Every trigonometric functions question from 5 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 6.4. 5 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/42/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Trigonometry past paper questions.
0580 Trigonometric Functions past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q14
cos x = 0.41. Find the reflex angle x.
Show solution
- Use the calculator first: cos⁻¹(0.41) = 65.79…° This angle is acute, not reflex.
- The cosine graph is symmetrical about 180°, so cos(360° − 65.79…°) is also 0.41.
- x = 360 − 65.79… = 294.20…
- 294.2° is between 180° and 360°, so it is reflex: 294.2°
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q12
(a) Sketch the graph of y = cos x for 0° ≤ x ≤ 360°.
(b) Solve the equation 3 cos x + 1 = 0 for 0° ≤ x ≤ 360°.
Show solution
- Mark the key points of y = cos x: (0°, 1), (90°, 0), (180°, −1), (270°, 0), (360°, 1).
- Join them with a smooth wave.
- Subtract 1 from both sides: 3 cos x = −1.
- Divide by 3: cos x = −1⁄3.
- x = cos⁻¹(−1⁄3) = 109.47…°
- The cosine graph is symmetrical about 180°, so the other solution is 360 − 109.47… = 250.52…°
- x = 109.5° or x = 250.5°
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q23
Two angles, A and B, are formed when two lines meet at a point, so A and B are on a straight line. Angle A is given by the equation 2 + 3 tan A = −0.5. Find the two possible values of angle B.
Show solution
- Solve for tan A first. Subtract 2 from both sides: 3 tan A = −2.5.
- Divide by 3: tan A = −5⁄6.
- A and B lie next to each other on a straight line, so B = 180° − A. This means tan B = −tan A = 5⁄6.
- B = tan⁻¹(5⁄6) = 39.80…°
- The tan graph repeats every 180°, so the other value is 39.8 + 180 = 219.8°.
- B = 39.8° or B = 219.8°
- Check: A = 180 − 39.8 = 140.2°, and tan 140.2° = −0.833 = −5⁄6 ✓
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q21
(a) Sketch the graph of y = cos x for 0° ≤ x ≤ 360°.
(b) Solve the equation 2 cos x + √3 = 0 for 0° ≤ x ≤ 360°.
Show solution
- Mark the key points of y = cos x: (0°, 1), (90°, 0), (180°, −1), (270°, 0), (360°, 1).
- Join them with a smooth wave shape.
- Rearrange: 2 cos x = −√3, so cos x = −√3⁄2.
- Ignore the minus sign first: cos 30° = √3⁄2, so the reference angle is 30°.
- cos is negative in the second and third quadrants (between 90° and 270°).
- x = 180 − 30 = 150° and x = 180 + 30 = 210°.
- x = 150° or x = 210°
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q17
(a) Sketch the graph of y = cos x for 0° ≤ x ≤ 360°. (b) cos x° = 1⁄√2 and x is a reflex angle. Find the value of x.
Show solution
- Mark the key points of y = cos x: (0°, 1), (90°, 0), (180°, −1), (270°, 0) and (360°, 1).
- Join them with a smooth wave.
- cos 45° = 1⁄√2, so 45° is one solution, but it is acute, not reflex.
- The cosine graph is symmetrical about 180°, so the other solution is 360 − 45 = 315°.
- 315° is between 180° and 360°, so it is reflex: x = 315
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