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A-LevelMathematicsTrigonometry9709/22 · Feb/Mar 2022 · Q47 Marks

Show that sin 2θ cot θ – cos 2θ = 1.

  1. (a)Show that sin 2θ cot θ – cos 2θ = 1.[3]
  2. (b)Hence find the exact value of sin 1/6π cot 1/6π.[2]
  3. (c)Find the smallest positive value of θ (in radians) satisfying the equation sin 2θ cot θ – 3 cos 2θ = 1.[2]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

  1. [B1] Use at least two of sin 2θ = 2 sin θ cos θ, cos 2θ = cos² θ – sin² θ, cot θ = cos θ/sin θ [OE]
  2. [M1] Express LHS in terms of sin θ and cos θ only and attempt valid simplification
  3. [A1] Obtain cos² θ + sin² θ or equivalent and hence 1 [AG – necessary detail needed]
  4. [M1] Substitute θ = π/6 and show or imply sin(π/3)cot(π/6) = 1 + cos(π/3)
  5. [A1] Obtain 1 + √3 or exact equivalent
  6. [M1] Use the identity from part (a) to obtain -2 cos 2θ = 0 or equivalent [Or alternative starting again, using valid simplification and reaching single trigonometric function.]
  7. [A1] Obtain θ = π/4

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 7 marks.
Source
A-Level Mathematics, Feb/Mar 2022 Paper 2 Variant 2, question 4.

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