The maximum value of (sin x)(cos x) is :

$y =\sin x \cos x$

$\sin x$

$2 \sin x \cos x = \sin 2x$

so  $y =\frac{1}{2} \times 2 \times \sin x \cos x$

$y =\frac{1}{2} \sin 2 x$

$-1 \leq \sin 2 x \leq 1$

so  $-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}$     $\Rightarrow \frac{-1}{2} \leq y \leq \frac{1}{2}$

y_max = $\frac{1}{2}$