# Trigonometry

> Mathematics · CIE A-Level 9709
> Source: https://www.owlsprep.com/study/cie-9709-u1-trigonometry/

This module covers core trigonometry for CIE 9709 P1: trigonometric ratios for right triangles, exact values for special angles, the graphs of sine, cosine, and tangent, the two fundamental identities, inverse trigonometric functions, and solving simple trigonometric equations between 0° and 360°.

**Prerequisites:** [Right triangle geometry and Pythagoras' theorem](https://www.owlsprep.com/study/cie-9709-u1-triangle-geometry/); [Basic angle measurement and quadrants](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry-quadrants/)

## Learning objectives

- Calculate trigonometric ratios for acute angles in right triangles
- Recall and use exact trigonometric values for special angles
- Sketch and interpret the graphs of sine, cosine, and tangent, including their periods, ranges, and asymptotes
- Apply the fundamental trigonometric identities to simplify expressions and solve equations
- Use inverse trigonometric functions and their principal value ranges
- Find all solutions of simple trigonometric equations between 0° and 360°

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Trigonometry** appears **106 times** in the last 10 years — **14.9%** of all P1 questions (106 of 692).

Most-tested forms: Solving trig equations (74), Pythagorean trig identity (38), Trig graphs (sin, cos, tan) (21), Trig ratios in right triangles (20), Quotient identity (tan = sin/cos) (6).

## Trigonometric Ratios for Acute Angles

These right-triangle ratios are assumed knowledge from IGCSE and are recapped here only briefly. For an acute angle $\theta$, SOH-CAH-TOA defines $\sin\theta$, $\cos\theta$, and $\tan\theta$ from the opposite, adjacent, and hypotenuse sides. Given one ratio, Pythagoras' theorem finds the third side and hence the other two ratios.

> **SOH-CAH-TOA**
>
> SOH: **S**ine = **O**pposite / **H**ypotenuse  
> CAH: **C**osine = **A**djacent / **H**ypotenuse  
> TOA: **T**angent = **O**pposite / **A**djacent

## Exact Trigonometric Values for Special Angles

CIE exams regularly require exact values for $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$, which are derived from equilateral and isosceles right triangles.

| Angle (°) | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ |
| --- | --- | --- | --- |
| 0 | 0 | 1 | 0 |
| 30 | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ |
| 45 | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | 1 |
| 60 | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
| 90 | 1 | 0 | Undefined |

**Worked example:** Find the exact value of $\sin 30^\circ + \cos 60^\circ - \tan 45^\circ$.

1. Substitute the exact values from the table:
2. $$\sin 30^\circ = \frac{1}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 45^\circ = 1$$
3. Calculate the result:
4. $$\frac{1}{2} + \frac{1}{2} - 1 = 1 - 1 = 0$$

**Worked example:** Find the exact value of $\cos 150^\circ$.

1. Write $150^\circ$ using the related acute angle $30^\circ$:
2. $$150^\circ = 180^\circ - 30^\circ$$
3. Since $150^\circ$ is obtuse, its cosine is negative. Apply the related-angle rule $\cos(180^\circ - \theta) = -\cos\theta$:
4. $$\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ$$
5. Substitute the exact value of $\cos 30^\circ$:
6. $$-\cos 30^\circ = -\frac{\sqrt{3}}{2}$$

## Fundamental Trigonometric Identities

Two fundamental identities are used constantly in CIE P1 problems. They simplify expressions, prove other identities, and help solve trigonometric equations.

**Fundamental Trigonometric Identities** — Paper 1 relies on just two fundamental identities:  
1. Quotient identity: $\tan\theta = \frac{\sin\theta}{\cos\theta}$  
2. Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$

*Example:* If you know $\sin\theta$, the Pythagorean identity gives $\cos^2\theta = 1 - \sin^2\theta$, and the quotient identity then gives $\tan\theta$.

**Worked example:** Show that $\frac{1 - \cos^2\theta}{\tan\theta} = \sin\theta\cos\theta$.

1. Replace $1 - \cos^2\theta$ using the Pythagorean identity:
2. $$1 - \cos^2\theta = \sin^2\theta$$
3. Replace $\tan\theta$ using the quotient identity, then divide:
4. $$\frac{\sin^2\theta}{\tan\theta} = \sin^2\theta \div \frac{\sin\theta}{\cos\theta} = \sin^2\theta \cdot \frac{\cos\theta}{\sin\theta}$$
5. Cancel one factor of $\sin\theta$:
6. $$\sin^2\theta \cdot \frac{\cos\theta}{\sin\theta} = \sin\theta\cos\theta$$

## Graphs of Sine, Cosine, and Tangent

The graphs of $y = \sin x$ and $y = \cos x$ are smooth waves that repeat every $360^\circ$ (their period) and stay between $-1$ and $1$. The graph of $y = \tan x$ behaves differently: it repeats every $180^\circ$ and has vertical asymptotes wherever $\cos x = 0$, at $x = 90^\circ$ and $x = 270^\circ$ within one $0^\circ$ to $360^\circ$ cycle.

| Graph | Period | Range | Asymptotes (0° to 360°) |
| --- | --- | --- | --- |
| $y = \sin x$ | $360^\circ$ | $-1 \leq y \leq 1$ | None |
| $y = \cos x$ | $360^\circ$ | $-1 \leq y \leq 1$ | None |
| $y = \tan x$ | $180^\circ$ | all real values | $x = 90^\circ, 270^\circ$ |

The transformation rules from earlier in the course carry over directly. A number multiplying the whole function changes the amplitude (a vertical stretch), while a number multiplying $x$ changes the period (a horizontal stretch).

| Function | Effect on the graph |
| --- | --- |
| $y = 3\sin x$ | Amplitude $3$: the graph oscillates between $-3$ and $3$; the period stays $360^\circ$. |
| $y = \cos 2x$ | The $2$ halves the period to $180^\circ$; the range stays $-1 \leq y \leq 1$. |
| $y = 1 - \cos 2x$ | $\cos 2x$ reflected in the $x$-axis and shifted up by $1$; the range becomes $0 \leq y \leq 2$. |

**Worked example:** Describe how the graph of $y = 1 - \cos 2x$ is obtained from $y = \cos x$, and state its range.

1. The coefficient $2$ inside $\cos 2x$ halves the period:
2. $$\text{Period} = \frac{360^\circ}{2} = 180^\circ$$
3. The negative sign reflects the curve in the $x$-axis, and the $+1$ shifts it up by one unit.
4. Because $\cos 2x$ lies between $-1$ and $1$, the range of $1 - \cos 2x$ is:
5. $$-1 \leq \cos 2x \leq 1 \implies 0 \leq 1 - \cos 2x \leq 2$$

## Inverse Trigonometric Functions

The inverse trigonometric functions $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$ reverse the trig ratios: given a ratio, they return an angle. Because the trig functions repeat, each inverse returns only one angle, the principal value, which is the value your calculator displays.

| Inverse function | Principal value range |
| --- | --- |
| $\sin^{-1}x$ | $-90^\circ \leq \theta \leq 90^\circ$ |
| $\cos^{-1}x$ | $0^\circ \leq \theta \leq 180^\circ$ |
| $\tan^{-1}x$ | $-90^\circ < \theta < 90^\circ$ |

A calculator only ever gives this principal value. To find any other solutions in a wider interval such as $0^\circ$ to $360^\circ$, use the symmetry of the graph (or the ASTC rule), as shown in the next section.

## Solving Trigonometric Equations (0° to 360°)

Trigonometric equations usually have two solutions between $0^\circ$ and $360^\circ$. Each ratio is positive in two of the four quadrants and negative in the other two, so a horizontal line $y = k$ crosses the curve twice in each $360^\circ$ period, giving two solutions in most cases. Use the ASTC rule to locate them.

> **ASTC Rule**
>
> **A**ll ratios positive in Q1, **S**ine positive in Q2, **T**angent positive in Q3, **C**osine positive in Q4

**Worked example:** Solve $2\sin \theta + 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$.

1. Rearrange to isolate the trigonometric ratio:
2. $$2\sin\theta = -1 \implies \sin\theta = -\frac{1}{2}$$
3. Find the acute reference angle from the positive value:
4. $$\alpha = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$
5. Sine is negative in Q3 and Q4, so calculate solutions for both quadrants:
6. $$Q3: 180^\circ + 30^\circ = 210^\circ \\ Q4: 360^\circ - 30^\circ = 330^\circ$$
7. Both solutions lie in the required range, so the solutions are $\theta = 210^\circ$ and $\theta = 330^\circ$.

## Common pitfalls

- **Wrong:** Forgetting there are 2 solutions between 0° and 360° for most equations
  - Why it fails: Examiners penalize missing solutions, which is a very common mistake
  - Correct: Always use the ASTC rule to find all valid quadrants and all solutions
- **Wrong:** Treating $\sin^2 \theta$ as $\sin(\theta^2)$
  - Why it fails: The notation means $(\sin \theta)^2$, not sine of theta squared
  - Correct: Remember $\sin^n \theta = (\sin \theta)^n$ for any exponent $n$
- **Wrong:** Giving an approximate decimal when an exact answer is required
  - Why it fails: CIE awards zero marks for non-exact answers when exact values are requested
  - Correct: Always check the question wording, and leave answers in surd/fraction form
- **Wrong:** Incorrectly rearranging $\sin^2 \theta + \cos^2 \theta = 1$ to $\cos^2 \theta = 1 + \sin^2 \theta$
  - Why it fails: Simple sign error when moving terms across the equals sign
  - Correct: Double-check: $\cos^2 \theta = 1 - \sin^2 \theta$ and $\sin^2 \theta = 1 - \cos^2 \theta$
- **Wrong:** Always subtracting the reference angle from 180° for the second solution
  - Why it fails: The calculation for the second solution depends on which quadrant it lies in
  - Correct: Find the quadrants first with ASTC, then calculate angles as $180+\alpha$ (Q3) and $360-\alpha$ (Q4)

## Cheatsheet

| Concept | Key Result |
| --- | --- |
| Trig Ratios | SOH-CAH-TOA: sin=O/H, cos=A/H, tan=O/A |
| Exact Values | Memorize 0°, 30°, 45°, 60°, 90° from special triangles |
| Trig Graphs | $\sin x, \cos x$: period $360^\circ$, range $[-1,1]$; $\tan x$: period $180^\circ$, asymptotes $90^\circ, 270^\circ$ |
| Core Identities | $\tan\theta = \frac{\sin\theta}{\cos\theta}$; $\sin^2\theta + \cos^2\theta = 1$ |
| Inverse Trig | $\sin^{-1}x\in[-90^\circ,90^\circ]$, $\cos^{-1}x\in[0^\circ,180^\circ]$, $\tan^{-1}x\in(-90^\circ,90^\circ)$; calculator gives principal value |
| ASTC Rule | Q1: All +, Q2: Sin +, Q3: Tan +, Q4: Cos + |
| Solving Equations | Find reference angle, get solutions in all valid quadrants |

## What's next

Mastery of this core trigonometry is essential for all further pure mathematics topics in CIE 9709. The identities and equation-solving skills you learn here are used constantly in graph transformations, more advanced identities, calculus, and mechanics problems. CIE regularly includes multi-part questions that combine these foundational skills with other topics, so reinforcing this sub-topic will improve your performance across the entire exam.

- [Series](https://www.owlsprep.com/study/cie-9709-u1-series/)
- [Differentiation](https://www.owlsprep.com/study/cie-9709-u1-differentiation/)
- [Integration](https://www.owlsprep.com/study/cie-9709-u1-integration/)

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