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4PM1/01R Jan 2022, Question 10: The Addition Formulae

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 15 marks

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The question, in short

Work through a cubic factor, double-angle identities, a trigonometric equation reduced to that cubic, and deduce the cosine value of its solutions.

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

Part (a) · 2 marks

Apply the factor theorem to verify that 4x − 3 divides 16x³ + 11x − 15.

  1. 1f(3/4) = 16(3/4)³ + 11(3/4) − 15
  2. 2= 27/4 + 33/4 − 15 = 0
  3. 3So (4x − 3) is a factor

Answerf(3/4) = 0, so (4x − 3) is a factor

Part (b) · 5 marks

Derive the two double-angle identities for sin 2θ and cos 2θ using the supplied formulae.

  1. 1(i) sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ
  2. 2(ii) cos 2θ = cos(θ + θ) = cos²θ − sin²θ
  3. 3= cos²θ − (1 − cos²θ)
  4. 4= 2 cos²θ − 1

Answer(i) sin 2θ = 2 sin θ cos θ shown (ii) cos 2θ = 2 cos²θ − 1 shown

Part (c) · 4 marks

Substitute x = cos θ to transform the given trigonometric equation into 16x³ + 11x − 15 = 0.

  1. 127 cos θ(2 cos²θ − 1) + 19 sin θ(2 sin θ cos θ) − 15 = 0
  2. 254 cos³θ − 27 cos θ + 38(1 − cos²θ) cos θ − 15 = 0
  3. 354 cos³θ − 27 cos θ + 38 cos θ − 38 cos³θ − 15 = 0
  4. 416 cos³θ + 11 cos θ − 15 = 0
  5. 5With x = cos θ: 16x³ + 11x − 15 = 0

Answer16x³ + 11x − 15 = 0 shown

Part (d) · 4 marks

Use the earlier results to deduce that every solution of the trigonometric equation satisfies cos θ = 3/4.

  1. 116x³ + 11x − 15 = (4x − 3)(4x² + 3x + 5)
  2. 2For 4x² + 3x + 5: b² − 4ac = 9 − 80 < 0, so no real roots
  3. 3So the only solution is 4x − 3 = 0
  4. 4cos θ = 3/4

Answercos θ = 3/4 is the only solution, shown

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