Precalculus (10th Edition)

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I know that $\sin$ is an odd function, which means $f(-\theta)=-f(\theta).$ Also, I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Therefore $\sin{1^\circ}+\sin{2^\circ}+...+\sin{358^\circ}+\sin{359^\circ}\\=\sin{1^\circ}+\sin{2^\circ}+...+\sin{(358^\circ-360^\circ)}+\sin{(359^\circ-360^\circ)}\\=\sin{1^\circ}+\sin{2^\circ}+...+\sin{-2^\circ}+\sin{-1^\circ}\\=\sin{1^\circ}+\sin{2^\circ}+...-\sin{2^\circ}-\sin{1^\circ}=0$.