# The sine and cosine formulae

> Edexcel International GCSE Further Pure Mathematics · 4PM1 2016
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-the-sine-and-cosine-formulae/

This guide covers the three core trigonometric formulae for non-right-angled triangles required for Edexcel IGCSE Further Pure Math (4PM1): the sine rule, cosine rule, and ½ab sin C area formula, plus exam-focused problem-solving strategies.

**Prerequisites:** [Right-angled trigonometry (SOHCAHTOA)](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-right-angled-trigonometry/); [Exact trigonometric values for 30°, 45°, 60°](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s10-exact-trig-values/)

## Learning objectives

- Recall and apply the sine rule to find unknown sides and angles in triangles
- Use the given cosine rule to solve for missing dimensions in non-right-angled triangles
- Calculate the area of any triangle using the ½ab sin C formula from memory
- Identify and resolve the ambiguous case of the sine rule in exam problems

## The Sine Rule

**Sine Rule** — Relates the lengths of sides of any triangle to the sines of their opposite angles. This formula must be recalled for the exam, it is not provided on the formula sheet.

*Notation:* \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

The sine rule is used when you are given at least one pair of corresponding side and opposite angle, plus one additional side or angle. Always label your triangle consistently: side $a$ is opposite angle $A$, side $b$ opposite angle $B$, side $c$ opposite angle $C$.

**Worked example:** In triangle $ABC$, angle $A = 40^\circ$, side $a = 8$ cm, side $b = 10$ cm. Find angle $B$, correct to 1 decimal place.

1. Substitute known values into the sine rule formula:
2. $$\frac{8}{\sin 40^\circ} = \frac{10}{\sin B}$$
3. Rearrange to isolate $\sin B$:
4. $$\sin B = \frac{10 \times \sin 40^\circ}{8}$$
5. Calculate the right-hand side: $\sin B = \frac{10 \times 0.6428}{8} = 0.8035$
6. Take inverse sine to find $B$: $B = \arcsin(0.8035) = 53.5^\circ$ (1 d.p.)
7. Note: This is the ambiguous case, so a second valid solution is $B = 180^\circ - 53.5^\circ = 126.5^\circ$, check if this fits the problem context.

> **warning**
>
> The ambiguous case only occurs when you are given two sides and a non-included angle (SSA). Always check if the supplementary angle is a valid solution (sum of angles must not exceed 180°).

> **Exam tip:** If the question does not specify which triangle to use, give both possible solutions for the ambiguous case to earn full marks.

*Calculator:* allowed

## The Cosine Rule

**Cosine Rule** — Used to find unknown sides or angles in triangles where the sine rule cannot be applied. This formula is provided on the exam formula sheet, you do not need to memorize it.

*Notation:* a^2 = b^2 + c^2 - 2bc \cos A

Use the cosine rule in two scenarios: 1) You know two sides and the included angle between them, and need to find the third side; 2) You know all three sides of the triangle, and need to find any interior angle.

**Worked example:** In triangle $PQR$, sides $p = 5$ cm, $q = 7$ cm, included angle $R = 60^\circ$. Find the length of side $r$, correct to 2 significant figures.

1. Label the triangle: side $r$ is opposite angle $R$, so substitute into the cosine rule:
2. $$r^2 = p^2 + q^2 - 2pq \cos R$$
3. $$r^2 = 5^2 + 7^2 - 2(5)(7)\cos 60^\circ$$
4. Calculate each term: $r^2 = 25 + 49 - 70 \times 0.5 = 74 - 35 = 39$
5. Take square root: $r = \sqrt{39} = 6.2$ cm (2 s.f.)

> **Exam tip:** Rearrange the cosine rule to solve for angles directly if needed: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ to save time in exams.

*Calculator:* allowed

## Area of a Triangle: ½ab sin C

**Trigonometric Triangle Area Formula** — Calculates the area of any triangle when you know the lengths of two sides and the size of the included angle between them. This formula must be recalled for the exam, it is not provided on the formula sheet.

*Notation:* \text{Area} = \frac{1}{2}ab \sin C

This formula works for all triangles, not just right-angled ones. The angle $C$ must always be the included angle between sides $a$ and $b$ for the formula to work correctly.

**Worked example:** Find the area of triangle $XYZ$ where sides $x = 12$ cm, $z = 9$ cm, included angle $Y = 30^\circ$. Give your answer as an exact value.

1. Identify the two sides and included angle: sides $x$ and $z$ meet at angle $Y$, so substitute into the formula:
2. $$\text{Area} = \frac{1}{2} \times x \times z \times \sin Y$$
3. $$\text{Area} = \frac{1}{2} \times 12 \times 9 \times \sin 30^\circ$$
4. Use exact value $\sin 30^\circ = 0.5$:
5. $$\text{Area} = 54 \times 0.5 = 27 \text{ cm}^2$$

> **Exam tip:** If the question asks for an exact area, use the exact trigonometric values for 30°, 45°, 60° instead of decimal approximations to avoid losing marks.

*Calculator:* allowed

## Combining Formulae for Exam Problems

Most exam questions will require you to use multiple formulae in sequence to solve a full problem. Always start by labeling your triangle clearly, listing all known values, then selecting the correct formula for the first unknown you need to find.

**Worked example:** In triangle $ABC$, $AB = 10$ cm, $BC = 8$ cm, angle $BAC = 50^\circ$. Find the area of triangle $ABC$, correct to 1 decimal place.

1. First, use the sine rule to find angle $ACB$:
2. $$\frac{BC}{\sin BAC} = \frac{AB}{\sin ACB}$$
3. $$\frac{8}{\sin 50^\circ} = \frac{10}{\sin C} \implies \sin C = \frac{10 \times 0.7660}{8} = 0.9575$$
4. Two possible values for $C$: $73.2^\circ$ or $106.8^\circ$
5. First solution: angle $B = 180 - 50 -73.2 = 56.8^\circ$, area = $\frac{1}{2} \times 8 \times 10 \times \sin 56.8^\circ = 33.5$ cm²
6. Second solution: angle $B = 180 -50 -106.8 = 23.2^\circ$, area = $\frac{1}{2} \times 8 \times10 \times \sin23.2^\circ =15.8$ cm²

> **Exam tip:** Show all intermediate steps when combining formulae: examiners award marks for correct application of each individual formula even if your final answer is wrong.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the sine rule when only sides and no opposite angle pairs are known
  - Why it fails: The sine rule requires at least one matching side-angle pair to function, leading to incorrect equations if this condition is not met
  - Correct: Use the cosine rule instead when you have two sides and included angle, or all three sides
- **Wrong:** Forgetting to check for the ambiguous case when using the sine rule with SSA measurements
  - Why it fails: Two valid triangles can exist, leading to missing half the solution and losing marks
  - Correct: Always calculate the supplementary angle ($180^\circ -$ calculated angle) and verify if it is a valid solution (sum of angles < 180°)
- **Wrong:** Using a non-included angle in the ½ab sin C area formula
  - Why it fails: The formula only works if the angle is between the two sides used, leading to incorrect area values
  - Correct: If you do not have the included angle, use the sine/cosine rule to find it first before calculating area
- **Wrong:** Trying to memorize the cosine rule when it is provided on the formula sheet
  - Why it fails: Wastes valuable memorization capacity for other required formulae like the sine rule and area formula
  - Correct: Familiarize yourself with the formula sheet layout to quickly locate the cosine rule during exams
- **Wrong:** Mixing up side and opposite angle labels when substituting into formulae
  - Why it fails: This leads to incorrect values for sides and angles, even if you applied the formula structure correctly
  - Correct: Always label your triangle with side $a$ opposite angle $A$, $b$ opposite $B$, $c$ opposite $C$ before substituting any values

## Cheatsheet

| Formula | Use Case | Given on Formula Sheet? |
| --- | --- | --- |
| $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ | Find unknown side/angle, 1+ side-angle pair known | No |
| $a^2 = b^2 + c^2 - 2bc \cos A$ | Find 3rd side (2 sides + included angle) / find angle (all 3 sides known) | Yes |
| $\text{Area} = \frac{1}{2}ab \sin C$ | Calculate area, 2 sides + included angle known | No |

## What's next

Now that you have mastered the sine and cosine formulae, you can apply these skills to solve 2D and 3D trigonometry problems, the next sub-topic in the Edexcel IGCSE Further Pure Math syllabus. You will also use these formulae alongside coordinate geometry and vector problems later in the course. Practise past exam questions to familiarize yourself with how these formulae are tested in structured written papers, and make sure you can quickly recall the sine rule and area formula under timed conditions. Remember to always check for the ambiguous case in SSA problems, as this is a common mark-losing trap in exams.

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