Answer
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 $\sin(390^{\circ})=\sin(390^{\circ}-360^{\circ})=\sin(30^{o})=\frac{1}{2}.$
Work Step by Step
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 $\sin(390^{\circ})=\sin(390^{\circ}-360^{\circ})=\sin(30^{o})=\frac{1}{2}.$
