WAEC 2019 · Paper 2 · Q10✱✱

(a)
If $\sin p = \frac12$ and $\cos q = -\frac13$ , evaluate $\sin(p - q)$ , where $0^\circ \le p \le 90^\circ$ and $90^\circ \le q \le 180^\circ$ .
(b)
Using the trapezium rule with seven ordinates, evaluate $\displaystyle\int_1^4 \frac{2}{\sqrt{x + 3}}\,dx$ (2 d.p.).

Worked solution (try it first)

(a)

$p$ is acute, so $\cos p = \sqrt{1 - \frac14}$
$= \dfrac{\sqrt3}{2}$ .
$q$ is obtuse, so $\sin q$ is positive: $\sin q = \sqrt{1 - \frac19}$
$= \dfrac{2\sqrt2}{3}$ .
$\sin(p - q) = \sin p\cos q - \cos p\sin q$
$= \dfrac12\left(-\dfrac13\right) - \dfrac{\sqrt3}{2} \times \dfrac{2\sqrt2}{3}$ .
$= -\dfrac16 - \dfrac{2\sqrt6}{6}$
$= \dfrac{-1 - 2\sqrt6}{6}$ .

(b)

Seven ordinates means six strips, $h = 0.5$ .
The ordinates of $\dfrac{2}{\sqrt{x + 3}}$ at $x = 1, 1.5, \ldots, 4$ are $1,\ 0.9428,\ 0.8944,\ 0.8528,\ 0.8165,\ 0.7845,\ 0.7559$ .
First and last: $1.7559$ .
Twice the rest: $2 \times 4.2910 = 8.5820$ .
Trapezium rule: $\dfrac{0.5}{2}(1.7559 + 8.5820) = 2.5845$ , about $2.58$ .

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