1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the value of $\alpha$ to 3 decimal places. (b) Hence write down the minimum value of $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$. (c) Solve, for $0 < x < 2\pi$, the equation $7 \, ext{cos} \, x - 24 \, ext{sin} \, x = 10$ giving your answers to 2 decimal places.

A-Level Edexcel Maths Pure Differentiation: 1. (a) Express $7 \, ext{cos} \

1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Question 2

$1.-(a)-Express-$7-\,--ext{cos}-\,-x---24-\,--ext{sin}-\,-x$-in-the-form-$R-\,--ext{cos}-\,-(x-+-\alpha)$-where-$R->-0$-and-$0-<-\alpha-<-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 2-2011-Paper 4.png$

1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the val... show full transcript

Worked Solution & Example Answer:1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Step 1

Express $7 \cos x - 24 \sin x$ in the form $R \cos (x + \alpha)$

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Answer

To express the equation in the form ( R \cos(x + \alpha) ), we identify:

Set coefficients:
- $R \cos \alpha = 7$
- $R \sin \alpha = -24$
Calculate ( R ): $R = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25$
Determine ( \alpha ): $\tan(\alpha) = \frac{-24}{7} \Rightarrow \alpha = \tan^{-1}\left(-\frac{24}{7}\right) \approx -1.287$ (radians) But since $\alpha$ must be in $0 < \alpha < \frac{\pi}{2}$ , we take the positive angle that gives us the same tangent value, which is approximately $2.354$ . Thus, the answer to three decimal places is:
- $\alpha \approx 2.355$ .

Step 2

Hence write down the minimum value of $7 \cos x - 24 \sin x$

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Answer

The expression $7 \cos x - 24 \sin x$ can take on minimum values determined by ( -R ), therefore:

Minimum value = $-25$ .

Step 3

Solve, for $0 < x < 2\pi$, the equation $7 \cos x - 24 \sin x = 10$

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Answer

From the equation:

$7 \cos x - 24 \sin x = 10$

We can express this as:

$25 \cos(x + \alpha) = 10 \implies \cos(x + \alpha) = \frac{10}{25} = 0.4$

Now, calculating ( x + \alpha ):

First find the principal value: $x + \alpha = \cos^{-1}(0.4) \approx 1.159$
Now, considering the periodic nature of cosine, we have: $x + \alpha = 1.159 + 2k\pi, \quad \text{and} \quad x + \alpha = -1.159 + 2k\pi$

Thus, the values for $x$ in the specified range can be derived:

For $k=0$ : $x \approx 1.159 - 2.355 \approx -1.196 \quad (not \, valid)$ $x \approx 1.159 - 2.354 \approx 4.045 \quad (valid: \approx 4.05)$
For $k=1$ : $x \approx 1.159 + 2\pi \approx 7.443 \quad (not \, valid \, in \, given \, range)$ $x \approx -1.159 + 2\pi \approx 5.124 \quad (valid: \approx 5.12)$

Thus, the solutions are:

$x \approx 4.05$ and $x \approx 5.12$ to two decimal places.

1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Worked Solution & Example Answer:1. (a) Express $7 \, ext{cos} \, x - 24 \, ext{sin} \, x$ in the form $R \, ext{cos} \, (x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 4

Express $7 \cos x - 24 \sin x$ in the form $R \cos (x + \alpha)$

Hence write down the minimum value of $7 \cos x - 24 \sin x$

Solve, for $0 < x < 2\pi$, the equation $7 \cos x - 24 \sin x = 10$

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