IB DP Mathematics: Analysis and Approaches – Circular functions, trigonometric identities and equations 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the circular functions, trigonometric identities and equations unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 3.5–3.11. Questions 1–9 are for SL and HL; questions 10–12 are HL only (AHL sections 3.9–3.11). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 requires a GDC. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator-allowed answers exactly or to 3 significant figures. Use radians unless a question uses degrees.

Learn the methods first in the circular functions study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Find the exact value of:

(a) cos(7π/6) [1] (b) tan(5π/3) [1] (c) sin(−3π/4) [1]

2. (calculator-free) Given that sin θ = 3/7 and π/2 < θ < π, find the exact value of:

(a) cos θ [2] (b) tan θ [1] (c) sin 2θ [1] (d) cos 2θ [1]

3. (calculator-free) Given that cos α = p, where 0 < α < π/2, write each of the following in terms of p.

(a) cos(π + α) [1] (b) cos(−α) [1] (c) sin(2π − α) [1] (d) tan(π + α) [1]

4. (calculator allowed) In triangle ABC, AB = 12 cm, AC = 8 cm and angle ABC = 35°.

(a) Show that there are two possible values of angle ACB, and find them. [3] (b) Find the two possible lengths of BC. [3]

5. (calculator-free) Let f(x) = 3 sin(2(x − π/6)) + 1.

(a) Write down the amplitude and the period of f. [2] (b) Write down the range of f. [1] (c) Describe a sequence of transformations that maps the graph of y = sin x onto the graph of y = f(x). [2] (d) Find the smallest positive value of x at which f has a maximum. [2]

6. (calculator-free) Solve 2 cos 2x + √3 = 0 for 0 ≤ x ≤ π. [4]

7. (calculator-free) Solve 2sin²x − 3 cos x − 3 = 0 for 0 ≤ x ≤ 2π. [6]

8. (calculator-free) Solve sin 2x = √3 cos x for 0 ≤ x ≤ 2π. [4]

9. (calculator allowed) A wind turbine has its hub 40 m above the ground. The tip of one blade is 25 m from the hub and turns at a constant rate, making one revolution every 5 seconds. At t = 0 the tip is at its lowest point. The height of the tip, h metres, after t seconds is modelled by h(t) = p cos(qt) + r.

(a) Find the values of p, q and r. [3] (b) Find the height of the tip when t = 1.2. [2] (c) Find the length of time in each revolution for which the tip is more than 55 m above the ground. [4]

10. (HL only, calculator-free) Given that sec θ = −3 and π < θ < 3π/2:

(a) Find the exact value of tan θ. [2] (b) Find the exact value of cosec θ. [2] (c) Show that sec²θ + cosec²θ ≡ sec²θ cosec²θ for all θ where both sides are defined. [3]

11. (HL only, calculator-free)

(a) Use a compound angle identity to show that tan(π/12) = 2 − √3. [4] (b) Given that tan A = 1/2, find tan 2A. [2] (c) Show that arctan(1/2) + arctan(1/3) = π/4. [4]

12. (HL only, calculator-free) Let f(x) = arccos x.

(a) Write down the domain and range of f. [2] (b) Using cos(π − θ) = −cos θ, show that arccos(−x) = π − arccos x for −1 ≤ x ≤ 1. [3] (c) Hence solve arccos(−x) = 2 arccos x. [3]

Answers

1. (a) Reference angle π/6, third quadrant, cos negative: −√3/2 [1] (b) Reference angle π/3, fourth quadrant, tan negative: −√3 [1] (c) −3π/4 is the same position as 5π/4, third quadrant, sin negative: −√2/2 [1] Examiner insight: On Paper 1 a decimal such as −0.866 is not an exact value and does not earn the accuracy mark.

2. (a) cos²θ = 1 − 9/49 = 40/49 [1]; θ is obtuse so cos θ < 0: cos θ = −2√10/7 [1] (b) tan θ = (3/7)/(−2√10/7) = −3/(2√10) = −3√10/20 [1] (c) sin 2θ = 2(3/7)(−2√10/7) = −12√10/49 [1] (d) cos 2θ = 1 − 2(9/49) = 31/49 [1] Examiner insight: The positive root +2√10/7 loses the accuracy mark in (a), and the wrong sign then follows through to (b) and (c) but not back to (a).

3. (a) Half-turn changes the sign of cos: −p [1] (b) cos is even: p [1] (c) sin α = √(1 − p²), and sin(2π − α) = −sin α = −√(1 − p²) [1] (d) tan(π + α) = tan α = √(1 − p²)/p [1] Examiner insight: Writing sin α = √(1 − p²) needs the fact that α is acute; a ± answer in (c) or (d) is marked wrong.

4. (a) sin C / 12 = sin 35° / 8, so sin C = 0.8604 [1]. C = 59.4° [1] or 180° − 59.36° = 121°; 121° + 35° = 156° < 180°, so both are possible [1] (b) Angle BAC = 180° − 35° − C = 85.64° or 24.36° [1]. BC = 8 sin A / sin 35° [1], so BC = 13.9 cm or 5.75 cm [1] Examiner insight: The “show that” mark in (a) needs the angle-sum check written down; quoting two angles without it earns only the value marks.

5. (a) Amplitude 3 [1]; period 2π/2 = π [1] (b) −2 ≤ f(x) ≤ 4 [1] (c) Horizontal stretch by scale factor 1/2, then translation π/6 to the right [1]; vertical stretch by scale factor 3, then translation 1 up [1] (d) Maximum where 2(x − π/6) = π/2 [1], so x = π/4 + π/6 = 5π/12 [1] Examiner insight: In (c) a translation of π/3 to the right is wrong, because the shift comes from the factorised form 2(x − π/6); scale factors must be stated, not just “stretch”.

6. cos 2x = −√3/2 [1]. Since 0 ≤ x ≤ π, 0 ≤ 2x ≤ 2π [1]. 2x = 5π/6 or 7π/6 [1], so x = 5π/12, 7π/12 [1] Examiner insight: This is calculator-free, so 1.31 and 1.83 are not accepted; the accuracy mark needs the exact values 5π/12 and 7π/12.

7. Use sin²x = 1 − cos²x: 2 − 2cos²x − 3 cos x − 3 = 0 [1], so 2cos²x + 3 cos x + 1 = 0 [1]. (2 cos x + 1)(cos x + 1) = 0 [1], giving cos x = −1/2 or cos x = −1 [1]. cos x = −1/2 gives x = 2π/3, 4π/3 [1]; cos x = −1 gives x = π [1] Examiner insight: Each root of the quadratic carries its own accuracy mark, so missing x = π costs a mark even when the other two are correct.

8. 2 sin x cos x − √3 cos x = 0 [1], so cos x(2 sin x − √3) = 0 [1]. cos x = 0 gives x = π/2, 3π/2 [1]; sin x = √3/2 gives x = π/3, 2π/3 [1] Examiner insight: Dividing by cos x loses π/2 and 3π/2; factorising is the only route to full marks.

9. (a) Lowest point at t = 0 means h(0) = p + r = 15 and the maximum is 65, so r = 40 [1] and p = −25 [1]. Period 5 s, so q = 2π/5 (1.26 to 3 s.f.) [1] (b) h(1.2) = −25 cos(2π × 1.2/5) + 40 [1] = 38.4 m [1] (c) −25 cos(2πt/5) + 40 > 55 gives cos(2πt/5) < −0.6 [1]. In the first revolution, 2πt/5 = 2.214 or 2π − 2.214 = 4.069 [1], so t = 1.762 or 3.238 [1]. Time above 55 m = 3.238 − 1.762 = 1.48 s [1] Examiner insight: A GDC intersection gives these values quickly, but write the equation you solved and both t-values; a bare 1.48 risks the method marks.

10. (a) 1 + tan²θ = sec²θ = 9, so tan²θ = 8 [1]. In the third quadrant tan θ > 0: tan θ = 2√2 [1] (b) cos θ = −1/3, so sin θ = tan θ cos θ = −2√2/3 [1]; cosec θ = −3/(2√2) = −3√2/4 [1] (c) LHS = 1/cos²θ + 1/sin²θ [1] = (sin²θ + cos²θ)/(sin²θ cos²θ) [1] = 1/(sin²θ cos²θ) = sec²θ cosec²θ = RHS [1] Examiner insight: In a “show that” identity, work from one side only and finish with the other side; working on both sides at once can lose the final mark.

11. (a) π/12 = π/3 − π/4 [1]. tan(π/3 − π/4) = (√3 − 1)/(1 + √3 × 1) [1]. Multiply by (√3 − 1)/(√3 − 1): (√3 − 1)²/(3 − 1) [1] = (4 − 2√3)/2 = 2 − √3 [1] (b) tan 2A = 2(1/2)/(1 − 1/4) [1] = 4/3 [1] (c) Let A = arctan(1/2), B = arctan(1/3). tan(A + B) = (1/2 + 1/3)/(1 − 1/6) [1] = (5/6)/(5/6) = 1 [1]. 0 < 1/3 < 1/2 < 1, so 0 < A, B < π/4 and 0 < A + B < π/2 [1]. The only angle in that interval with tangent 1 is π/4, so A + B = π/4 [1] Examiner insight: tan(A + B) = 1 alone does not prove the result; the reasoning mark needs the interval for A + B, since 5π/4 also has tangent 1.

12. (a) Domain −1 ≤ x ≤ 1 [1]; range 0 ≤ f(x) ≤ π [1] (b) Let θ = arccos x, so 0 ≤ θ ≤ π and cos θ = x [1]. Then cos(π − θ) = −cos θ = −x [1]. Since 0 ≤ π − θ ≤ π, π − θ lies in the range of arccos, so arccos(−x) = π − θ = π − arccos x [1] (c) π − arccos x = 2 arccos x [1], so 3 arccos x = π and arccos x = π/3 [1], giving x = 1/2 [1] Examiner insight: “Hence” means you must use (b); solving by trial of values scores no method marks even if x = 1/2 is found.

Where marks are usually lost

• Decimal answers on calculator-free questions where exact values such as 5π/12 or −2√10/7 are required.
• The wrong sign for cos θ or tan θ because the quadrant in the question was ignored.
• Missing the second triangle in the ambiguous case, or not checking the angle sum.
• Not changing the interval for 2x or 3x before listing solutions.
• Dividing by cos x or sin x instead of factorising, as in question 8.
• Reading the horizontal translation from sin(2x − π/3) as π/3 instead of π/6.
• Period taken as b rather than 2π/b in a model.
• Rounding intermediate values in a model before the final step, which changes the third significant figure.
• (HL only) Proving an inverse-function result without stating the range that makes the angle unique.
• (HL only) Sign slips in compound angle expansions, especially cos(A + B).

Next steps

• Recap the key facts in the circular functions revision notes.
• Re-learn any weak topic in the circular functions study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AA exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).