A2 October 2020 Paper 2 Q4
4.
| Scheme | Marks | AO |
|---|---|---|
| \((\cos\theta + \mathrm{i}\sin\theta)^7 = \cos^7\theta + \dbinom{7}{1}\cos^6\theta(\mathrm{i}\sin\theta) + \dbinom{7}{2}\cos^5\theta(\mathrm{i}\sin\theta)^2 + \ldots\) Some simplification may be done at this stage e.g. \(c^7 + 7c^6\mathrm{i}s - 21c^5s^2 - 35c^4\mathrm{i}s^3 + 35c^3s^4 + 21c^2\mathrm{i}s^5 - 7cs^6 - \mathrm{i}s^7\) | M1 | 1.1b |
| \(\mathrm{i}\sin 7\theta = {}^7\mathrm{C}_1c^6\mathrm{i}s + {}^7\mathrm{C}_3c^4\mathrm{i}^3s^3 + {}^7\mathrm{C}_5c^2\mathrm{i}^5s^5 + \mathrm{i}^7s^7\) or \(= 7c^6\mathrm{i}s + 35c^4\mathrm{i}^3s^3 + 21c^2\mathrm{i}^5s^5 + \mathrm{i}^7s^7\) | M1 | 2.1 |
| \(\sin 7\theta = 7c^6s - 35c^4s^3 + 21c^2s^5 - s^7\) | A1 | 1.1b |
| \(= 7\left(1 - s^2\right)^3s - 35\left(1 - s^2\right)^2s^3 + 21\left(1 - s^2\right)s^5 - s^7\) \(= 7\left(1 - 3s^2 + 3s^4 - s^6\right)s - 35\left(1 - 2s^2 + s^4\right)s^3 + 21\left(1 - s^2\right)s^5 - s^7\) | M1 | 2.1 |
| \(\left\{7s - 21s^3 + 21s^5 - 7s^7 - 35s^3 + 70s^5 - 35s^7 + 21s^5 - 21s^7 - s^7\right\}\) leading to \(\sin 7\theta = 7\sin\theta - 56\sin^3\theta + 112\sin^5\theta - 64\sin^7\theta\,*\) | A1* | 1.1b |
| (5) |
Notes
(a)
M1: Attempts to expand \((\cos\theta + \mathrm{i}\sin\theta)^7\) including a recognisable attempt at binomial coefficients
Some simplification may be done at this stage. (May only see imaginary terms)
M1: Identifies imaginary terms with \(\sin 7\theta\)
A1: Correct expression with coefficients evaluated and i’s dealt with correctly
M1: Replaces \(\cos^2\theta\) with \(1 - \sin^2\theta\) and applies the expansions of \(\left(1 - \sin^2\theta\right)^2\) and \(\left(1 - \sin^2\theta\right)^3\) to their expression
A1*: Reaches the printed answer with no errors and expansion of brackets seen.
| Scheme | Marks | AO |
|---|---|---|
| \(1 + \sin 7\theta = 0 \Rightarrow \sin 7\theta = -1\) | M1 | 3.1a |
| \(7\theta = -450, -90, 270, 630, \ldots\) or \(7\theta = -\dfrac{5\pi}{2}, -\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{7\pi}{2}, \ldots\) | A1 | 1.1b |
| \(\theta = -\dfrac{450}{7}, -\dfrac{90}{7}, \dfrac{270}{7}, \dfrac{630}{7}, \ldots \Rightarrow \sin\theta = \ldots\) or \(\theta = -\dfrac{5\pi}{14}, -\dfrac{\pi}{14}, \dfrac{3\pi}{14}, \dfrac{7\pi}{14}, \ldots \Rightarrow \sin\theta = \ldots\) | M1 | 2.2a |
| \(x = \sin\theta = -0.901, -0.223, 0.623, 1\) | A1 A1 | 1.1b 2.3 |
| (5) | ||
| (10 marks) |
Notes
(b)
M1: Makes the connection with part (a) and realises the need to solve \(\sin 7\theta = -1\)
A1: At least one correct value for \(7\theta\)
M1: Divides by 7 and deduces that \(x\) values are found by finding at least one value for \(\sin\theta\)
A1: Awrt 2 correct values for \(x\)
A1: Awrt all 4 \(x\) values correct and no extras