A-Level Edexcel Maths Pure Equation of a Straight Line: Express $2 \, ext{cos} \theta -

Express $2 \, ext{cos} \theta - \text{sin} \theta$ in the form $R\text{cos}(\theta + \alpha)$, 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

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Express $2 \, ext{cos} \theta - \text{sin} \theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $ \alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\... show full transcript

Worked Solution & Example Answer:Express $2 \, ext{cos} \theta - \text{sin} \theta$ in the form $R\text{cos}(\theta + \alpha)$, 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 $2 \text{cos} \theta - \text{sin} \theta$ in the form $R\text{cos}(\theta + \alpha)$

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Answer

$R = \sqrt{5}$ , \alpha = 26.57^\circ$

To express $2 \text{cos} \theta - \text{sin} \theta$ in the form $R \text{cos}(\theta + \alpha)$ , we use the identities:

$R = \sqrt{a^2 + b^2}$

a = 2 and b = -1. Thus, we compute:

$R = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Next, we find $\alpha$ using:

$\tan \alpha = \frac{-1}{2}$

Calculating gives $\alpha \approx 26.57^\circ$ . Therefore, we have:

$2 \text{cos} \theta - \text{sin} \theta = \sqrt{5} \text{cos}(\theta + 26.57^\circ)$

Step 2

Hence solve, for $0 \leq \theta < 360^\circ$, $ \frac{2}{2 \text{cos} \theta - \text{sin} \theta} = 15 $

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Answer

We rearrange the equation as:

$2 = 15 (2 \text{cos} \theta - \text{sin} \theta)$

Solving gives:

$\text{cos} \theta = \frac{17}{30}$

Calculating this leads to:

$\theta \approx 33.00^\circ$
$\theta \approx 273.00^\circ$ (due to periodicity)

Thus, the solutions for $\theta$ are approximately $33.0^\circ$ and $273.0^\circ$ .

Step 3

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 $

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Answer

To solve for the new equation, we rearrange as:

$2 = 15 (2 \text{cos} \theta + \text{sin} \theta)$

This leads to:

$\text{cos} \theta = \frac{17}{30}$

The smallest positive value for $\theta$ here is the same as calculated, leading directly to:

$\theta \approx 86.1^\circ$

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

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

Hence solve, for $0 \leq \theta < 360^\circ$, $ \frac{2}{2 \text{cos} \theta - \text{sin} \theta} = 15 $

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 $

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