3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$. Give the exact value of $R$ and give the value of $R$ to 2 decimal places. (b) Hence solve, for $0 < \theta < 360^\circ$, $$ \frac{2}{2 \text{cos} \, \theta - \text{sin} \, \theta} = 15 $$ Give your answers to one decimal place. (c) Use your solutions to parts (a) and (b) to deduce the smallest positive value of $\theta$ for which $$ \frac{2}{2 \text{cos} \, \theta + \text{sin} \, \theta} = 15 $$ Give your answer to one decimal place.

A-Level Edexcel Maths Pure Modelling with Functions: 3. (a) Express $2 \, ext{cos} \

3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Question 4

$3.-(a)-Express-$2-\,--ext{cos}-\,--heta----ext{sin}-\,--heta$-in-the-form-$R-\,--ext{cos}-(-heta-+-\alpha)$,-where-$R$-and-$\alpha$-are-constants,-$R->-0$-and-$0-<-\alpha-<-90^\circ$-Edexcel-A-Level Maths Pure-Question 4-2016-Paper 3.png$

3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \... show full transcript

Worked Solution & Example Answer:3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Step 1

Express $2 \cos \theta - \sin \theta$ in the form $R \cos (\theta + \alpha)$

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Answer

To express $2 \cos \theta - \sin \theta$ in the form $R \cos (\theta + \alpha)$ , we match it to the standard form. Looking for constants, we need:

Find $R$ : $R = \sqrt{a^2 + b^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$ Thus, the exact value of $R$ is $\sqrt{5}$ .
Find $\alpha$ : Using the relationship: $\tan \alpha = \frac{-1}{2} \implies \alpha = \tan^{-1}(-0.5) \approx 26.57^\circ$

Thus, $2 \cos \theta - \sin \theta = \sqrt{5} \cos \left(\theta + 26.57^\circ \right)$ , with the value of $R \approx 2.24$ to two decimal places.

Step 2

Solve for $0 < \theta < 360^\circ$

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Answer

Starting from the equation:

$\frac{2}{2 \cos \theta - \sin \theta} = 15$

Multiplying both sides by the denominator yields:

$2 = 15 (2 \cos \theta - \sin \theta)$ \Rightarrow 15 \sin \theta - 30 \cos \theta = -2 $$

Rearranging gives:

$15 \sin \theta - 30 \cos \theta = -2$

This can now be manipulated based on the earlier form found. Solving yields: $\theta \approx 33.00^\circ \text{ and } 273.29^\circ$

Step 3

Deduce the smallest positive value of $\theta$

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Answer

From the solutions in part (b), we have:

$\theta_1 \approx 33.00^\circ$
$\theta_2 \approx 273.29^\circ$

The smallest positive value of $\theta$ for which $\frac{2}{2 \cos \theta + \sin \theta} = 15$ is the first solution:

Thus, the answer is $33.0^\circ$ .

3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Worked Solution & Example Answer:3. (a) Express $2 \, ext{cos} \, heta - ext{sin} \, heta$ in the form $R \, ext{cos} ( heta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 3

Express $2 \cos \theta - \sin \theta$ in the form $R \cos (\theta + \alpha)$

Solve for $0 < \theta < 360^\circ$

Deduce the smallest positive value of $\theta$

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