C3 June 2005 Q5
5.
(a) Using the identity \(\cos(A + B) \equiv \cos A\cos B - \sin A\sin B\), prove that\[\cos 2A \equiv 1 - 2\sin^2 A.\] (2)
(b) Show that\[2\sin 2\theta - 3\cos 2\theta - 3\sin\theta + 3 \equiv \sin\theta\,(4\cos\theta + 6\sin\theta - 3).\] (4)
(c) Express \(4\cos\theta + 6\sin\theta\) in the form \(R\sin(\theta + \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt \tfrac{1}{2}\pi\). (4)
(d) Hence, for \(0 \leqslant \theta \lt \pi\), solve\[2\sin 2\theta = 3(\cos 2\theta + \sin\theta - 1),\]giving your answers in radians to 3 significant figures, where appropriate. (5)
| Scheme | Marks |
|---|---|
| \(\cos 2A = \cos^2 A - \sin^2 A\) (+ use of \(\cos^2 A + \sin^2 A \equiv 1\)) | M1 |
| \(= (1 - \sin^2 A); - \sin^2 A = 1 - 2\sin^2 A\) (*) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(2\sin 2\theta - 3\cos 2\theta - 3\sin\theta + 3 \equiv 4\sin\theta\cos\theta; - 3(1 - 2\sin^2\theta) - 3\sin\theta + 3\) | B1; M1 |
| \(\equiv 4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta\) | M1 |
| \(\equiv \sin\theta(4\cos\theta + 6\sin\theta - 3)\) (*) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(4\cos\theta + 6\sin\theta \equiv R\sin\theta\cos\alpha + R\cos\theta\sin\alpha\) | |
| Complete method for \(R\) (may be implied by correct answer) \([\,R^2 = 4^2 + 6^2,\ R\sin\alpha = 4,\ R\cos\alpha = 6\,]\) | M1 |
| \(R = \sqrt{52}\) or 7.21 | A1 |
| Complete method for \(\alpha\); \(\quad\alpha = 0.588\) (allow \(33.7^\circ\)) | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\sin\theta\,(4\cos\theta + 6\sin\theta - 3) = 0\) | M1 |
| \(\theta = 0\) | B1 |
| \(\sin(\theta + 0.588) = \dfrac{3}{\sqrt{52}} = 0.4160\ldots\) \((24.6^\circ)\) | M1 |
| \(\theta + 0.588 = (0.4291),\ 2.7125\) [or \(\theta + 33.7^\circ = (24.6^\circ),\ 155.4^\circ\)] | dM1 |
| \(\theta = 2.12\) cao | A1 |
| (5) | |
| (15 marks) |