A2 June 2023 Paper 1 Q12

Notes

B1: seen at any stage

M1: (2nd) expansion of \(\left(z - \frac{1}{z}\right)^5\)

A1: (1st) correct expansion with \(z^n\), \(\frac{1}{z^n}\) terms paired and factorised

M1: (3rd) FT their expansion

A1: (2nd) i must appear in each term

A1: (3rd) www

Alternative method 1

Scheme	Marks
Considering \(\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)^5\)	M1
\(32\mathrm{i}\sin^5\theta\)	B1
\(= \mathrm{e}^{5\mathrm{i}\theta} - 5\mathrm{e}^{3\mathrm{i}\theta} + 10\mathrm{e}^{\mathrm{i}\theta} - 10\mathrm{e}^{-\mathrm{i}\theta} + 5\mathrm{e}^{-3\mathrm{i}\theta} - \mathrm{e}^{-5\mathrm{i}\theta}\)	M1
\(= \mathrm{e}^{5\mathrm{i}\theta} - \mathrm{e}^{-5\mathrm{i}\theta} - 5\left(\mathrm{e}^{3\mathrm{i}\theta} - \mathrm{e}^{-3\mathrm{i}\theta}\right) + 10\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)\)	A1
Using \(\mathrm{e}^{\mathrm{i}n\theta} - \mathrm{e}^{-\mathrm{i}n\theta} = 2\mathrm{i}\sin n\theta\) with sum of terms	M1
\(= 2\mathrm{i}\sin 5\theta - 10\mathrm{i}\sin 3\theta + 20\mathrm{i}\sin\theta\)	A1
\(\Rightarrow \sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta\)	A1

M1: (2nd) expansion of \(\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)^5\)

A1: (1st) correct expansion with \(\mathrm{e}^{\mathrm{i}\theta}\), \(\mathrm{e}^{-\mathrm{i}\theta}\) terms paired and factorised

M1: (3rd) FT their expansion (corrected from the printed mark scheme, which has \(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta} = 2\mathrm{i}\sin n\theta\); the identity is \(\mathrm{e}^{\mathrm{i}n\theta} - \mathrm{e}^{-\mathrm{i}n\theta} = 2\mathrm{i}\sin n\theta\))

A1: (2nd) i must appear in each term

A1: (3rd) www

Alternative method 2

Scheme	Marks
Equating Im components of \((\cos\theta + \mathrm{i}\sin\theta)^5\) \(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\)	B1
\(= 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta\)	M1*
\(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\)	A1
Equating Im components of \((\cos\theta + \mathrm{i}\sin\theta)^3\) \(\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta\)	M1*
\(\sin^3\theta = \dfrac{3}{4}\sin\theta - \dfrac{1}{4}\sin 3\theta\)	A1
\(\Rightarrow 16\sin^5\theta = \sin 5\theta + 20\left(\frac{3}{4}\sin\theta - \frac{1}{4}\sin 3\theta\right) - 5\sin\theta\)	M1dep
\(\Rightarrow \sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta\)	A1

B1: using binomial expansion and de Moivre’s theorem

M1*: (1st) correct expression for \(\sin 5\theta\) and substituting in \(\cos^2\theta = 1 - \sin^2\theta\)

A1: (1st) or \(16\sin^5\theta = \sin 5\theta + 20\sin^3\theta - 5\sin\theta\) or any correct rearrangement

M1*: (2nd) using binomial expansion and de Moivre’s theorem; correct expression for \(\sin 3\theta\) in terms of \(\sin\theta\) and \(\cos\theta\)

A1: (2nd) or \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\) or any correct rearrangement

M1dep: substituting expression for \(\sin^3\theta\) into \(\sin^5\theta\).

A1: (3rd) www