(a) Copy and complete the table for the relation \(y = 2 \cos 2x -

MATHEMATICS
WAEC 1995

(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).

x30°60°90°120°150°180°
\(y = 2\cos 2x - 1\)1.001.0

(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)

x30°60°90°120°150°180°
\(y = 2\cos 2x - 1\)1.00-2-3-201.0

(b)

(c)

x30°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) \(2 \cos 2x + \frac{1}{2} = 0\)

\(2 \cos 2x + \frac{1}{2} - 1\frac{1}{2} = -1\frac{1}{2}\)

\(2 \cos 2x - 1 = -1\frac{1}{2}\)

The values of x at \(2 \cos 2x + \frac{1}{2} = 0\) are x = 51° 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°.



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