Answer
$\dfrac{1}{2}$
Work Step by Step
Recall:
$\sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)$
Hence,
$\sin(70^\circ)\cos(40^\circ)-\cos(70^\circ)\sin(40^\circ)\\
=\sin(70^\circ-40^\circ)\\
=\sin(30^\circ)\\
=\frac{1}{2}.$
