AQA A-Level Mathematics: E: Trigonometry (7357) – 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 E: Trigonometry (E1 to E9) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Section E is Paper 1 content and can also be assessed on Papers 2 and 3. A calculator is allowed throughout 7357, but where a question asks for an exact answer or says “show that”, give full working.

Learn the content first with the Trigonometry study guide and the Trigonometry revision notes. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/.

Questions

1. A sector of a circle has angle 1.5 radians and area 27 cm².

(a) Find the radius of the circle. [2] (b) Find the perimeter of the sector. [1]

2. Find the exact value of:

(a) sin(7π/4) [1] (b) cosec(2π/3) [1] (c) Use a compound angle formula to show that tan(π/12) = 2 − √3. [3]

3. (a) State the range of y = arccos x. [1] (b) Find the exact value of arccos(cos(5π/3)). [2] (c) For 0 ≤ x ≤ 2π, state the values of x for which sec x is undefined, and state the range of sec x. [2]

4. In triangle PQR, PQ = 9 cm, PR = 6 cm and angle QPR = 2π/3 radians.

(a) Show that QR = 3√19 cm. [2] (b) Find the exact area of the triangle. [2] (c) Calculate angle PQR, in degrees to 1 decimal place. [2]

5. (a) Show that, when θ is small, (4 cos θ − 4 + 3θ sin 2θ) / (θ tan θ) ≈ 4. [3] (b) Calculate the exact value of the expression when θ = 0.1, and find the percentage error in the approximation. [2]

6. Given that sec θ = −3 and π/2 < θ < π, find the exact value of:

(a) tan θ [2] (b) sin 2θ and cos 2θ [3]

7. Solve 3 tan²x + 5 sec x + 1 = 0 for 0° ≤ x < 360°. [5]

8. Solve 2 cos(3x − 20°) = 1 for 0° ≤ x ≤ 180°, giving your answers to 1 decimal place. [5]

9. (a) Prove that sin 3θ ≡ 3 sin θ − 4 sin³θ. [4] (b) Hence solve sin 3θ = 2 sin θ for 0 ≤ θ < 2π, giving exact answers. [4]

10. (a) Express √5 cos x + 2 sin x in the form R cos(x − α), where R > 0 and 0 < α < π/2. Give α to 3 decimal places. [3] (b) Find the maximum value of 1 / (4 + √5 cos x + 2 sin x) and the value of x in 0 ≤ x < 2π where it occurs. [2] (c) Solve √5 cos x + 2 sin x = 2 for 0 ≤ x < 2π. [4]

11. The height above the ground of a seat on an observation wheel is modelled by h = 22 − 20 cos(πt/12) metres, t minutes after the seat passes the lowest point.

(a) State the maximum height and the first time it is reached. [2] (b) State the time for one revolution. [1] (c) For how long in each revolution is the seat more than 32 m above the ground? [4]

12. A particle is in equilibrium under three forces: its weight of 20 N acting vertically downwards, a horizontal force of 12 N acting to the left, and a force of P N acting to the right at an angle α above the horizontal.

(a) Resolve horizontally and vertically to form two equations in P and α. [2] (b) Find P, in exact form and to 3 significant figures, and find α to 1 decimal place. [4]

Answers

1. (a) ½r²(1.5) = 27 [1], so r² = 36 and r = 6 cm [1]. (b) Arc = 6 × 1.5 = 9 cm, so perimeter = 9 + 6 + 6 = 21 cm [1]. Examiner insight: the perimeter of a sector includes both radii; giving the arc length alone loses the mark.

2. (a) −√2/2 [1] (b) 1 / sin(2π/3) = 2/√3 = 2√3/3 [1] (c) tan(π/3 − π/4) = (tan π/3 − tan π/4) / (1 + tan π/3 tan π/4) [1] = (√3 − 1) / (1 + √3) [1] = (√3 − 1)² / (3 − 1) = (4 − 2√3)/2 = 2 − √3 [1] Examiner insight: a calculator decimal for tan(π/12) earns nothing on a “show that”; the compound angle step and the rationalising step must both be written.

3. (a) 0 ≤ y ≤ π [1] (b) cos(5π/3) = ½ [1], and arccos(½) = π/3 [1]. (c) Undefined at x = π/2 and x = 3π/2 [1]; range sec x ≤ −1 or sec x ≥ 1 [1]. Examiner insight: answering 5π/3 in (b) scores at most the first mark, because arccos only returns values in [0, π].

4. (a) QR² = 9² + 6² − 2(9)(6) cos(2π/3) [1] = 117 + 54 = 171, so QR = √171 = 3√19 cm [1]. (b) Area = ½(9)(6) sin(2π/3) [1] = 27√3/2 cm² [1]. (c) sin Q = 6 sin(2π/3) / (3√19) [1] = 0.3974…, so Q = 23.4° [1]. Examiner insight: in (a) the substitution with cos(2π/3) = −½ must be visible; writing “QR² = 171” with no line before it does not show the given result.

5. (a) 4 cos θ − 4 ≈ 4(1 − θ²/2) − 4 = −2θ² [1] 3θ sin 2θ ≈ 3θ(2θ) = 6θ², and θ tan θ ≈ θ² [1] (−2θ² + 6θ²) / θ² = 4 [1] (b) Exact value at θ = 0.1: 3.9485 (5 s.f.) [1]; percentage error = (4 − 3.9485)/3.9485 × 100 = 1.30% [1]. Examiner insight: the small angle results only hold in radians, and sin 2θ must become 2θ, not θ; either slip loses the method mark.

6. (a) tan²θ = sec²θ − 1 = 8 [1]; θ is in the second quadrant where tan is negative, so tan θ = −2√2 [1]. (b) cos θ = −1/3 and sin θ = 2√2/3 [1] sin 2θ = 2(2√2/3)(−1/3) = −4√2/9 [1] cos 2θ = 2(−1/3)² − 1 = −7/9 [1] Examiner insight: the sign mark depends on a reason; state the quadrant rather than just choosing the negative root.

7. 3(sec²x − 1) + 5 sec x + 1 = 0 [1] 3 sec²x + 5 sec x − 2 = 0 [1] (3 sec x − 1)(sec x + 2) = 0; sec x = 1/3 has no solutions because |sec x| ≥ 1 [1] sec x = −2, so cos x = −½ [1] x = 120°, 240° [1] Examiner insight: an answer that tries to solve cos x = 3 without saying it is impossible usually loses the rejection mark.

8. cos(3x − 20°) = ½ [1] Interval: −20° ≤ 3x − 20° ≤ 520° [1] 3x − 20° = 60°, 300°, 420° [1] x = 26.7°, 106.7°, 146.7° [1] for any two correct, [1] for all three with no extras. Examiner insight: the final accuracy mark is lost for an extra answer outside 0° to 180° as well as for a missing one, so check every value against the interval.

9. (a) sin 3θ = 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) 3 sin θ − 4 sin³θ = 2 sin θ, so sin θ(1 − 4 sin²θ) = 0 [1] sin θ = 0: θ = 0, π [1] sin θ = ½: θ = π/6, 5π/6 [1] sin θ = −½: θ = 7π/6, 11π/6 [1] Examiner insight: “Hence” means the result from (a) must be used; dividing by sin θ throws away 0 and π and costs a mark.

10. (a) R cos α = √5 and R sin α = 2 [1] R = √(5 + 4) = 3 [1] tan α = 2/√5, so α = 0.730 [1] (b) Least value of the denominator is 4 − 3 = 1, so the maximum is 1 [1], when x − α = π, so x = 3.87 [1]. (c) cos(x − α) = 2/3 [1] x − α lies in [−0.730, 5.553), so x − α = 0.8411 or 2π − 0.8411 = 5.4421 [1] x = 1.57 (exactly π/2) [1] x = 6.17 [1] Examiner insight: in (b), the maximum of a reciprocal comes from the minimum of the denominator; giving 1/7 is a common route to zero marks.

11. (a) Maximum when cos(πt/12) = −1: h = 42 m [1], first at t = 12 minutes [1]. (b) Period = 2π ÷ (π/12) = 24 minutes [1]. (c) 22 − 20 cos(πt/12) > 32 gives cos(πt/12) < −½ [1] For 0 ≤ t ≤ 24: 2π/3 < πt/12 < 4π/3 [1] 8 < t < 16 [1] So the seat is above 32 m for 8 minutes [1]. Examiner insight: reversing the inequality when dividing by −20 is the key step; finding only the boundary times without the interval loses the final two marks.

12. (a) Horizontally: P cos α = 12 [1]; vertically: P sin α = 20 [1]. (b) P² = 12² + 20² = 544 [1], so P = 4√34 = 23.3 N [1] tan α = 20/12 = 5/3 [1], so α = 59.0° [1] Examiner insight: squaring and adding uses sin²α + cos²α ≡ 1; say so, and give the force with units.

Where marks are usually lost

• Answering in degrees when the interval is in radians, or the reverse.
• Missing solutions in a multiple angle question because the interval was not changed first.
• Dividing by sin θ or cos θ instead of factorising.
• Not rejecting impossible values such as sec x = 1/3, or rejecting them without a reason.
• Choosing a sign for tan θ or sin θ without naming the quadrant.
• Using the principal value of arccos or arcsin when it lies outside the question’s interval.
• Skipping the identity step in a “show that” or “prove” question.
• For a cos x + b sin x in R cos(x − α) form, using tan α = a/b instead of b/a.
• In context questions, giving a boundary time instead of the length of the interval asked for.

Next steps

• Recap with the Trigonometry revision notes.
• Re-read any weak section in the Trigonometry study guide.
• Practise differentiating trigonometric functions with the Differentiation practice questions.
• Course hub: /boards/aqa/a-level/mathematics/.
• Printable checklist: /checklists/aqa/a-level/mathematics/.
• All free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.6 E: Trigonometry (E1 to E9), and Appendix B: mathematical formulae and identities.