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A-LevelMathematicsTrigonometry9709/23 · Oct/Nov 2024 · Q711 Marks

Prove that cos(θ+30°)cos(θ+60°) = ¼√3 - ½sin 2θ.

  1. (a)Prove that cos(θ+30°)cos(θ+60°) = ¼√3 - ½sin 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 ½√3.[3]

Mark scheme

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

  1. [4] State (cosθcos 30 - sinθsin 30)(cosθcos 60 – sinθsin 60) B1
  2. Expand and use correct exact values M1
  3. Obtain ¼√3(cos²θ+sin²θ) - ½sinθcosθ or similarly simplified equivalent A1
  4. Conclude ¼√3 - ½sin 2θ A1 AG – necessary detail needed.
  5. [4] Use identity to obtain value for sin4α *M1
  6. Obtain sin4α=¼√3-3 or 0.466... A1
  7. Show correct process to obtain one value of α DM1
  8. Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between 0° and 90°.
  9. [3] Substitute θ = -10 to obtain cos 20cos50=¼√3-½sin(-20) B1
  10. Substitute θ = 10 to obtain cos 40cos70=¼√3-½sin 20 B1
  11. Add and confirm ½√3 with clear indication that sin(-20) = -sin 20 B1 AG – necessary detail needed.
  12. Alternative solution for Question 7(c)
  13. Rewrite as sin 70cos 50+ sin 50cos 70 or cos 20sin 40+ cos 40sin 20 or sin 70 sin 40+ cos 70 cos 40 B1
  14. Obtain sin120 or sin 60 or cos30 B1
  15. Confirm ½√3 B1 AG – necessary detail needed.

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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 3, question 7.

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