WAEC 2022 · Paper 2 · Q9✱

(a)
Copy and complete the table of values for $y = 3\sin x + 7\cos x$ for $0^\circ \le x \le 180^\circ$ .

$x$ 0° 20° 40° 60° 80° 100° 120° 140° 160° 180°

$y$ 7.0 4.2 −0.9

Model answer

$x$ 0° 20° 40° 60° 80° 100° 120° 140° 160° 180°

$y$ 7.0 7.6 7.3 6.1 4.2 1.7 −0.9 −3.4 −5.6 −7.0

For example, at $x = 20^\circ$ : $y = 3(0.342) + 7(0.940) = 7.6$ (1 d.p.).
(b)
Using a scale of 2 cm to $20^\circ$ on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, draw the graph for $0^\circ \le x \le 180^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to $20^\circ$ , 2 cm to 2 units. The curve rises slightly to about 7.6 near $23^\circ$ , then falls to $-7$ at $180^\circ$ .

For (c): (i) at $x = 150^\circ$ , $y \approx −4.6$ ; (ii) $y > 0$ to the left of the crossing, $0^\circ \le x < 113^\circ$ .
(c)
Using the graph, find the: (i) value of $y$ when $x = 150^\circ$ ; (ii) range of values of $x$ for which $y > 0$ .

$x$	0°	20°	40°	60°	80°	100°	120°	140°	160°	180°
$y$	7.0	7.6	7.3	6.1	4.2	1.7	−0.9	−3.4	−5.6	−7.0

x is in degrees. y = 3 sin x + 7 cos x.

Worked solution (try it first)

In degree mode, to 1 decimal place.
For example, $x = 20^\circ$ : $3(0.342) + 7(0.940) = 1.03 + 6.58$
$\approx 7.6$ .
$x = 160^\circ$ : $3(0.342) + 7(-0.940) \approx -5.6$ .
The full row is $7.0, 7.6, 7.3, 6.1, 4.2, 1.7, -0.9, -3.4, -5.6, -7.0$ .

$y > 0$ where the curve is above the $x$ -axis.
It crosses the axis at about $113^\circ$ , so $y > 0$ for $0^\circ \le x < 113^\circ$ .