C3 June 2011 Q6
6.
(a) Prove that \[\frac{1}{\sin 2\theta} - \frac{\cos 2\theta}{\sin 2\theta} = \tan\theta, \quad \theta \neq 90n^\circ,\ n \in \mathbb{Z}\] (4)
(b) Hence, or otherwise,
(i) show that \(\tan 15^\circ = 2 - \sqrt{3}\), (3)
(ii) solve, for \(0 \lt x \lt 360^\circ\), \[\operatorname{cosec} 4x - \cot 4x = 1\] (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{\sin 2\theta} - \dfrac{\cos 2\theta}{\sin 2\theta} = \dfrac{1 - \cos 2\theta}{\sin 2\theta}\) | M1 |
| \(= \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}\) | M1A1 |
| \(= \dfrac{\sin\theta}{\cos\theta} = \tan\theta\) cso | A1* |
| (4) |
| Scheme | Marks |
|---|---|
| (i) \(\tan 15^\circ = \dfrac{1}{\sin 30^\circ} - \dfrac{\cos 30^\circ}{\sin 30^\circ}\) | M1 |
| \(\tan 15^\circ = \dfrac{1}{\frac{1}{2}} - \dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}\) cso | dM1 A1* |
| (3) | |
| (ii) \(\tan 2x = 1\) | M1 |
| \(2x = 45^\circ\) | A1 |
| \(2x = 45^\circ + 180^\circ\) | M1 |
| \(x = 22.5^\circ, 112.5^\circ, 202.5^\circ, 292.5^\circ\) | A1 (any two) A1 |
| (5) | |
| (12 marks) |
Alt for (b)(i)
| Scheme | Marks |
|---|---|
| \(\tan 15^\circ = \tan(60^\circ - 45^\circ)\) or \(\tan(45^\circ - 30^\circ)\) | |
| \(\tan 15^\circ = \dfrac{\tan 60 - \tan 45}{1 + \tan 60\tan 45}\) or \(\dfrac{\tan 45 - \tan 30}{1 + \tan 45\tan 30}\) | M1 |
| \(\tan 15^\circ = \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}}\) or \(\dfrac{1 - \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}}\) | M1 |
| Rationalises to produce \(\tan 15^\circ = 2 - \sqrt{3}\) | A1* |