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Date	November Example questions	Marks available	3	Reference code	EXN.1.AHL.TZ0.12
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Expand	Question number	12	Adapted from	N/A

Question

Use the binomial theorem to expand ${(\cos θ + i \sin θ)}^{4}$ . Give your answer in the form $a + b i$ where $a$ and $b$ are expressed in terms of $\sin θ$ and $\cos θ$ .

[3]

Use de Moivre’s theorem and the result from part (a) to show that $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ .

[5]

Use the identity from part (b) to show that the quadratic equation $x^{2} - 6 x + 1 = 0$ has roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ .

[5]

Hence find the exact value of ${cot}^{2} \frac{3 π}{8}$ .

[4]

Deduce a quadratic equation with integer coefficients, having roots ${cosec}^{2} \frac{π}{8}$ and ${cosec}^{2} \frac{3 π}{8}$ .

[3]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses the binomial theorem on ${(\cos θ + i \sin θ)}^{4}$ M1

$= {}_{4}C_{0} \cos^{4} θ + {}_{4}C_{1} \cos^{3} θ (i \sin θ) + {}_{4}C_{2} \cos^{2} θ (i^{2} \sin^{2} θ) + {}_{4}C_{3} \cos θ (i^{3} \sin^{3} θ) + {}_{4}C_{4} (i^{4} \sin^{4} θ)$ A1

$= (\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ A1

[3 marks]

(using de Moivre’s theorem with $n = 4$ gives) $\cos 4 θ + i \sin 4 θ$ (A1)

equates both the real and imaginary parts of $\cos 4 θ + i \sin 4 θ$ and $(\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ M1

$\cos 4 θ = \cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ$ and $\sin 4 θ = 4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ$

recognizes that $\cot 4 θ = \frac{\cos 4 θ}{\sin 4 θ}$ (A1)

substitutes for $\sin 4 θ$ and $\cos 4 θ$ into $\frac{\cos 4 θ}{\sin 4 θ}$ M1

$\cot 4 θ = \frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}$

divides the numerator and denominator by $\sin^{4} θ$ to obtain

$\cot 4 θ = \frac{\frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{\sin^{4} θ}}{\frac{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}{\sin^{4} θ}}$ A1

$\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ AG

[5 marks]

setting $\cot 4 θ = 0$ and putting $x = \cot^{2} θ$ in the numerator of $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ gives $x^{2} - 6 x + 1 = 0$ M1

attempts to solve $\cot 4 θ = 0$ for $θ$ M1

$4 θ = \frac{π}{2}, \frac{3 π}{2}, \dots (4 θ = \frac{1}{2} (2 n + 1) π, n = 0, 1, \dots)$ (A1)

$θ = \frac{π}{8}, \frac{3 π}{8}$ A1

Note: Do not award the final A1 if solutions other than $θ = \frac{π}{8}, \frac{3 π}{8}$ are listed.

finding the roots of $\cot 4 θ = 0 (θ = \frac{π}{8}, \frac{3 π}{8})$ corresponds to finding the roots of $x^{2} - 6 x + 1 = 0$ where $x = \cot^{2} θ$ R1