## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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