A Level Mathematics: Pure Mathematics 3 Trigonometry, Vectors and Complex Numbers — Practice Questions (Cambridge 9709)

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 — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Each question practises a skill tested in the June 2025 Paper 32. After each answer there is a tip and the real question to try next.

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Questions

1. (a) Write 8 sin θ + 15 cos θ as R cos(θ − α), with R > 0 and 0 < α < ½π, stating α to 4 decimal places. (b) Hence solve the equation 8 sin 2x + 15 cos 2x = 10 for 0 < x < π. [6]

2. Solve the equation tan 2x = 4 cot x for 0° < x < 180°. [4]

3. By evaluating a suitable function at x = 1.2 and x = 1.4, show that the equation x = 1 + e⁻ˣ has a root in the interval 1.2 < x < 1.4. [2]

4. Starting with x₁ = 1.25, use the iteration xₙ₊₁ = 1 + e^(−xₙ) to find the root of x = 1 + e⁻ˣ correct to 2 decimal places, showing every iterate to 4 decimal places. [3]

5. The points A and B have position vectors i − 2j + 4k and 3i + j + 2k relative to the origin O. (a) Write down a vector equation of the line AB. (b) Determine whether each of the points C(7, 7, −2) and D(5, 4, 1) lies on this line. [4]

6. The points P, Q and R have coordinates (2, 1, −1), (4, 3, 0) and (5, 1, 3). (a) Calculate the exact value of cos QPR by means of a scalar product. (b) Use your answer to part (a) to find the area of triangle PQR in exact form. [5]

7. Form a quartic equation in x and use it to find both square roots of 1 − 4√3 i, giving each as x + iy with x and y exact real numbers. [5]

8. The complex number z satisfies |z − 6 − 8i| = 3. Find the least possible value of |z|. [2]

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Answers

1. (a) R = √(8² + 15²) = 17 [1]. R cos(θ − α) = R cos θ cos α + R sin θ sin α, so R sin α = 8 and R cos α = 15, giving tan α = 8/15 [1] and α = 0.4900 [1]. (b) 17 cos(2x − 0.4900) = 10, so cos(2x − 0.4900) = 10/17 and 2x − 0.4900 = ±0.9419 + 2kπ [1]. Since 0 < 2x < 2π: 2x = 0.4900 + 0.9419 = 1.4319, or 2x = 0.4900 − 0.9419 + 2π = 5.8313 [1]. So x = 0.716 or x = 2.92 [1].

Tip: write the new range for 2x before solving, so that you look for every solution in 0 < 2x < 2π and not just the first one.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 7.

2. Write in sines and cosines: sin 2x/cos 2x = 4 cos x/sin x, so sin 2x sin x = 4 cos x cos 2x, and with sin 2x = 2 sin x cos x this gives 2 sin² x cos x = 4 cos x cos 2x [1]. Factorise: cos x (2 sin² x − 4 cos 2x) = 0, so cos x = 0, giving x = 90° (check: tan 180° = 0 and cot 90° = 0) [1]. Otherwise, using cos 2x = 1 − 2 sin² x: 2 sin² x = 4 − 8 sin² x, so sin² x = 2/5 and sin x = 0.6325 (sin x > 0 in this range) [1]. x = 39.2°, 90° or 140.8° [1].

Tip: do not divide by an expression such as cos x that can be zero; factorise it out instead. Rewriting everything in tan x here would lose the solution x = 90°, where tan x is undefined but the original equation still holds.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 4.

3. Let f(x) = x − 1 − e⁻ˣ. f(1.2) = 0.2 − e^(−1.2) = −0.101 and f(1.4) = 0.4 − e^(−1.4) = 0.153 [1]. The sign changes and f is continuous, so there is a root between 1.2 and 1.4 [1].

Tip: rearrange to f(x) = 0 first, give the values you calculated, and state the conclusion in words.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 6(b).

4. Iterates: 1.25, 1.2865, 1.2762, 1.2791, 1.2783, 1.2785, 1.2785 [2]. The root is 1.28 to 2 decimal places [1].

Tip: keep going until two successive iterates agree when rounded to the accuracy asked for, and use the full calculator value each time, not the rounded one.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 6(c).

5. (a) AB = (3i + j + 2k) − (i − 2j + 4k) = 2i + 3j − 2k [1]. r = i − 2j + 4k + λ(2i + 3j − 2k) [1]. (b) For C, the x-component gives 1 + 2λ = 7, so λ = 3; then y = −2 + 9 = 7 and z = 4 − 6 = −2, which match, so C lies on the line [1]. For D, 1 + 2λ = 5 gives λ = 2; then y = 4 matches but z = 4 − 4 = 0 ≠ 1, so D does not lie on the line [1].

Tip: find λ from one component, then check it in all the other components. One matching component is not enough.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 9(a).

6. (a) PQ = (2, 2, 1) and PR = (3, 0, 4) [1]. PQ · PR = 6 + 0 + 4 = 10, |PQ| = 3 and |PR| = 5 [1]. cos QPR = 10/15 = 2/3 [1]. (b) sin QPR = √(1 − 4/9) = √5/3 [1]. Area = ½ × 3 × 5 × √5/3 = 5√5/2 [1].

Tip: both vectors must start at the vertex of the angle (here P). Using QP instead of PQ changes the sign of the scalar product and gives the wrong angle.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 9.

7. Let (x + iy)² = 1 − 4√3 i. Comparing real and imaginary parts: x² − y² = 1 and 2xy = −4√3 [1]. So y = −2√3/x and x² − 12/x² = 1, giving x⁴ − x² − 12 = 0 [1]. (x² − 4)(x² + 3) = 0, and x is real, so x² = 4 and x = ±2 [1]. Then y = ∓√3 [1]. The square roots are 2 − √3 i and −2 + √3 i [1].

Tip: match the sign of y to each value of x using 2xy, and reject x² = −3 because x must be real.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 5.

8. The points lie on a circle with centre 6 + 8i and radius 3. The centre is √(6² + 8²) = 10 from the origin [1]. The least value of |z| is 10 − 3 = 7 [1].

Tip: |z − a| = r is a circle with centre a. The nearest point to the origin lies on the line through the origin and the centre.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 3.

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Where marks are usually lost

• α given in degrees or to too few decimal places when radians to 4 decimal places were asked for.
• Only one solution found after solving for 2x, because the range was not doubled.
• A point tested on a line using only one component.
• Vectors for an angle not both taken from the vertex.
• Signs of x and y mixed up when square roots of a complex number are paired.