Edexcel C3 2010 January — Question 3 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Year2010
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.3 This is a standard C3 harmonic form question with routine steps: expand R cos(x+α), equate coefficients to find R and α, then solve a simple equation of form R cos(x+α)=4. The method is well-practiced and requires no novel insight, making it slightly easier than average for A-level.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

3. (a) Express \(5 \cos x - 3 \sin x\) in the form \(R \cos ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
(b) Hence, or otherwise, solve the equation $$5 \cos x - 3 \sin x = 4$$ for \(0 \leqslant x < 2 \pi\), giving your answers to 2 decimal places.

Question 3:
Part (a):
AnswerMarks Guidance
WorkingMark Guidance
\(5\cos x - 3\sin x = R\cos x\cos\alpha - R\sin x\sin\alpha\)
Equate \(\cos x\): \(5 = R\cos\alpha\)
Equate \(\sin x\): \(3 = R\sin\alpha\)
\(R = \sqrt{5^2+3^2} = \sqrt{34}\ \{= 5.83095...\}\)M1; A1 \(R^2=5^2+3^2\); \(\sqrt{34}\) or awrt 5.8
\(\tan\alpha = \frac{3}{5} \Rightarrow \alpha = 0.5404195003...^c\)M1 \(\tan\alpha = \pm\frac{3}{5}\) or \(\tan\alpha = \pm\frac{5}{3}\) or \(\sin\alpha = \pm\frac{3}{\text{their }R}\) or \(\cos\alpha = \pm\frac{5}{\text{their }R}\)
A1\(\alpha=\) awrt 0.54 or \(\alpha=\) awrt \(0.17\pi\) or \(\alpha = \frac{\pi}{\text{awrt }5.8}\)
Hence, \(5\cos x - 3\sin x = \sqrt{34}\cos(x+0.5404)\)
Total: (4)
Part (b):
AnswerMarks Guidance
WorkingMark Guidance
\(5\cos x - 3\sin x = 4\)
\(\sqrt{34}\cos(x+0.5404) = 4\)
\(\cos(x+0.5404) = \frac{4}{\sqrt{34}}\ \{=0.68599...\}\)M1 \(\cos(x \pm \text{their }\alpha) = \frac{4}{\text{their }R}\)
\((x+0.5404) = 0.814826916...^c\)M1 For applying \(\cos^{-1}\left(\frac{4}{\text{their }R}\right)\)
\(x = 0.2744...^c\)A1 awrt \(0.27^c\)
\((x+0.5404) = 2\pi - 0.814826916...^c\ \{=5.468358...^c\}\)ddM1 \(2\pi - \text{their }0.8148\)
\(x = 4.9279...^c\)A1 awrt \(4.93^c\)
Hence \(x = \{0.27, 4.93\}\)
Total: (5)
[9]
*Note: If there are any EXTRA solutions inside the range \(0 \leq x < 2\pi\), then withhold the final accuracy mark if the candidate would otherwise score all 5 marks. Also ignore EXTRA solutions outside the range \(0 \leq x < 2\pi\).*
## Question 3:

### Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $5\cos x - 3\sin x = R\cos x\cos\alpha - R\sin x\sin\alpha$ | | |
| Equate $\cos x$: $5 = R\cos\alpha$ | | |
| Equate $\sin x$: $3 = R\sin\alpha$ | | |
| $R = \sqrt{5^2+3^2} = \sqrt{34}\ \{= 5.83095...\}$ | M1; A1 | $R^2=5^2+3^2$; $\sqrt{34}$ or awrt 5.8 |
| $\tan\alpha = \frac{3}{5} \Rightarrow \alpha = 0.5404195003...^c$ | M1 | $\tan\alpha = \pm\frac{3}{5}$ or $\tan\alpha = \pm\frac{5}{3}$ or $\sin\alpha = \pm\frac{3}{\text{their }R}$ or $\cos\alpha = \pm\frac{5}{\text{their }R}$ |
| | A1 | $\alpha=$ awrt 0.54 or $\alpha=$ awrt $0.17\pi$ or $\alpha = \frac{\pi}{\text{awrt }5.8}$ |
| Hence, $5\cos x - 3\sin x = \sqrt{34}\cos(x+0.5404)$ | | |

**Total: (4)**

### Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $5\cos x - 3\sin x = 4$ | | |
| $\sqrt{34}\cos(x+0.5404) = 4$ | | |
| $\cos(x+0.5404) = \frac{4}{\sqrt{34}}\ \{=0.68599...\}$ | M1 | $\cos(x \pm \text{their }\alpha) = \frac{4}{\text{their }R}$ |
| $(x+0.5404) = 0.814826916...^c$ | M1 | For applying $\cos^{-1}\left(\frac{4}{\text{their }R}\right)$ |
| $x = 0.2744...^c$ | A1 | awrt $0.27^c$ |
| $(x+0.5404) = 2\pi - 0.814826916...^c\ \{=5.468358...^c\}$ | ddM1 | $2\pi - \text{their }0.8148$ |
| $x = 4.9279...^c$ | A1 | awrt $4.93^c$ |
| Hence $x = \{0.27, 4.93\}$ | | |

**Total: (5)**

**[9]**

*Note: If there are any EXTRA solutions inside the range $0 \leq x < 2\pi$, then withhold the final accuracy mark if the candidate would otherwise score all 5 marks. Also ignore EXTRA solutions outside the range $0 \leq x < 2\pi$.*

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3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos ( x + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$.\\
(b) Hence, or otherwise, solve the equation

$$5 \cos x - 3 \sin x = 4$$

for $0 \leqslant x < 2 \pi$, giving your answers to 2 decimal places.\\

\hfill \mbox{\textit{Edexcel C3 2010 Q3 [9]}}