(a) (i) Expand \( (\cos \theta + \text{i} \sin \theta)^5 \) using the binomial theorem:

\( (\cos \theta + \text{i} \sin \theta)^5 = \sum_{k=0}^5 \binom{5}{k} (\cos \theta)^{5-k} (\text{i} \sin \theta)^k \)

\( = \cos^5 \theta + 5 \cos^4 \theta (\text{i} \sin \theta) + 10 \cos^3 \theta (\text{i} \sin \theta)^2 + 10 \cos^2 \theta (\text{i} \sin \theta)^3 + 5 \cos \theta (\text{i} \sin \theta)^4 + (\text{i} \sin \theta)^5 \). A1 (for correct binomial coefficients)

\( = \cos^5 \theta + 5 \text{i} \cos^4 \theta \sin \theta – 10 \cos^3 \theta \sin^2 \theta – 10 \text{i} \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + \text{i} \sin^5 \theta \). A1

[2 marks]

(a) (ii) By De Moivre’s theorem: \( (\cos \theta + \text{i} \sin \theta)^5 = \cos 5\theta + \text{i} \sin 5\theta \). M1

Equate imaginary parts of the expansion from (i):

\( \sin 5\theta = 5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta \). M1 A1

[2 marks]

(a) (iii) Equate real parts from (i):

\( \cos 5\theta = \cos^5 \theta – 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta \). A1

[2 marks]

(b) Solve \( z^5 – 1 = 0 \), where \( z = r (\cos \alpha + \text{i} \sin \alpha) \):

\( z^5 = 1 \Rightarrow (r \text{cis} \alpha)^5 = 1 \text{cis} 0 \). M1

\( r^5 = 1 \Rightarrow r = 1 \). A1

\( \text{cis} 5\alpha = \text{cis} 0 \Rightarrow 5\alpha = 0 + 360k, k \in \mathbb{Z} \Rightarrow \alpha = 72k \). M1

Smallest positive \(\alpha\): \( \alpha = 72^\circ \) (for \( k = 1 \)). A1

Note: Award M1A0 if final answer is in radians.

[4 marks]

(c) Use \( \sin 5\alpha = 0 \) (since \( 5 \cdot 72^\circ = 360^\circ \)) and the result from (a)(ii): M1

\( 0 = 5 \cos^4 \alpha \sin \alpha – 10 \cos^2 \alpha \sin^3 \alpha + \sin^5 \alpha \). A1

Since \( \sin \alpha \neq 0 \), divide by \( \sin \alpha \):

\( 0 = 5 \cos^4 \alpha – 10 \cos^2 \alpha \sin^2 \alpha + \sin^4 \alpha \). M1

Substitute \( \cos^2 \alpha = 1 – \sin^2 \alpha \):

\( 0 = 5 (1 – \sin^2 \alpha)^2 – 10 (1 – \sin^2 \alpha) \sin^2 \alpha + \sin^4 \alpha \). A1

Expand and simplify: \( 0 = 5 (1 – 2 \sin^2 \alpha + \sin^4 \alpha) – 10 \sin^2 \alpha + 10 \sin^4 \alpha + \sin^4 \alpha \).

\( 0 = 5 – 10 \sin^2 \alpha + 5 \sin^4 \alpha – 10 \sin^2 \alpha + 10 \sin^4 \alpha + \sin^4 \alpha \).

\( 0 = 16 \sin^4 \alpha – 20 \sin^2 \alpha + 5 \). AG

[4 marks]

(d) Solve the quadratic in \( u = \sin^2 \alpha \): \( 16u^2 – 20u + 5 = 0 \). M1

\( u = \frac{20 \pm \sqrt{400 – 320}}{32} = \frac{20 \pm \sqrt{80}}{32} \). A1

\( \sin \alpha = \pm \sqrt{\frac{20 \pm \sqrt{80}}{32}} = \pm \sqrt{\frac{10 \pm 2 \sqrt{5}}{16}} = \pm \frac{\sqrt{10 \pm 2 \sqrt{5}}}{4} \). A1

Since \( \alpha = 72^\circ > 60^\circ \), \( \sin 72^\circ > \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \), choose positive signs: R1

\( \sin 72^\circ = \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \). A1

Note: Award A1 regardless of signs. Accept equivalent forms with integral denominator.

[5 marks]

Total [19 marks]