Moderate -0.3 This is a straightforward compound angle question requiring expansion of cos(θ-60°) using the addition formula, rearranging to form tan θ = k, then solving within the given range. It's slightly easier than average as it follows a standard template with no conceptual surprises, though it does require careful algebraic manipulation and knowledge of the CAST diagram.
Use correct \(\cos(A-B)\) formula to get \(\cos\theta\cos 60 + \sin\theta\sin 60 = 3\sin\theta\)
Step 2
M1
Substitute \(\cos 60 = \frac{1}{2}\), \(\sin 60 = \frac{\sqrt{3}}{2}\) to obtain equation in \(\tan\theta\) (or \(\cos\theta\) or \(\sin\theta\)). Allow \(\frac{1}{2}\) and \(\frac{\sqrt{3}}{2}\) interchanged. e.g. \(\frac{1}{2} + \frac{\sqrt{3}}{2}\tan\theta = 3\tan\theta\)
Step 3
A1
Obtain \(\tan\theta = \frac{1}{6-\sqrt{3}}\) or \(\tan\theta = \frac{6+\sqrt{3}}{33}\) or \(0.2343\), or \(\cos\theta = \frac{3\frac{\sqrt{3}}{2}}{\sqrt{10-3\sqrt{3}}}\) or \(0.9736\) or \(\sin\theta = \frac{\frac{1}{2}}{\sqrt{10-3\sqrt{3}}}\) or \(0.2281\)
Step 4
A1
Obtain answer e.g. \(\theta = 13.2°\). May be more accurate, allow value rounding to \(13.2°\). \(\theta = 13.1867°\)
Step 5
A1 FT
Obtain second answer e.g. \(\theta = 193.2°\) and no others in given interval. FT on previous \(\theta\), must have scored M1. Note if \(\theta\) is negative (e.g. \(-13.2\)): \(-13.2+180=166.8\) A0 but \(-13.2+360=346.8\) A1 FT. Ignore answers outside given interval. Treat answers in radians as misread: \(0.23015\), \(3.3717\).
Alternative method (using \(R\cos(\theta\pm\alpha)\)):
Answer
Marks
Guidance
Step 1
B1
Use correct \(\cos(A-B)\) formula to get \(\cos\theta\cos 60 + \sin\theta\sin 60 = 3\sin\theta\)
Step 2
M1
Correct method for finding \(\tan\alpha\) from \(p\cos\theta + q\sin\theta = 0\): \(\tan\alpha = \pm\frac{q}{p}\)
Step 3
A1
Correct value of \(\alpha\): \(76.8°\) or \(1.34\) radians (may be more accurate)
Step 4
A1
Obtain answer e.g. \(\theta = 13.2°\). Allow value rounding to \(13.2°\), \(\theta = 13.1867°\)
Step 5
A1 FT
Obtain second answer e.g. \(\theta = 193.2°\) and no others in given interval. Same FT conditions as above.
Total: 5 marks
## Question 2:
**Step 1** | B1 | Use correct $\cos(A-B)$ formula to get $\cos\theta\cos 60 + \sin\theta\sin 60 = 3\sin\theta$
**Step 2** | M1 | Substitute $\cos 60 = \frac{1}{2}$, $\sin 60 = \frac{\sqrt{3}}{2}$ to obtain equation in $\tan\theta$ (or $\cos\theta$ or $\sin\theta$). Allow $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ interchanged. e.g. $\frac{1}{2} + \frac{\sqrt{3}}{2}\tan\theta = 3\tan\theta$
**Step 3** | A1 | Obtain $\tan\theta = \frac{1}{6-\sqrt{3}}$ or $\tan\theta = \frac{6+\sqrt{3}}{33}$ or $0.2343$, or $\cos\theta = \frac{3\frac{\sqrt{3}}{2}}{\sqrt{10-3\sqrt{3}}}$ or $0.9736$ or $\sin\theta = \frac{\frac{1}{2}}{\sqrt{10-3\sqrt{3}}}$ or $0.2281$
**Step 4** | A1 | Obtain answer e.g. $\theta = 13.2°$. May be more accurate, allow value rounding to $13.2°$. $\theta = 13.1867°$
**Step 5** | A1 FT | Obtain second answer e.g. $\theta = 193.2°$ and no others in given interval. FT on previous $\theta$, must have scored M1. Note if $\theta$ is negative (e.g. $-13.2$): $-13.2+180=166.8$ A0 but $-13.2+360=346.8$ A1 FT. Ignore answers outside given interval. Treat answers in radians as misread: $0.23015$, $3.3717$.
---
**Alternative method** (using $R\cos(\theta\pm\alpha)$):
**Step 1** | B1 | Use correct $\cos(A-B)$ formula to get $\cos\theta\cos 60 + \sin\theta\sin 60 = 3\sin\theta$
**Step 2** | M1 | Correct method for finding $\tan\alpha$ from $p\cos\theta + q\sin\theta = 0$: $\tan\alpha = \pm\frac{q}{p}$
**Step 3** | A1 | Correct value of $\alpha$: $76.8°$ or $1.34$ radians (may be more accurate)
**Step 4** | A1 | Obtain answer e.g. $\theta = 13.2°$. Allow value rounding to $13.2°$, $\theta = 13.1867°$
**Step 5** | A1 FT | Obtain second answer e.g. $\theta = 193.2°$ and no others in given interval. Same FT conditions as above.
**Total: 5 marks**
---