Precalculus (10th Edition)

We know that $\sin$ has a period of $360^{ \circ}$, hence using this first we try and find a value where the argument is between $0^{\circ}$ and $360^{\circ}$ and then evaluate the funcion. Therefore $\sin(390^{\circ})=\sin(390^{\circ}-360^{\circ})=\sin(30^{o})=\frac{1}{2}.$
We know that $\sin$ has a period of $360^{ \circ}$, hence using this first we try and find a value where the argument is between $0^{\circ}$ and $360^{\circ}$ and then evaluate the function. Therefore $\sin(390^{\circ})=\sin(390^{\circ}-360^{\circ})=\sin(30^{o})=\frac{1}{2}.$