We have, cos x = - $\frac{1}{3}$ , x in quadrant III since, x is in III quadrant ⇒ π < x < $\frac{3\pi}{2}$ ⇒ $\frac{\pi}{2}$ < $\frac{x}{2}$ < $\frac{3\pi}{4}$ ⇒ $\frac{x}{2}$ lies in II quadrant ⇒ sin $\frac{x}{2}$ > 0 , cos $\frac{x}{2}$ < 0 , tan $\frac{x}{2}$ < 0 , cos $\frac{x}{2}$ = ± $\sqrt{\frac{1+\cos x}{2}}$ = ± $\sqrt{\frac{1-\frac{1}{3}}{2}}$ = + $\frac{1}{\sqrt{3}}$ ⇒ cos $\frac{x}{2}$ = - $\frac{1}{\sqrt{3}}$ Now , sin $\frac{x}{2}$ = ± $\sqrt{\frac{1-\cos x}{2}}$ = ± $\sqrt{\frac{1+\frac{1}{3}}{2}}$ = ± $\sqrt{\frac{2}{3}}$ = $\sqrt{\frac{2}{3}}$ and tan $\frac{x}{2}$ = $\frac{\frac{\sin x}{2}}{\frac{\cos x}{2}}$ = - $\sqrt{2}$ Hence, sin $\frac{x}{2}$ = $\sqrt{\frac{2}{3}}$ or $\frac{\sqrt{6}}{3}$ , cos $\frac{x}{2}$ = $-\frac{\sqrt{3}}{3}$ , tan $\frac{x}{2}$ = - $\sqrt{2}$