Figure 1 shows an oscilloscope screen. The curve shown on the screen satisfies the equation y = \sqrt{3} cos x + sin x. (a) Express the equation of the curve in the form y = R sin(x + \alpha), where R and \alpha are constants, R > 0 and 0 < \alpha < \frac{\pi}{2}. (b) Find the values of x, 0 < x < 2\pi, for which y = 1.

A-Level Edexcel Maths Pure Polynomials: Figure 1 shows an oscilloscope s

Figure 1 shows an oscilloscope screen - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

Question 7

Figure 1 shows an oscilloscope screen. The curve shown on the screen satisfies the equation y = \sqrt{3} cos x + sin x. (a) Express the equation of the curve in t... show full transcript

Worked Solution & Example Answer:Figure 1 shows an oscilloscope screen - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

Step 1

Express the equation of the curve in the form y = R sin(x + α)

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Answer

To express the equation ( y = \sqrt{3} cos x + sin x ) in the form ( y = R sin(x + \alpha) ), we identify ( R ) and ( \alpha ) using the following relationships:

Calculate ( R ): $R^2 = (\sqrt{3})^2 + (1)^2 = 3 + 1 = 4 \Rightarrow R = 2$
Determine ( \alpha ): $\tan \alpha = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{\pi}{6}$ .

Thus, the equation can be written as: $y = 2 \sin(x + \frac{\pi}{6})$ .

Step 2

Find the values of x, 0 < x < 2π, for which y = 1

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Answer

To find the values of ( x ) for which ( y = 1 ), we set up the equation:

$1 = 2 \sin(x + \frac{\pi}{6})$

From this, we can isolate ( \sin(x + \frac{\pi}{6}) ): $\sin(x + \frac{\pi}{6}) = \frac{1}{2}$

The general solutions for this equation are given by:

( x + \frac{\pi}{6} = \frac{\pi}{6} + 2k\pi ) (First quadrant)
( x + \frac{\pi}{6} = \frac{5\pi}{6} + 2k\pi ) (Second quadrant)

For the first case: $x = 0 + 2k\pi - \frac{\pi}{6} \Rightarrow x = -\frac{\pi}{6} + 2k\pi$ This does not satisfy ( 0 < x < 2\pi ).

For the second case: $x = \frac{5\pi}{6} - \frac{\pi}{6} \Rightarrow x = \frac{4\pi}{6} = \frac{2\pi}{3}$

Now including both solutions within the range:

For k=0: ( x = \frac{2\pi}{3} )
For k=1: ( x = 2\pi - \frac{\pi}{6} + 2\pi = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} )

Thus, the valid values of ( x ) are: $x = \frac{2\pi}{3}, \frac{11\pi}{6}$ .

Figure 1 shows an oscilloscope screen - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

Worked Solution & Example Answer:Figure 1 shows an oscilloscope screen - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

Express the equation of the curve in the form y = R sin(x + α)

Find the values of x, 0 < x < 2π, for which y = 1

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