4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the value of $\alpha$ to 3 decimal places. (b) $p(\theta) = \frac{4}{12 + 6\text{cos}(\theta) + 8\text{sin}(\theta)} \quad 0 \leq \theta \leq 2\pi$. Calculate (i) the maximum value of $p(\theta)$, (ii) the value of $\theta$ at which the maximum occurs.

A-Level Edexcel Maths Pure Trigonometric Functions: 4. (a) Express $6 \, ext{cos} \

4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1

Question 25

$4.-(a)-Express-$6-\,--ext{cos}-\,--heta-+-8-\,--ext{sin}-\,--heta$-in-the-form-$R-\text{cos}(\theta---\alpha)$,-where-$R->-0$-and-$0-<-\alpha-<-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 25-2013-Paper 1.png$

4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. G... show full transcript

Worked Solution & Example Answer:4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1

Step 1

Express $6 \, \text{cos} \, \theta + 8 \, \text{sin} \, \theta$ in the form $R \text{cos}(\theta - \alpha)$

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Answer

To express $6 \, \text{cos} \, \theta + 8 \, \text{sin} \, \theta$ in the form $R \text{cos}(\theta - \alpha)$ , we first find $R$ using the Pythagorean theorem:

$R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10.$

Next, we calculate $\alpha$ using the tangent function:

$\tan(\alpha) = \frac{8}{6} = \frac{4}{3}.$

Thus, we find:

$$\alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927 \text{ (to 3 decimal places)}.$

Step 2

Calculate (i) the maximum value of $p(\theta)$

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Answer

To find the maximum value of $p(\theta)$ , we rewrite the function:

$p(\theta) = \frac{4}{12 + 10\cos(\theta - \alpha)}$

The maximum value occurs when the denominator is minimized, which happens when $\cos(\theta - \alpha) = -1$ :

$\text{Minimum } 12 + 10(-1) = 2.$

Thus, the maximum value of $p(\theta)$ is:

$p(\theta) = \frac{4}{2} = 2.$

Step 3

Calculate (ii) the value of $\theta$ at which the maximum occurs

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Answer

The maximum occurs when:

$\cos(\theta - \alpha) = -1.$

This results in:

$\theta - \alpha = \pi \\ \Rightarrow \theta = \pi + \alpha.$

Substituting the value of $\alpha$ :

$\theta = \pi + 0.927 \approx 4.07.$

4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1

Worked Solution & Example Answer:4. (a) Express $6 \, ext{cos} \, heta + 8 \, ext{sin} \, heta$ in the form $R \text{cos}(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1

Express $6 \, \text{cos} \, \theta + 8 \, \text{sin} \, \theta$ in the form $R \text{cos}(\theta - \alpha)$

Calculate (i) the maximum value of $p(\theta)$

Calculate (ii) the value of $\theta$ at which the maximum occurs

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