Answer
$\frac{1}{2}$
Work Step by Step
RECALL:
For any integer $k$, $\sin{\theta} = \sin{(\theta+180^ok)}$.
We know that $\sin$ has a period of $360^o$, hence first we try and find a value where the argument is between $0^{o}$ and $360^{o}$.
Therefore, using the property above gives:
$\sin(390^{o})
\\=\sin{(30^o+360^o)}
\\=\sin{(30^{o}+180^o\cdot2)}
\\=\sin{30^o}
\\=\frac{1}{2}$
