Edexcel IAL - Further Pure Mathematics 3- 1.1 Definition, Graphs and Properties of Hyperbolic Functions- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 1.1 Definition, Graphs and Properties of Hyperbolic Functions -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 1.1 Definition, Graphs and Properties of Hyperbolic Functions -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 1.1 Definition, Graphs and Properties of Hyperbolic Functions
Hyperbolic Functions
The hyperbolic functions are defined using exponential functions. They are analogous to the trigonometric functions but arise naturally in problems involving growth, decay, and certain differential equations.
Definitions in Terms of Exponentials
For any real value of \( x \), the six hyperbolic functions are defined as follows:
| Hyperbolic Function | Definition in Terms of Exponentials |
|---|---|
| \( \sinh x \) | \( \dfrac{1}{2}(e^x – e^{-x}) \) |
| \( \cosh x \) | \( \dfrac{1}{2}(e^x + e^{-x}) \) |
| \( \tanh x \) | \( \dfrac{\sinh x}{\cosh x} = \dfrac{e^x – e^{-x}}{e^x + e^{-x}} \) |
| \( \text{sech x }\) | \( \dfrac{1}{\cosh x} = \dfrac{2}{e^x + e^{-x}} \) |
| \( \text{csch x} \) | \( \dfrac{1}{\sinh x} = \dfrac{2}{e^x – e^{-x}} \) |
| \( \coth x \) | \( \dfrac{\cosh x}{\sinh x} \) |
Basic Properties of Hyperbolic Functions
Some important properties are:
\( \cosh x \geq 1 \) for all \( x \)
\( \sinh x \) is an odd function
\( \cosh x \) is an even function
\( \tanh x \) lies between \( -1 \) and \( 1 \)
Graphs of Hyperbolic Functions
The shapes of the graphs are important:
| Function | Key Graphical Properties | Graph |
|---|---|---|
| \( y = \cosh x \) | Minimum value of \( 1 \) at \( x = 0 \); curve is symmetric about the \( y \)-axis |  |
| \( y = \sinh x \) | Passes through the origin; increases rapidly as \( x \to \pm\infty \) |  |
| \( y = \tanh x \) | Horizontal asymptotes at \( y = 1 \) and \( y = -1 \) |  |
Important Hyperbolic Identities
Using the exponential definitions, the following identities can be derived and used:
\( \cosh^2 x – \sinh^2 x = 1 \)
\( \cosh^2 x + \sinh^2 x = \cosh 2x \)
\( 1 – \tanh^2 x = \mathrm{sech^2 x}\)
Solving Equations of the Form \( a\cosh x + b\sinh x = c \)
Such equations are usually solved by rewriting the hyperbolic functions in exponential form and letting \( e^x = t \).
Example :
Show that \( \cosh^2 x – \sinh^2 x = 1 \).
▶️ Answer/Explanation
\( \cosh x = \dfrac{e^x + e^{-x}}{2}, \quad \sinh x = \dfrac{e^x – e^{-x}}{2} \)
\( \cosh^2 x – \sinh^2 x = \dfrac{(e^x + e^{-x})^2 – (e^x – e^{-x})^2}{4} \)
\( = \dfrac{4}{4} = 1 \)
Conclusion: The identity is proved.
Example :
Show that \( \cosh^2 x + \sinh^2 x = \cosh 2x \).
▶️ Answer/Explanation
Using exponential forms:
\( \cosh^2 x + \sinh^2 x = \dfrac{(e^x + e^{-x})^2 + (e^x – e^{-x})^2}{4} \)
\( = \dfrac{2(e^{2x} + e^{-2x})}{4} = \dfrac{e^{2x} + e^{-2x}}{2} \)
\( = \cosh 2x \)
Conclusion: The identity is verified.
Example :
Solve the equation \( 3\cosh x + 4\sinh x = 5 \).
▶️ Answer/Explanation
Write in exponential form:
\( 3\left(\dfrac{e^x + e^{-x}}{2}\right) + 4\left(\dfrac{e^x – e^{-x}}{2}\right) = 5 \)
\( \dfrac{7e^x – e^{-x}}{2} = 5 \)
\( 7e^x – e^{-x} = 10 \)
Let \( e^x = t \), then \( e^{-x} = \dfrac{1}{t} \)
\( 7t – \dfrac{1}{t} = 10 \)
\( 7t^2 – 10t – 1 = 0 \)
\( t = \dfrac{10 \pm \sqrt{128}}{14} \)
Since \( t = e^x > 0 \), only the positive root is valid.
Conclusion: The solution is \( x = \ln t \), where \( t = \dfrac{10 + \sqrt{128}}{14} \).