Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 1 Trigonometry – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 1.5 Trigonometry of Pure Mathematics 1 in the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The section is examined on Paper 1 (1 hour 50 minutes, 75 marks), which is compulsory for AS Level and A Level. A scientific calculator is allowed on Paper 1, but where a question asks for an exact value, find it without a decimal. Give other answers to 3 significant figures, or angles in degrees to 1 decimal place, unless told otherwise.

Course hub: Cambridge A Level Mathematics. Printable list: 9709 checklist. Learn the content in the study guide and recap with the revision notes. For mixed Paper 1 questions, try the Pure Mathematics 1 mixed practice. For quadratic factorising, see the quadratics practice.

Questions

1. Find the exact value of each of the following.

(a) tan 315° [1] (b) sin(4π/3) [1] (c) cos 480° [1]

2. It is given that sin θ = −1/3, where 180° < θ < 270°. Find the exact values of cos θ and tan θ. [3]

3. Give each answer exactly, in radians.

(a) Find cos⁻¹(−1/2). [1] (b) Find tan⁻¹(−√3). [1] (c) Find cos⁻¹(cos(−π/3)), explaining why the answer is not −π/3. [2]

4. The function f is defined by f(x) = a − b cos x for 0 ≤ x ≤ 2π, where a and b are positive constants. The range of f is −1 ≤ f(x) ≤ 7.

(a) Find the values of a and b. [2] (b) Sketch the graph of y = f(x), showing the coordinates of the end points and of the maximum point. [2] (c) Solve the equation f(x) = 5, giving your answers exactly. [3]

5. (a) State the period of y = tan 2x, in radians. [1] (b) Solve tan 2x = 3 for 0 ≤ x ≤ π. [3]

6. (a) Prove the identity (1 − cos θ)(1 + cos θ) / (sin θ cos θ) ≡ tan θ. [3] (b) Hence solve (1 − cos θ)(1 + cos θ) / (sin θ cos θ) = 2.5 for 0° < θ < 360°. [2]

7. Solve the equation 6 sin²x + cos x = 5 for 0° ≤ x ≤ 360°. [5]

8. Solve the equation 4 sin x cos x = 3 cos²x for 0 ≤ x ≤ 2π. [5]

9. Solve the equation 5 cos(2x − 40°) = 2 for 0° ≤ x ≤ 180°. [4]

10. The function f is defined by f(x) = 3 + 2 sin x for −π/2 ≤ x ≤ π/2.

(a) State the range of f. [1] (b) Explain why f has an inverse. [1] (c) Find an expression for f⁻¹(x), and state the domain of f⁻¹. [3] (d) Solve the equation f(x) = 4, giving your answer exactly. [2]

11. (a) Show that the equation 4 tan x sin x = 15 can be written as 4cos²x + 15 cos x − 4 = 0. [3] (b) Hence solve 4 tan x sin x = 15 for 0° ≤ x ≤ 360°. [3] (c) Use your answer to part (b) to solve 4 tan 2y sin 2y = 15 for 0° ≤ y ≤ 180°. [2]

Answers

1. (a) 315° is in the fourth quadrant with related angle 45°, and tan is negative there: −1 [1]. (b) 4π/3 = 240°, third quadrant, related angle 60°, sin negative: −√3/2 [1]. (c) 480° − 360° = 120°, second quadrant, related 60°, cos negative: −1/2 [1]. Examiner insight: each mark here is for the exact value only; a decimal such as −0.866 earns nothing, even if it is correct to 3 s.f.

2. cos²θ = 1 − (−1/3)² = 8/9 [1]. θ is in the third quadrant, so cos θ < 0: cos θ = −2√2/3 [1]. tan θ = (−1/3) ÷ (−2√2/3) = 1/(2√2) = √2/4 [1]. Examiner insight: the sign mark depends on stating or using the quadrant; −2√2/3 given with no reason may still score, but +2√2/3 loses that mark and any tan value built on it.

3. (a) 2π/3 [1]. (b) −π/3 [1]. (c) cos(−π/3) = cos(π/3) = 1/2, so cos⁻¹(1/2) = π/3 [1]. The principal value of cos⁻¹ must lie between 0 and π, and −π/3 is outside that range [1]. Examiner insight: the explanation mark needs the range 0 ≤ cos⁻¹x ≤ π stated; “because cos is even” alone does not explain why −π/3 is rejected.

4. (a) The least value is a − b and the greatest is a + b, so a − b = −1 and a + b = 7 [1]. a = 3, b = 4 [1]. (b) The curve y = 3 − 4 cos x starts at a minimum, rises to a maximum and falls back, with end points (0, −1) and (2π, −1) [1] and maximum point (π, 7) [1]. (c) 3 − 4 cos x = 5, so cos x = −1/2 [1]. x = 2π/3 [1] and x = 2π − 2π/3 = 4π/3, so x = 2π/3, 4π/3 [1]. Examiner insight: in (b) a curve drawn the wrong way up (starting at a maximum, like +4 cos x) loses the shape mark even if the labelled values are right; in (c) the domain is in radians, so answers of 120° and 240° lose the accuracy mark.

5. (a) Period = π/2 [1]. (b) 0 ≤ 2x ≤ 2π. tan⁻¹3 = 1.2490, so 2x = 1.2490 [1] or 2x = 1.2490 + π = 4.3906 [1]. x = 0.625, 2.20 [1]. Examiner insight: the second value of 2x comes from adding π, not 2π; adding 2π gives 2x = 7.53, which is outside 0 to 2π, and the solution x = 2.20 is then lost.

6. (a) (1 − cos θ)(1 + cos θ) = 1 − cos²θ [1] = sin²θ [1]. So the left side is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ [1]. (b) tan θ = 2.5, so θ = 68.2° [1] or 68.2° + 180° = 248.2°, so θ = 68.2°, 248.2° [1]. Examiner insight: in a “prove” question the final mark needs a complete chain from the left side to tan θ, with no step skipped; in (b) “hence” means you must use tan θ = 2.5, not start again.

7. 6(1 − cos²x) + cos x = 5 [1], so 6cos²x − cos x − 1 = 0 [1]. (3 cos x + 1)(2 cos x − 1) = 0, giving cos x = 1/2 or cos x = −1/3 [1]. cos x = 1/2: x = 60°, 300° [1]. cos x = −1/3: x = 109.5°, 250.5° [1]. Examiner insight: the first method mark is for replacing sin²x with 1 − cos²x; forming a quadratic in cos x and then solving it correctly usually earns the next two marks even if a later angle is wrong.

8. cos x(4 sin x − 3 cos x) = 0 [1]. cos x = 0 gives x = π/2, 3π/2 [1]. Otherwise 4 sin x = 3 cos x, so tan x = 3/4 [1]. x = 0.644 [1] and 0.644 + π = 3.79 [1]. Examiner insight: dividing both sides by cos x at the start loses π/2 and 3π/2, and so loses a mark; an answer with only 0.644 and 3.79 scores at most 3 out of 5.

9. cos(2x − 40°) = 0.4, and −40° ≤ 2x − 40° ≤ 320° [1]. cos⁻¹0.4 = 66.42°, so 2x − 40° = 66.42° or 360° − 66.42° = 293.58° [1]. x = (66.42° + 40°)/2 = 53.2° [1] and x = (293.58° + 40°)/2 = 166.8° [1]. Examiner insight: −66.42° is outside the new interval (it is below −40°) and would give x = −13.2°, which is not in 0° ≤ x ≤ 180°. The final accuracy mark needs both values in range and no extra solutions inside the interval.

10. (a) 1 ≤ f(x) ≤ 5 [1]. (b) f is one-one on −π/2 ≤ x ≤ π/2 (sin x is increasing there), so it has an inverse [1]. (c) y = 3 + 2 sin x, so sin x = (y − 3)/2 [1]. f⁻¹(x) = sin⁻¹((x − 3)/2) [1]. The domain of f⁻¹ is the range of f: 1 ≤ x ≤ 5 [1]. (d) 3 + 2 sin x = 4, so sin x = 1/2 [1]. π/6 is in the domain and 5π/6 is not, so x = π/6 [1]. Examiner insight: in (d) the answer x = π/6, 5π/6 loses the final mark because 5π/6 is outside the stated domain of f; always test each value against the domain.

11. (a) 4 (sin x/cos x) sin x = 15, so 4 sin²x = 15 cos x [1]. 4(1 − cos²x) = 15 cos x [1]. 4 − 4cos²x − 15 cos x = 0, so 4cos²x + 15 cos x − 4 = 0 [1]. (b) (4 cos x − 1)(cos x + 4) = 0, so cos x = 1/4 [1]. cos x = −4 has no solutions, because −1 ≤ cos x ≤ 1 [1]. x = 75.5°, 284.5° [1]. (c) 2y = 75.52° or 284.48° (as 0° ≤ 2y ≤ 360°) [1]. y = 37.8°, 142.2° [1]. Examiner insight: (a) is a given answer, so every line of working must be shown, including the step where tan x becomes sin x/cos x; the final mark is lost if the last line jumps to the printed equation.

Where marks are usually lost

• Giving only the calculator’s principal value (Questions 5, 7, 9) and missing the other solutions in the interval.
• Forgetting to change the interval for 2x or 2x − 40°, then stopping at 360°.
• Leaving the answer as a value of 2x or 2y instead of dividing back.
• Degrees in a radian question (Questions 4 and 8), or the reverse.
• Cancelling cos x in Question 8 and losing π/2 and 3π/2.
• Not rejecting cos x = −4 with a reason, or not rejecting a value outside a function’s domain (Question 10).
• Writing a cos value with the wrong sign because the quadrant was not checked (Question 2).
• In a “show that” or “prove” part, skipping the line where sin² is replaced by 1 − cos².
• Rounding cos⁻¹0.4 or tan⁻¹3 to 1 decimal place before adding or dividing, which can change the final digit.

Next steps

• Recap with the trigonometry revision notes.
• Review the method in the trigonometry study guide.
• Course hub: Cambridge A Level Mathematics.
• Printable 9709 checklist.
• Take the AS 10-minute diagnostic or use the 9709 self-check bank.
• Book a free trial class.

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Section 1.5 Trigonometry, in 1 Pure Mathematics 1 (for Paper 1).