Answer
$\cos\theta$ = - $\frac{2\sqrt 2}{3}$
Work Step by Step
We know from first Pythagorean identity that-
$\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$
As $\theta$ terminates in Q II, Therefore $\cos\theta$ will be negative-
$\cos\theta$ = - $\sqrt (1-\sin^{2}\theta)$
substitute the given value of $\sin\theta$-
$\cos\theta$ = - $\sqrt (1-(\frac{1}{3})^{2})$
$\cos\theta$ = - $\sqrt (1-\frac{1}{9})$
$\cos\theta$ = - $\sqrt (\frac{9 - 1}{9})$ = $\sqrt (\frac{8}{9})$
$\cos\theta$ = - $\frac{2\sqrt 2}{3}$
