Cambridge IGCSE Mathematics 0580: Exact trigonometric values and trigonometric functions – Study Guide

This study guide teaches subtopics E6.3 Exact trigonometric values and E6.4 Trigonometric functions from topic 6, Trigonometry, of the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. Both subtopics are Extended only: the syllabus lists C6.3 and C6.4 as “Extended content only”, so they are not tested on the Core papers (Papers 1 and 3). Extended candidates meet them on Paper 2, which is non-calculator, and Paper 4, which is a calculator paper. Paper 2 is where exact values matter most, because you cannot look a value up.

Use it alongside the course hub and the printable 0580 checklist. The trigonometry study guide covers the rest of topic 6, and the trigonometry revision notes hold the condensed exact-value table. Afterwards, test yourself with the trigonometry practice questions and the Extended diagnostic.

What this unit covers

Syllabus section	What you must be able to do	Tier
E6.3.1	Know the exact values of sin x and cos x for x = 0°, 30°, 45°, 60° and 90°	Extended only
E6.3.2	Know the exact values of tan x for x = 0°, 30°, 45° and 60°	Extended only
E6.4.1	Recognise, sketch and interpret the graphs of y = sin x, y = cos x and y = tan x for 0° ⩽ x ⩽ 360°	Extended only
E6.4.2	Solve trigonometric equations involving sin x, cos x or tan x for 0° ⩽ x ⩽ 360°	Extended only

The exact values are not printed in the List of formulas on Paper 2 or Paper 4. The list gives only the sine rule, the cosine rule and the area formula for trigonometry, so you must learn the values below.

E6.3 Exact trigonometric values

Where the values come from

Two triangles generate every value.

The 45° triangle. Take a right-angled isosceles triangle with the two shorter sides equal to 1. By Pythagoras the hypotenuse is √(1² + 1²) = √2. Both acute angles are 45°.

sin 45° = opposite/hypotenuse = 1/√2 = √2/2
cos 45° = adjacent/hypotenuse = 1/√2 = √2/2
tan 45° = opposite/adjacent   = 1/1  = 1

The 30°–60° triangle. Take an equilateral triangle with sides 2 and cut it in half down a line of symmetry. Each half is a right-angled triangle with hypotenuse 2, short side 1 and height √(2² − 1²) = √3. The angles are 30°, 60° and 90°. The side of length 1 is opposite the 30° angle; the side √3 is opposite the 60° angle.

sin 30° = 1/2        cos 30° = √3/2       tan 30° = 1/√3 = √3/3
sin 60° = √3/2       cos 60° = 1/2        tan 60° = √3/1 = √3

0° and 90°. Picture the right-angled triangle with hypotenuse 1 as the angle shrinks to 0°: the opposite side shrinks to 0 and the adjacent side becomes 1. So sin 0° = 0, cos 0° = 1 and tan 0° = 0. As the angle grows to 90° the roles swap: sin 90° = 1 and cos 90° = 0. The syllabus does not ask for tan 90°, because tan 90° has no value (you would be dividing by cos 90° = 0).

The values you must know

x	0°	30°	45°	60°	90°
sin x	0	1/2	√2/2	√3/2	1
cos x	1	√3/2	√2/2	1/2	0
tan x	0	√3/3	1	√3	(not required)

Patterns that help you recall it:

• Reading sin x from 0° to 90°, the values are √0/2, √1/2, √2/2, √3/2, √4/2.
• The cos row is the sin row reversed, because cos x = sin(90° − x).
• tan x = sin x ÷ cos x, so tan 60° = (√3/2) ÷ (1/2) = √3.

1/√2 and √2/2 are equal, as are 1/√3 and √3/3. Rationalising the denominator is part of E1.18 Surds, so either form is acceptable unless a question asks for a rationalised denominator.

Worked example 1: evaluating an expression (non-calculator)

Find the exact value of tan 60° × cos 30° + 4 sin² 45°.

tan 60° = √3,   cos 30° = √3/2,   sin 45° = √2/2

tan 60° × cos 30° = √3 × √3/2 = 3/2
sin² 45° = (√2/2)² = 2/4 = 1/2,   so 4 sin² 45° = 2

Total = 3/2 + 2 = 7/2

Note that sin² 45° means (sin 45°)², not sin(45²).

Worked example 2: exact sides in a right-angled triangle (non-calculator)

(a) In triangle PQR, angle Q = 90°, angle P = 60° and PR = 12 cm. Find the exact lengths of QR and PQ.

QR is opposite angle P:  QR = 12 sin 60° = 12 × √3/2 = 6√3 cm
PQ is adjacent to P:     PQ = 12 cos 60° = 12 × 1/2  = 6 cm

(b) In another right-angled triangle, the side adjacent to a 30° angle is 9 cm. Find the exact length of the side opposite the 30° angle.

opposite = 9 tan 30° = 9 × 1/√3 = 9/√3
Rationalise: 9/√3 × √3/√3 = 9√3/3 = 3√3 cm

Worked example 3: finding an angle from an exact ratio

A right-angled triangle has a side of 5 cm opposite angle θ and a side of 5√3 cm adjacent to θ. Find θ without a calculator.

tan θ = 5 / (5√3) = 1/√3
1/√3 = √3/3, which is tan 30°
θ = 30°

If a ratio simplifies to 1/2, √2/2, √3/2, 1/√3, 1 or √3, look for a standard angle.

E6.4 Trigonometric functions

The three graphs from 0° to 360°

Learn these key points.

x	0°	90°	180°	270°	360°
sin x	0	1	0	−1	0
cos x	1	0	−1	0	1
tan x	0	no value	0	no value	0

y = sin x starts at the origin, rises to a maximum of 1 at x = 90°, falls through 0 at 180° to a minimum of −1 at 270°, and returns to 0 at 360°. It is a smooth wave that repeats every 360°.

y = cos x has the same wave shape shifted along by 90°. It starts at its maximum of 1 when x = 0°, crosses the x-axis at 90°, reaches −1 at 180°, crosses again at 270° and is back at 1 at 360°.

y = tan x is a different shape. It passes through (0°, 0), (180°, 0) and (360°, 0) and increases steeply on each branch. It has vertical asymptotes at x = 90° and x = 270°: the curve gets closer and closer to these lines without meeting them. Unlike sin and cos, tan x can take any value, and its pattern repeats every 180°.

The syllabus says a sketch does not need to be to scale, but it must be freehand, show the important features (intercepts, turning points, symmetry) with values marked on the axes, have labelled axes and interact correctly with the axes, for example by tending towards an asymptote. For these graphs that means:

• mark 90°, 180°, 270° and 360° on the x-axis, and 1 and −1 on the y-axis for sin and cos;
• show the turning points of sin and cos at the right x-values;
• draw the asymptotes of tan x as dashed vertical lines at 90° and 270°, with the branches approaching but not touching them.

Interpreting the graphs: symmetry and signs

The graphs show which angles share a value. You use these facts to solve equations.

sin:  sin(180° − x) = sin x       symmetric about x = 90°
cos:  cos(360° − x) = cos x       symmetric about x = 180°
tan:  tan(x + 180°) = tan x       repeats every 180°

Reading the graphs for sign:

Interval	sin x	cos x	tan x
0° < x < 90°	+	+	+
90° < x < 180°	+	−	−
180° < x < 270°	−	−	+
270° < x < 360°	−	+	−

Worked example 4: using symmetry. Given that sin 40° ≈ 0.643, find the other angle between 0° and 360° with a sine of about 0.643, and the two angles with a sine of about −0.643.

sin(180° − 40°) = sin 40°, so the other angle is 140°.
The graph of sin x is below the axis between 180° and 360°.
By symmetry, the negative values occur 40° past 180° and 40° before 360°:
x = 180° + 40° = 220°   and   x = 360° − 40° = 320°

How many solutions? Draw the horizontal line y = k across the graph and count crossings. For 0° ⩽ x ⩽ 360°:

• sin x = k or cos x = k with −1 < k < 1 (k ≠ 0) has two solutions;
• sin x = 1 has one solution (90°); cos x = 1 has two (0° and 360°);
• sin x = 0 has three solutions (0°, 180°, 360°) because both ends of the range are included;
• sin x = k or cos x = k with k > 1 or k < −1 has no solutions;
• tan x = k (k ≠ 0) always has two solutions, one on each branch that crosses the line.

Solving trigonometric equations

Method

1. Rearrange so that sin x, cos x or tan x is on its own.
2. Find one angle: from the exact-value table on Paper 2, or with sin⁻¹, cos⁻¹ or tan⁻¹ on Paper 4.
3. Use the graph’s symmetry to find every other angle in 0° ⩽ x ⩽ 360°.
4. Check each answer lies in the range, and give angles to 1 decimal place unless they are exact.

The second-solution rules, all from the graphs above:

sin x = k:  x and 180° − x
cos x = k:  x and 360° − x
tan x = k:  x and x + 180°

A negative calculator angle is outside the range. Apply the rule to the calculator’s value, then add 360° (or 180° for tan) to any angle that is still negative. For sin x = −0.4: sin⁻¹(−0.4) = −23.6°, so x = 180° − (−23.6°) = 203.6° and x = −23.6° + 360° = 336.4°.

Worked example 5: non-calculator equation with a sine

Solve 2 sin x + 1 = 0 for 0° ⩽ x ⩽ 360°.

2 sin x = −1
sin x = −1/2
The related acute angle is 30° (since sin 30° = 1/2).
sin x is negative between 180° and 360°.
x = 180° + 30° = 210°
x = 360° − 30° = 330°

x = 210° or 330°

Worked example 6: non-calculator equation with a tangent

Solve √3 tan x = 1 for 0° ⩽ x ⩽ 360°.

tan x = 1/√3
tan 30° = 1/√3, so x = 30°
tan repeats every 180°: x = 30° + 180° = 210°

x = 30° or 210°

Worked example 7: calculator equation with a sine (Paper 4 style)

Solve 5 sin x = 2 for 0° ⩽ x ⩽ 360°.

sin x = 2/5 = 0.4
x = sin⁻¹(0.4) = 23.578...°
Second solution: 180° − 23.578...° = 156.421...°

x = 23.6° or 156.4° (1 d.p.). Subtract the unrounded value, then round.

Worked example 8: calculator equation with a negative tangent

Solve 3 tan x + 7 = 0 for 0° ⩽ x ⩽ 360°.

tan x = −7/3
tan⁻¹(−7/3) = −66.801...°   (outside the range)
Add 180°:  −66.801...° + 180° = 113.198...°
Add 180° again: 113.198...° + 180° = 293.198...°

x = 113.2° or 293.2° (1 d.p.). A quick check with the sign table: tan x is negative between 90° and 180° and between 270° and 360°, and both answers fall in those intervals.

Common errors

• Swapping sin 30° and sin 60°. Sketch the half-equilateral triangle: the side of length 1 is opposite the 30° angle.
• Writing tan 30° = √3 instead of 1/√3. tan 30° is less than 1, because the opposite side is shorter than the adjacent side.
• Giving 0.866 or 0.707 when the question asks for an exact value. The syllabus says exact answers may need surd form.
• Stopping at the calculator’s angle. Most equations in this range have two solutions, and the graph is how you find the second one.
• Using 180° − x for a cosine equation. That rule belongs to sine; for cosine use 360° − x.
• Leaving a negative calculator angle such as −66.8° as an answer when the range is 0° ⩽ x ⩽ 360°.
• Drawing tan x crossing the lines x = 90° or x = 270°, or joining the branches across an asymptote.

Next steps

Practise the equation and exact-value questions in the Extended trigonometry practice set. If you sit Core papers, the Core diagnostic covers your content instead.

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027, Subject content, section 6 Trigonometry: E6.3 Exact trigonometric values and E6.4 Trigonometric functions (Extended subject content).