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4PM1/02R Jun 2024, Question 11: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 12 marks

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The question, in short

Use a given formula to prove a double-angle identity for cos 2A, then apply it to a squared expression, and finally locate the exact stationary point P of a trig curve on 0 ≤ x ≤ π/6.

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

Part (a) · 2 marks

Starting from the formula on page 2, derive the identity cos 2A = 2cos²A − 1.

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

Answercos 2A = 2cos²A − 1 shown

Part (b) · 3 marks

Using part a, prove that (2cos²A − 1)² equals (cos 4A + 1)/2.

  1. 1(2cos²A − 1)² = (cos 2A)²
  2. 2cos 4A = 2cos²2A − 1, so cos²2A = (cos 4A + 1)/2
  3. 3(2cos²A − 1)² = (cos 4A + 1)/2

Answer(2cos²A − 1)² = (cos 4A + 1)/2 shown

Part (c) · 7 marks

For y = (sin 2x)/2 + (2cos²x − 1)²/2 + 1/8, find exact coordinates of the stationary point P where 0 ≤ x ≤ π/6.

  1. 1y = (sin 2x)/2 + (cos 4x)/4 + 3/8
  2. 2dy/dx = cos 2x − sin 4x
  3. 3= cos 2x − 2sin 2x cos 2x = cos 2x(1 − 2sin 2x) = 0
  4. 4In 0 ≤ x ≤ π/6, cos 2x ≠ 0, so sin 2x = 1/2, x = π/12
  5. 5y = (1/2)(1/2) + (1/4)(1/2) + 3/8 = 3/4

AnswerP(π/12, 3/4)

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