(a) Copy and complete the table for the relation (y = 2 cos 2x – 1).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| (y = 2cos 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).
Explanation
(a)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| (y = 2cos 2x – 1) | 1.0 | 0 | -2 | -3 | -2 | 0 | 1.0 |
(b)
(c)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| (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} – 1frac{1}{2} = -1frac{1}{2})
(2 cos 2x – 1 = -1frac{1}{2})
The values of x at (2 cos 2x + frac{1}{2} = 0) are x = 51° and 129°.
(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° and 132°.