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4PM1/02 Jun 2025, Question 11: Solving Trig Equations

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 15 marks

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The question, in short

Prove a trigonometric identity in cos⁴θ and sin⁴θ, then apply it to solve an equation for θ in a given interval and evaluate a definite integral using calculus.

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

Part (a) · 4 marks

Prove that the difference of the fourth powers of cos θ and sin θ equals cos 2θ.

  1. 1cos⁴θ − sin⁴θ = (cos²θ − sin²θ)(cos²θ + sin²θ)
  2. 2cos²θ + sin²θ = 1
  3. 3= cos²θ − sin²θ = cos 2θ

Answercos⁴θ − sin⁴θ = cos 2θ shown

Part (b) · 7 marks

Use the identity to solve the given equation for θ between 0 and π, giving answers to 2 decimal places.

  1. 1RHS = 2cos(2θ + π/4) by (a)
  2. 28cos²(2θ + π/4) − 2cos(2θ + π/4) − 3 = 0
  3. 3(4cos(2θ + π/4) − 3)(2cos(2θ + π/4) + 1) = 0
  4. 4cos(2θ + π/4) = 3/4 or −1/2
  5. 5In [π/4, 9π/4): 2θ + π/4 = 2π/3, 4π/3, 2π − 0.7227…, 2π + 0.7227…
  6. 6θ = 0.65, 1.70, 2.39, 3.11

Answerθ = 0.65, 1.70, 2.39, 3.11

Part (c) · 4 marks

By calculus, evaluate the definite integral from π/16 to π/8, expressing the result as a − b√2 with rational a and b.

  1. 1cos⁴2x − sin⁴2x = cos 4x
  2. 2∫ (cos 4x − 8sin 4x) dx = (1/4)sin 4x + 2cos 4x
  3. 3= (1/4)[sin(π/2) − sin(π/4)] + 2[cos(π/2) − cos(π/4)]
  4. 4= (1/4)(1 − √2/2) + 2(0 − √2/2) = 1/4 − 9√2/8

Answer1/4 − 9√2/8

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