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x + (-e^{-x})}{2} = \frac{e^x - e^{-x}}{2} = \sinh x$$ $$\mathbf{\frac{d}{dx}(\cosh x) = \sinh x}$$
- Derivative of \(\tanh x\): $$\mathbf{\frac{d}{dx}(\tanh x) = \mathrm{sech}^2 x}$$
Analogy/Memory Aid: Compare this to standard trigonometry:
| Trigonometric | Hyperbolic |
|---|---|
| \(\frac{d}{dx}(\sin x) = \cos x\) | \(\frac{d}{dx}(\sinh x) = \cosh x\) |
| \(\frac{d}{dx}(\cos x) = -\sin x\) (Note the minus!) | \(\frac{d}{dx}(\cosh x) = \sinh x\) (NO minus!) |
Encouragement: The fact that there are fewer negative signs in hyperbolic differentiation usually makes it easier to manage!
5.2 Differentiation of Inverse Hyperbolic Functions
The results for the derivatives of the inverse functions are given in the formulae booklet and will be used frequently:
- $$\mathbf{\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{x^2 + 1}}}$$
- $$\mathbf{\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2 - 1}}}$$
- $$\mathbf{\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1 - x^2}}$$
5.3 Applications to Integration
Integration involving hyperbolic functions often relies on recognising that the integral is the derivative of an inverse function. You must be able to use the standard integrals associated with these inverse functions:
Standard Integrals:
- $$\mathbf{\int \frac{1}{\sqrt{x^2 + a^2}} dx = \sinh^{-1} \left(\frac{x}{a}\right) + C}$$
- $$\mathbf{\int \frac{1}{\sqrt{x^2 - a^2}} dx = \cosh^{-1} \left(\frac{x}{a}\right) + C}$$
- $$\mathbf{\int \frac{1}{a^2 - x^2} dx = \frac{1}{a} \tanh^{-1} \left(\frac{x}{a}\right) + C} \quad (\text{for } |x| < |a|)$$
Common Mistake to Avoid: Confusing the integration formula for \(\sqrt{a^2 - x^2}\) (which is \(\sin^{-1}\)) with \(\sqrt{x^2 - a^2}\) (which is \(\cosh^{-1}\)). Note the order of \(x^2\) and \(a^2\) is critical!
Key Takeaway: Differentiation of \(\cosh x\) gives positive \(\sinh x\). Integration often leads to the logarithmic forms of the inverse functions.
Section 6: Solving Equations of the Form \(a \sinh x + b \cosh x = c\)
When solving an equation that mixes \(\sinh x\) and \(\cosh x\), the most reliable method is to revert to their exponential definitions. This transforms the problem into a quadratic equation in \(e^x\).
Step-by-Step Method
Suppose you need to solve \(2 \sinh x + 3 \cosh x = 4\).
- Substitute Definitions: Replace \(\sinh x\) and \(\cosh x\): $$2 \left(\frac{e^x - e^{-x}}{2}\right) + 3 \left(\frac{e^x + e^{-x}}{2}\right) = 4$$
- Clear Denominators: Multiply the entire equation by 2: $$2(e^x - e^{-x}) + 3(e^x + e^{-x}) = 8$$
- Expand and Group \(e^x\) and \(e^{-x}\) terms: $$(2e^x - 2e^{-x}) + (3e^x + 3e^{-x}) = 8$$ $$5e^x + e^{-x} = 8$$
- Form a Quadratic in \(e^x\): Multiply the whole equation by \(e^x\). Since \(e^x \cdot e^{-x} = 1\), we get: $$5(e^x)^2 + 1 = 8e^x$$ Rearrange into standard quadratic form (Let \(u = e^x\)): $$\mathbf{5u^2 - 8u + 1 = 0}$$
- Solve the Quadratic: Use the quadratic formula to find the value(s) of \(u\): $$u = \frac{8 \pm \sqrt{64 - 4(5)(1)}}{10} = \frac{8 \pm \sqrt{44}}{10} = \frac{8 \pm 2\sqrt{11}}{10}$$ $$u = \frac{4 \pm \sqrt{11}}{5}$$
- Solve for \(x\): Remember \(u = e^x\). Since \(e^x\) must be positive, and both solutions are positive here (since \(\sqrt{11} \approx 3.3\)), both are valid: $$x = \ln \left(\frac{4 + \sqrt{11}}{5}\right) \quad \text{and} \quad x = \ln \left(\frac{4 - \sqrt{11}}{5}\right)$$
Important Note: If the quadratic step gives you a negative value for \(u\), that root must be rejected, as \(e^x\) cannot be negative!
Key Takeaway: To solve mixed hyperbolic equations, always convert to exponential form, form a quadratic in \(e^x\), and use logarithms for the final answer.