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0606/22 Feb/Mar 2025, Question 8: Solving Trig Equations

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 7 marks

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The question, in short

Prove a trigonometric identity involving sin θ and tan²θ, then use the result to solve an equation in 3x over the range −180° ≤ x ≤ 180°.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Show that sin θ tan²θ divided by 1 + tan²θ simplifies to sin³θ.

  1. 1Numerator: sin θ tan²θ = sin θ × sin²θ/cos²θ
  2. 2Denominator: 1 + tan²θ = sec²θ = 1/cos²θ
  3. 3(sin θ × sin²θ/cos²θ) ÷ (1/cos²θ) = sin θ × sin²θ = sin³θ

Answer(sin θ tan²θ)/(1 + tan²θ) = sin³θ shown

Part (b) · 5 marks

Using part a, solve sin³3x = 1/8 for −180° ≤ x ≤ 180°.

  1. 1By (a): sin³3x = 1/8, so sin 3x = 1/2
  2. 2−540° ≤ 3x ≤ 540°
  3. 33x = −330°, −210°, 30°, 150°, 390°, 510°
  4. 4x = −110°, −70°, 10°, 50°, 130°, 170°

Answerx = −110°, −70°, 10°, 50°, 130°, 170°

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Also searched as: 0606/22/F/M/25 · 0606_m25_qp_22 · 0606 m25 22 · question 8.
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