OCR C4 — Question 2 7 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.3 This is a standard C4 harmonic form question with routine application of the R cos(θ+α) method followed by a straightforward equation solve. The technique is well-practiced at this level, requiring only algebraic manipulation to find R and α, then solving a basic trigonometric equation. Slightly above average difficulty due to the two-part nature and need for careful angle work, but still a textbook exercise with no novel insight required.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

2 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < { } _ { 2 } ^ { 1 } \pi\).-
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(4\cos\theta - \sin\theta = R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\)
\(\Rightarrow R\cos\alpha = 4,\ R\sin\alpha = 1\)M1 correct pairs
\(\Rightarrow R^2 = 1^2 + 4^2 = 17,\ R = \sqrt{17} = 4.123\)B1 \(R = \sqrt{17} = 4.123\)
\(\tan\alpha = \frac{1}{4} \Rightarrow \alpha = 0.245\)M1 \(\tan\alpha = \frac{1}{4}\) o.e.
A1\(\alpha = 0.245\)
\(\sqrt{17}\cos(\theta + 0.245) = 3\)
\(\Rightarrow \cos(\theta + 0.245) = \frac{3}{\sqrt{17}}\)
\(\Rightarrow \theta + 0.245 = 0.756,\ 5.527\)M1 \(\theta + 0.245 = \arccos\frac{3}{\sqrt{17}}\), ft their \(R\), \(\alpha\) for method
\(\Rightarrow \theta = 0.511,\ 5.282\)A1A1 [7] penalise extra solutions in the range \((-1)\)
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\cos\theta - \sin\theta = R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$ | | |
| $\Rightarrow R\cos\alpha = 4,\ R\sin\alpha = 1$ | M1 | correct pairs |
| $\Rightarrow R^2 = 1^2 + 4^2 = 17,\ R = \sqrt{17} = 4.123$ | B1 | $R = \sqrt{17} = 4.123$ |
| $\tan\alpha = \frac{1}{4} \Rightarrow \alpha = 0.245$ | M1 | $\tan\alpha = \frac{1}{4}$ o.e. |
| | A1 | $\alpha = 0.245$ |
| $\sqrt{17}\cos(\theta + 0.245) = 3$ | | |
| $\Rightarrow \cos(\theta + 0.245) = \frac{3}{\sqrt{17}}$ | | |
| $\Rightarrow \theta + 0.245 = 0.756,\ 5.527$ | M1 | $\theta + 0.245 = \arccos\frac{3}{\sqrt{17}}$, ft their $R$, $\alpha$ for method |
| $\Rightarrow \theta = 0.511,\ 5.282$ | A1A1 [7] | penalise extra solutions in the range $(-1)$ |

---
2 Express $4 \cos \theta - \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < { } _ { 2 } ^ { 1 } \pi$.-\\
Hence solve the equation $4 \cos \theta - \sin \theta = 3$, for $0 \leqslant \theta \leqslant 2 \pi$.

\hfill \mbox{\textit{OCR C4  Q2 [7]}}