Answer
$\dfrac{1}{2}$
Work Step by Step
We know that $\cos{\theta}$ has a period of $360^{\circ}$ (hence $\cos{(\theta+360^0)}=\cos{\theta}$), thus first we try and find a value where the argument is between $-180^{\circ}$ and $180^{\circ}$. T
$\cos(420^{\circ})=\cos (60^{\circ}+360^{\circ})=\cos (60^{\circ})=\dfrac{1}{2}$
