(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = 2\cos 2x - 1\)
1.0
0
1.0
(b) Using a scale of 2cm = 30^\circ on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0^\circ \leq x \leq 180^\circ\).
(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 \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = 2\cos 2x - 1\)
1.0
0
-2
-3
-2
0
1.0
(b)
(c)
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = \frac{1}{180} (x - 360)\)
-2.0
-1.83
-1.67
-1.5
-1.33
-1.17
-1.0
(d)(i) \(2 \cos 2x + \frac{1}{2} = 0\)
\(2 \cos 2x + \frac{1}{2} - 1\frac{1}{2} = -1\frac{1}{2}\)
\(2 \cos 2x - 1 = -1\frac{1}{2}\)
The values of x at \(2 \cos 2x + \frac{1}{2} = 0\) are x = 51^\circ and 129^\circ.
(ii) \(2 \cos 2x - \frac{x}{180} + 1 = 0\)
\(2 \cos 2x - 1 = \frac{1}{180} (x - 360)\)
The values of x is where the two graphs intersect; which are at x = 54^\circ and 132^\circ.
Source: Myschool Past Questions Archive (Record #54528)