June 2022 Paper 2 Q1
1 Express \(\cos\theta + \sqrt{3}\sin\theta\) in the form \(R\cos(\theta - \alpha)\), where \(R\) and \(\alpha\) are exact values to be determined. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(R^2 = 1^2 + \sqrt{3}^2\) | M1 | 1.1 |
| \(\tan\alpha = \frac{\sqrt{3}}{1}\) or \(\sin\alpha = \frac{\sqrt{3}}{2}\) or \(\cos\alpha = \frac{1}{2}\) soi | M1 | 1.1 |
| \(R = 2\) or \(\alpha = \frac{\pi}{3}\) or \(\alpha = 60^\circ\) seen | A1 | 1.1 |
| \(2\cos\left(\theta - \frac{\pi}{3}\right)\) or \(2\cos(\theta - 60^\circ)\) isw | A1 | 1.1 |
| [4] |
Notes
M1: may be implied by correct answer
M1: may see eg \(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right)\)
may be implied by correct answer
Alternatively
| Scheme | Marks |
|---|---|
| \(\cos\theta + \sqrt{3}\sin\theta = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\) so \(1 = R\cos\alpha\) and \(\sqrt{3} = R\sin\alpha\) | M1 |
| \(\dfrac{1}{\cos\alpha} = \dfrac{\sqrt{3}}{\sin\alpha}\) | M1 |
| \(\alpha = \frac{\pi}{3}\) or \(\alpha = 60^\circ\) seen | A1 |
| \(2\cos\left(\theta - \frac{\pi}{3}\right)\) or \(2\cos(\theta - 60^\circ)\) isw | A1 |
M1: for equating coefficients
M1: for eliminating \(R\)