Further Trigonometry

Welcome to Further Trigonometry!

In standard A Level Mathematics, you learned how to juggle trigonometric identities and solve equations using various rules. In Further Pure Mathematics 1 (FP1), we take this a step further by introducing a "universal" tool: the t-formulae. Think of these as a secret code that allows you to transform tricky trigonometric expressions into much simpler algebraic ones. Whether you're proving an identity or solving a complex equation, the t-formulae are your new best friends!

1. What are the t-formulae?

The core idea of this chapter is a single substitution. We define a new variable, \(t\), based on the tangent of a half-angle.

We let:
\(t = \tan\left(\frac{\theta}{2}\right)\)

By using this substitution, we can express \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\) entirely in terms of \(t\). This is incredibly powerful because it turns a trig problem into an algebra problem.

The Three Essential Formulae

You must learn (and be able to derive) these three results:

1. \(\sin \theta = \frac{2t}{1+t^2}\)
2. \(\cos \theta = \frac{1-t^2}{1+t^2}\)
3. \(\tan \theta = \frac{2t}{1-t^2}\)

Quick Review: Notice that the denominators for \(\sin \theta\) and \(\cos \theta\) are exactly the same (\(1+t^2\)). This makes them easier to remember! Don't worry if this seems tricky at first—once you use them a few times, they become second nature.

How do we get these? (The Derivation)

You might be asked to show where these come from. We use the standard double angle formulae from your core A Level studies:

We know that \(\tan(2A) = \frac{2\tan A}{1-\tan^2 A}\).
If we let \(A = \frac{\theta}{2}\), then \(2A = \theta\). Substituting these in gives us the formula for \(\tan \theta\) directly!

To find \(\sin \theta\) and \(\cos \theta\), we can imagine a right-angled triangle where the "opposite" side is \(2t\) and the "adjacent" side is \(1-t^2\). By using Pythagoras' Theorem, the hypotenuse becomes \(1+t^2\). From there, you just use "SOH CAH TOA" to find the others.

Key Takeaway: The t-formulae act as a bridge between the world of trigonometry and the world of algebra.

2. Applying t-formulae to Identities

One common exam task is to prove that one trigonometric expression is equivalent to another. When the expression involves both "whole" angles (like \(\theta\)) and "half" angles (like \(\frac{\theta}{2}\)), the t-formulae are the perfect tool.

Step-by-Step Process:

1. Substitute: Replace every instance of \(\sin \theta\), \(\cos \theta\), or \(\tan \theta\) with their \(t\) equivalents.
2. Simplify: Multiply through to get rid of the "fractions within fractions" (complex fractions).
3. Factorise: Often, the numerator and denominator will have common factors that cancel out.
4. Convert back: Once simplified, remember that \(t\) is just \(\tan\left(\frac{\theta}{2}\right)\).

Example Analogy: Proving identities with t-formulae is like untangling a knot. By converting everything to \(t\), you're essentially laying the string out flat so you can see exactly where the loops go.

Common Mistake: Forgetting that \(\sec \theta\), \(\csc \theta\), and \(\cot \theta\) are just the reciprocals!
If \(\sin \theta = \frac{2t}{1+t^2}\), then \(\csc \theta = \frac{1+t^2}{2t}\). Just flip the fraction!

3. Solving Trigonometric Equations

The syllabus specifically mentions solving equations of the form:
\(a \cos x + b \sin x = c\)

While you might have used the \(R\cos(x \pm \alpha)\) method in standard math, the t-formulae provide a robust alternative, especially when you need to find all solutions in a specific range.

The Process:

1. Replace \(\cos x\) with \(\frac{1-t^2}{1+t^2}\) and \(\sin x\) with \(\frac{2t}{1+t^2}\).
2. Multiply the entire equation by \((1+t^2)\) to clear the denominators.
3. You will be left with a quadratic equation in \(t\) (e.g., \(At^2 + Bt + C = 0\)).
4. Solve for \(t\) using the quadratic formula or factorisation.
5. Finally, solve \(\tan\left(\frac{x}{2}\right) = t\) to find your values for \(x\).

Did you know?
There is one "hidden" solution to watch out for! Because \(\tan(90^\circ)\) is undefined, the t-substitution doesn't "see" solutions where \(\frac{x}{2} = 90^\circ\) (which means \(x = 180^\circ\)). Always check if \(x = 180^\circ\) works by plugging it back into the original equation at the very end!

Key Takeaway: Using \(t = \tan(x/2)\) turns a trigonometric equation into a quadratic equation, which we already know how to solve easily.

4. Summary and Memory Aids

To succeed in this chapter, focus on mastering the "algebraic heavy lifting." Here is a quick cheat sheet for your revision:

The "Triangle" Memory Trick

If you forget the formulae, draw a right-angled triangle with:
- Opposite: \(2t\)
- Adjacent: \(1-t^2\)
- Hypotenuse: \(1+t^2\)
From this triangle, you can derive \(\sin\), \(\cos\), and \(\tan\) using basic SohCahToa.

Quick Review Box:

Prerequisites: You should be comfortable with basic trig identities and solving quadratic equations.
Range Check: When solving \(\tan(x/2) = t\), if the range for \(x\) is \(0 < x < 360\), your range for \(x/2\) is \(0 < x/2 < 180\). Don't forget to adjust your calculator search range!
Check Your Work: Always pick a value from your final answer and plug it back into the original equation to ensure it works.

Don't be discouraged if the algebra looks messy! Further Maths is all about staying calm through long calculations. You've got this!