A2 June 2022 Paper 1 Q14
14
(a) Find \(\left(3 - \mathrm{e}^{2\mathrm{i}\theta}\right)\left(3 - \mathrm{e}^{-2\mathrm{i}\theta}\right)\) in terms of \(\cos 2\theta\). [2]
(b) Hence show that the sum of the infinite series \[\sin\theta + \frac{1}{3}\sin 3\theta + \frac{1}{9}\sin 5\theta + \frac{1}{27}\sin 7\theta + \ldots\] can be expressed as \(\dfrac{6\sin\theta}{5 - 3\cos 2\theta}\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\left(3 - e^{2i\theta}\right)\left(3 - e^{-2i\theta}\right) = 9 - 3\left(e^{2i\theta} + e^{-2i\theta}\right) + 1 =\) \(= 10 - 6\cos 2\theta\) | M1 A1 | 1.1 1.1 |
| [2] |
Notes
M1: For expanding correctly
| Scheme | Marks | AO |
|---|---|---|
| let \(S = \sin\theta + \frac{1}{3}\sin 3\theta + \frac{1}{9}\sin 5\theta + \frac{1}{27}\sin 7\theta + \ldots\) and \(C = \cos\theta + \frac{1}{3}\cos 3\theta + \frac{1}{9}\cos 5\theta + \frac{1}{27}\cos 7\theta + \ldots\) \(C + iS = \mathrm{e}^{i\theta} + \dfrac{1}{3}\mathrm{e}^{3i\theta} + \dfrac{1}{9}\mathrm{e}^{5i\theta} + \dfrac{1}{27}\mathrm{e}^{7i\theta} + \ldots\) | M1 | 2.1 |
| \(= \dfrac{e^{i\theta}}{1 - \frac{1}{3}\mathrm{e}^{2i\theta}}\) | M1 A1 | 2.1 2.2a |
| \(= \dfrac{3\mathrm{e}^{i\theta}}{3 - \mathrm{e}^{2i\theta}} = \dfrac{3\mathrm{e}^{i\theta}\left(3 - \mathrm{e}^{-2i\theta}\right)}{10 - 6\cos 2\theta}\) | M1* | 3.1a |
| \(= \dfrac{9(\cos\theta + i\sin\theta) - 3(\cos\theta - i\sin\theta)}{10 - 6\cos 2\theta}\) | M1dep* | 2.1 |
| \(S = \dfrac{9\sin\theta + 3\sin\theta}{10 - 6\cos 2\theta} = \dfrac{6\sin\theta}{5 - 3\cos 2\theta}\) | A1 | 2.2a |
| [6] |
Notes
M1: At least 2 terms of C + iS soi by correct GP formula
M1: sum to infinity of GP formula for their series (which must be geometric)
A1: oe
M1*: multiply numerator and denominator by \(3 - \mathrm{e}^{-2i\theta}\)
M1dep*: \(\mathrm{e}^{i\theta} = \cos\theta + i\sin\theta\) used when denominator has been simplified to a real expression
A1: AG