Using de Moivre's theorem, show that tan 5θ = (5 tan θ - 10 tan³ θ + tan⁵ θ) / (1 - 10 tan² θ + 5 tan⁴ θ). Hence show that the equation x² – 10x + 5 = 0 has roots tan²(π/5) and tan²(2π/5). Deduce a quadratic equation, with integer coefficients, having roots sec²(π/5) and sec²(2π/5).

Mark scheme

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

1. [B1] (a) (cosθ + isinθ)⁵ = cos5θ + isin5θ
2. [M1A1] (a) (c+is)⁵ = c⁵ + 5c⁴si –10c³s² –10ic²s³ + 5cs⁴ + is⁵
3. [M1] (a) tan 5θ = (Im(cos5θ + isin5θ)) / (Re(cos5θ + isin5θ)) Divide numerator and denominator by c⁵
4. [A1] (a) = (5 tan θ - 10 tan³ θ + tan⁵ θ) / (1 - 10 tan² θ + 5 tan⁴ θ) (AG)
5. [B1] (b) tan 5θ = 0 ⇒ 5θ = nπ ⇒ θ = nπ/5 for n = 0, 1, 2, 3, 4
6. [B1] (b) t⁵ − 10t³ + 5t = 0 has roots tan(nπ/5) and factorise t(t⁴ - 10t² + 5) = 0
7. [M1] (b) Identify x = t² and form x² - 10x + 5 = 0
8. [A1] (b) Explain why tan²(3π/5) = tan²(2π/5) and tan²(4π/5) = tan²(π/5) (AG)
9. [M1] (c) sec² α = 1 + tan² α
10. [M1] (c) y = 1 + x ⇒ x = y - 1 ⇒ (y − 1)² − 10(y − 1) + 5 = 0
11. [A1] (c) ⇒ y² − 12y + 16 = 0