JAMB 2003 · UME · Q28

If $\frac\pi2 \le \theta \le 2\pi$ , find the maximum value of $f(\theta) = \dfrac{4}{6 + 2\cos\theta}$ .

Worked solution (try it first)

The top is fixed at 4, so the fraction is largest when the bottom, $6 + 2\cos\theta$ , is smallest.
$\cos\theta$ is smallest ( $-1$ ) at $\theta = \pi$ , which is inside the range.
The bottom is then $6 - 2 = 4$ .
Sine and cosine are both over the hypotenuse, 5, so find it by Pythagoras first.
$1\frac13$ (option A) is $\frac43 = \frac{1}{\tan\theta}$ , not $\sin\theta + \cos\theta$ .