Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 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 2.3, Trigonometry, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.3 is part of Pure Mathematics 2 and is examined on Paper 2, the AS Level Pure Mathematics route; the same outcomes appear as section 3.3 for Paper 3. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Where a question asks for an exact value, the working must be exact throughout. Otherwise give angles in degrees to 1 decimal place and other answers to 3 significant figures.

Related: the study guide, the revision notes, the Pure Mathematics 1 trigonometry practice for the Paper 1 basics, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

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

(a) cosec 225° [1] (b) sec(2π/3) [1] (c) cot(7π/6) [1]

2. It is given that tan θ = −√5/2, where 90° < θ < 180°. Find the exact values of sec θ and cosec θ. [3]

3.

(a) Sketch the graph of y = cosec x for 0 < x < 2π. State the equations of the asymptotes and the coordinates of the turning points. [3] (b) State the set of values of k for which the equation cosec x = k has no solutions. [1]

4.

(a) Prove the identity (1 + sec θ)/(tan θ + sin θ) ≡ cosec θ. [3] (b) Hence solve the equation (1 + sec θ)/(tan θ + sin θ) = 4 for 0° < θ < 360°. [2]

5. Solve the equation 3 cot²θ + 5 cosec θ + 1 = 0 for 0° < θ < 360°. [5]

6.

(a) By writing 15° as 60° − 45°, show that tan 15° = 2 − √3. [3] (b) Use the formula for tan 2A and the result of part (a) to show that tan 30° = 1/√3. [2]

7. Solve the equation 2 sin(θ − 30°) = 3 cos(θ + 60°) for 0° ≤ θ ≤ 360°. [5]

8. Solve the equation tan 2θ = 3 tan θ for 0° < θ < 180°. [4]

9.

(a) Express 3 cos θ − 2 sin θ in the form R cos(θ + α), where R > 0 and 0° < α < 90°. Give the exact value of R and the value of α correct to 2 decimal places. [3] (b) Hence solve the equation 3 cos 2x − 2 sin 2x = 1.5 for 0° ≤ x ≤ 180°. [4] (c) Find the exact value of the greatest value of 1/(5 + 3 cos θ − 2 sin θ). [2]

10.

(a) Prove the identity sec²θ + cosec²θ ≡ 4 cosec²2θ. [3] (b) Hence solve the equation sec²θ + cosec²θ = 6 for 0° < θ < 90°. [3] (c) Explain why the equation sec²θ + cosec²θ = 3 has no solutions. [1]

11. The angle A is such that sec A = 3 and 270° < A < 360°. Find the exact value of

(a) sin A [2] (b) sin 2A and cos 2A [3] (c) tan(A + 45°), giving your answer in the form (p√2 + q)/7, where p and q are integers. [3]

12.

(a) By writing 3θ as 2θ + θ, show that sin 3θ ≡ 3 sin θ − 4 sin³θ. [4] (b) Show that the equation 2 sin 3θ + 2 cos 2θ = 3 can be written as 8s³ + 4s² − 6s + 1 = 0, where s = sin θ. [2] (c) Show that (2s − 1) is a factor of 8s³ + 4s² − 6s + 1. Hence solve 2 sin 3θ + 2 cos 2θ = 3 for 0° < θ < 180°. [5]

Answers

1. (a) sin 225° = −√2/2, so cosec 225° = −√2 [1] (b) cos(2π/3) = −1/2, so sec(2π/3) = −2 [1] (c) tan(7π/6) = tan(π/6) = 1/√3, so cot(7π/6) = √3 [1] Examiner insight: each mark is for the exact value with the correct sign; a decimal such as −1.414 is not exact and earns nothing.

2. sec²θ = 1 + tan²θ = 9/4 [1]. θ is in the second quadrant, where cos θ < 0, so sec θ = −3/2 [1]. cot θ = −2/√5, so cosec²θ = 1 + 4/5 = 9/5, and sin θ > 0 there, so cosec θ = 3/√5 = 3√5/5 [1] Examiner insight: a positive sec θ loses the second mark; the sign must come from the quadrant, and cosec θ is positive because sin θ is.

3. (a) Two U-shaped branches: one above the x-axis with minimum point (π/2, 1), one below with maximum point (3π/2, −1) [1]. Asymptotes x = 0, x = π, x = 2π [1]. Correct shape, with the curve approaching each asymptote and no part between y = −1 and y = 1 [1] (b) −1 < k < 1 [1] Examiner insight: in (b), including k = ±1 in the set loses the mark, because cosec x = 1 and cosec x = −1 each have a solution.

4. (a) LHS = (1 + 1/cos θ)/(sin θ/cos θ + sin θ) [1]. Multiply top and bottom by cos θ: (cos θ + 1)/(sin θ + sin θ cos θ) = (1 + cos θ)/(sin θ(1 + cos θ)) [1]. Cancel (1 + cos θ): 1/sin θ = cosec θ [1] (b) cosec θ = 4, so sin θ = 1/4 [1]. θ = 14.5°, 165.5° [1] Examiner insight: the final mark in (a) needs the factor sin θ(1 + cos θ) shown before cancelling; in (b), “hence” means you must use cosec θ = 4.

5. 3(cosec²θ − 1) + 5 cosec θ + 1 = 0, so 3cosec²θ + 5 cosec θ − 2 = 0 [1]. (3 cosec θ − 1)(cosec θ + 2) = 0, so cosec θ = 1/3 or cosec θ = −2 [1]. cosec θ = 1/3 has no solutions, because |cosec θ| ≥ 1 [1]. cosec θ = −2 gives sin θ = −1/2, so θ = 210° [1] and θ = 330° [1] Examiner insight: the rejection mark needs a reason; simply leaving out cosec θ = 1/3 does not earn it.

6. (a) tan(60° − 45°) = (tan 60° − tan 45°)/(1 + tan 60° tan 45°) = (√3 − 1)/(1 + √3) [1]. Multiply top and bottom by (√3 − 1): (√3 − 1)²/(3 − 1) [1] = (4 − 2√3)/2 = 2 − √3 [1] (b) tan 30° = 2(2 − √3)/(1 − (2 − √3)²) = (4 − 2√3)/(1 − (7 − 4√3)) = (4 − 2√3)/(4√3 − 6) [1]. Since 4√3 − 6 = 2√3(2 − √3), this is 2(2 − √3)/(2√3(2 − √3)) = 1/√3 [1] Examiner insight: both parts are “show that” with the answer given, so a calculator check of tan 15° earns no marks; the rationalising step must be written out.

7. 2 sin(θ − 30°) = 2(sin θ(√3/2) − cos θ(1/2)) = √3 sin θ − cos θ [1]. 3 cos(θ + 60°) = 3(cos θ(1/2) − sin θ(√3/2)) = (3/2) cos θ − (3√3/2) sin θ [1]. Collect: (5√3/2) sin θ = (5/2) cos θ, so tan θ = 1/√3 [1]. θ = 30° [1] and θ = 210° [1] Examiner insight: the first two marks are for correct expansions with exact values of sin and cos of 30° and 60°; a sign slip in cos(θ + 60°) loses the second mark and the answers.

8. 2 tan θ/(1 − tan²θ) = 3 tan θ, so 2 tan θ = 3 tan θ − 3tan³θ [1]. tan θ(3tan²θ − 1) = 0 [1]. tan θ = 0 gives θ = 0° or 180°, both outside 0° < θ < 180°; so tan²θ = 1/3, tan θ = ±1/√3 [1]. θ = 30°, 150° [1] Examiner insight: dividing by tan θ at the start is only acceptable here if you also say why tan θ = 0 gives no solutions in the open interval; otherwise the second mark is lost.

9. (a) R cos(θ + α) = R cos θ cos α − R sin θ sin α, so R cos α = 3 and R sin α = 2 [1]. R = √13 [1]. tan α = 2/3, α = 33.69° [1] (b) √13 cos(2x + 33.69°) = 1.5, so cos(2x + 33.69°) = 1.5/√13 [1]. The range is 33.69° ≤ 2x + 33.69° ≤ 393.69°, so 2x + 33.69° = 65.42° or 294.58° [1]. 2x = 31.73° or 260.89° [1]. x = 15.9°, 130.4° [1] (c) The least value of 5 + √13 cos(θ + α) is 5 − √13 [1]. Greatest value = 1/(5 − √13) = (5 + √13)/12 [1] Examiner insight: in (c) the greatest value of a reciprocal comes from the least value of the denominator; using 5 + √13 gives the least value and loses both marks.

10. (a) sec²θ + cosec²θ = 1/cos²θ + 1/sin²θ = (sin²θ + cos²θ)/(sin²θ cos²θ) [1] = 1/(sin²θ cos²θ) [1] = 4/(2 sin θ cos θ)² = 4/sin²2θ = 4 cosec²2θ [1] (b) 4 cosec²2θ = 6, so sin²2θ = 2/3; sin 2θ > 0 for 0° < 2θ < 180°, so sin 2θ = √(2/3) [1]. 2θ = 54.74° or 125.26° [1]. θ = 27.4°, 62.6° [1] (c) |cosec 2θ| ≥ 1, so 4 cosec²2θ ≥ 4 > 3 and no solution exists [1] Examiner insight: in (b), missing 2θ = 180° − 54.74° is the usual way to lose a mark; always work in the doubled interval.

11. (a) cos A = 1/3, so sin²A = 1 − 1/9 = 8/9 [1]. A is in the fourth quadrant, so sin A = −2√2/3 [1] (b) sin 2A = 2(−2√2/3)(1/3) [1] = −4√2/9 [1]. cos 2A = 2(1/3)² − 1 = −7/9 [1] (c) tan A = −2√2, so tan(A + 45°) = (−2√2 + 1)/(1 + 2√2) [1]. Multiply top and bottom by (2√2 − 1): (1 − 2√2)(2√2 − 1)/(8 − 1) [1] = −(9 − 4√2)/7 = (4√2 − 9)/7 [1] Examiner insight: follow-through is allowed in (b) and (c) from a wrong sign of sin A, but only if the method is correct; the accuracy marks are still lost.

12. (a) sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ [1] = 2 sin θ cos²θ + (1 − 2sin²θ) sin θ [1] = 2 sin θ(1 − sin²θ) + sin θ − 2sin³θ [1] = 3 sin θ − 4 sin³θ [1] (b) 2(3s − 4s³) + 2(1 − 2s²) = 3 [1]. 6s − 8s³ + 2 − 4s² − 3 = 0, so 8s³ + 4s² − 6s + 1 = 0 [1] (c) With s = 1/2: 1 + 1 − 3 + 1 = 0, so (2s − 1) is a factor [1]. Dividing gives 8s³ + 4s² − 6s + 1 = (2s − 1)(4s² + 4s − 1) [1]. 4s² + 4s − 1 = 0 gives s = (−1 ± √2)/2; s = (−1 − √2)/2 = −1.207 is rejected because sin θ ≥ −1, so sin θ = 0.2071 [1]. sin θ = 1/2: θ = 30°, 150° [1]. sin θ = 0.2071: θ = 12.0°, 168.0° [1] Examiner insight: (a) needs cos²θ replaced by 1 − sin²θ in a written line; in (c), stating f(1/2) = 0 without the substituted values does not earn the factor-theorem mark.

Where marks are usually lost

• Signs of sec, cosec or cot not taken from the quadrant (Questions 2 and 11).
• A root such as cosec θ = 1/3 dropped with no reason, or kept and “solved” (Question 5).
• Errors in the sign between the terms of cos(A ± B) or in the denominator of tan(A ± B) (Questions 6 and 7).
• Cancelling tan θ or sin θ without checking whether it could be zero (Question 8).
• Rounding α to 33.7° too early, so the answer to Question 9(b) ends in the wrong decimal.
• Not doubling the interval before solving for 2θ or 2x + α (Questions 9 and 10).
• Using the wrong extreme of the denominator for a reciprocal (Question 9(c)).
• “Show that” and “prove” parts with a skipped Pythagorean or rationalising step.
• Solutions from an impossible quadratic root, such as sin θ = −1.207, left in the final answer.

Next steps

• Trigonometry revision notes
• Trigonometry study guide
• Pure Mathematics 2 practice set and the Pure 3 trigonometry, vectors and complex numbers practice
• A Level Mathematics hub
• Printable 9709 checklist
• AS diagnostic, the 9709 self-check bank, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.3 Trigonometry.