A2 October 2020 Paper 1 Q5
5 By expanding \(\left(z^2 + \dfrac{1}{z^2}\right)^3\), where \(z = \mathrm{e}^{\mathrm{i}\theta}\), show that \(4\cos^3 2\theta = \cos 6\theta + 3\cos 2\theta\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\left(z^2 + \dfrac{1}{z^2}\right)^3 = z^6 + 3z^2 + \dfrac{3}{z^2} + \dfrac{1}{z^6}\) | M1 A1 | 1.1 1.1 |
| \(z^2 + \dfrac{1}{z^2} = 2\cos 2\theta \Rightarrow \left(z^2 + \dfrac{1}{z^2}\right)^3 = 8\cos^3 2\theta\) and \(z^2 + \dfrac{1}{z^2} = 2\cos 2\theta\) and \(z^6 + \dfrac{1}{z^6} = 2\cos 6\theta\) and \(\left(z^2 + \dfrac{1}{z^2}\right)^3 = \left(z^6 + \dfrac{1}{z^6}\right) + 3\left(z^2 + \dfrac{1}{z^2}\right)\) | M1 | 3.1a |
| \(\Rightarrow 8\cos^3 2\theta = 2\cos 6\theta + 6\cos 2\theta\) | A1 | 2.1 |
| \(\Rightarrow 4\cos^3 2\theta = \cos 6\theta + 3\cos 2\theta\) AG | A1 | 1.1 |
| [5] |
Notes
M1: Use of binomial
M1: use of De Moivre for any one term