Use de Moivre's theorem to express cot 7θ in terms of cot θ. Use the equation cot 7θ = 0 to show that the roots of the equation x⁶ − 21x⁴ + 35x² − 7 = 0 are cot²((kπ)/14) for k = 1, 3, 5, 9, 11, 13, and deduce that cot²(π/14) cot²(3π/14) cot²(5π/14) = 7.

Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

1. [M1] (c+is)⁷ = c⁷ +7c⁶ (is)+...+(is)⁷
2. [A1] cos 7θ = c⁷-21c⁵s² +35c³s⁴ -7cs⁶
3. [M1] sin 7θ = 7c⁶s-35c⁴s³ + 21c²s⁵ - s⁷
4. [A1] cot 7θ = (cot⁷θ - 21cot⁵θ + 35cot³θ - 7cotθ) / (7cot⁶θ - 35cot⁴θ + 21cot²θ - 1)
5. [M1A1] cot7θ=0 and cotθ ≠ 0⇒ x⁶ −21x⁴ + 35x² − 7 = 0,where x = cot θ and θ = kπ/14 where k = 1, 3, 5, 9, 11, 13
6. [M1A1] Product of roots cot(π/14) cot(3π/14) cot(5π/14) cot(9π/14) cot(11π/14) cot(13π/14) = -7
7. [M1] But cot(13π/14) = -cot(π/14), cot(11π/14) = -cot(3π/14), cot(9π/14) = -cot(5π/14)
8. [A1] Hence cot²(π/14) cot²(3π/14) cot²(5π/14) = 7. (AG)(Penultimate line must be seen.)
9. SC Award B1 if product of roots mentioned, without proper pairing seen.