C3 January 2009 Q6
6.
(a)
(i) By writing \(3\theta = (2\theta + \theta)\), show that\[\sin 3\theta = 3\sin\theta - 4\sin^3\theta.\] (4)
(ii) Hence, or otherwise, for \(0 < \theta < \dfrac{\pi}{3}\), solve\[8\sin^3\theta - 6\sin\theta + 1 = 0.\]Give your answers in terms of \(\pi\). (5)
(b) Using \(\sin(\theta - \alpha) = \sin\theta\cos\alpha - \cos\theta\sin\alpha\), or otherwise, show that\[\sin 15^\circ = \frac{1}{4}(\sqrt{6} - \sqrt{2}).\] (4)
| Scheme | Marks |
|---|---|
| (i) \(\sin 3\theta = \sin(2\theta + \theta)\) \(= \sin 2\theta\cos\theta + \cos 2\theta\sin\theta\) \(= 2\sin\theta\cos\theta.\cos\theta + (1 - 2\sin^2\theta)\sin\theta\) | M1 A1 |
| \(= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta\) | M1 |
| \(= 3\sin\theta - 4\sin^3\theta\ \ \ast\) cso | A1 |
| (4) | |
| (ii) \(8\sin^3\theta - 6\sin\theta + 1 = 0\) \(-2\sin 3\theta + 1 = 0\) | M1 A1 |
| \(\sin 3\theta = \dfrac{1}{2}\) | M1 |
| \(3\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\) \(\theta = \dfrac{\pi}{18}, \dfrac{5\pi}{18}\) | A1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\sin 15^\circ = \sin(60^\circ - 45^\circ) = \sin 60^\circ\cos 45^\circ - \cos 60^\circ\sin 45^\circ\) | M1 |
| \(= \dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}\times\dfrac{1}{\sqrt{2}}\) | M1 A1 |
| \(= \dfrac{1}{4}\sqrt{6} - \dfrac{1}{4}\sqrt{2} = \dfrac{1}{4}(\sqrt{6} - \sqrt{2})\ \ \ast\) cso | A1 |
| (4) | |
| (13 marks) |
Alternatives to (b) ①
| \(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ\) | M1 |
| \(= \dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}}\times\dfrac{1}{2}\) | M1 A1 |
| \(= \dfrac{1}{4}\sqrt{6} - \dfrac{1}{4}\sqrt{2} = \dfrac{1}{4}(\sqrt{6} - \sqrt{2})\ \ \ast\) cso | A1 (4) |
Alternatives to (b) ②
| Using \(\cos 2\theta = 1 - 2\sin^2\theta\), \(\cos 30^\circ = 1 - 2\sin^2 15^\circ\) \(2\sin^2 15^\circ = 1 - \cos 30^\circ = 1 - \dfrac{\sqrt{3}}{2}\) | |
| \(\sin^2 15^\circ = \dfrac{2 - \sqrt{3}}{4}\) | M1 A1 |
| \(\left(\dfrac{1}{4}(\sqrt{6} - \sqrt{2})\right)^2 = \dfrac{1}{16}(6 + 2 - 2\sqrt{12}) = \dfrac{2 - \sqrt{3}}{4}\) | M1 |
| Hence \(\sin 15^\circ = \dfrac{1}{4}(\sqrt{6} - \sqrt{2})\ \ \ast\) cso | A1 (4) |