Express \(2 \sin \theta - 3 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 < \alpha < \frac{1}{2}\pi\).
Hence write down the greatest and least possible values of \(1 + 2 \sin \theta - 3 \cos \theta\). [6]
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Answer: \(2\sin\theta - 3\cos\theta = R\sin(\theta - \alpha)\)\(= R\sin\theta\cos\alpha - R\cos\theta\sin\alpha\)
\(\Rightarrow\) Rcosα =2, Rsinα=3
\(\Rightarrow R^2 = 2^2 + 3^2 = 13, R = \sqrt{13}\)
tan α = 3/2.
\(\Rightarrow \alpha = 0.983\)
minimum \(1 - \sqrt{13}\), maximum \(1 + \sqrt{13}\)
Answer Marks
M1 correct pairs
B1 \(R = \sqrt{13}\) or 3.61 or better
M1 A1 0.98 or better accept multiples of π that round to 0.98
B1 B1 or -2.61, 4.61 or better
condone wrong sign at this stage correct division, ft from first M1 radians only accept multiples of π that round to 0.98 allow B1, B1ft for l-√R and 1+√R for their R to 2dp or better
Total: [6] Question 4(i):
Answer: \(x = 2\sin\theta, y = \cos 2\theta\)When \(\theta = \pi/3\), \(x = 2\sin\pi/3 = \sqrt{3}\)
\(y = \cos 2\pi/3 = -\frac{1}{2}\)
Answer Marks
B1 \(x = \sqrt{3}\)
B1 \(y = -\frac{1}{2}\)
exact only (isw all dec answers following exact ans )
Total: [2] Question 4 (EITHER):
Answer: \(\frac{dv}{d\theta} = 2\cos\theta, \frac{dy}{d\theta} = -2\sin 2\theta\)\(\Rightarrow \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) used
\(\frac{dy}{dx} = \frac{-\sin 2\theta}{\cos\theta}\)
\(= \frac{-\sin 2\pi/3}{-\cos\pi/3} = \frac{-\sqrt{3}/2}{-1/2} = -\sqrt{3}\)
Answer Marks
M1 \(\frac{dy}{dx} = (\frac{dy}{d\theta})/(\frac{dx}{d\theta})\) used
A1 any correct equivalent form
A1 exact www
....... ..............................................
M1 A1 A1 exact www
ft their derivatives if right way up (condone one further minor slip if intention clear) condone poor notation can isw if incorrect simplification
Total: [5] Question 4 (OR):
Answer: expressing \(y\) in terms of \(x\), \(y=1-x^2/2\)\(\frac{dy}{dx} = -x\) or \(-2\sin\theta\)
\(= -\sqrt{3}\)
Answer Marks
M1 A1 A1 exact www
Total: [3] Question 4(ii):
Answer: \(y = 1 - 2\sin^2\theta = 1 - 2(x/2)^2 = 1 - \frac{1}{2}x^2\)
Answer Marks
M1A1 or reference to (i) if used there
[2] for M1, need correct trig identity and attempt to substitute for x allow SC B1 for y=cos 2arcsin(x/2) or equivalent
Total: [2]
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**Answer:** $2\sin\theta - 3\cos\theta = R\sin(\theta - \alpha)$
$= R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$
$\Rightarrow$ Rcosα =2, Rsinα=3
$\Rightarrow R^2 = 2^2 + 3^2 = 13, R = \sqrt{13}$
tan α = 3/2.
$\Rightarrow \alpha = 0.983$
minimum $1 - \sqrt{13}$, maximum $1 + \sqrt{13}$
| M1 | correct pairs |
| B1 | $R = \sqrt{13}$ or 3.61 or better |
| M1 | |
| A1 | 0.98 or better accept multiples of π that round to 0.98 |
| B1 B1 | or -2.61, 4.61 or better |
| | **condone wrong sign at this stage** **correct division, ft from first M1 radians only accept multiples of π that round to 0.98** |
| | **allow B1, B1ft for l-√R and 1+√R for their R to 2dp or better** |
**Total: [6]**
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## Question 4(i):
**Answer:** $x = 2\sin\theta, y = \cos 2\theta$
When $\theta = \pi/3$, $x = 2\sin\pi/3 = \sqrt{3}$
$y = \cos 2\pi/3 = -\frac{1}{2}$
| B1 | $x = \sqrt{3}$ |
| B1 | $y = -\frac{1}{2}$ |
| | exact only (isw all dec answers following exact ans ) |
**Total: [2]**
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## Question 4 (EITHER):
**Answer:** $\frac{dv}{d\theta} = 2\cos\theta, \frac{dy}{d\theta} = -2\sin 2\theta$
$\Rightarrow \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ used
$\frac{dy}{dx} = \frac{-\sin 2\theta}{\cos\theta}$
$= \frac{-\sin 2\pi/3}{-\cos\pi/3} = \frac{-\sqrt{3}/2}{-1/2} = -\sqrt{3}$
| M1 | $\frac{dy}{dx} = (\frac{dy}{d\theta})/(\frac{dx}{d\theta})$ used |
| A1 | any correct equivalent form |
| A1 | exact www |
| ....... | .............................................. |
| M1 | |
| A1 | |
| A1 | exact www |
| | ft their derivatives if right way up (condone one further minor slip if intention clear) condone poor notation can isw if incorrect simplification |
**Total: [5]**
---
## Question 4 (OR):
**Answer:** expressing $y$ in terms of $x$, $y=1-x^2/2$
$\frac{dy}{dx} = -x$ or $-2\sin\theta$
$= -\sqrt{3}$
| M1 | |
| A1 | |
| A1 | exact www |
**Total: [3]**
---
## Question 4(ii):
**Answer:** $y = 1 - 2\sin^2\theta = 1 - 2(x/2)^2 = 1 - \frac{1}{2}x^2$
| M1A1 | or reference to (i) if used there |
| [2] | for M1, need correct trig identity and attempt to substitute for x allow SC B1 for y=cos 2arcsin(x/2) or equivalent |
**Total: [2]**
---
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Express $2 \sin \theta - 3 \cos \theta$ in the form $R \sin(\theta - \alpha)$, where $R$ and $\alpha$ are constants to be determined, and $0 < \alpha < \frac{1}{2}\pi$.
Hence write down the greatest and least possible values of $1 + 2 \sin \theta - 3 \cos \theta$. [6]
\hfill \mbox{\textit{OCR MEI C4 2011 Q3 [6]}}