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A-LevelMathematicsTrigonometry9709/21 · Oct/Nov 2024 · Q711 Marks

Prove that cos(θ+30°)cos(θ+60°) = (1/4)√3 - (1/2)sin 2θ.

  1. (a)Prove that cos(θ+30°)cos(θ+60°) = (1/4)√3 - (1/2)sin 2θ.[4]
  2. (b)Solve the equation 5 cos(2α+30°)cos(2α+60°) = 1 for 0° < α < 90°.[4]
  3. (c)Show that the exact value of cos 20° cos 50° + cos 40° cos 70° is (1/2)√3.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

  1. [B1] State (cosθ cos 30 - sinθ sin 30)(cosθ cos 60 - sinθ sin 60)
  2. [M1] Expand and use correct exact values
  3. [A1] Obtain (1/4)√3(cos²θ+sin²θ) - sinθ cosθ or similarly simplified equivalent
  4. [A1] Conclude (1/4)√3 - (1/2)sin 2θ AG – necessary detail needed.
  5. [*M1] Use identity to obtain value for sin 4α
  6. [A1] Obtain sin 4α = (5√3-4)/10 or 0.466...
  7. [DM1] Show correct process to obtain one value of α
  8. [A1] Obtain 6.9 and 38.1
  9. [B1] Substitute θ = -10 to obtain cos 20cos 50 = (1/4)√3 - (1/2)sin(-20)
  10. [B1] Substitute θ = 10 to obtain cos 40 cos 70 = (1/4)√3 - (1/2)sin(20)
  11. [B1] Add and confirm (1/2)√3 with clear indication that sin(-20) = -sin 20

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 11 marks.
Source
A-Level Mathematics, Oct/Nov 2024 Paper 2 Variant 1, question 7.

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