WAEC 2019 · Paper 2 · Q9

(a)
Copy and complete the table of values for $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .

$x$ $0^\circ$ $60^\circ$ $120^\circ$ $180^\circ$ $240^\circ$ $300^\circ$ $360^\circ$

$y$ $2.0$ $-3.6$

Model answer

$x$ 0° 60° 120° 180° 240° 300° 360°

$y$ 2.0 3.6 1.6 −2.0 −3.6 −1.6 2.0

For example, at $x = 60^\circ$ : $y = 2(0.5) + 3(0.866) = 3.6$ (1 d.p.).
(b)
Using a scale of 2 cm to $60^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). With points only every $60^\circ$ , draw a smooth wave: it peaks at about $3.6$ near $56^\circ$ and dips to about $-3.6$ near $236^\circ$ .

For (c): (i) draw $y = -1$ : $x \approx 162.4^\circ$ and $310.2^\circ$ . (ii) At $x = 342^\circ$ , $y \approx 1.0$ .
(c)(i)
Using the graph, solve $2\cos x + 3\sin x = -1$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Using the graph, find, correct to one decimal place, the value of $y$ when $x = 342^\circ$ .

$x$	0°	60°	120°	180°	240°	300°	360°
$y$	2.0	3.6	1.6	−2.0	−3.6	−1.6	2.0

x in degrees. The line y = −1 gives the solutions of (c)(i).

Worked solution (try it first)

Draw the line $y = -1$ and read down from where it meets the curve: $x \approx 162^\circ$ and $x \approx 310^\circ$ .

Read up from $x = 342^\circ$ to the curve and across: $y \approx 1.0$ .
(By calculation, $2\cos 342^\circ + 3\sin 342^\circ \approx 1.90 - 0.93$
$= 0.97$ .)