C3 January 2011 Q1
1.
(a) Express \(7\cos x - 24\sin x\) in the form \(R\cos(x + \alpha)\) where \(R \gt 0\) and \(0 \lt \alpha \lt \frac{\pi}{2}\). Give the value of \(\alpha\) to 3 decimal places. (3)
(b) Hence write down the minimum value of \(7\cos x - 24\sin x\). (1)
(c) Solve, for \(0 \leqslant x \lt 2\pi\), the equation\[7\cos x - 24\sin x = 10\]giving your answers to 2 decimal places. (5)
| Scheme | Marks |
|---|---|
| \(7\cos x - 24\sin x = R\cos(x + \alpha)\) \(7\cos x - 24\sin x = R\cos x\cos\alpha - R\sin x\sin\alpha\) | |
| Equate \(\cos x\): \(7 = R\cos\alpha\) Equate \(\sin x\): \(24 = R\sin\alpha\) | |
| \(R = \sqrt{7^2 + 24^2};= 25\) | B1 |
| \(\tan\alpha = \tfrac{24}{7} \Rightarrow \alpha = 1.287002218\ldots^{c}\) | M1 A1 |
| Hence, \(7\cos x - 24\sin x = 25\cos(x + 1.287)\) | |
| (3) |
Notes
B1: \(R = 25\)
M1: \(\tan\alpha = \tfrac{24}{7}\) or \(\tan\alpha = \tfrac{7}{24}\)
A1: awrt 1.287
| Scheme | Marks |
|---|---|
| Minimum value \(= \underline{-25}\) | B1ft |
| (1) |
Notes
B1ft: \(-25\) or \(-R\)
| Scheme | Marks |
|---|---|
| \(7\cos x - 24\sin x = 10\) \(25\cos(x + 1.287) = 10\) | |
| \(\cos(x + 1.287) = \dfrac{10}{25}\) | M1 |
| \(\text{PV} = 1.159279481\ldots^{c}\) or \(66.42182152\ldots^\circ\) | M1 |
| So, \(x + 1.287 = \left\{1.159279\ldots^{c}, 5.123906\ldots^{c}, 7.442465\ldots^{c}\right\}\) | M1 |
| gives, \(x = \{3.836906\ldots, 6.155465\ldots\}\) | A1 A1 |
| (5) | |
| (9 marks) |
Notes
M1: \(\cos(x \pm \text{their } \alpha) = \dfrac{10}{(\text{their } R)}\)
M1: For applying \(\cos^{-1}\left(\dfrac{10}{\text{their } R}\right)\)
M1: either \(2\pi\) + or \(-\) their \(\text{PV}^{c}\) or \(360^\circ\) + or \(-\) their \(\text{PV}^\circ\)
A1: awrt 3.84 OR 6.16
A1: awrt 3.84 AND 6.16