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4PM1/02 Jun 2020, Question 10: Solving Trig Equations

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 16 marks

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The question, in short

Prove two cosine sum-to-product identities, solve a trig equation in θ, prove a triple-cosine identity, then integrate a cosine product.

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

Part (a) · 2 marks

Show the sum of cos(A+B) and cos(A−B) equals 2cos A cos B.

  1. 1cos(A + B) = cos A cos B − sin A sin B
  2. 2cos(A − B) = cos A cos B + sin A sin B
  3. 3Adding: 2cos A cos B

Answercos(A + B) + cos(A − B) = 2cos A cos B shown

Part (b) · 3 marks

Deduce the sum-to-product formula for cos P + cos Q using suitable substitutions.

  1. 1Let A + B = P and A − B = Q
  2. 2A = (P + Q)/2, B = (P − Q)/2
  3. 3Substitute into (a): cos P + cos Q = 2cos((P + Q)/2)cos((P − Q)/2)

Answercos P + cos Q = 2cos((P + Q)/2)cos((P − Q)/2) shown

Part (c) · 4 marks

Find all θ in 0 to π/2 satisfying cos 5θ + cos 7θ = 0, giving answers as multiples of π.

  1. 1cos 5θ + cos 7θ = 2cos 6θ cos θ = 0
  2. 2cos 6θ = 0 ⇒ 6θ = π/2, 3π/2, 5π/2 ⇒ θ = π/12, π/4, 5π/12
  3. 3cos θ = 0 ⇒ θ = π/2

Answerθ = π/12, π/4, 5π/12, π/2

Part (d) · 3 marks

Show cos 8x + 2cos 6x + cos 4x simplifies to 4cos 6x cos²x.

  1. 1(cos 8x + cos 6x) + (cos 6x + cos 4x) = 2cos 7x cos x + 2cos 5x cos x
  2. 2= 2cos x(cos 7x + cos 5x)
  3. 3= 2cos x × 2cos 6x cos x
  4. 4= 4cos 6x cos²x

Answercos 8x + 2cos 6x + cos 4x = 4cos 6x cos²x shown

Part (e) · 4 marks

Evaluate exactly the integral from 0 to π/3 of cos 6x cos²x using calculus.

  1. 1cos 6x cos²x = (1/4)(cos 8x + 2cos 6x + cos 4x)
  2. 2∫ = (1/4)[(1/8)sin 8x + (1/3)sin 6x + (1/4)sin 4x] from 0 to π/3
  3. 3= (1/4)[(1/8)sin(8π/3) + (1/3)sin 2π + (1/4)sin(4π/3)]
  4. 4= (1/4)[√3/16 + 0 − √3/8] = −√3/64

Answer−√3/64

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