Learn9MA0Paper 1 · Pure Mathematics 1TrigonometryExact values and the unit circle

Trigonometry · Subtopic 5 of 13

Exact Values and the Unit Circle

How sine, cosine and tangent are defined for any angle using the unit circle, the CAST rule for signs, and the exact values at $30^\circ$ , $45^\circ$ and $60^\circ$ that you need without a calculator.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define $\sin\theta$ , $\cos\theta$ and $\tan\theta$ for any angle using the unit circle.
Use the CAST diagram to find the sign of a ratio in each quadrant.
Find exact values at $30^\circ$ , $45^\circ$ , $60^\circ$ and related angles, in degrees and radians.
Given one ratio, find the others exactly using a right-angled triangle.

Pause and think

In a right-angled triangle, every angle other than the right angle is less than $90^\circ$ . So what could $\sin150^\circ$ possibly mean?

Section 2 of 14

Core idea

Start hereExact answers without a calculator

A calculator says $\sin60^\circ=0.8660254\ldots$ , a decimal that never ends. But in an exam you're often asked for the exact value: $\dfrac{\sqrt3}2$ . Where does that come from? From cutting an equilateral triangle in half. Two simple triangles give you exact sin, cos and tan for $30^\circ$ , $45^\circ$ and $60^\circ$ , and a circle extends them to every angle there is.

The big idea

The half-square ( $1,1,\sqrt2$ ) and half-equilateral ( $1,\sqrt3,2$ ) triangles give exact trig values for $30^\circ,45^\circ,60^\circ$ . On the unit circle, the point at angle $\theta$ is $(\cos\theta,\sin\theta)$ , which defines sin and cos for any angle; the quadrant fixes the sign.

1. The ratios

\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad\cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad\tan\theta=\frac{\text{opp}}{\text{adj}}=\frac{\sin\theta}{\cos\theta}

opp: the side opposite the angle $\theta$ .
adj: the side next to $\theta$ (not the hypotenuse).
hyp: the hypotenuse, opposite the right angle.

Why $\tan=\frac{\sin}{\cos}$ : $\dfrac{\sin\theta}{\cos\theta}=\dfrac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}}=\dfrac{\text{opp}}{\text{adj}}$ .

2. The two special triangles

The two special triangles

Half an equilateral triangle of side

2

gives sides

1,\sqrt3,2

with angles

30^\circ

and

60^\circ

. Half a unit square gives sides

1,1,\sqrt2

with

45^\circ

. Read any exact value off these, e.g.

\sin60^\circ=\dfrac{\sqrt3}{2}

\tan45^\circ=1

Building the triangles

$\text{Square side }1,\text{ cut on the diagonal: angles }45^\circ,45^\circ,90^\circ$
The diagonal splits the right angle equally.
$\text{hyp}=\sqrt{1^2+1^2}=\sqrt2$
Pythagoras.
$\text{Equilateral side }2,\text{ cut in half: angles }30^\circ,60^\circ,90^\circ$
The cut bisects the $60^\circ$ top angle and the base.
$\text{height}=\sqrt{2^2-1^2}=\sqrt3$
Pythagoras with hypotenuse $2$ and half-base $1$ .

Exact values

\begin{array}{c|ccc}&30^\circ&45^\circ&60^\circ\\\hline\sin&\tfrac12&\tfrac{\sqrt2}2&\tfrac{\sqrt3}2\\\cos&\tfrac{\sqrt3}2&\tfrac{\sqrt2}2&\tfrac12\\\tan&\tfrac1{\sqrt3}&1&\sqrt3\end{array}

TrickPattern

\sin0^\circ,30^\circ,45^\circ,60^\circ,90^\circ=\frac{\sqrt0}2,\frac{\sqrt1}2,\frac{\sqrt2}2,\frac{\sqrt3}2,\frac{\sqrt4}2

. Cosine is the same list backwards.

3. Any angle: the unit circle

The unit circle: the point is (cos θ, sin θ)

θ (degrees) = 0

The height of the point is

\sin\theta

and the distance across is

\cos\theta

. In the second quadrant the height is still positive but the across is negative, so

\sin

is positive and

\cos

is negative.

Draw a circle of radius $1$ centred at the origin. Measure $\theta$ anticlockwise from the positive $x$ -axis. Since the hypotenuse is $1$ , the point's coordinates are $(\cos\theta,\sin\theta)$ . This works for $\theta=200^\circ$ or $-30^\circ$ , angles no triangle could hold.

Which ratios are positive? (CAST)

Drawing diagram…

Going anticlockwise: All, Sin, Tan, Cos ("All Students Take Calculus"). The sign comes from whether $x$ (cos) and $y$ (sin) are positive.

Worked example $\sin150^\circ$ , $\cos240^\circ$ , $\tan315^\circ$

Find each exactly.

Goal: Exact values with correct signs.
Strategy: Reference angle (acute angle to the $x$ -axis) → table value → sign from the quadrant.

$150^\circ:\ \text{ref }30^\circ,\ \text{Q2 (sin positive)}\Rightarrow\sin150^\circ=\tfrac12$
$180-150=30$ .
$240^\circ:\ \text{ref }60^\circ,\ \text{Q3 (cos negative)}\Rightarrow\cos240^\circ=-\tfrac12$
$240-180=60$ .
$315^\circ:\ \text{ref }45^\circ,\ \text{Q4 (tan negative)}\Rightarrow\tan315^\circ=-1$
$360-315=45$ .

Answer

\frac12

-\frac12

-1

✓ CheckCalculator:

\sin150^\circ=0.5

✓,

\cos240^\circ=-0.5

✓,

\tan315^\circ=-1

✓.

Worked exampleFrom one ratio to the others

$\sin\theta=\frac35$ and $\theta$ is obtuse. Find $\cos\theta$ and $\tan\theta$ .

Goal: Exact values.
Strategy: Build the triangle for the size; use the quadrant for the sign.

$\text{adj}=\sqrt{5^2-3^2}=4$
opp $3$ , hyp $5$ .
Obtuse: second quadrant, where cos and tan are negative.
$\cos\theta=-\tfrac45,\qquad\tan\theta=-\tfrac34$
Apply the signs.

Answer

\cos\theta=-\frac45

\tan\theta=-\frac34

✓ Check

\sin^2+\cos^2=\frac9{25}+\frac{16}{25}=1

✓.

Pause and think

Find exactly $\cos\frac{5\pi}6$ and $\sin\left(-\frac\pi3\right)$ .

Summary

Triangles $1,1,\sqrt2$ and $1,\sqrt3,2$ .
Unit circle point $(\cos\theta,\sin\theta)$ .
Reference angle + CAST sign.

Unit circle: Radius-1 circle at the origin.
Reference angle: Acute angle to the x-axis.

HistorySine comes from a mistranslation. Indian mathematicians called the half-chord of a circle jya-ardha, shortened to jya. Arabic scholars wrote it as jiba, without vowels: jb. Later Latin translators read jb as jaib, meaning "fold" or "pocket", and translated it as sinus. That is where our word "sine" comes from.

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