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

1 $\frac{1}{2}$ $\frac{1}{4}$ $\sqrt{2}$

$\frac{1}{2}$

$y =\sin x \cos x$ $\sin x$ $2 \sin x \cos x = \sin 2x$ 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  $-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}$     $\Rightarrow \frac{-1}{2} \leq y \leq \frac{1}{2}$ ymax = $\frac{1}{2}$