Use de Moivre's theorem to show that cos 4θ = 8 cos⁴ θ – 8 cos² θ + 1.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.

1. [M1] Proves initial result: (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ = c⁴ + 4c³(is) + 6c²(is)² + 4c(is)³ + (is)⁴
2. [M1] cos 4θ = c⁴ – 6c²(1 – c²) + (1 – c²)² = 8 cos⁴ θ – 8 cos² θ + 1 (AG)
3. [A1] cos 4θ = 8 cos⁴ θ – 8 cos² θ + 1 (AG)
4. [B1] Verifies any two cases (MS solution uses θ = 4π/7, 12π/7, 6π/7, 20π/7 values which differ from QP values, but shows identity cos 4θ = -cos 3θ for those. Example: cos(4*4π/7) = -cos(3*4π/7))
5. [B1] Verifies remaining two cases (MS solution continues with its own set of angles, e.g. cos(4*(12π/7)) = -cos(3*(12π/7)); cos(4*4π) = 1 = -(-1) = -cos 3π)
6. [M1] Shows roots of equation to be as given: 8cos⁴θ – 8 cos²θ + 1 = – (4cos³θ – 3 cosθ) ⇒ 8c⁴ + 4c³ – 8c² – 3c + 1 = 0 (*)
7. [A1] ⇒ cos ⅛π, cos ⅜π, cos ⅝π, –1 are the roots. (AG)
8. [M1A1] States sum of roots of equation: Sum of roots of (*) are -4/8 = -1/2. Result (AG)