WAEC 2025 · Paper 2 · Q8✱✱

(a)
Complete the table of values for $y = 2\cos2x - 1$ for $x = 0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 180^\circ$ .
Separate values with commas, e.g. 3, −2
(b)
Draw the graph of $y = 2\cos2x - 1$ for $0^\circ \le x \le 180^\circ$ . On the same axes, draw $y = \frac{1}{180}(x - 360)$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve starts at 1, falls to $-3$ at $90^\circ$ and rises back to 1 at $180^\circ$ . The line $y = \frac{1}{180}(x - 360)$ runs from $(0^\circ, -2)$ to $(180^\circ, -1)$ .

For (c): $2\cos 2x + \frac12 = 0$ is $2\cos 2x - 1 = -1.5$ , so draw $y = -1.5$ (not the sloping line): $x \approx 52.2^\circ$ and $127.8^\circ$ . The sloping line would solve a different equation; it meets the curve at $x \approx 55.1^\circ$ and $131.1^\circ$ .
(c)
Use the graph to find the values of $x$ for which $2\cos2x + \frac12 = 0$ .
Separate values with commas, e.g. 3, −2

x is in degrees. For (c), 2cos 2x + ½ = 0 is where the curve meets y = −1.5.

Worked solution (try it first)

Plot the seven points and join them with a smooth curve.
The line $y = \frac{1}{180}(x - 360)$ passes through $(0, -2)$ and $(180, -1)$ : draw it with a ruler.

Rearrange to match the curve: $2\cos 2x + \frac12 = 0$ is $2\cos 2x - 1 = -1\frac12$ .
Draw $y = -1.5$ and read where it meets the curve: $x \approx 52^\circ$ and $x \approx 128^\circ$ .