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

A-Level Edexcel Maths Pure Basic Trigonometry: Express $6 \, \cos \theta + 8 \,

Express $6 \, \cos \theta + 8 \, \sin \theta$ in the form $R \cos(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

Question 26

$Express-$6-\,-\cos-\theta-+-8-\,-\sin-\theta$-in-the-form-$R-\cos(\theta---\alpha)$,-where-$R->-0$-and-$0-<-\alpha-<-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 26-2013-Paper 1.png$

Express $6 \, \cos \theta + 8 \, \sin \theta$ in the form $R \cos(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the value of \alpha$ to 3... show full transcript

Worked Solution & Example Answer:Express $6 \, \cos \theta + 8 \, \sin \theta$ in the form $R \cos(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

Step 1

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

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Answer

To express $6 \cos \theta + 8 \sin \theta$ in the form $R \cos(\theta - \alpha)$ , we need to find $R$ and $\alpha$ . Using Pythagoras' theorem:

$R^2 = 6^2 + 8^2 = 36 + 64 = 100 \Rightarrow R = 10.$
To find $\alpha$ , we use:

$$\tan \alpha = \frac{8}{6} \Rightarrow \alpha = \arctan\left(\frac{8}{6}\right) \approx 0.927. $Thus, we can express$ 6 \cos \theta + 8 \sin \theta$ as:

$10 \cos(\theta - 0.927).$

Step 2

Calculate $p(\theta)$

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Answer

Now substituting $\alpha$ into $p(\theta)$ gives:

$p(\theta) = \frac{4}{12 + 10 \cos \theta}.$
We find the maximum of this function. To do this, we first find the critical points. The maximum value occurs when $\cos \theta$ is minimized. Since the minimum value of $\cos \theta$ is -1, we substitute:

$p(\theta) = \frac{4}{12 + 10(-1)} = \frac{4}{2} = 2.$
Thus, the maximum value of $p(\theta)$ is $2$ .

Step 3

Find the value of $\theta$ at which the maximum occurs

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Answer

The maximum occurs when:

$\cos \theta = -1 \Rightarrow \theta = \pi.$
Thus, the value of $\theta$ at which the maximum occurs is $\theta \approx 3.142$ .

Express $6 \, \cos \theta + 8 \, \sin \theta$ in the form $R \cos(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

Worked Solution & Example Answer:Express $6 \, \cos \theta + 8 \, \sin \theta$ in the form $R \cos(\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

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

Calculate $p(\theta)$

Find the value of $\theta$ at which the maximum occurs

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