A2 June 2022 Paper 2 Q9
9 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) oe | B1 | 2.1 |
| \(\mathrm{e}^{4\mathrm{i}\theta} = \cos 4\theta + \mathrm{i}\sin 4\theta\) | M1 | 1.1 |
| \(\mathrm{Re}\left(\mathrm{e}^{4\mathrm{i}\theta}\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right)^4\right)\) \(= \mathrm{Re}\left((\cos 4\theta + \mathrm{i}\sin 4\theta)(2\cos\theta)^4\right)\) \(= \cos 4\theta \times (2\cos\theta)^4\) \(= 16\cos 4\theta\cos^4\theta\) (so \(a = 16\)) | A1 | 2.2a |
| [3] |
Notes
M1: Use of DMT
A1: Must equate terms to complete demonstration
| Scheme | Marks | AO |
|---|---|---|
| DR \(\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right)^4 = \left(\mathrm{e}^{\mathrm{i}\theta}\right)^4 + 4\left(\mathrm{e}^{\mathrm{i}\theta}\right)^3\mathrm{e}^{-\mathrm{i}\theta}\) \(+ 6\left(\mathrm{e}^{\mathrm{i}\theta}\right)^2\left(\mathrm{e}^{-\mathrm{i}\theta}\right)^2 + 4\mathrm{e}^{\mathrm{i}\theta}\left(\mathrm{e}^{-\mathrm{i}\theta}\right)^3 + \left(\mathrm{e}^{-\mathrm{i}\theta}\right)^4\) | *M1 | 3.1a |
| \(\therefore \mathrm{e}^{4\mathrm{i}\theta}\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right)^4\) \(= \mathrm{e}^{8\mathrm{i}\theta} + 4\mathrm{e}^{6\mathrm{i}\theta} + 6\mathrm{e}^{4\mathrm{i}\theta} + 4\mathrm{e}^{2\mathrm{i}\theta} + 1\) \(\mathrm{e}^{8\mathrm{i}\theta} = \cos 8\theta + \mathrm{i}\sin 8\theta\) or \(\mathrm{Re}\left(\mathrm{e}^{8\mathrm{i}\theta}\right) = \cos 8\theta\) etc | dep*M1 | 1.1 |
| \(\therefore 16\cos 4\theta\cos^4\theta =\) \(\cos 8\theta + 4\cos 6\theta + 6\cos 4\theta + 4\cos 2\theta + 1\) | dep*M1 | 1.1 |
| \(\displaystyle \begin{aligned} &\theta = \frac{\pi}{12} \Rightarrow 16\cos\frac{\pi}{3}\cos^4\frac{\pi}{12} = \\[6pt] &\cos\frac{2\pi}{3} + 4\cos\frac{\pi}{2} + 6\cos\frac{\pi}{3} + 4\cos\frac{\pi}{6} + 1 \end{aligned}\) | dep*M1 | 3.1a |
| \(\displaystyle \begin{aligned} &\Rightarrow \cos\frac{\pi}{12} = \\[6pt] &\sqrt[4]{\dfrac{1}{8}\left(-\dfrac{1}{2} + 4 \times 0 + 6 \times \dfrac{1}{2} + 4 \times \dfrac{\sqrt{3}}{2} + 1\right)} \end{aligned}\) | dep*M1 | 1.1 |
| \(= \sqrt[4]{\dfrac{1}{8}\left(\dfrac{7}{2} + 2\sqrt{3}\right)} = \sqrt[4]{\dfrac{1}{16}\left(7 + 4\sqrt{3}\right)}\) \(= \dfrac{1}{\sqrt[4]{16}}\sqrt[4]{7 + 4\sqrt{3}} = \dfrac{1}{2}\sqrt[4]{7 + 4\sqrt{3}}\) | A1 | 2.2a |
| [6] |
Notes
*M1: Expands correct brackets using binomial theorem. Terms can be unsimplified but must have correct numerical coefficients.
eg \(\left(\mathrm{e}^{2\mathrm{i}\theta} + 1\right)^4\) or \(\left(z + z^{-1}\right)^4\) or \(\left(\dfrac{\sqrt{3}}{2} + 1 + \dfrac{1}{2}\mathrm{i}\right)^4\) etc
if expansion seen in 9(a), must be used in 9(b) to gain mark here
dep*M1: Use of Euler’s formula to convert exponential form to trigonometric form
dep*M1: Taking real parts
dep*M1: Choice of \(\theta\) soi and substituted into identity with their 16.
dep*M1: Gives correct numerical values to all \(\cos\dfrac{n\pi}{6}\) terms. Also dependent on use of Euler’s formula and choice of \(\theta\).
A1: So \(b = 7\) and \(c = 4\) (can be embedded). cao.