Answer
$0$
Work Step by Step
We know that $\sin (360-x)=-\sin x$
$\implies \sin x+\sin(360-x)=0$
Re-arranging the terms of the expression yields
$=(\sin 1^{\circ}+\sin 359^{\circ})+(\sin 2^{\circ}+\sin 358^{\circ})+(\sin 3^{\circ}+\sin 357^{\circ})....+(\sin 179^{\circ}+\sin 181^{\circ})+\sin{180^\circ}\\
=0+0+0+...+0\\
=0$
