a. Copy and complete the table for the relation y = 2cos2x - 1
x
0º
30º
60º
90º
120º
150º
180º
y = 2cos 2x - 1
1.0
0
1.0
b. Using a scale of 2cm = 30º on the x - axis and 2cm = 1 unit on the y - axis, draw the graph of y = 2cos2x - 1 for 0º ≤ x ≤ 180º
c. On the same axes, draw the graph of y = \(\frac{1}{180}\)(x - 360)
d. Use your graphs to find the values of x for which 2cos2x + \(\frac{1}{2}\) = 0
Verified Correct Answer: .
Step-by-Step Mathematical Solution
a.
x
0º
30º
60º
90º
120º
150º
180º
cos2x
1
0.5
-0.5
-1.0
-0.5
0.5
1
2cos2x
2
1.0
-1.0
-2.0
-1.0
1.0
2
y = 2cos2x - 1
1.0
0
-2.0
-3.0
-2.0
0
1.0
b See graph above
c. See table and graph above
d. 2cos2x + \(\frac{1}{2}\) = 0
Substract 1\(\frac{1}{2}\) from both sides
2cos2x + \(\frac{1}{2}\) - 1\(\frac{1}{2}\) = 0 - 1\(\frac{1}{2}\)
From the graph, look for the value of x corresponding to where y = + 1\(\frac{1}{2}\). This occurs at x = 51º and 129º
Source: WAEC / JAMB / NECO Classroom Archive (Record #75950)