Welcome to the World of Hyperbolic Functions!

In your GCSE and A-level studies, you’ve become very familiar with trigonometric functions like \(\sin\), \(\cos\), and \(\tan\). These are often called "circular functions" because they relate to coordinates on a circle.

In this chapter, we are going to meet their cousins: the hyperbolic functions. Don't worry if this seems a bit strange at first! While they have similar names (\(\sinh\), \(\cosh\), and \(\tanh\)), they are actually built using the exponential function \(e^x\). You’ll find they have some very "circular-like" properties but with a few interesting twists.

Why learn this? Hyperbolic functions aren't just abstract math; they describe the shape of a hanging power line or a heavy necklace (a curve called a catenary) and are used extensively in engineering and special relativity!

1. The Definitions: What are they?

Instead of using a unit circle, these functions are based on a hyperbola. We define the three main functions using \(e^x\) and \(e^{-x}\). You need to memorize these three definitions:

  • Hyperbolic Sine: \(\sinh x = \frac{e^x - e^{-x}}{2}\) (pronounced "shine" or "sin-itch")
  • Hyperbolic Cosine: \(\cosh x = \frac{e^x + e^{-x}}{2}\) (pronounced "cosh")
  • Hyperbolic Tangent: \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}\) (pronounced "tansh" or "tan-itch")

Quick Tip: Notice that \(\sinh x\) has a minus sign in the numerator, while \(\cosh x\) has a plus sign. You can remember this because \(\cosh\) is the "heavier" one (addition) and creates that hanging chain shape!

Key Takeaway:

Hyperbolic functions are just combinations of \(e^x\) and \(e^{-x}\). If you ever get stuck, you can always replace the hyperbolic term with its \(e\) definition to solve a problem.

2. Sketching the Graphs

Visualizing these functions helps you understand how they behave. Let's look at their shapes:

The Graph of \(y = \sinh x\)

The graph of \(\sinh x\) looks a bit like \(y = x^3\). It starts at the origin \((0,0)\) and goes off to infinity in both directions. It is an odd function, meaning it has rotational symmetry about the origin.

The Graph of \(y = \cosh x\)

This is the "hanging chain" curve.

  • It never goes below \(y = 1\).
  • The y-intercept is always \((0, 1)\) because \(\frac{e^0 + e^0}{2} = \frac{1+1}{2} = 1\).
  • It is an even function (symmetrical across the y-axis), just like the regular \(\cos x\).

The Graph of \(y = \tanh x\)

This graph is "trapped" between two horizontal asymptotes: \(y = 1\) and \(y = -1\). It passes through the origin and looks like a stretched "S" shape. It will never actually reach 1 or -1, no matter how large \(x\) gets!

Common Mistake to Avoid: Many students accidentally draw \(\cosh x\) starting at the origin. Remember: \(\cosh 0 = 1\) and \(\sinh 0 = 0\).

3. Hyperbolic Identities

Just like \(\cos^2 x + \sin^2 x \equiv 1\), hyperbolic functions have their own rules. However, there is a tiny change in the signs!

The fundamental identity you must know is:
\(\cosh^2 x - \sinh^2 x \equiv 1\)

Did you know? This identity is why they are called "hyperbolic." The equation for a standard hyperbola is \(x^2 - y^2 = 1\). If you let \(x = \cosh t\) and \(y = \sinh t\), they fit this equation perfectly!

How to prove it:

If you are asked to prove this, just plug in the exponential definitions:
\(\left(\frac{e^x + e^{-x}}{2}\right)^2 - \left(\frac{e^x - e^{-x}}{2}\right)^2\)
When you expand the brackets, the \(e^{2x}\) and \(e^{-2x}\) terms cancel out, leaving you with exactly 1.

4. Inverse Hyperbolic Functions

If we want to find the value of \(x\) for a given \(\sinh x\), we use the inverse functions: \(\text{arsinh } x\), \(\text{arcosh } x\), and \(\text{artanh } x\). (Some calculators use \(\sinh^{-1} x\)).

Important Note on \(\text{arcosh } x\): Because the \(\cosh x\) graph is U-shaped, it isn't "one-to-one" (two different \(x\) values can give the same \(y\)). To have an inverse, we only look at the positive side where \(x \geq 0\). Therefore, \(\text{arcosh } x\) is only defined for \(x \geq 1\).

Logarithmic Forms

Because the original functions are made of \(e^x\), the inverses are made of natural logs (\(\ln\)). You should be able to use these formulas:

  • \(\text{arsinh } x = \ln(x + \sqrt{x^2 + 1})\) for all \(x\)
  • \(\text{arcosh } x = \ln(x + \sqrt{x^2 - 1})\) for \(x \geq 1\)
  • \(\text{artanh } x = \frac{1}{2} \ln(\frac{1+x}{1-x})\) for \(|x| < 1\)

Don't worry if these look intimidating! You can actually derive them. For example, to find \(\text{arsinh } x\), you set \(y = \sinh x\), replace \(\sinh x\) with its exponential definition, and solve for \(x\) using the quadratic formula. It’s a very common exam task!

5. Step-by-Step: Solving a Hyperbolic Equation

Example: Solve \(2\sinh x + \cosh x = 1\)

Step 1: Replace with exponentials.
\(2(\frac{e^x - e^{-x}}{2}) + (\frac{e^x + e^{-x}}{2}) = 1\)

Step 2: Simplify the fractions.
\((e^x - e^{-x}) + \frac{1}{2}e^x + \frac{1}{2}e^{-x} = 1\)
Multiply everything by 2 to clear the fraction: \(2e^x - 2e^{-x} + e^x + e^{-x} = 2\)
\(3e^x - e^{-x} = 2\)

Step 3: Turn it into a quadratic.
Multiply the whole equation by \(e^x\) (remember \(e^x \cdot e^{-x} = 1\)):
\(3e^{2x} - 1 = 2e^x\)
\(3(e^x)^2 - 2e^x - 1 = 0\)

Step 4: Solve the quadratic.
Let \(u = e^x\): \(3u^2 - 2u - 1 = 0\)
Factorizing gives \((3u + 1)(u - 1) = 0\), so \(u = 1\) or \(u = -1/3\).

Step 5: Find \(x\).
Since \(e^x = u\), we have \(e^x = 1\) (which means \(x = 0\)).
We also have \(e^x = -1/3\), but since \(e^x\) is always positive, there are no real solutions for this part.
Final Answer: \(x = 0\)

Quick Review Box

Key Identities:
\(\tanh x \equiv \frac{\sinh x}{\cosh x}\)
\(\cosh^2 x - \sinh^2 x \equiv 1\)

Key Values:
\(\sinh 0 = 0\)
\(\cosh 0 = 1\)
\(\tanh 0 = 0\)

The "Big Secret": If a problem looks too hard, convert everything to \(e^x\) and \(e^{-x}\). It almost always turns into a quadratic equation you can solve!