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4PM1/02R Jun 2019, Question 10: Solving Trig Equations

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

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The question, in short

Use double-angle identities to derive expressions for cos 2A and cos 4A, solve a trigonometric equation in θ, and integrate f(A) = 4cos⁴A − 4cos²A + 1.

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

Part (a) · 2 marks

Use the cos(A + B) formula to show that cos 2A equals 2cos²A − 1.

  1. 1cos(A + A) = cos A cos A − sin A sin A
  2. 2cos 2A = cos²A − sin²A
  3. 3= cos²A − (1 − cos²A)
  4. 4= 2cos²A − 1

Answercos 2A = 2cos²A − 1 shown

Part (b) · 4 marks

Show that cos 4A can be written as 8cos⁴A − 8cos²A + 1.

  1. 1cos 4A = 2cos²2A − 1
  2. 2= 2(2cos²A − 1)² − 1
  3. 3= 2(4cos⁴A − 4cos²A + 1) − 1
  4. 4= 8cos⁴A − 8cos²A + 1

Answercos 4A = 8cos⁴A − 8cos²A + 1 shown

Part (c) · 5 marks

Solve cos²(θ/4 + π/24)[cos²(θ/4 + π/24) − 1] = −1/16 for 0 ≤ θ < 2π, giving answers in terms of π.

  1. 1Let x = θ/4 + π/24: cos²x(cos²x − 1) = cos⁴x − cos²x = (cos 4x − 1)/8
  2. 24x = θ + π/6
  3. 3(cos(θ + π/6) − 1)/8 = −1/16 ⇒ cos(θ + π/6) = 1/2
  4. 4θ + π/6 = π/3, 5π/3 (in range π/6 ≤ θ + π/6 < 13π/6)
  5. 5θ = π/6, 3π/2

Answerθ = π/6, 3π/2

Part (d) · 4 marks

Using calculus, find the exact value of the integral of f(A) from π/6 to π/2 in the form aπ − b√c.

  1. 1From (b): 4cos⁴A − 4cos²A = (cos 4A − 1)/2, so f(A) = (cos 4A)/2 + 1/2
  2. 2∫ f(A) dA = sin 4A/8 + A/2
  3. 3[sin 4A/8 + A/2] from π/6 to π/2 = (0 + π/4) − (√3/16 + π/12)
  4. 4= π/6 − √3/16

Answerπ/6 − √3/16

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