Moderate -0.3 This is a standard harmonic form question with routine application of the R cos(θ+α) formula (R=√10, tan α=3), followed by a straightforward observation that R cos(θ+α)=4 is impossible since maximum value is √10≈3.16<4. The technique is well-practiced and requires no novel insight, making it slightly easier than average.
1 Express \(\cos \theta - 3 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\).
Hence show that the equation \(\cos \theta - 3 \sin \theta = 4\) has no solution.
Or 3.2 or better, not \(\pm\sqrt{10}\) unless \(+\sqrt{10}\) chosen
\(\tan\alpha = 3 \Rightarrow \alpha = 1.249\)
M1 A1
ft their pairs (condone sign errors but division must be correct way round); A1 for 1.249 or better (accept 1.25), with no errors seen in method for angle
Maximum value of \(\cos\theta - 3\sin\theta\) is \(\sqrt{10} < 4\)
B1
Or equivalent convincing numerical statement e.g. \(\frac{4}{\sqrt{10}} > 1\). May be embedded in attempt at solution. Do not accept general statements. Dependent on first B1. SC: If \(\cos\alpha=1, \sin\alpha=3 \Rightarrow \tan\alpha=3\) scores M0A0B1M1A1B1 (max 4/6). Note \(R=\sqrt{10}\) and \(\tan\alpha=3\) with no wrong working could score full marks
# Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta - 3\sin\theta = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)$ $\Rightarrow 1 = R\cos\alpha, \; 3 = R\sin\alpha$ | M1 A1 | Correct pairs. Condone sign errors for M mark (so accept $R\sin\alpha = -3$) |
| $R^2 = 1^2 + 3^2 = 10 \Rightarrow R = \sqrt{10}$ | B1 | Or 3.2 or better, not $\pm\sqrt{10}$ unless $+\sqrt{10}$ chosen |
| $\tan\alpha = 3 \Rightarrow \alpha = 1.249$ | M1 A1 | ft their pairs (condone sign errors but division must be correct way round); A1 for 1.249 or better (accept 1.25), with no errors seen in method for angle |
| Maximum value of $\cos\theta - 3\sin\theta$ is $\sqrt{10} < 4$ | B1 | Or equivalent convincing numerical statement e.g. $\frac{4}{\sqrt{10}} > 1$. May be embedded in attempt at solution. Do not accept general statements. **Dependent on first B1**. SC: If $\cos\alpha=1, \sin\alpha=3 \Rightarrow \tan\alpha=3$ scores M0A0B1M1A1B1 (max 4/6). Note $R=\sqrt{10}$ and $\tan\alpha=3$ with no wrong working could score full marks |
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