A2 June 2019 Paper 2 Q4
4. The infinite series C and S are defined by
\[\mathrm{C} = \cos\theta + \frac{1}{2}\cos 5\theta + \frac{1}{4}\cos 9\theta + \frac{1}{8}\cos 13\theta + \ldots\]\[\mathrm{S} = \sin\theta + \frac{1}{2}\sin 5\theta + \frac{1}{4}\sin 9\theta + \frac{1}{8}\sin 13\theta + \ldots\]Given that the series C and S are both convergent,
Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{C} + \mathrm{iS} = \cos\theta + \mathrm{i}\sin\theta + \dfrac{1}{2}(\cos 5\theta + \mathrm{i}\sin 5\theta)\left(+\dfrac{1}{4}(\cos 9\theta + \mathrm{i}\sin 9\theta) + \ldots\right)\) | M1 | 1.1b |
| \(= \mathrm{e}^{\mathrm{i}\theta} + \dfrac{1}{2}\mathrm{e}^{5\mathrm{i}\theta}\left(+\dfrac{1}{4}\mathrm{e}^{9\mathrm{i}\theta} + \ldots\right)\) | A1 | 2.1 |
| \(\mathrm{C} + \mathrm{iS} = \dfrac{\mathrm{e}^{\mathrm{i}\theta}}{1 - \frac{1}{2}\mathrm{e}^{4\mathrm{i}\theta}}\) | M1 | 3.1a |
| \(= \dfrac{2\mathrm{e}^{\mathrm{i}\theta}}{2 - \mathrm{e}^{4\mathrm{i}\theta}}\) * | A1* | 1.1b |
| (4) |
Notes
Way 1
M1: Combines the two series by pairing the multiples of \(\theta\) (At least up to \(5\theta\))
A1: Converts to Euler form correctly (At least up to \(5\theta\))
M1: Recognises that C + iS is a convergent geometric series and uses the sum to infinity of a GP
A1*: Reaches the printed answer with no errors
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{C} + \mathrm{iS} = \cos\theta + \mathrm{i}\sin\theta + \dfrac{1}{2}(\cos 5\theta + \mathrm{i}\sin 5\theta)\left(+\dfrac{1}{4}(\cos 9\theta + \mathrm{i}\sin 9\theta) + \ldots\right)\) | M1 | 1.1b |
| \(\mathrm{C} + \mathrm{iS} = \cos\theta + \mathrm{i}\sin\theta + \dfrac{1}{2}(\cos\theta + \mathrm{i}\sin\theta)^5\left(+\dfrac{1}{4}(\cos\theta + \mathrm{i}\sin\theta)^9 + \ldots\right)\) | A1 | 2.1 |
| \(\mathrm{C} + \mathrm{iS} = \dfrac{\cos\theta + \mathrm{i}\sin\theta}{1 - \frac{1}{2}(\cos\theta + \mathrm{i}\sin\theta)^4} = \dfrac{\mathrm{e}^{\mathrm{i}\theta}}{1 - \frac{1}{2}\mathrm{e}^{4\mathrm{i}\theta}}\) | M1 | 3.1a |
| \(= \dfrac{2\mathrm{e}^{\mathrm{i}\theta}}{2 - \mathrm{e}^{4\mathrm{i}\theta}}\) * | A1* | 1.1b |
| (4) |
M1: Combines the two series by pairing the multiples of \(\theta\) (At least up to \(5\theta\))
A1: Converts to power form correctly (At least up to \(5\theta\))
M1: Recognises that C + iS is a convergent geometric series and uses the sum to infinity of a GP
A1*: Reaches the printed answer with no errors
Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2\mathrm{e}^{\mathrm{i}\theta}}{2 - \mathrm{e}^{4\mathrm{i}\theta}} \times \dfrac{2 - \mathrm{e}^{-4\mathrm{i}\theta}}{2 - \mathrm{e}^{-4\mathrm{i}\theta}}\) | M1 | 3.1a |
| \(\dfrac{4\mathrm{e}^{\mathrm{i}\theta} - 2\mathrm{e}^{-3\mathrm{i}\theta}}{4 - 2\mathrm{e}^{-4\mathrm{i}\theta} - 2\mathrm{e}^{4\mathrm{i}\theta} + 1}\) | A1 | 1.1b |
| \(\dfrac{4\cos\theta + 4\mathrm{i}\sin\theta - 2\cos 3\theta + 2\mathrm{i}\sin 3\theta}{5 - 2\cos 4\theta + 2\mathrm{i}\sin 4\theta - 2\cos 4\theta - 2\mathrm{i}\sin 4\theta}\) Dependent on the first M | dM1 | 2.1 |
| \(S = \dfrac{4\sin\theta + 2\sin 3\theta}{5 - 4\cos 4\theta}\) * | A1* | 1.1b |
| (4) | ||
| (8 marks) |
Notes
Way 1
M1: Multiplies numerator and denominator by \(2 - \mathrm{e}^{-4\mathrm{i}\theta}\)
A1: Correct fraction in terms of exponentials
dM1: Converts back to trigonometric form
A1*: Reaches the printed answer with no errors
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2\mathrm{e}^{\mathrm{i}\theta}}{2 - \mathrm{e}^{4\mathrm{i}\theta}} = \dfrac{2(\cos\theta + \mathrm{i}\sin\theta)}{2 - (\cos 4\theta + \mathrm{i}\sin 4\theta)} \times \dfrac{2 - (\cos 4\theta - \mathrm{i}\sin 4\theta)}{2 - (\cos 4\theta - \mathrm{i}\sin 4\theta)}\) | M1 | 3.1a |
| \(\dfrac{4\cos\theta + 4\mathrm{i}\sin\theta - 2\cos\theta\cos 4\theta - 2\sin\theta\sin 4\theta + 2\mathrm{i}\sin 4\theta\cos\theta - 2\mathrm{i}\sin\theta\cos 4\theta}{4 + \cos^2 4\theta + \sin^2 4\theta - 4\cos 4\theta}\) | A1 | 1.1b |
| \(\dfrac{4\cos\theta + 4\mathrm{i}\sin\theta - 2\cos 3\theta + 2\mathrm{i}\sin 3\theta}{5 - 2\cos 4\theta + 2\mathrm{i}\sin 4\theta - 2\cos 4\theta - 2\mathrm{i}\sin 4\theta}\) Dependent on the first M | dM1 | 2.1 |
| \(S = \dfrac{4\sin\theta + 2\sin 3\theta}{5 - 4\cos 4\theta}\) * | A1* | 1.1b |
M1: Converts back to trigonometric form and realises the need to make the denominator real and multiplies numerator and denominator by the complex conjugate of the denominator which is correct for their fraction
A1: Correct fraction in terms of trigonometric functions
dM1: Uses the correct addition formula to obtain \(\sin 3\theta\) in the numerator
A1*: Reaches the printed answer with no errors