Answer only one of the following two alternatives. EITHER

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

1. [B1] Uses de Moivre's theorem and binomial theorem. (cos θ + i sin θ)³ = cos 3θ + i sin 3θ and cos³ θ + 3i cos² θ sin θ – 3 cos θ sin² θ – i sin³ θ
2. [M1A1M1M1A0] Award B0M1A1M1M1A0 i.e.4/6 (If line 1 missing, or no reference made to cos3θ and sin3θ being real and imaginary parts.)
3. [M1] Equates real and imaginary parts. sin 3θ = 3 cos² θ sin θ – sin³ θ cos 3θ = cos³ θ – 3 cos θ sin² θ
4. [M1] Uses tan A = (sin A)/(cos A) . tan 3θ = (3 cos² θ sin θ – sin³ θ)/(cos³ θ – 3 cos θ sin² θ)
5. [A1] ∴ tan 3θ = (3 tan θ - tan³ θ)/(1 - 3 tan² θ) (*) (AG)
6. [B1] States roots of tan 3θ = 1 between 0 and π . 3θ = π/4, 5π/4, 9π/4
7. [B1] (all) θ = π/12, 5π/12, 3π/4
8. [M1] Put tan 3θ = 1 in (*) ⇒ t³ - 3t² - 3t + 1 = 0
9. [A1] (one) Roots are tan π/12, tan 5π/12, tan 3π/4
10. [A1] (all)
11. [M1] Obtains cubic equation and solves. (t+1)(t² - 4t + 1) = 0 ⇒ t = -1 , 2±√3
12. [A1] (one) Evaluates each root. tan 3π/4 = -1 tan π/12 = 2-√3
13. [A1] (all) tan 5π/12 = 2+√3