OxfordAQA International AS and A-level Mathematics: Compound and double angle formulae and R-form (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 follow the second part of content area P2.4 (Trigonometry) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level Unit P2, so it is A2 content and not examined at International AS. Angles are in degrees unless a question uses π.

Links: course hub, checklist, study guide, revision notes and the free 10-minute diagnostics.

Questions

1. Using tan 165° = tan(120° + 45°), show that tan 165° = √3 − 2. [3]

2. A is obtuse with sin A = 20/29. B is acute with cos B = 5/13. Find the exact values of sin(A − B) and cos(A + B). [4]

3. Show that sin(x + 60°) + sin(x − 60°) ≡ sin x. [3]

4. Given that cos x = −5/13 and 90° < x < 180°, find the exact value of tan 2x. [3]

5.

(a) Prove that cos⁴x − sin⁴x ≡ cos 2x. [2] (b) Prove that sin 3x ≡ 3 sin x − 4 sin³x. [4]

6. Show that cot θ − tan θ ≡ 2 cot 2θ. [3]

7.

(a) Express √5 sin x − 2 cos x in the form R sin(x − α), where R > 0 and 0° < α < 90°. Give α to 1 decimal place. [3] (b) Solve √5 sin x − 2 cos x = 1.5 for 0° ≤ x ≤ 360°, giving answers to 1 decimal place. [3] (c) Find the greatest value of 12/(5 − √5 sin x + 2 cos x). [2]

8.

(a) Express cos x − sin x in the form R cos(x + α), where R > 0 and 0 < α < π/2. [3] (b) Solve cos x − sin x = 1 for 0 ≤ x ≤ 2π, giving exact answers. [3]

9. Solve 2 cos 2θ = 1 + 3 cos θ for 0° ≤ θ ≤ 360°, giving answers to 1 decimal place. [5]

10. Solve tan 2x = 3 tan x for −180° ≤ x ≤ 180°. [5]

11. Find the exact value of ∫ (sin x + 2 cos x)² dx from 0 to π/3. [6]

12.

(a) Prove that cos(A + B) cos(A − B) ≡ cos²A − sin²B. [3] (b) Hence find the exact value of cos 75° cos 15°. [2] (c) Solve cos(x + 30°) cos(x − 30°) = 1/8 for 0° ≤ x ≤ 360°, giving answers to 1 decimal place. [3]

Answers

1. tan 165° = (tan 120° + tan 45°)/(1 − tan 120° tan 45°) [1] = (1 − √3)/(1 + √3) [1]. Multiply top and bottom by (1 − √3): (1 − √3)²/(1 − 3) = (4 − 2√3)/(−2) = √3 − 2 [1] [3] Examiner insight: in a “show that” question, write the formula, the substituted values and the rationalising step. A result copied from memory has no method to give credit to.

2. A is obtuse, so cos A = −21/29 [1]. B is acute, so sin B = 12/13 [1]. sin(A − B) = (20/29)(5/13) − (−21/29)(12/13) = 100/377 + 252/377 = 352/377 [1]. cos(A + B) = (−21/29)(5/13) − (20/29)(12/13) = −105/377 − 240/377 = −345/377 [1] [4] Examiner insight: the sign of cos A is the main risk. A short note (“A obtuse, so cos A < 0”) lets a reader see that the sign was a decision, not a slip.

3. sin(x + 60°) = sin x cos 60° + cos x sin 60° and sin(x − 60°) = sin x cos 60° − cos x sin 60° [1]. Adding, the cos x sin 60° terms cancel [1]. This leaves 2 sin x cos 60° = 2 sin x (1/2) = sin x [1] [3] Examiner insight: write cos 60° = 1/2 explicitly on the last line. Finishing on the target expression, with the exact value used, completes the proof.

4. x is obtuse, so sin x = 12/13 [1]. tan x = −12/5 [1]. tan 2x = 2(−12/5)/(1 − 144/25) = (−24/5)/(−119/25) = 120/119 [1] [3] Examiner insight: the exact answer is a fraction. A decimal approximation of an angle, then of tan 2x, would not match the “exact” instruction.

5. (a) cos⁴x − sin⁴x = (cos²x − sin²x)(cos²x + sin²x) [1] = cos 2x × 1 = cos 2x [1] (b) sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x [1]. = 2 sin x cos²x + (1 − 2 sin²x) sin x [1]. Replace cos²x with 1 − sin²x: 2 sin x − 2 sin³x + sin x − 2 sin³x [1]. = 3 sin x − 4 sin³x [1] Examiner insight: in part (b), choose a form of cos 2x and a replacement for cos²x that leave only sin x. Mixing sin and cos terms makes the final step hard to reach.

6. cot θ − tan θ = cos θ/sin θ − sin θ/cos θ = (cos²θ − sin²θ)/(sin θ cos θ) [1]. The numerator is cos 2θ and the denominator is (1/2) sin 2θ [1]. So the expression is 2 cos 2θ/sin 2θ = 2 cot 2θ [1] [3] Examiner insight: name the double angle formulae as you use them. The reader then sees why sin θ cos θ became half of sin 2θ.

7. (a) r sin(x − α) = r cos α sin x − r sin α cos x, so r cos α = √5 and r sin α = 2 [1]. R = √(5 + 4) = 3 [1]. tan α = 2/√5, so α = 41.8°, giving 3 sin(x − 41.8°) [1] (b) 3 sin(x − 41.8°) = 1.5, so sin(x − 41.8°) = 0.5 [1]. x − 41.8° = 30° or 150° [1]. x = 71.8° or 191.8° [1] (c) The denominator is 5 − 3 sin(x − 41.8°), whose least value is 5 − 3 = 2 [1]. The greatest value of the fraction is 12/2 = 6 [1] Examiner insight: in part (c), the greatest value comes from the least positive denominator. Check that the denominator never reaches zero (here it ranges from 2 to 8) before taking its least value.

8. (a) r cos(x + α) = r cos α cos x − r sin α sin x, so r cos α = 1 and r sin α = 1 [1]. R = √2 [1]. tan α = 1, so α = π/4 and the expression is √2 cos(x + π/4) [1] (b) √2 cos(x + π/4) = 1, so cos(x + π/4) = 1/√2 [1]. For 0 ≤ x ≤ 2π, π/4 ≤ x + π/4 ≤ 9π/4, so x + π/4 = π/4, 7π/4 or 9π/4 [1]. x = 0, 3π/2 or 2π [1] Examiner insight: write the shifted interval for x + π/4 before listing angles. It shows which solutions to keep, including those at the ends of the interval.

9. 2 cos 2θ = 4cos²θ − 2, so 4cos²θ − 2 = 1 + 3 cos θ [1]. 4cos²θ − 3 cos θ − 3 = 0 [1]. cos θ = (3 ± √57)/8 [1]. (3 + √57)/8 ≈ 1.32 is greater than 1, so reject it [1]. cos θ = (3 − √57)/8 ≈ −0.569, so θ = 124.7° or 235.3° [1] [5] Examiner insight: rejecting a root with a reason (it is greater than 1) is part of the solution. Listing both roots as if both gave angles misses this.

10. 2 tan x/(1 − tan²x) = 3 tan x. Multiply by (1 − tan²x): 2t = 3t(1 − t²) with t = tan x [1]. t(3t² − 1) = 0 [1]. t = 0: x = −180°, 0°, 180° [1]. t = 1/√3: x = 30° or −150° [1]. t = −1/√3: x = −30° or 150° [1] [5] Examiner insight: the root t = 0 disappears if you divide by tan x. Factorising keeps it, and the interval then asks for three values from it.

11. (sin x + 2 cos x)² = sin²x + 4 sin x cos x + 4cos²x [1]. Use sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2 and 4 sin x cos x = 2 sin 2x [1]: the integrand is 5/2 + (3/2) cos 2x + 2 sin 2x [1]. Integral: (5/2)x + (3/4) sin 2x − cos 2x [1]. At π/3: 5π/6 + 3√3/8 + 1/2. At 0: −1 [1]. Difference: 5π/6 + 3√3/8 + 3/2 [1] [6] Examiner insight: the lower limit contributes −cos 0 = −1, which turns into +1 when subtracted. Substituting both limits on separate lines guards against that sign.

12. (a) (cos A cos B − sin A sin B)(cos A cos B + sin A sin B) [1] = cos²A cos²B − sin²A sin²B [1]. Replace cos²B = 1 − sin²B and sin²A = 1 − cos²A: cos²A − cos²A sin²B − sin²B + cos²A sin²B = cos²A − sin²B [1] (b) A + B = 75° and A − B = 15°, so A = 45° and B = 30° [1]. cos²45° − sin²30° = 1/2 − 1/4 = 1/4 [1] (c) cos²x − sin²30° = 1/8, so cos²x = 3/8 [1]. cos x = ±0.612 [1]. x = 52.2°, 127.8°, 232.2°, 307.8° [1] Examiner insight: “hence” in part (b) points to the identity from part (a), so choose A and B to fit it. Evaluating cos 75° and cos 15° separately ignores that instruction.

Where marks are usually lost

• Using the wrong sign in cos(A ± B), or in the sign of cos A for an obtuse angle in question 2.
• Giving a decimal where the question asks for an exact value, as in questions 1, 2, 4 and 12(b).
• Dividing by tan x in question 10, or by sin x in an equation like it, and dropping solutions.
• Working on both sides of an identity at once instead of starting from one side.
• Not showing the substitution of an exact value, such as cos 60° = 1/2 in question 3.
• Choosing a cos 2x form that leaves two functions, so the equation in question 9 cannot be factorised.
• Writing an R-form that leaves α obtuse, or expanding the form with the wrong sign when matching coefficients.
• Forgetting to shift the interval for x − α or x + α before reading off solutions.
• Leaving out roots when the interval is longer than 0° to 360°, or mis-handling the negative angles in question 10.
• Losing the lower-limit term in a definite integral involving cos 2x.

Next steps

• Condense the topic with the revision notes.
• Read the full explanations in the study guide.
• Return to the course hub and tick off the checklist.
• Take the free 10-minute diagnostics to find what to revise next.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: content area P2.4 Trigonometry (compound angle and double angle formulae, the forms r cos(θ ± α) and r sin(θ ± α), identities, integration and equations), International A-level Unit P2.