A2 June 2024 Paper 1 Q13

AQACurrent spec9 marksDe Moivre's Theorem

(a) Use de Moivre’s theorem to show that\[\cos 3\theta = 4\cos^3\theta - 3\cos\theta\] [3 marks]

(b) Use de Moivre’s theorem to express \(\sin 3\theta\) in terms of \(\sin\theta\) [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Expands \((\cos\theta + \mathrm{i}\sin\theta)^3\) PI correct real part	M1	1.1a
Equates real parts and uses \(\sin^2\theta = 1 - \cos^2\theta\)	M1	1.1a
Completes a reasoned argument using de Moivre’s theorem to show \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\) AG	R1	2.1
	(3)

Typical solution

\[\begin{aligned}\cos 3\theta + \mathrm{i}\sin 3\theta &= (\cos\theta + \mathrm{i}\sin\theta)^3 \\ &= \cos^3\theta + 3\mathrm{i}\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - \mathrm{i}\sin^3\theta\end{aligned}\]

Equating real parts

\[\begin{aligned}\cos 3\theta &= \cos^3\theta - 3\cos\theta\sin^2\theta \\ &= \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) \\ &= 4\cos^3\theta - 3\cos\theta\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Equates imaginary parts and uses \(\cos^2\theta = 1 - \sin^2\theta\)	M1	1.1a
Obtains \(3\sin\theta - 4\sin^3\theta\)	A1	1.1b
	(2)

Typical solution

Equating imaginary parts

\[\begin{aligned}\sin 3\theta &= 3\cos^2\theta\sin\theta - \sin^3\theta \\ &= 3(1 - \sin^2\theta)\sin\theta - \sin^3\theta \\ &= 3\sin\theta - 4\sin^3\theta\end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Substitutes expressions for \(\cos 3\theta\) and their \(\sin 3\theta\) into \(\cot 3\theta = \dfrac{\cos 3\theta}{\sin 3\theta}\) or \(\tan 3\theta = \dfrac{\sin 3\theta}{\cos 3\theta}\)	B1F	1.1b
Manipulates their rational function of \(\sin\theta\) and \(\cos\theta\) to obtain at least one instance of \(\cot\theta\) or \(\tan\theta\)	M1	3.1a
Manipulates their rational function of \(\sin\theta\) and \(\cos\theta\) to obtain only \(\cot\theta\) (and \(\operatorname{cosec}\theta\)) terms	M1	3.1a
Completes a reasoned argument from the final or intermediate results in parts (a) and (b) to show \(\cot 3\theta = \dfrac{\cot^3\theta - 3\cot\theta}{3\cot^2\theta - 1}\) AG	R1	2.1
	(4)
	(9 marks)

Typical solution

\[\begin{aligned}\cot 3\theta &= \frac{\cos 3\theta}{\sin 3\theta} = \frac{4\cos^3\theta - 3\cos\theta}{3\sin\theta - 4\sin^3\theta} \\ &= \frac{4\left(\dfrac{\cos^3\theta}{\sin^3\theta}\right) - 3\left(\dfrac{\cos\theta}{\sin^3\theta}\right)}{3\left(\dfrac{\sin\theta}{\sin^3\theta}\right) - 4\left(\dfrac{\sin^3\theta}{\sin^3\theta}\right)} \\ &= \frac{4\cot^3\theta - 3\cot\theta\operatorname{cosec}^2\theta}{3\operatorname{cosec}^2\theta - 4} \\ &= \frac{4\cot^3\theta - 3\cot\theta(1 + \cot^2\theta)}{3(1 + \cot^2\theta) - 4}\end{aligned}\]\[\cot 3\theta = \frac{\cot^3\theta - 3\cot\theta}{3\cot^2\theta - 1}\]