Answer
$0$
Work Step by Step
Notice that $130=180-50, 220=180+40, 310=360-50$, we have:
$sin40^\circ+sin130^\circ+sin220^\circ+sin310^\circ=sin40^\circ+sin(180-50)^\circ+sin(180+40)^\circ+sin(360-50)^\circ=sin40^\circ+sin50^\circ-sin40^\circ-sin50^\circ=0$
