Pearson Edexcel International Advanced Level Mathematics: Further trigonometry and R-formulae (YMA01) – 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 Unit P3 section 2 (topics 2.1 to 2.3) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification, Issue 3, April 2019. The specification lists P3 as compulsory for IAL Mathematics and Pure Mathematics only; IAS weighting is N/A. Learn the content first in the study guide and the revision notes. Course hub: Edexcel A Level Mathematics. Printable checklist. To gauge your level beforehand, use a free 10-minute diagnostic.

Give angles in degrees to 1 decimal place unless told otherwise. Use radians where the question uses π. Keep unrounded values in your working until the final answer.

Questions

1. Find the exact value of

(a) sec(5π/6), [1] (b) cot 300°, [1] (c) arccos(−√3/2), in radians. [1]

2. (a) State the domain and the range of y = arccos x, with the range in radians. [2] (b) Find the exact value of arcsin(sin(11π/6)). [1] (c) Find the exact value of sin(arctan(5/12)). [2]

3. (a) Solve 3 cosec θ = 5 for 0 ≤ θ < 360°. [2] (b) Solve cot 2x = 2 for 0 < x < 180°. [3]

4. Solve 3 sec² θ − 5 tan θ − 5 = 0 for −180° < θ ≤ 180°. [6]

5. (a) Prove that tan θ + cot θ ≡ 2 cosec 2θ. [3] (b) Hence solve tan θ + cot θ = 5 for 0 < θ < 180°. [3]

6. A is acute with sin A = 7/25. B is obtuse with sin B = 5/13.

(a) Find cos A and cos B. [2] (b) Find the exact value of cos(A − B). [3]

7. (a) Show that tan(x + 45°) ≡ (1 + tan x)/(1 − tan x). [2] (b) Hence solve tan(x + 45°) + tan x = 0 for 0 < x < 180°, giving each answer to 1 d.p. [4]

8. Solve cos 2θ + 3 cos θ + 2 = 0 for 0 ≤ θ < 360°. [5]

9. The angle θ is obtuse and sin θ = 4/5. Find the exact value of

(a) cos 2θ, [2] (b) tan 2θ, [2] (c) cos(θ/2). [2]

10. f(θ) = 9 cos θ − 4 sin θ.

(a) Write f(θ) in the form R cos(θ + α), with R > 0 and 0 < α < 90°. Give R as a surd and α to 1 d.p. [3] (b) Solve f(θ) = 6 for 0 ≤ θ < 360°. [4] (c) Find the least value of 12 + f(θ) and the smallest positive θ at which it occurs. [3]

11. (a) Write √3 sin x − cos x in the form R sin(x − α), with R > 0 and 0 < α < π/2. [3] (b) Solve √3 sin x − cos x = 1 for 0 ≤ x ≤ 2π. [3] (c) Explain why √3 sin x − cos x = 3 has no solutions. [1]

12. g(x) = 4 cos² x + 3 sin x cos x.

(a) Show that g(x) ≡ 2 + 2 cos 2x + (3/2) sin 2x. [3] (b) Write 2 cos 2x + (3/2) sin 2x in the form R cos(2x − α), with R > 0 and 0 < α < 90°. Give α to 2 d.p. [3] (c) Hence solve g(x) = 3 for 0 ≤ x < 180°. [3] (d) State the greatest value of g(x) and the smallest positive x at which it occurs. [1]

Answers

1. (a) cos(5π/6) = −√3/2, so sec(5π/6) = −2/√3 = −2√3/3 [1] [1] (b) cot 300° = cos 300°/sin 300° = (1/2)/(−√3/2) = −√3/3 [1] [1] (c) arccos lies between 0 and π, and cos(5π/6) = −√3/2, so 5π/6 [1] [1] Examiner insight: Reciprocal values are best written with a rational denominator, and the sign should be checked against the quadrant before you write the final answer.

2. (a) Domain −1 ≤ x ≤ 1 [1]. Range 0 ≤ y ≤ π [1] [2] (b) sin(11π/6) = −1/2. The answer must lie in the range of arcsin, so it is −π/6 [1] [1] (c) Let α = arctan(5/12). A right-angled triangle has opposite 5, adjacent 12, hypotenuse 13 [1]. sin α = 5/13 [1] [2] Examiner insight: A bare 11π/6 is a common answer to part (b). State the range of the inverse function when you use it, because that is where the credit is decided.

3. (a) sin θ = 3/5 [1]. θ = 36.9°, 143.1° [1] [2] (b) tan 2x = 1/2 [1]. 0 < 2x < 360°, so 2x = 26.6° or 206.6° [1]. x = 13.3°, 103.3° [1] [3] Examiner insight: In part (b), write the interval for 2x before solving. If you stop at 2x = 26.6° you give only one of the two solutions that the question requires.

4. Replace sec² θ by 1 + tan² θ: 3 + 3 tan² θ − 5 tan θ − 5 = 0 [1] 3 tan² θ − 5 tan θ − 2 = 0 [1] (3 tan θ + 1)(tan θ − 2) = 0 [1] tan θ = 2 or tan θ = −1/3 [1] tan θ = 2: θ = 63.4° and 63.4° − 180° = −116.6° [1] tan θ = −1/3: θ = −18.4° and 161.6° [1] Solutions: θ = −116.6°, −18.4°, 63.4°, 161.6° [6] Examiner insight: The interval runs from −180° to 180°, so each tangent value gives one positive and one negative angle. Check each root against the interval before you finish.

5. (a) LHS = sin θ/cos θ + cos θ/sin θ = (sin² θ + cos² θ)/(sin θ cos θ) [1] = 1/(sin θ cos θ) [1] = 2/(2 sin θ cos θ) = 2/sin 2θ = 2 cosec 2θ = RHS [1] [3] (b) 2 cosec 2θ = 5, so sin 2θ = 2/5 [1] 0 < 2θ < 360°, so 2θ = 23.6° or 156.4° [1] θ = 11.8°, 78.2° [1] [3] Examiner insight: A proof is judged on every line, so keep the stages visible and finish with the target expression. Part (b) is meant to use part (a), and a solution that goes back to tan and cot is much longer.

6. (a) cos A = 24/25 [1] (A acute). cos B = −12/13 [1] (B obtuse, so cosine is negative) [2] (b) cos(A − B) = cos A cos B + sin A sin B [1] = (24/25)(−12/13) + (7/25)(5/13) [1] = (−288 + 35)/325 = −253/325 [1] [3] Examiner insight: Show the substitution as well as the answer. The sign of cos B is the usual slip, and the working shows where it came from.

7. (a) tan(x + 45°) = (tan x + tan 45°)/(1 − tan x tan 45°) [1] tan 45° = 1, so this is (1 + tan x)/(1 − tan x) [1] [2] (b) Let t = tan x. (1 + t)/(1 − t) + t = 0 [1] 1 + t + t − t² = 0, so t² − 2t − 1 = 0 [1] t = 1 ± √2 [1] x = arctan(1 + √2) = 67.5° and x = 180° + arctan(1 − √2) = 67.5°, 157.5° [1] [4] Examiner insight: In part (b) the 45° identity from part (a) is meant to be used, so state it before substituting. Because the interval is 0 to 180°, a negative tangent value has to be converted into an angle in that interval.

8. cos 2θ = 2 cos² θ − 1, giving 2 cos² θ − 1 + 3 cos θ + 2 = 0 [1] 2 cos² θ + 3 cos θ + 1 = 0 [1] (2 cos θ + 1)(cos θ + 1) = 0 [1] cos θ = −1/2: θ = 120°, 240° [1] cos θ = −1: θ = 180° [1] [5] Examiner insight: Choosing 2 cos² θ − 1 keeps one function in the quadratic. The root cos θ = −1 is easy to forget, so check each factor gives its own solutions.

9. (a) cos 2θ = 1 − 2 sin² θ = 1 − 2(16/25) [1] = −7/25 [1] [2] (b) θ is obtuse, so cos θ = −3/5 and tan θ = −4/3 [1] tan 2θ = 2(−4/3)/(1 − 16/9) = (−8/3)/(−7/9) = 24/7 [1] [2] (c) cos θ = 2cos²(θ/2) − 1, so cos²(θ/2) = (1 + cos θ)/2 = 1/5 [1] 90° < θ < 180°, so θ/2 is acute and cos(θ/2) is positive: √5/5 [1] [2] Examiner insight: For a half angle, a reason for the sign of the square root is part of a complete answer. Your answer to part (b) can be checked against tan 2θ = sin 2θ/cos 2θ.

10. (a) R cos(θ + α) = R cos α cos θ − R sin α sin θ, so R cos α = 9 and R sin α = 4 [1] R = √(81 + 16) = √97 [1] tan α = 4/9, so α = 24.0° [1] [3] (b) cos(θ + α) = 6/√97 [1] θ + 24.0° = 52.5° or 307.5° (range 24.0° ≤ θ + α < 384.0°) [1] θ = 28.5° [1] θ = 283.6° [1] [4] (c) Least value of f is −√97, so least value of 12 + f is 12 − √97 ≈ 2.15 [1] This occurs when cos(θ + α) = −1, so θ + 24.0° = 180° [1] θ = 156.0° [1] [3] Examiner insight: Write the two matching equations for R cos α and R sin α before dividing. In part (b) the interval for θ + α is longer than 360°, and the second solution depends on seeing that.

11. (a) R sin(x − α) = R sin x cos α − R cos x sin α, so R cos α = √3 and R sin α = 1 [1] R = √(3 + 1) = 2 [1] tan α = 1/√3, so α = π/6 [1] [3] (b) sin(x − π/6) = 1/2 [1] −π/6 ≤ x − π/6 ≤ 11π/6, so x − π/6 = π/6 or 5π/6 [1] x = π/3, π [1] [3] (c) The greatest value of 2 sin(x − π/6) is 2, which is less than 3, so the equation has no solutions [1] [1] Examiner insight: In part (c), compare 3 with the greatest value R. A statement that mentions R and the range of sine is a complete explanation.

12. (a) cos² x = (1 + cos 2x)/2, so 4 cos² x = 2 + 2 cos 2x [1] sin x cos x = (1/2) sin 2x, so 3 sin x cos x = (3/2) sin 2x [1] g(x) = 2 + 2 cos 2x + (3/2) sin 2x [1] [3] (b) R cos α = 2 and R sin α = 3/2 [1] R = √(4 + 9/4) = 5/2 [1] tan α = 3/4, so α = 36.87° [1] [3] (c) g(x) = 3 gives (5/2) cos(2x − 36.87°) = 1, so cos(2x − 36.87°) = 2/5 [1] −36.87° ≤ 2x − 36.87° < 323.13°, so 2x − 36.87° = ±66.42° [1] 2x = 103.29° or 330.45°, so x = 51.6°, 165.2° [1] [3] (d) Greatest value of g is 2 + 5/2 = 9/2, first at 2x = 36.87°, so x = 18.4° [1] [1] Examiner insight: Keep α to at least two decimal places in the working, because rounding it early can change the last digit of x. A shown identity needs both the cos² and the sin x cos x conversions stated.

Where marks are usually lost

• Finding only one value for 2θ or 2x and not listing the second solution in the doubled interval.
• Giving arcsin(sin x) as x when x lies outside the range of arcsin.
• Taking the wrong sign for cos B or sin B when B is obtuse.
• Using sec² θ = 1 + tan² θ with a slip in the constant, so the quadratic is wrong from the start.
• Dividing by sin θ or cos θ and losing solutions.
• Choosing the wrong form of cos 2θ, which leaves a mixed equation in sin and cos.
• Writing R sin α and R cos α the wrong way round in an R-form, which gives α as 90° minus the correct value.
• Rounding α before solving, which shifts the final angle.
• Leaving a proof without a closing line, or working on both sides at once.

Next steps

• Review the revision notes for any question you missed.
• Reread the study guide for the matching worked examples.
• Return to the Edexcel A Level Mathematics hub and tick off the topic on the printable checklist.
• Use the free 10-minute diagnostics to choose what to revise next.
• Book a free trial class at /trial/.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited: Unit P3, section 2 (topics 2.1 to 2.3).