A2 October 2020 Paper 1 Q12
12
(a) Given that \(z = \cos\theta + \mathrm{i}\sin\theta\), express \(z^n + \dfrac{1}{z^n}\) and \(z^n - \dfrac{1}{z^n}\) in simplified trigonometric form. [2]
(b) By considering \(\left(z + \dfrac{1}{z}\right)^3\left(z - \dfrac{1}{z}\right)^3\), find constants \(A\) and \(B\) such that\[\sin^3\theta\cos^3\theta = A\sin 6\theta + B\sin 2\theta.\] [6]
| Scheme | Marks | AO |
|---|---|---|
| \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) | B1 | 1.1 |
| \(z^n - \dfrac{1}{z^n} = 2\mathrm{i}\sin n\theta\) | B1 | 1.1 |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(\left(z + \dfrac{1}{z}\right)^3\left(z - \dfrac{1}{z}\right)^3 = -64\mathrm{i}\cos^3\theta\sin^3\theta\) | B1 | 2.1 |
| \(\left(z + \dfrac{1}{z}\right)^3\left(z - \dfrac{1}{z}\right)^3 = \left(z^3 + 3z + \dfrac{3}{z} + \dfrac{1}{z^3}\right)\left(z^3 - 3z + \dfrac{3}{z} - \dfrac{1}{z^3}\right)\) | M1 M1 | 2.1 1.1 |
| \(= z^6 - 3z^2 + \dfrac{3}{z^2} - \dfrac{1}{z^6}\) | A1 | 1.1 |
| \(= 2\mathrm{i}\sin 6\theta - 6\mathrm{i}\sin 2\theta\) | M1 | 2.1 |
| \(\Rightarrow \cos^3\theta\sin^3\theta = -\dfrac{1}{32}\sin 6\theta + \dfrac{3}{32}\sin 2\theta\) | A1 | 2.2a |
| [6] |
Notes
M1 M1: binomial expansions oe; expanding the whole expression
or \(\left[\left(z + \dfrac{1}{z}\right)\left(z - \dfrac{1}{z}\right)\right]^3\) etc
award second M1 if changes into trig and makes some attempt at using the addition formulae