WAEC 2014 · Paper 2 · Q7

(a)

Copy and complete the table of values for the relation $y = 2\sin x + 1$ .

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$	$270^\circ$
$y$	$1.0$		$2.7$					$0.0$	$-0.7$

Model answer

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$	$270^\circ$
$y$	$1.0$	$2.0$	$2.7$	$3.0$	$2.7$	$2.0$	$1.0$	$0.0$	$-0.7$	$-1.0$

For example $x = 30^\circ$ : $2(0.5) + 1 = 2.0$ , and $x = 270^\circ$ : $2(-1) + 1 = -1.0$ .

(b)
Using scales of 2 cm to $30^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 2\sin x + 1$ for $0^\circ \le x \le 270^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve rises from 1 to its maximum 3 at $90^\circ$ and falls to $-1$ at $270^\circ$ .

For (c): $\sin x = \frac14$ means $2\sin x + 1 = 1.5$ , so draw $y = 1.5$ : it meets the curve at $x \approx 14.5^\circ$ and $165.5^\circ$ .
(c)
Use the graph to find the values of $x$ for which $\sin x = \frac14$ .
Separate values with commas, e.g. 3, −2

x in degrees. The line y = 1.5 gives sin x = ¼.

Worked solution (try it first)

The graph is of $2\sin x + 1$ , so change the equation to match: $\sin x = \frac14$ gives $2\sin x + 1 = 2 \times \frac14 + 1 = 1.5$ .
Draw the line $y = 1.5$ and read down from where it crosses the curve: $x \approx 15^\circ$ and $x \approx 165^\circ$ .