Let z = cos θ + i sin θ. Show that zⁿ + 1/zⁿ = 2 cos nθ and zⁿ - 1/zⁿ = 2i sin nθ. By considering (z - 1/z)⁴ (z + 1/z)², show that sin⁴ θ cos² θ = 1/32 (cos 6θ – 2 cos 4θ – cos 2θ + 2). Hence find the exact value of ∫₀^(π/2) sin⁴ θ cos² θ dθ.

Mark scheme

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

1. [M1A1] zⁿ = cos nθ + i sin nθ (by De Moivre's Theorem). z⁻ⁿ = cos(-nθ) + i sin(-nθ) = cos nθ - i sin nθ. Adding: zⁿ + 1/zⁿ = 2 cos nθ. Subtracting: zⁿ - 1/zⁿ = 2i sin nθ. (AG)
2. [M1] Consider (z - 1/z)⁴ (z + 1/z)². Using the identities, this is (2i sin θ)⁴ (2 cos θ)².
3. [M1] (2i sin θ)⁴ = 16i⁴ sin⁴ θ = 16 sin⁴ θ. (2 cos θ)² = 4 cos² θ. Product is 64 sin⁴ θ cos² θ.
4. [M1] Expand (z - 1/z)⁴ = z⁴ - 4z² + 6 - 4/z² + 1/z⁴. (Or use (z-1/z)² = z²-2+1/z² then square it)
5. [M1] Expand (z + 1/z)² = z² + 2 + 1/z².
6. [M1A1] Multiply the expanded terms: (z⁴ - 4z² + 6 - 4/z² + 1/z⁴)(z² + 2 + 1/z²) = (z⁶ + 1/z⁶) - 2(z⁴ + 1/z⁴) - (z² + 1/z²) + 4.
7. [M1A1] Substitute back using zⁿ + 1/zⁿ = 2 cos nθ: = 2 cos 6θ - 2(2 cos 4θ) - (2 cos 2θ) + 4 = 2 cos 6θ - 4 cos 4θ - 2 cos 2θ + 4. Equating: 64 sin⁴ θ cos² θ = 2 cos 6θ - 4 cos 4θ - 2 cos 2θ + 4. [A1] Divide by 64: sin⁴ θ cos² θ = 1/32 (cos 6θ – 2 cos 4θ – cos 2θ + 2). (AG)
8. ALT METHOD for part 2: [M1, A1, A1] Expand and group each bracket separately and sub trig. [M1, M1] Multiply together two brackets and manipulate trig expressions. [M1A1] LHS = RHS.
9. [M1A1] ∫₀^(π/2) sin⁴ θ cos² θ dθ = ∫₀^(π/2) 1/32 (cos 6θ - 2 cos 4θ - cos 2θ + 2) dθ. Integrate term by term: [1/6 sin 6θ - 1/2 sin 4θ - 1/2 sin 2θ + 2θ].
10. [A1] Evaluate from 0 to π/2: 1/32 [1/6 sin 3π - 1/2 sin 2π - 1/2 sin π + 2(π/2)] - [0] = 1/32 (0 - 0 - 0 + π) = π/32.