FP2 June 2006 Q6
6.
(a) Use de Moivre’s theorem to show that \[\sin 5\theta = \sin\theta(16\cos^4\theta - 12\cos^2\theta + 1).\] (5)
(b) Hence, or otherwise, solve, for \(0 \leqslant \theta \lt \pi\) \[\sin 5\theta + \cos\theta\sin 2\theta = 0.\] (6)
| Scheme | Marks |
|---|---|
| In this solution \(\cos\theta = c\) and \(\sin\theta = s\) | |
| \(\cos 5\theta + \mathrm{i}\sin 5\theta = (c + \mathrm{i}s)^5\) | M1 |
| \((= c^5 + 5c^4\,\mathrm{i}s + 10c^3(\mathrm{i}s)^2 + 10c^2(\mathrm{i}s)^3 + 5c(\mathrm{i}s)^4 + (\mathrm{i}s)^5)\) | |
| \(\sin 5\theta = 5c^4s - 10c^2s^3 + s^5\) | M1 A1 |
| \(= 5c^4s - 10c^2(1 - c^2)s + (1 - c^2)^2s \qquad s^2 = 1 - c^2\) | M1 |
| \(= s(16c^4 - 12c^2 + 1)\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\sin\theta(16\cos^4\theta - 12\cos^2\theta + 1) + 2\cos^2\theta\sin\theta = 0\) | M1 |
| \(\sin\theta = 0 \Rightarrow \theta = 0\) | B1 |
| \(16c^4 - 10c^2 + 1 = (8c^2 - 1)(2c^2 - 1) = 0\) | M1 |
| \(c = \pm\dfrac{1}{2\sqrt{2}},\ c = \pm\dfrac{1}{\sqrt{2}}\) any two | A1 |
| \(\theta \approx 1.21, 1.93;\ \theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}\) any two | A1 |
| all four accept awrt 0.79, 1.21, 1.93, 2.36 Ignore any solutions out of range. | A1 |
| (6) | |
| (11 marks) |