OxfordAQA International AS and A-level Mathematics: Trigonometric identities and solving trigonometric equations (9660) – 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 the second part of content area PP1.2 in the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: the tangent identity, the Pythagorean identity sin²θ + cos²θ = 1, and solving simple trigonometric equations in a given interval in degrees or radians. The specification states this content is part of International AS, Unit PSM1 (section 3.2, PP1). Read the study guide or the revision notes first. The course hub and checklist show the wider course, and the free 10-minute diagnostics help you pick what to practise next.

Round angles in degrees to 1 decimal place and radians to 3 significant figures unless a question says otherwise. Show all working.

Questions

1. Solve cos θ = −0.65 for 0° ≤ θ ≤ 360°. [3]

2. Given that sin A = 20/29 and A is obtuse, find the exact values of cos A and tan A. [3]

3. Show that (sin θ + cos θ)² + (sin θ − cos θ)² = 2 for all values of θ. [3]

4. Solve tan θ = −1.6 for −180° ≤ θ ≤ 180°. [3]

5. Solve sin(θ + 25°) = −0.3 for 0° ≤ θ ≤ 360°. [4]

6. Solve cos(x − π/4) = 0.6 for 0 ≤ x ≤ 2π, giving x in radians. [4]

7. A student is asked to solve 2 sin θ cos θ = cos θ for 0° ≤ θ ≤ 360°. They divide both sides by cos θ and obtain θ = 30° and 150° only.

(a) Find all the solutions of the equation in the interval. [3] (b) Explain the student’s error. [1]

8. Solve 6cos²θ + sin θ = 5 for 0° ≤ θ ≤ 360°. [6]

9. Solve 2cos²x + 3 sin x = 3 for 0 ≤ x ≤ 2π, giving exact answers in terms of π. [6]

10. (a) Show that sin θ tan θ = (1 − cos²θ) / cos θ. [2] (b) Hence solve sin θ tan θ = 1.5 for 0° ≤ θ ≤ 360°. [4]

11. The depth d metres of water at a harbour entrance t hours after midnight is modelled by d = 6 + 2.5 sin(30t)°, for 0 ≤ t ≤ 12.

(a) State the greatest and least depths predicted by the model. [2] (b) Find the times when the depth is 4.5 metres, giving t to 2 decimal places. [4] (c) Explain why the model gives no time when the depth is 9 metres. [1]

12. (a) Solve 2 sin x = 3 cos x for −π ≤ x ≤ π. [4] (b) Given that tan θ = 3/4 and 180° < θ < 270°, find the exact values of sin θ and cos θ. [4]

Answers

1. Reference angle: cos⁻¹(0.65) = 49.46° [1]. Cosine is negative in the second and third quadrants, so θ = 180° − 49.46° = 130.54° [1] and θ = 180° + 49.46° = 229.46° [1]. θ = 130.5° and 229.5° [3] Examiner insight: Giving 130.5° only is the common slip; show the symmetry step that gives the second answer, so the working makes the source of each solution clear.

2. cos²A = 1 − (20/29)² = 1 − 400/841 = 441/841 [1]. A is obtuse, so cosine is negative: cos A = −21/29 [1]. tan A = (20/29) ÷ (−21/29) = −20/21 [1]. cos A = −21/29, tan A = −20/21 [3] Examiner insight: Taking the positive root is the usual way to lose accuracy here; write down the sign rule you are using before you take the square root.

3. Expand: (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ − 2 sin θ cos θ + cos²θ) [1]. The cross terms cancel, leaving 2 sin²θ + 2 cos²θ [1]. This equals 2(sin²θ + cos²θ) = 2 × 1 = 2 [1]. Examiner insight: In a “show that”, quote the identity you use at the step where you use it; an unexplained jump to 2 leaves the key reasoning unseen.

4. Principal value: tan⁻¹(−1.6) = −57.99° [1]. Add 180°: θ = 122.01° [1]. Subtracting 180° gives −237.99°, which is outside the interval, so there are no further solutions [1]. θ = −58.0° and 122.0° [3] Examiner insight: With a negative interval start, check both directions from the principal value; stating that you have checked the end of the range protects against a missing or surplus answer.

5. Let u = θ + 25°; then 25° ≤ u ≤ 385° [1]. sin⁻¹(0.3) = 17.46°, so u = 180° + 17.46° = 197.46° or u = 360° − 17.46° = 342.54° [1]. Subtract 25°: θ = 172.46° and 317.54° [1]. The next possible value, u = 557.46°, is outside the interval [1]. θ = 172.5° and 317.5° [4] Examiner insight: Writing the interval for the bracket first is the habit that stops an out-of-range solution being kept; do it before you touch the inverse function.

6. Let u = x − π/4; then −π/4 ≤ u ≤ 7π/4 [1]. cos⁻¹(0.6) = 0.9273 [1]. Cosine is symmetric, so u = 0.9273 or u = 2π − 0.9273 = 5.3559; the value u = −0.9273 is below −π/4 (about −0.785) so is rejected [1]. Add π/4: x = 1.7127 and 6.1413 [1]. x = 1.71 and 6.14 [4] Examiner insight: Comparing −0.9273 with −π/4 numerically shows why that root is excluded; a solution list with no sign of this check is easy to doubt.

7. (a) Rearrange: 2 sin θ cos θ − cos θ = 0, so cos θ (2 sin θ − 1) = 0 [1]. cos θ = 0 gives θ = 90° and 270° [1]. sin θ = 1/2 gives θ = 30° and 150° [1]. θ = 30°, 90°, 150°, 270° (b) Dividing by cos θ removes the solutions where cos θ = 0, namely 90° and 270° [1]. Examiner insight: Moving everything to one side and factorising keeps every solution; if you cancel a function that can be zero, state the case you set aside and solve it separately.

8. Replace sin²θ: 6cos²θ = 6(1 − sin²θ), so 6 − 6 sin²θ + sin θ = 5 [1]. Rearrange: 6 sin²θ − sin θ − 1 = 0 [1]. Factorise: (3 sin θ + 1)(2 sin θ − 1) = 0 [1]. sin θ = 1/2 gives θ = 30° and 150° [1]. sin θ = −1/3 has reference angle 19.47°, so θ = 180° + 19.47° = 199.5° [1] and θ = 360° − 19.47° = 340.5° [1]. θ = 30°, 150°, 199.5°, 340.5° [6] Examiner insight: Here the squared term is cos²θ, so you must substitute for it using the identity, not for sin θ; choose the function that leaves a single type in the quadratic.

9. Replace cos²x by 1 − sin²x: 2 − 2 sin²x + 3 sin x = 3 [1]. So 2 sin²x − 3 sin x + 1 = 0 [1]. Factorise: (2 sin x − 1)(sin x − 1) = 0 [1]. sin x = 1 gives x = π/2 [1]. sin x = 1/2 has reference angle π/6, with sine positive in the first and second quadrants, so x = π/6 [1] and x = π − π/6 = 5π/6 [1]. x = π/6, π/2, 5π/6 [6] Examiner insight: When the question asks for exact values, a decimal answer risks losing credit; keep the multiples of π through to the end.

10. (a) sin θ tan θ = sin θ × sin θ / cos θ = sin²θ / cos θ [1]. Then sin²θ = 1 − cos²θ, so the expression is (1 − cos²θ) / cos θ [1]. (b) The equation becomes (1 − cos²θ) / cos θ = 3/2 [1]. Multiply by 2 cos θ: 2 − 2cos²θ = 3 cos θ, so 2cos²θ + 3 cos θ − 2 = 0 [1]. Factorise: (2 cos θ − 1)(cos θ + 2) = 0, and cos θ = −2 is impossible, so cos θ = 1/2 [1]. θ = 60° and 300° [1]. θ = 60° and 300° [4] Examiner insight: “Hence” tells you to use part (a); starting again from the original equation without it may still work but wastes time and can leave the intended link unseen.

11. (a) The sine term lies between −1 and 1, so d lies between 6 − 2.5 and 6 + 2.5 [1]. Greatest depth 8.5 m, least depth 3.5 m [1] (b) 6 + 2.5 sin(30t)° = 4.5 gives sin(30t)° = −0.6 [1]. Here 0° ≤ 30t ≤ 360°, and the reference angle is 36.87°, so 30t = 180 + 36.87 = 216.87 [1] and 30t = 360 − 36.87 = 323.13 [1]. Divide by 30: t = 7.23 and 10.77 [1] (c) d = 9 requires sin(30t)° = 1.2, which is impossible because sine never exceeds 1 (equivalently, 9 is above the greatest depth of 8.5 m) [1]. Examiner insight: In a context question, check the interval for 30t before solving and give the final answer for t, not for the angle.

12. (a) Divide by cos x: 2 tan x = 3, so tan x = 1.5 [1]. tan⁻¹(1.5) = 0.9828 [1]. Subtract π for the second solution: 0.9828 − π = −2.1588 [1]. Both lie in −π ≤ x ≤ π [1]. x = 0.983 and −2.16 (b) sin θ = (3/4) cos θ, so (9/16)cos²θ + cos²θ = 1 [1]. Then (25/16)cos²θ = 1 and cos²θ = 16/25 [1]. In the third quadrant both sine and cosine are negative, so cos θ = −4/5 [1] and sin θ = (3/4)(−4/5) = −3/5 [1]. sin θ = −3/5, cos θ = −4/5 Examiner insight: For (a), the interval runs from −π, so subtract π from the first solution rather than adding it; for (b) the stated quadrant is the only thing that fixes the signs.

Where marks are usually lost

• Stopping after the principal value and missing the second solution in the interval.
• Choosing the wrong quadrant symmetry for a negative value on the right-hand side.
• Not changing the interval when the angle is shifted, as in questions 5 and 6.
• Cancelling a factor common to both sides instead of factorising, as in question 7.
• Substituting for the wrong squared function, so the quadratic still has two different trigonometric functions.
• Leaving an impossible root such as cos θ = −2 in the working without rejecting it.
• Rounding in the middle of a calculation and shifting the final digit.
• Giving decimals when exact answers in terms of π are requested.
• Mixing degrees and radians within one question.

Next steps

• Re-read the study guide for any method you found slow.
• Use the revision notes for a final recap.
• Tick off the unit on the printable checklist and browse the course hub.
• Take the free 10-minute diagnostics to find your next weak area.
• Book a free trial class if you want to work through these with a teacher.

Official syllabus

These questions follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA (International AS exams from May/June 2018, A-level from May/June 2019), content area PP1.2 Trigonometry.