Answer
$-\frac{\sqrt 3}{2}$
Work Step by Step
As the period of cosine function is $360^{\circ}$,
$\cos(-510^{\circ})=\cos(-510^{\circ}+360^{\circ})$
$=\cos(-150^{\circ})=\cos 150^{\circ}$
(As $\cos (-x)=\cos x$ )
$=\cos(180^{\circ}-30^{\circ})=-\cos 30^{\circ}$
( As $\cos(180^{\circ}-x)=-\cos x$ )
$=-\frac{\sqrt 3}{2}$
