Markscheme:

Part (a):
Using binomial theorem on \( (\cos \theta + i \sin \theta)^4 \): (M1)

\( = \binom{4}{0}\cos^4\theta + \binom{4}{1}\cos^3\theta(i\sin\theta) + \binom{4}{2}\cos^2\theta(i\sin\theta)^2 + \binom{4}{3}\cos\theta(i\sin\theta)^3 + \binom{4}{4}(i\sin\theta)^4 \) (A1)

\( = (\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta) + i(4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta) \) (A1)

[3 marks]

Part (b):
By de Moivre's theorem: \( \cos 4\theta + i\sin 4\theta \) (A1)

Equating real and imaginary parts: (M1)
\( \cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta \)
\( \sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta \)

\( \cot 4\theta = \frac{\cos 4\theta}{\sin 4\theta} \) (A1)

Dividing numerator and denominator by \( \sin^4\theta \): (M1)

\( \cot 4\theta = \frac{\cot^4\theta - 6\cot^2\theta + 1}{4\cot^3\theta - 4\cot\theta} \) (A1)

[5 marks]

Part (c):
Setting \( \cot 4\theta = 0 \) and letting \( x = \cot^2\theta \): (M1)
Numerator: \( x^2 - 6x + 1 = 0 \) (M1)

\( \cot 4\theta = 0 \) when \( 4\theta = \frac{\pi}{2}, \frac{3\pi}{2}, ... \) (A1)
\( \theta = \frac{\pi}{8}, \frac{3\pi}{8} \) (A1)

Therefore roots are \( \cot^2\frac{\pi}{8} \) and \( \cot^2\frac{3\pi}{8} \). (R1)

[5 marks]

Part (d):
Solving \( x^2 - 6x + 1 = 0 \): (M1)
\( x = 3 \pm 2\sqrt{2} \) (A1)

Since \( \cot^2\frac{\pi}{8} > \cot^2\frac{3\pi}{8} \): (R1)
\( \cot^2\frac{3\pi}{8} = 3 - 2\sqrt{2} \) (A1)

[4 marks]

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