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8 (a) 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

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8-(a)-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.png

8 (a) 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 full transcript

Worked Solution & Example Answer:8 (a) 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 map the graph of y=sin⁡xy = \sin x onto the graph of y=3sin⁡x−3cos⁡x+4y = \sqrt{3} \sin x - 3 \cos x + 4, we can follow these steps:

  1. Identify the coefficients: We start with y=3sin⁡x−3cos⁡xy = \sqrt{3} \sin x - 3 \cos x. We can rewrite this in the form y=Rsin⁡(x+α)y = R \sin (x + \alpha) by finding RR and α\alpha. Using the identities, we find:

    • R=(3)2+(−3)2=3+9=12=23R = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}.
    • tan⁡α=−33=−3⇒α=−π3\tan \alpha = \frac{-3}{\sqrt{3}} = -\sqrt{3} \Rightarrow \alpha = -\frac{\pi}{3}.

  2. Translation: The graph is translated vertically. The constant +4+4 indicates a vertical translation upwards by 4 units.

  3. Final transformation: Therefore, we express the transformation as:

    • Start with y=sin⁡xy = \sin x.
    • Stretch vertically by a factor of 232\sqrt{3}: y=23sin⁡(x+π3)y = 2\sqrt{3} \sin(x + \frac{\pi}{3}).
    • Finally, translate the graph upwards by 4 units: y=23sin⁡(x+π3)+4y = 2\sqrt{3} \sin(x + \frac{\pi}{3}) + 4.

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 determine the least value of the expression 13sin⁡x−3cos⁡x+4\frac{1}{\sqrt{3} \sin x - 3 \cos x + 4}, we need to find the maximum of the denominator 3sin⁡x−3cos⁡x+4\sqrt{3} \sin x - 3 \cos x + 4:

  1. Find the derivative: By setting the derivative satisfactory for the critical points, we can analyze the maximum.

  2. Equation manipulation: Setting tan⁡x=33\tan x = \frac{3}{\sqrt{3}} gives critical points. Upon evaluating inside this, we can find that:

    • The least value is calculated when sin⁡x=−1\sin x = -1 and cos⁡x=−12\cos x = -\frac{1}{2}, leading to a value of 2−32\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

The maximum of the denominator 3sin⁡x−3cos⁡x+4\sqrt{3} \sin x - 3 \cos x + 4 can be evaluated similarly:

  1. Investigation for maximum: Check in critical angles to establish where functions approach their limits.
  2. Conclusion: Through evaluation, the greatest value will be derived from expressions found from graph behavior which we relate back through analysis of the oscillations of sine and cosine, giving the final answer as 2+32\frac{2 + \sqrt{3}}{2}.

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