A-LevelFurther MathematicsComplex numbers9231/12 · May/June 2015 · Q811 Marks

By considering ∑_{r=1}^{n} z²ʳ⁻¹, where z = cos θ + i sin θ, show that, if sin θ ≠ 0, ∑_{r=1}^{n} sin (2r − 1)θ = sin² nθ / sin θ Deduce that ∑_{r=1}^{n} (2r − 1) cos [(2r − 1)π / 2n] = − cosec (π/2n) cot (π/2n).

Mark scheme

Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.

[M1A1] ∑z²ʳ⁻¹ = z(1 - [z²ⁿ]) / (1 - z²) = (z - z²ⁿ⁺¹) / (1 - z²)
[M1A1] = 1 - (cos 2nθ + i sin 2nθ) / (1 - (cos 2θ + i sin 2θ))
[A1] = 1 - cos 2nθ - i sin 2nθ / (-2i sin θ)
[M1A1] Equating imaginary parts: ∑sin (2r – 1)θ = 2 sin² nθ / 2 sin θ = sin² nθ / sin θ (AG)
[M1A1] Differentiating: ∑(2r – 1) cos (2r−1)θ = 2n sin nθ cos nθ cosec θ – sin² nθ cosec θ cot θ
[dM1] Putting θ = π/2n
[A1A1] ∑(2r – 1) cos [(2r−1)π / 2n] = − cosec (π/2n) cot (π/2n). (AG)

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Topic: Complex numbers, A-Level Further Mathematics (syllabus 9231). Structured question, 11 marks.
Source: A-Level Further Mathematics, May/June 2015 Paper 1 Variant 2, question 8.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content