4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the exact value of $R$ and give the value of $\alpha$ in radians to 3 decimal places. (b) Show that the equation $$5\cot2x - 3\csc2x = 2$$ can be rewritten in the form $$5\cos2x - 2\sin2x = c$$ where $c$ is a positive constant to be determined. (c) Hence or otherwise, solve, for $0 \leq x < \pi$, $$5\cot2x - 3\csc2x = 2$$ giving your answers to 2 decimal places. (Solutions based entirely on graphical or numerical methods are not acceptable.)

A-Level Edexcel Maths Pure Equation of a Straight Line: 4. (a) Write $5 ext{cos }\theta

4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Question 6

$4.-(a)-Write-$5-ext{cos-}\theta---2-ext{sin-}\theta$-in-the-form-$R\text{cos}(\theta-+-\alpha)$,-where-$R$-and-$\alpha$-are-constants,-$R->-0$-and-$0-<-\alpha-<-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 6-2017-Paper 4.png$

4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \fra... show full transcript

Worked Solution & Example Answer:4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Step 1

Write $5\text{cos }\theta - 2\text{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$

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Answer

To express $5\text{cos }\theta - 2\text{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$ , we can use the identity:

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

where (a = 5) and (b = -2). Thus,

$R = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}$

Next, we find the angle $\alpha$ using:

$\tan \alpha = \frac{b}{a} = \frac{-2}{5}$

So,

$\alpha = \tan^{-1}\left(\frac{-2}{5}\right) \approx -0.3803 \, \text{(in radians)}.$

To fit into the given range $0 < \alpha < \frac{\pi}{2}$ , we convert:

$\alpha = \pi + (-0.3803) \approx 2.7613 ext{ radians, rounding to three decimal places gives } 2.761.$

Step 2

Show that the equation can be rewritten

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Answer

To rewrite the equation $5\cot2x - 3\csc2x = 2$ , we substitute in the trigonometric identities where:

$\cot2x = \frac{\cos2x}{\sin2x} \text{ and } \csc2x = \frac{1}{\sin2x}.$

Thus, the equation becomes:

$5\frac{\cos2x}{\sin2x} - 3\frac{1}{\sin2x} = 2.$

Multiplying the entire equation by $\sin2x$ leads to:

$5\cos2x - 3 = 2\sin2x.$

Rearranging gives:

$5\cos2x - 2\sin2x = 3$

This matches the required form, establishing that $c = 3$ .

Step 3

Solve for $0 \leq x < \pi$

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Answer

Using the equation derived, $5\cos2x - 2\sin2x = 3$ , we can express $\sin2x$ in terms of $\cos2x$ to find our solutions:

$\sin2x = \frac{5}{2}\cos2x - \frac{3}{2}.$

Next, we can substitute $\sin2x = \sqrt{1 - \cos^2 2x}$ into the equation, further manipulating it to find:

$satisfying$ 5(\sqrt{1 - \cos^2 2x}) - 2\cos^2 2x = 3$$

Solving for $x$ in the specified interval gives:

$x \approx 0.30 \, \text{and} \, 2.46 \, \text{ (to two decimal places)}.$

4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Worked Solution & Example Answer:4. (a) Write $5 ext{cos }\theta - 2 ext{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 6 - 2017 - Paper 4

Write $5\text{cos }\theta - 2\text{sin }\theta$ in the form $R\text{cos}(\theta + \alpha)$

Show that the equation can be rewritten

Solve for $0 \leq x < \pi$

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