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A-LevelMathematicsTrigonometry9709/32 · May/June 2019 · Q35 Marks

Showing all necessary working, solve the equation cot 2θ = 2 tan θ for 0° < θ < 180°.

  1. Showing all necessary working, solve the equation cot 2θ = 2 tan θ for 0° < θ < 180°.[5]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [*M1] Use correct trig formulae to obtain an equation in tan θ or equivalent (e.g. all in sin θ or all in cos θ) (1-tan²θ)/(2tanθ) = 2tanθ. Allow (cot²θ-1)/(2cotθ) = 2/cotθ
  2. [A1] Obtain a correct simplified equation (5tan²θ = 1 or sin²θ = 1/6 or cos²θ = 5/6)
  3. [DM1] Solve for θ (Dependent on the first M1)
  4. [A1] Obtain answer 24.1° (or 155.9°) (One correct in range to at least 3 sf)
  5. [A1] Obtain second answer (FT 180° – their 24.1° and no others in range. Correct to at least 3 sf. Accept 156° but not 156.0. Ignore values outside range. If working in tanθ or cosθ need to be considering both square roots to score the second A1. Mark 0.421, 2.72 as a MR, so A0A1)

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source
A-Level Mathematics, May/June 2019 Paper 3 Variant 2, question 3.

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