Trigonometry

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

1. Trigonometric Ratios for Acute Angles★☆☆☆☆⏱ 3 min

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

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

📐 Worked Example

Find the exact value of $cos 15 0^{\circ}$ .

1
Write $15 0^{\circ}$ using the related acute angle $3 0^{\circ}$ :
2
$150^\circ = 180^\circ - 30^\circ$
3
Since $15 0^{\circ}$ is obtuse, its cosine is negative. Apply the related-angle rule $cos (18 0^{\circ} - θ) = - cos θ$ :
4
$\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ$
5
Substitute the exact value of $cos 3 0^{\circ}$ :
6
$-\cos 30^\circ = -\frac{\sqrt{3}}{2}$

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

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

📘 Definition

Fundamental Trigonometric Identities

Paper 1 relies on just two fundamental identities:

Quotient identity: $tan θ = \frac{s i n θ}{c o s θ}$
Pythagorean identity: $sin^{2} θ + cos^{2} θ = 1$

Example:

If you know $sin θ$ , the Pythagorean identity gives $cos^{2} θ = 1 - sin^{2} θ$ , and the quotient identity then gives $tan θ$ .

📐 Worked Example

Show that $\frac{1 - c o s ^{2} θ}{t a n θ} = sin θ cos θ$ .

1
Replace $1 - cos^{2} θ$ using the Pythagorean identity:
2
$1 - \cos^2\theta = \sin^2\theta$
3
Replace $tan θ$ 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 $sin θ$ :
6
$\sin^2\theta \cdot \frac{\cos\theta}{\sin\theta} = \sin\theta\cos\theta$

4. Graphs of Sine, Cosine, and Tangent★★☆☆☆⏱ 6 min

The graphs of $y = sin x$ and $y = cos x$ are smooth waves that repeat every $36 0^{\circ}$ (their period) and stay between $- 1$ and $1$ . The graph of $y = tan x$ behaves differently: it repeats every $18 0^{\circ}$ and has vertical asymptotes wherever $cos x = 0$ , at $x = 9 0^{\circ}$ and $x = 27 0^{\circ}$ within one $0^{\circ}$ to $36 0^{\circ}$ cycle.

Graph	Period	Range	Asymptotes (0° to 360°)
$y = sin x$	$36 0^{\circ}$	$- 1 \leq y \leq 1$	None
$y = cos x$	$36 0^{\circ}$	$- 1 \leq y \leq 1$	None
$y = tan x$	$18 0^{\circ}$	all real values	$x = 9 0^{\circ}, 27 0^{\circ}$

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 $36 0^{\circ}$ .
$y = cos 2 x$	The $2$ halves the period to $18 0^{\circ}$ ; the range stays $- 1 \leq y \leq 1$ .
$y = 1 - cos 2 x$	$cos 2 x$ 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 $y = 1 - cos 2 x$ is obtained from $y = cos x$ , and state its range.

1
The coefficient $2$ inside $cos 2 x$ 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 $cos 2 x$ lies between $- 1$ and $1$ , the range of $1 - cos 2 x$ is:
5
$-1 \leq \cos 2x \leq 1 \implies 0 \leq 1 - \cos 2x \leq 2$

5. Inverse Trigonometric Functions★★★☆☆⏱ 4 min

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
$sin^{- 1} x$	$- 9 0^{\circ} \leq θ \leq 9 0^{\circ}$
$cos^{- 1} x$	$0^{\circ} \leq θ \leq 18 0^{\circ}$
$tan^{- 1} x$	$- 9 0^{\circ} < θ < 9 0^{\circ}$

A calculator only ever gives this principal value. To find any other solutions in a wider interval such as $0^{\circ}$ to $36 0^{\circ}$ , use the symmetry of the graph (or the ASTC rule), as shown in the next section.

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

Trigonometric equations usually have two solutions between $0^{\circ}$ and $36 0^{\circ}$ . 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 $36 0^{\circ}$ period, giving two solutions in most cases. Use the ASTC rule to locate them.

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

7. 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)

8. 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
Trig Graphs	$sin x, cos x$ : period $36 0^{\circ}$ , range $[- 1, 1]$ ; $tan x$ : period $18 0^{\circ}$ , asymptotes $9 0^{\circ}, 27 0^{\circ}$
Core Identities	$tan θ = \frac{s i n θ}{c o s θ}$ ; $sin^{2} θ + cos^{2} θ = 1$
Inverse Trig	$sin^{- 1} x \in [- 9 0^{\circ}, 9 0^{\circ}]$ , $cos^{- 1} x \in [0^{\circ}, 18 0^{\circ}]$ , $tan^{- 1} x \in (- 9 0^{\circ}, 9 0^{\circ})$ ; 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

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.