3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$. (b) Hence, or otherwise, solve the equation $5 \cos x - 3 \sin x = 4$ for $0 \leq x < 2\pi$, giving your answers to 2 decimal places.

A-Level Edexcel Maths Pure Trigonometric Functions: 3. (a) Express $5 \cos x

3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Question 4

$3.-(a)-Express-$5-\cos-x---3-\sin-x$-in-the-form-$R-\cos(x-+-\alpha)$,-where-$R->-0$-and-$0-<-\alpha-<-\frac{1}{2}-\pi$-Edexcel-A-Level Maths Pure-Question 4-2010-Paper 2.png$

3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$. (b) Hence, or otherwise, solve the equatio... show full transcript

Worked Solution & Example Answer:3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Step 1

Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$

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Answer

To express the equation in the desired form, we begin by comparing it with the formula:

$R \cos(x + \alpha) = R \cos x \cos \alpha - R \sin x \sin \alpha.$

First, equate the coefficients:

$R \cos \alpha = 5$
$R \sin \alpha = -3$

Next, we calculate $R$ using the Pythagorean identity:

$R = \sqrt{(5^2) + (-3^2)} = \sqrt{25 + 9} = \sqrt{34}.$

Now we can find \alpha:

$\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{-3}{5} \implies \alpha = \tan^{-1}\left(-\frac{3}{5}\right).$

Thus, we can express it as:

$5 \cos x - 3 \sin x = \sqrt{34} \cos\left(x + \tan^{-1}\left(-\frac{3}{5}\right)\right).$

Step 2

Solve the equation $5 \cos x - 3 \sin x = 4$

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Answer

Using the result from part (a), we solve:

$\sqrt{34} \cos\left(x + \tan^{-1}\left(-\frac{3}{5}\right)\right) = 4.$

This simplifies to:

$\cos\left(x + \tan^{-1}\left(-\frac{3}{5}\right)\right) = \frac{4}{\sqrt{34}}.$

Calculate \tan^{-1}\left(-\frac{3}{5}\right):

Let \alpha = \tan^{-1}( -0.5404 ) \approx -0.5404.

Thus, we rewrite the equation as:

$x - 0.5404 = \cos^{-1}\left(\frac{4}{\sqrt{34}}\right).$

Calculating gives \cos^{-1}(0.6859) which yields the principal solution:

$x = 0.5404 + 0.8150.$

This results in:

First solution: $x \approx 1.3554$ (already in range).

Finding the second solution: $x = 2\pi - (0.5404 + 0.8150) \approx 6.0288 - 0.5404 = 5.4884.$

Finally, rounding off gives the answers: $x \approx 1.36, 5.49.$

3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Worked Solution & Example Answer:3. (a) Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Express $5 \cos x - 3 \sin x$ in the form $R \cos(x + \alpha)$

Solve the equation $5 \cos x - 3 \sin x = 4$

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