Trigonometry

Mathematics· Pure Mathematics 1 Unit 5: Trigonometry· 20 min read

1. Trigonometric Ratios for Acute Angles★☆☆☆☆⏱ 5 min

📘 Definition

Trigonometric Ratios

$sin θ$ , $cos θ$ , $tan θ$

For an acute angle $θ$ in a right-angled triangle, ratios of sides are defined relative to $θ$ as opposite, adjacent, and hypotenuse.

Example:

In a 3-4-5 triangle with $θ$ opposite the side of length 3, $sin θ = 3/5$ .

📐 Worked Example

Find $tan θ$ and $cos θ$ if $sin θ = 5/13$ and $θ$ is acute.

1
Use the definition of sine to assign side lengths. $sin θ = \frac{opposite}{hypotenuse}$ , so opposite = 5, hypotenuse = 13.
2
Use Pythagoras' theorem to find the adjacent side:
3
$(\text{adjacent})^2 + 5^2 = 13^2 \\ (\text{adjacent})^2 = 169 - 25 = 144 \\ \text{adjacent} = 12$
4
Calculate the required ratios using SOH-CAH-TOA:
5
$\cos\theta = \frac{12}{13}, \quad \tan\theta = \frac{5}{12}$

2. Exact Trigonometric Values for Special Angles★★☆☆☆⏱ 6 min

CIE exams regularly require exact values for $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ , which are derived from equilateral and isosceles right triangles.

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

📐 Worked Example

Find the exact value of $sin 3 0^{\circ} + cos 6 0^{\circ} - tan 4 5^{\circ}$ .

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
$\frac{1}{2} + \frac{1}{2} - 1 = 1 - 1 = 0$

3. Fundamental Pythagorean Identities★★★☆☆⏱ 7 min

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

📘 Definition

Pythagorean Trigonometric Identities

Two identities derived from Pythagoras' theorem on the unit circle:

$sin^{2} θ + cos^{2} θ = 1$
$1 + tan^{2} θ = \frac{1}{c o s ^{2} θ}$

Example:

If you know $sin θ$ , you can use the first identity to find $cos^{2} θ = 1 - sin^{2} θ$ .

📐 Worked Example

Simplify $(1 + tan^{2} θ) cos^{2} θ$ .

1
Substitute the second Pythagorean identity into the expression:
2
$\left(\frac{1}{\cos^2 \theta}\right)\cos^2 \theta$
3
Cancel the common term, assuming $cos θ \neq = 0$ :
4
$\left(\frac{1}{\cancel{\cos^2 \theta}}\right)\cancel{\cos^2 \theta} = 1$

4. Solving Trigonometric Equations (0° to 360°)★★★★☆⏱ 7 min

Trigonometric equations usually have 2 solutions between 0° and 360°, because each ratio is positive in one quadrant and negative in another. Use the ASTC rule to find all solutions.

📐 Worked Example

Solve $2 sin θ + 1 = 0$ for $0^{\circ} \leq θ \leq 36 0^{\circ}$ .

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
$\alpha = \arcsin\left(\frac{1}{2}\right) = 30^\circ$
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 $θ = 21 0^{\circ}$ and $θ = 33 0^{\circ}$ .

5. Common Pitfalls

Wrong move:

Forgetting there are 2 solutions between 0° and 360° for most equations

Why:

Examiners penalize missing solutions, which is a very common mistake

Correct move:

Always use the ASTC rule to find all valid quadrants and all solutions

Wrong move:

Treating $sin^{2} θ$ as $sin (θ^{2})$

Why:

The notation means $(sin θ)^{2}$ , not sine of theta squared

Correct move:

Remember $sin^{n} θ = (sin θ)^{n}$ for any exponent $n$

Wrong move:

Giving an approximate decimal when an exact answer is required

Why:

CIE awards zero marks for non-exact answers when exact values are requested

Correct move:

Always check the question wording, and leave answers in surd/fraction form

Wrong move:

Incorrectly rearranging $sin^{2} θ + cos^{2} θ = 1$ to $cos^{2} θ = 1 + sin^{2} θ$

Why:

Simple sign error when moving terms across the equals sign

Correct move:

Double-check: $cos^{2} θ = 1 - sin^{2} θ$ and $sin^{2} θ = 1 - cos^{2} θ$

Wrong move:

Always subtracting the reference angle from 180° for the second solution

Why:

The calculation for the second solution depends on which quadrant it lies in

Correct move:

Find the quadrants first with ASTC, then calculate angles as $180 + α$ (Q3) and $360 - α$ (Q4)

6. Quick Reference 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
Pythagorean IDs	$sin^{2} θ + cos^{2} θ = 1$ ; $1 + tan^{2} θ = 1/ cos^{2} θ$
ASTC Rule	Q1: All +, Q2: Sin +, Q3: Tan +, Q4: Cos +
Solving Equations	Find reference angle, get solutions in all valid quadrants

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Solve 2-term trigonometric equation
2022 · 1
Simplify expression with identities
2021 · 1
Find exact trigonometric value

Going deeper

syllabusOfficial CIE 9709 SyllabusFull unit learning outcomes

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.