# Trigonometry

> Edexcel International A-Level Mathematics · IAL Maths P2 (WMA12)
> Source: https://www.owlsprep.com/study/edexcel-ial-math-p2-trigonometry/

This guide covers the core trigonometric identities and equation-solving skills required for Edexcel IAL Pure Maths 2 (P2) Trigonometry, strictly aligned to the 2018 IAS specification.

**Prerequisites:** [P1 Trigonometric graphs and basic equation solving](https://www.owlsprep.com/study/edexcel-ial-math-p1-basic-trigonometry/); Solving linear and quadratic algebraic equations

## Learning objectives

- Recall and apply the two core P2 trigonometric identities: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$
- Solve linear trigonometric equations with transformed arguments in given intervals
- Solve quadratic trigonometric equations using the $\text{sin}^2\theta + \text{cos}^2\theta = 1$ identity
- Avoid common exam pitfalls when answering P2 trigonometry questions

## Core P2 Trigonometric Identities

**P2 Required Trigonometric Identities** — Two identities that you must memorize (they are not provided in the exam formula booklet) for all P2 trigonometry problems:

$$1. \\ \text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta} \quad (\text{valid when } \text{cos}\theta \neq 0)$$

$$2. \\ \text{sin}^2\theta + \text{cos}^2\theta = 1$$

You can rearrange the second identity to substitute for squared trigonometric ratios: $\text{sin}^2\theta = 1 - \text{cos}^2\theta$ or $\text{cos}^2\theta = 1 - \text{sin}^2\theta$, which is critical for solving quadratic trig equations later.

**Worked example:** Given $\text{sin}\theta = \frac{5}{13}$ and $\theta$ is an acute angle, find the exact values of $\text{cos}\theta$ and $\text{tan}\theta$.

1. Step 1: Use the $\text{sin}^2\theta + \text{cos}^2\theta = 1$ identity, substitute $\text{sin}\theta = \frac{5}{13}$:

   $$(\frac{5}{13})^2 + \text{cos}^2\theta = 1$$
2. Rearrange to solve for $\text{cos}^2\theta$:

   $$\text{cos}^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$$
3. Since $\theta$ is acute, $\text{cos}\theta$ is positive, so $\text{cos}\theta = \frac{12}{13}$
4. Step 2: Use the $\text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta}$ identity:

   $$\text{tan}\theta = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$$

*Calculator:* allowed

## Solving Trig Equations with Transformed Arguments

P2 trig equations include a single linear transformation of the angle (argument), e.g. $x + 30^\text{o}$, $2x$, $2(x+\frac{\text{π}}{6})$. Follow this 4-step method to solve them correctly:

1. Substitute $u$ for the transformed argument, and adjust the given interval by applying the same transformation to the upper and lower bounds
2. Find all solutions for $u$ in the adjusted interval
3. Rearrange to solve for $x$ from each valid $u$ value
4. Verify all $x$ solutions fall within the original given interval

> **tip**
>
> Always adjust the interval first: this is the most frequently skipped step that leads to missing or extra solutions.

**Worked example:** Solve $\text{sin}(x+\frac{\text{π}}{2}) = \frac{3}{4}$ for $0 < x < 2\text{π}$. Give your answers to 2 decimal places.

1. Step 1: Let $u = x + \frac{\text{π}}{2}$. Adjust the interval: $0 + \frac{\text{π}}{2} < u < 2\text{π} + \frac{\text{π}}{2} → \frac{\text{π}}{2} < u < \frac{5\text{π}}{2}$
2. Step 2: Solve $\text{sin}u = 0.75$. Principal solution: $u = \text{arcsin}(0.75) ≈ 0.848$ rad, which is below the lower interval bound. Next valid solutions:

   $$u = \text{π} - 0.848 ≈ 2.294 \text{ rad}, \quad u = 2\text{π} + 0.848 ≈ 7.131 \text{ rad}$$
3. Step 3: Solve for $x$: $x = u - \frac{\text{π}}{2}$

   $$x ≈ 2.294 - 1.571 = 0.72, \quad x ≈ 7.131 - 1.571 = 5.56$$
4. Step 4: Verify both 0.72 and 5.56 fall within $0 < x < 2\text{π}$ (≈6.28), so they are valid solutions.

*Calculator:* allowed

## Solving Quadratic Trigonometric Equations

Quadratic trig equations include a mix of a squared trig ratio and a linear trig ratio, e.g. $6\text{cos}^2x + \text{sin}x - 5 = 0$. Use the $\text{sin}^2\theta + \text{cos}^2\theta = 1$ identity to rewrite the equation as a quadratic in a single trig ratio, then solve as normal.

> **warning**
>
> Never divide both sides of an equation by $\text{sin}x$ or $\text{cos}x$: this eliminates all solutions where the ratio equals zero, leading to lost marks. Always factor instead.

**Worked example:** Solve $6\text{cos}^2x + \text{sin}x - 5 = 0$ for $0 ≤ x < 360^\text{o}$. Give non-exact answers to 1 decimal place.

1. Step 1: Replace $\text{cos}^2x$ with $1 - \text{sin}^2x$ to get a quadratic in $\text{sin}x$:

   $$6(1 - \text{sin}^2x) + \text{sin}x -5 = 0 → 6\text{sin}^2x - \text{sin}x - 1 = 0$$
2. Step 2: Let $y = \text{sin}x$, solve the quadratic equation:

   $$(3y + 1)(2y - 1) = 0 → y = -\frac{1}{3} \text{ or } y = \frac{1}{2}$$
3. Step 3: Solve $\text{sin}x = \frac{1}{2}$: solutions in interval are $x = 30^\text{o}, 150^\text{o}$
4. Step 4: Solve $\text{sin}x = -\frac{1}{3}$: principal solution ≈ -19.47°, so valid solutions in interval are $x = 180 + 19.47 = 199.5^\text{o}, 360 - 19.47 = 340.5^\text{o}$
5. Final solutions: $30^\text{o}, 150^\text{o}, 199.5^\text{o}, 340.5^\text{o}$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to adjust the interval when solving equations with transformed arguments
  - Why it fails: You will miss valid solutions or include solutions outside the original interval, losing up to half the marks for the question
  - Correct: First substitute $u$ for the transformed argument, apply the same transformation to the original interval bounds to get the valid range for $u$ before solving
- **Wrong:** Using P3 trig identities (sec, cosec, double angle, compound angle) to solve P2 equations
  - Why it fails: These are not required for P2, and you risk making unnecessary errors or wasting time in the exam
  - Correct: Only use the two core P2 identities ($\text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta}$ and $\text{sin}^2\theta + \text{cos}^2\theta = 1$) for all P2 trig problems
- **Wrong:** Rejecting negative trig ratio solutions without checking the interval
  - Why it fails: Angles in quadrants 2, 3 and 4 have negative sine or cosine values, so you will eliminate up to half of your valid solutions
  - Correct: Find all solutions for the trig ratio in the adjusted interval, including those where the ratio is negative, before solving for $x$
- **Wrong:** Dividing both sides of a trig equation by $\text{sin}x$ or $\text{cos}x$ to simplify
  - Why it fails: This removes all solutions where the divided ratio equals zero, leading to lost marks
  - Correct: Rearrange all terms to one side of the equation and factor out the common trig ratio instead of dividing

## Cheatsheet

| Concept | Formula/Method | Key Exam Note |
| --- | --- | --- |
| Core Identity 1 | $\text{sin}^2\theta + \text{cos}^2\theta = 1$ | Must memorize, not in formula booklet |
| Core Identity 2 | $\text{tan}\theta = \frac{\text{sin}\theta}{\text{cos}\theta}$ | Valid only when $\text{cos}\theta ≠ 0$ |
| Transformed Argument Equations | 1. Adjust interval for $u$ 2. Solve for $u$ 3. Solve for $x$ | Adjust interval first to avoid missing solutions |
| Quadratic Trig Equations | Replace squared ratio with identity → solve quadratic → solve trig equations | Never divide by a trig ratio, factor instead |

## What's next

Now that you have mastered P2 trigonometry, you are ready to progress to more advanced trigonometric content in Pure Mathematics 3 (P3), including reciprocal trigonometric ratios, compound and double angle identities, and R-form expressions. These skills build directly on the core identities and equation-solving methods you learned here, so make sure you are confident solving all P2-style trig equations before moving on. You can also practice applying these skills to P2 past paper questions to familiarize yourself with Edexcel's exam phrasing and mark scheme expectations, and ensure you are avoiding the common pitfalls outlined in this guide.

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