In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Given that $$2 \, ext{sin}(x - 60^ ext{o}) = \text{cos}(x - 30^ ext{o})$$ show that $$\tan{x} = 3 \sqrt{3}$$ (b) Hence or otherwise solve, for $0 \leq x < 180^ ext{o}$ $$2 \text{sin} 2\theta = \text{cos}(2\theta + 30^ ext{o})$$ giving your answers to one decimal place.

A-Level Edexcel Maths Pure Applications of Differentiation: In this question you must show a

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Question 15

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Given that $$2 \, ext{sin}(... show full transcript

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Step 1

Given that 2 sin(x - 60°) = cos(x - 30°) show that tan x = 3 √3

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Answer

To prove that $\tan{x} = 3\sqrt{3}$ based on the equation, we start by applying the compound angle formulas:

Expand the equation: $2 \text{sin}(x - 60^ ext{o}) = 2 \left( \text{sin}(x) \text{cos}(60^ ext{o}) - \text{cos}(x) \text{sin}(60^ ext{o}) \right)$ This gives: $2 \text{sin}(x) \cdot \frac{1}{2} - 2 \text{cos}(x) \cdot \frac{\sqrt{3}}{2}$ Thus, $\text{sin}(x) - \sqrt{3} \text{cos}(x)$
Simplify: $\text{sin}(x) - \sqrt{3} \text{cos}(x) = \text{cos}(x - 30^ ext{o})$ Expanding the right side gives us: $\text{cos}(x) \cdot \text{cos}(30^ ext{o}) + \text{sin}(x) \cdot \text{sin}(30^ ext{o})$ leading to: $\text{cos}(x) \cdot \frac{\sqrt{3}}{2} + \text{sin}(x) \cdot \frac{1}{2}$
Setting the equations equal to each other: $\text{sin}(x) - \sqrt{3} \text{cos}(x) = \text{cos}(x) \cdot \frac{\sqrt{3}}{2} + \text{sin}(x) \cdot \frac{1}{2}$ Rearranging gives: $\text{sin}(x) - \text{sin}(x) \cdot \frac{1}{2} = \text{cos}(x) \cdot \frac{\sqrt{3}}{2} + \sqrt{3} \text{cos}(x)$ This simplifies to: $\frac{1}{2} \text{sin}(x) = \left(\sqrt{3} + \frac{\sqrt{3}}{2}\right) \text{cos}(x)$
Divide both sides by cos(x) and simplify: $\tan{x} = 3\sqrt{3}$ Thus, showing the required relation.

Step 2

Hence or otherwise solve, for 0 ≤ x < 180° 2 sin 2θ = cos(2θ + 30°)

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Answer

To solve the equation:

Rewrite the equation: $2 \text{sin}(2\theta) = \text{cos}(2\theta + 30^ ext{o})$
Use the cosine angle addition formula: Rewrite the right side: $\text{cos}(2\theta + 30^ ext{o}) = \text{cos}(2\theta) \text{cos}(30^ ext{o}) - \text{sin}(2\theta) \text{sin}(30^ ext{o})$ This results in: $\text{cos}(2\theta) \cdot \frac{\sqrt{3}}{2} - \text{sin}(2\theta) \cdot \frac{1}{2}$
Set the equations equal: Thus, we have: $2 \text{sin}(2\theta) = \text{cos}(2\theta) \cdot \frac{\sqrt{3}}{2} - \text{sin}(2\theta) \cdot \frac{1}{2}$ Which simplifies to: $2\text{sin}(2\theta) + \frac{1}{2}\text{sin}(2\theta) = \frac{\sqrt{3}}{2} \text{cos}(2\theta)$ or: $\frac{5}{2} \text{sin}(2\theta) = \frac{\sqrt{3}}{2} \text{cos}(2\theta)$ Thus: $\tan(2\theta) = \frac{\sqrt{3}}{5}$
Find angles: Solve for $2\theta$ : $2\theta = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)$ This gives a solution approximately: $2\theta \approx 0.346$ Thus, $\theta \approx 0.173\text{o}$
Use periodicity for other solutions: Add $heta = 180^ ext{o}$ to find more values.
Final values: Convert to one decimal place for the final answer.

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Given that 2 sin(x - 60°) = cos(x - 30°) show that tan x = 3 √3

Hence or otherwise solve, for 0 ≤ x < 180° 2 sin 2θ = cos(2θ + 30°)

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