(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\). x 0° 30° 60° 90° 120° 150° 180° \(y = 2\cos 2x - 1\) 1.0 0 1.0 (b) Using a scale of ...
Question 1 Report
(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| \(y = 2\cos 2x - 1\) | 1.0 | 0 | 1.0 |
(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0° \leq x \leq 180°\).
(c) On the same axis, draw the graph of \(y = \frac{1}{180} (x - 360)\)
(d) Use your graphs to find the : (i) values of x for which \(2 \cos 2x + \frac{1}{2} = 0\); (ii) roots of the equation \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
Answer Details
(a) For each value of x, evaluate \(y=2\cos 2x-1\).
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=2\cos2x-1\) | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.
For the straight line, the plotting values are:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=\dfrac{x-360}{180}\) | -2.00 | -1.83 | -1.67 | -1.50 | -1.33 | -1.17 | -1.00 |
(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]
(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]
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