Mastering Hyperbolic Functions for Further Maths

Hyperbolic functions are a cornerstone of the Further Maths curriculum, appearing frequently in calculus, differential equations, and complex analysis. While they share names and structural similarities with circular trigonometric functions, they are defined using exponential functions, making them essential for modelling physical phenomena like catenary curves and wave propagation.

In this guide, we will explore the definitions of $\sinh x$, $\cosh x$, and $\tanh x$, master the core identities, and learn how to apply Osborne's Rule to derive new relationships. Understanding these functions is not just about memorising formulas; it is about recognising the underlying exponential behaviour that governs them.

Defining the Hyperbolic Functions

Unlike circular functions which relate to the unit circle $x^2 + y^2 = 1$, hyperbolic functions relate to the hyperbola $x^2 - y^2 = 1$. They are defined using the exponential function $e^x$ and its reciprocal $e^{-x}$:

$$\cosh x = \frac{e^x + e^{-x}}{2}$$ $$\sinh x = \frac{e^x - e^{-x}}{2}$$ $$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

These definitions allow us to treat hyperbolic functions as algebraic expressions. For instance, $\cosh x$ is an even function, meaning $\cosh(-x) = \cosh x$, while $\sinh x$ is an odd function, meaning $\sinh(-x) = -\sinh x$.

The Fundamental Identity and Osborne's Rule

One of the most important identities is the hyperbolic Pythagorean identity. Starting from the definitions:

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

Thus, $\cosh^2 x - \sinh^2 x = 1$.

To derive other identities, we use Osborne's Rule. This rule states that you can convert any trigonometric identity into a hyperbolic one by replacing the trig functions with their hyperbolic counterparts and changing the sign of any term containing the product of two sines (or $\sinh$ functions). For example, since $\cos^2 x + \sin^2 x = 1$, replacing $\sin^2 x$ with $\sinh^2 x$ and flipping the sign gives $\cosh^2 x - \sinh^2 x = 1$.

Worked Example 1: Solving Equations

Solve the equation $2\cosh x + \sinh x = 2$.

Step 1: Substitute the exponential definitions. $$2\left(\frac{e^x + e^{-x}}{2}\right) + \left(\frac{e^x - e^{-x}}{2}\right) = 2$$

Step 2: Simplify the expression. $$e^x + e^{-x} + 0.5e^x - 0.5e^{-x} = 2$$ $$1.5e^x + 0.5e^{-x} = 2$$

Step 3: Multiply by $2e^x$ to form a quadratic in $e^x$. $$3e^{2x} + 1 = 4e^x \implies 3(e^x)^2 - 4(e^x) + 1 = 0$$

Step 4: Factorise. $$(3e^x - 1)(e^x - 1) = 0$$ So, $e^x = 1/3$ or $e^x = 1$.

Step 5: Solve for $x$. $x = \ln(1/3) = -\ln 3$ or $x = \ln(1) = 0$.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are logarithmic in form. For example, to find $\text{arsinh } x$, let $y = \text{arsinh } x$, so $x = \sinh y = \frac{e^y - e^{-y}}{2}$. Multiplying by $2e^y$ gives $e^{2y} - 2xe^y - 1 = 0$. Solving this quadratic for $e^y$ using the quadratic formula and taking the natural log yields:

$$\text{arsinh } x = \ln(x + \sqrt{x^2 + 1})$$

Worked Example 2: Proving an Identity

Prove that $\cosh(2x) = 1 + 2\sinh^2 x$.

Step 1: Start with the definition of $\cosh(2x)$. $$\cosh(2x) = \frac{e^{2x} + e^{-2x}}{2}$$

Step 2: Use the identity $\cosh^2 x - \sinh^2 x = 1$, so $\cosh^2 x = 1 + \sinh^2 x$. $$\cosh(2x) = \cosh^2 x + \sinh^2 x$$

Step 3: Substitute $\cosh^2 x = 1 + \sinh^2 x$. $$\cosh(2x) = (1 + \sinh^2 x) + \sinh^2 x = 1 + 2\sinh^2 x$$

Common Mistakes

1. Sign Errors: Forgetting to flip the sign when using Osborne's Rule is the most common error. Always check the $\cosh^2 x - \sinh^2 x = 1$ identity.
2. Domain Confusion: Unlike $\sin x$ and $\cos x$, $\cosh x$ is always $\ge 1$. If you find $\cosh x = 0.5$, you have made a calculation error.
3. Algebraic slips: When solving equations, students often forget to multiply the constant term by the exponential factor when clearing denominators.

FAQ

Are hyperbolic functions periodic? No, they are not periodic. They are based on exponential growth and decay, unlike circular functions which repeat.

What is the difference between $\text{sech } x$ and $\text{csch } x$? $\text{sech } x = 1/\cosh x$ and $\text{csch } x = 1/\sinh x$. They are the reciprocals of the primary hyperbolic functions.

Do I need to memorise the logarithmic forms of inverse functions? These are usually provided in the formula booklet, but you must be able to derive them using the exponential definitions.

Conclusion

Hyperbolic functions are a powerful tool in your mathematical arsenal. By mastering their exponential definitions and the application of identities, you can simplify complex problems with ease. Ready to see these concepts in motion? Visit MathInstructor AI to generate a free, narrated animated lesson on hyperbolic functions and visualise these curves in real-time.