Use de Moivre's theorem to show that tan 5θ = (5t - 10t³ + t⁵) / (1 - 10t² + 5t⁴), where t = tan θ. Deduce that the roots of the equation t⁴ – 10t² + 5 = 0 are ±tan(π/5) and ±tan(2π/5). Hence show that tan(π/5) tan(2π/5) = √5.

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

1. [M1] (cos θ + i sin θ)⁵ = cos 5θ + i sin 5θ
2. [A1] = c⁵ + 5c⁴(is) + 10c³(is)² + 10c²(is)³ + 5c(is)⁴ + (is)⁵
3. [M1A1] tan 5θ = (5c⁴s - 10c²s³ + s⁵) / (c⁵ - 10c³s² + 5cs⁴) = ... = (5t - 10t³ + t⁵) / (1 - 10t² + 5t⁴) (AG)
4. [M1] tan 5θ = 0 ⇒ t(t⁴ – 10t² + 5) = 0 (Allow non-factorised form.)
5. [A1] Rejecting t = 0, which implies θ = kπ, roots of t⁴ – 10t² + 5 = 0 are: ±tan(π/5) and ±tan(2π/5), since 5θ = kπ and k = ±1, ±2.
6. [A1] (Deduction of roots and conditions)
7. [M1A1] Product of roots of u²-10u+5=0 where u=t² gives tan²(π/5) tan²(2π/5) = 5. Hence tan(π/5) tan(2π/5) = √5. (S.C. Award B1 if result deduced from information given in the question.)