Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$. Fully justify your answer. Show that the least value of $\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}$ is $\frac{2 - \sqrt{3}}{2}$. Find the greatest value of \(\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}\.

A-Level AQA Maths Pure Trigonometric Functions: Determine a sequence of transfor

Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2

Question 8

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Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$. Fully justify your answer. Sho... show full transcript

Worked Solution & Example Answer:Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2

Step 1

Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$

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Answer

To transform the graph of $y = \sin x$ into $y = \sqrt{3} \sin x - 3 \cos x + 4$ , we follow these steps:

Identify the amplitude and the resulting transformations: We compare the given function with $R \sin(\alpha x + \phi) + d$ format. Here, we find:
- $R = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$
- The angle ( \alpha ) can be calculated using the identities from trigonometry:
  - ( \tan \alpha = -\frac{-3}{\sqrt{3}} = \sqrt{3} o \alpha = \frac{\pi}{3} )
Translate the function: The equation can now be reformulated:
- $y = 2 \sin(\frac{x}{3} - \phi) + 4$ (Translation by +4 units in the y-direction).
- The graph is stretched vertically by a factor of $2\sqrt{3}$ .
Final transformation description:
- Stretch in the y-direction by a factor of $2\sqrt{3}$ and translate upwards by 4 units.

Step 2

Show that the least value of $\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}$ is $\frac{2 - \sqrt{3}}{2}$

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Answer

To find the least value of (f(x) = \sqrt{3} \sin x - 3 \cos x + 4), we need to determine when (f(x)) reaches its maximum:

Set up the function: [ f(x) = \sqrt{3} \sin x - 3 \cos x + 4 ]
Determine the critical points: Differentiate (f(x)) with respect to (x) to find local maxima and minima.
Use the range of sine and cosine functions: Calculate: [ \text{Max value of } f(x) = 4 + R = 4 + 2\sqrt{3} ]
Find the least value of (\frac{1}{f(x)}): The least value occurs when (f(x)) is at its maximum, hence: [ \text{Least value} = \frac{1}{\text{Max value}} = \frac{1}{(4 + 2\sqrt{3})} = \frac{2 - \sqrt{3}}{2} ]

Step 3

Find the greatest value of $\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}$

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Answer

Using the previous findings:

From the earlier step, we know that the greatest value of ( \sqrt{3} \sin x - 3 \cos x + 4 ) is (4 + 2\sqrt{3}).
Therefore, the greatest value of (\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}) is: [ \frac{1}{4 + 2\sqrt{3}} ]

Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2

Worked Solution & Example Answer:Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$ - AQA - A-Level Maths Pure - Question 8 - 2018 - Paper 2

Determine a sequence of transformations which maps the graph of $y = \sin x$ onto the graph of $y = \sqrt{3} \sin x - 3 \cos x + 4$

Show that the least value of \(\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}\) is \(\frac{2 - \sqrt{3}}{2}\)

Find the greatest value of \(\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}\)

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