Hyperbolic Functions (Further Pure Mathematics 2, Section 2.1)
Hey there! Welcome to the world of Hyperbolic Functions. Don't worry if the name sounds intimidating; these functions are simply partners to the trigonometric (circular) functions you already know (sin, cos, tan), but defined using the exponential function \(e^x\). They are essential in physics, engineering (especially modeling hanging cables—the famous catenary curve!), and integration methods in Further Maths.
In this chapter, we will master their definitions, identities, graphs, and the powerful logarithmic forms of their inverses.
1. Definitions of Hyperbolic Functions
The hyperbolic functions are built directly from the exponential functions \(e^x\) and \(e^{-x}\).
a) The Core Definitions: \(\cosh x\) and \(\sinh x\)
The two fundamental hyperbolic functions are hyperbolic cosine (\(\cosh x\), pronounced cosh) and hyperbolic sine (\(\sinh x\), pronounced shine or sinh).
1. Hyperbolic Cosine (\(\cosh x\)):
The definition of \(\cosh x\) is the average of \(e^x\) and \(e^{-x}\). \[ \cosh x = \frac{e^x + e^{-x}}{2} \]
Memory Aid: Since \(\cos x\) is an Even function, \(\cosh x\) is also an Even function. The formula uses the plus sign, making it symmetric about the y-axis.
2. Hyperbolic Sine (\(\sinh x\)):
The definition of \(\sinh x\) is half the difference between \(e^x\) and \(e^{-x}\). \[ \sinh x = \frac{e^x - e^{-x}}{2} \]
Memory Aid: Since \(\sin x\) is an Odd function, \(\sinh x\) is also an Odd function. The formula uses the minus sign, making it symmetric about the origin.
b) Other Hyperbolic Functions
Just like standard trig, the other four hyperbolic functions are defined based on the relationship between \(\sinh x\) and \(\cosh x\):
- Hyperbolic Tangent: \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\)
- Hyperbolic Cotangent: \(\text{coth } x = \frac{1}{\tanh x} = \frac{\cosh x}{\sinh x}\)
- Hyperbolic Secant: \(\text{sech } x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}\)
- Hyperbolic Cosecant: \(\text{cosech } x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}\)
\(\cosh x = \frac{e^x + e^{-x}}{2}\)
\(\sinh x = \frac{e^x - e^{-x}}{2}\)
2. Hyperbolic Identities
The identities are similar to trigonometric identities, but be extremely careful: the signs are often flipped!
a) The Fundamental Identity (The Hyperbolic Pythagorean Theorem)
The most important identity connects \(\cosh x\) and \(\sinh x\):
\[ \cosh^2 x - \sinh^2 x = 1 \]
Proof (You must be able to prove this):
- Start with the definitions: \[ \cosh^2 x - \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2 \]
- Expand the numerators: \[ = \frac{1}{4} \left[ (e^{2x} + 2e^x e^{-x} + e^{-2x}) - (e^{2x} - 2e^x e^{-x} + e^{-2x}) \right] \]
- Simplify the middle terms (\(e^x e^{-x} = e^0 = 1\)): \[ = \frac{1}{4} \left[ (e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x}) \right] \]
- Distribute the minus sign and cancel terms: \[ = \frac{1}{4} \left[ e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x} \right] \]
- The remaining terms are \(2 + 2 = 4\): \[ = \frac{1}{4} (4) = 1 \]
Key Takeaway: Remember that in regular trig, \(\cos^2 x + \sin^2 x = 1\). In hyperbolic functions, the sign changes: \(\cosh^2 x - \sinh^2 x = 1\). This is often the source of calculation mistakes, so highlight it in your notes!
b) Derived Identities
We can derive other identities by dividing the fundamental identity by \(\cosh^2 x\) or \(\sinh^2 x\):
- Dividing by \(\cosh^2 x\): \[ \frac{\cosh^2 x}{\cosh^2 x} - \frac{\sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} \] Identity 2: \(1 - \tanh^2 x = \text{sech}^2 x\) (Note the sign difference from \(1 + \tan^2 x = \sec^2 x\))
- Dividing by \(\sinh^2 x\): \[ \frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\sinh^2 x} = \frac{1}{\sinh^2 x} \] Identity 3: \(\coth^2 x - 1 = \text{cosech}^2 x\) (Note the sign difference from \(1 + \cot^2 x = \csc^2 x\))
c) Double Angle Identities
These follow the structure of standard trigonometric double angles but again, watch the signs!
- Hyperbolic Sine: \(\sinh 2x = 2 \sinh x \cosh x\) (Same as trig)
- Hyperbolic Cosine: \(\cosh 2x = \cosh^2 x + \sinh^2 x\) (Note: This is PLUS, unlike \(\cos 2x = \cos^2 x - \sin^2 x\))
We can rearrange \(\cosh 2x\) using the fundamental identity: \[ \cosh 2x = 2 \cosh^2 x - 1 \] \[ \cosh 2x = 1 + 2 \sinh^2 x \]
Tip for Struggling Students: If you forget the hyperbolic identities, try quickly rewriting the trig identity and remember to flip the signs for any term that originally contained an odd power of sine. Since \(\cosh x\) is purely positive, it tends to keep the positive sign when it dominates the identity.
3. Graphs and Key Properties
Sketching these graphs is a required skill. You must know the domain, range, and asymptotic behavior.
a) Graph of \(y = \cosh x\)
- Shape: Looks like a parabola but is actually the shape formed by a uniform heavy chain hanging freely between two supports—this is called the Catenary curve.
- Domain: \(x \in \mathbb{R}\)
- Range: \(y \ge 1\) (Since \(\frac{e^x + e^{-x}}{2}\) is always positive, and its minimum occurs at \(x=0\), where \(\cosh 0 = 1\)).
- Symmetry: Even function (symmetric about the y-axis).
Did you know? Architects use the catenary shape inverted (an arch) because it distributes weight perfectly, avoiding bending stress. The Gateway Arch in St. Louis, USA, is an inverted catenary!
b) Graph of \(y = \sinh x\)
- Shape: An S-shape, similar to \(y = x^3\) but growing exponentially.
- Domain: \(x \in \mathbb{R}\)
- Range: \(y \in \mathbb{R}\)
- Symmetry: Odd function (symmetric about the origin), passes through (0, 0).
c) Graph of \(y = \tanh x\)
- Shape: Similar to \(\tan^{-1} x\).
- Domain: \(x \in \mathbb{R}\)
- Range: \(-1 < y < 1\)
- Asymptotes: Horizontal asymptotes at \(y = 1\) (as \(x \to \infty\)) and \(y = -1\) (as \(x \to -\infty\)).
Why the asymptotes for \(\tanh x\)?
As \(x \to \infty\): \[ \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \] Divide top and bottom by \(e^x\): \[ \tanh x = \frac{1 - e^{-2x}}{1 + e^{-2x}} \] As \(x \to \infty\), \(e^{-2x} \to 0\), so \(\tanh x \to \frac{1 - 0}{1 + 0} = 1\). This confirms the asymptote \(y=1\).
4. Inverse Hyperbolic Functions
Since \(\sinh x\) and \(\tanh x\) are one-to-one (monotonic), their inverses are straightforward. However, \(\cosh x\) is many-to-one (since it's even), so its domain must be restricted to \(x \ge 0\) to ensure a valid inverse.
a) Definitions
The inverse functions are denoted as \(\sinh^{-1} x\), \(\cosh^{-1} x\), and \(\tanh^{-1} x\). They answer the question: "What value of \(x\) gives me this output?"
b) Deriving the Logarithmic Forms (Area Functions)
The key requirement of the syllabus is to derive and use the logarithmic forms of these inverse functions. These forms are often called area functions because they relate to the area under a hyperbola.
We will derive \(\sinh^{-1} x\) step-by-step. The process is similar for \(\cosh^{-1} x\) and \(\tanh^{-1} x\).
Step-by-Step Derivation of \(\sinh^{-1} x\)
Let \(y = \sinh^{-1} x\). By definition, this means \(x = \sinh y\).
Step 1: Substitute the exponential definition. \[ x = \frac{e^y - e^{-y}}{2} \]
Step 2: Clear the fraction and eliminate the negative exponent.
Multiply by 2:
\[ 2x = e^y - e^{-y} \]Multiply the entire equation by \(e^y\) (This is the crucial algebraic trick to form a quadratic):
\[ 2x e^y = e^{2y} - e^0 \] \[ 2x e^y = e^{2y} - 1 \]Step 3: Rearrange into a quadratic equation in terms of \(e^y\).
\[ (e^y)^2 - (2x) e^y - 1 = 0 \]Step 4: Solve the quadratic using the quadratic formula.
Let \(E = e^y\). Then \(E^2 - (2x)E - 1 = 0\). Using \(E = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-2x, c=-1\):
\[ e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)} \] \[ e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \] \[ e^y = \frac{2x \pm 2\sqrt{x^2 + 1}}{2} \] \[ e^y = x \pm \sqrt{x^2 + 1} \]Step 5: Apply domain restrictions.
Since \(e^y\) must be positive, we check the two solutions:
Since \(\sqrt{x^2 + 1}\) is always greater than \(x\), the negative root \(x - \sqrt{x^2 + 1}\) is negative. Therefore, we must choose the positive root.
\[ e^y = x + \sqrt{x^2 + 1} \]Step 6: Take the natural logarithm to find \(y\).
\[ y = \ln(x + \sqrt{x^2 + 1}) \]Thus, the logarithmic form of \(\sinh^{-1} x\) is:
\[ \sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) \quad \text{for } x \in \mathbb{R} \]
c) Summary of Logarithmic Forms (MF19 Formulas)
You must be able to use these, and derive the first three (or recognize the derivation process):
- \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) \quad (\text{for all } x)\)
- \(\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}) \quad (\text{for } x \ge 1)\)
- \(\tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) \quad (\text{for } |x| < 1)\)
Common Mistake Alert!
When solving for \(\cosh^{-1} x\), the quadratic step gives you two positive solutions for \(e^y\). However, because we restricted the domain of \(\cosh x\) to \(y \ge 0\), we must select the solution that yields \(y \ge 0\). For \(\cosh^{-1} x\), the convention is to always take the positive root: \(e^y = x + \sqrt{x^2 - 1}\).
Example Use: Solving an Equation
Problem: Solve \(\cosh x = 3\).
Solution (using the logarithmic form):
Since \(x = \cosh^{-1} 3\), we use the logarithmic form:
\[ x = \ln(3 + \sqrt{3^2 - 1}) \] \[ x = \ln(3 + \sqrt{8}) \] \[ x = \ln(3 + 2\sqrt{2}) \]
Because the graph of \(\cosh x\) is symmetric, there is also a negative solution, which is \(\ln(3 - 2\sqrt{2})\), but the principal value derived from the positive branch is sufficient unless otherwise stated.
The logarithmic forms turn potentially complex hyperbolic equations into standard logarithmic equations, making them much easier to solve!
Key Takeaways for Hyperbolic Functions
1. Definitions are Exponential: \(\cosh\) (plus) and \(\sinh\) (minus) are the foundation.
2. Identities are Opposite: The fundamental identity is \(\cosh^2 x - \sinh^2 x = 1\) (notice the minus sign!).
3. Logarithmic Forms are Essential: Understand the derivation process (using \(e^y\) and the quadratic formula) to find the logarithmic forms of the inverse functions.