sin x = $\frac{1}{4}$ , x in quadrant II ⇒ $\frac{\pi}{2}$ < x < π ⇒ $\frac{\pi}{4}$ < $\frac{x}{2}$ < $\frac{\pi}{2}$ ⇒ $\frac{x}{2}$ lies in first quadrant ⇒ sin $\frac{x}{2}$ > 0 , cos $\frac{x}{2}$ > 0 , tan $\frac{x}{2}$ > 0 Also , $\cos^2$ x = 1 - $\sin^2 x$ ⇒ $\cos^2 x$ = 1 - $\left(\frac{1}{4}\right)^2$ = 1 - $\frac{1}{16}$ = $\frac{15}{16}$ ⇒ cos x = ± $\frac{\sqrt{15}}{4}$ ⇒ cos x = - $\frac{\sqrt{15}}{4}$ (Since x lies in II quadrant) cos $\frac{x}{2}$ = ± $\sqrt{\frac{1+\cos x}{2}}$ = ± $\sqrt{\frac{1-\frac{\sqrt{15}}{4}}{2}}$ = ± $\sqrt{\frac{4-\sqrt{15}}{8}}$ = ± $\frac{\sqrt{8-2\sqrt{15}}}{4}$ = $\frac{\sqrt{8-2\sqrt{15}}}{4}$ sin $\frac{x}{2}$ = ± $\sqrt{\frac{1-\cos x}{2}}$ = ± $\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}$ = ± $\sqrt{\frac{4+\sqrt{15}}{8}}$ = $\frac{\sqrt{8+2\sqrt{15}}}{4}$ tan $\frac{x}{2}$ = $\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}$ = $\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}$ × $\frac{\sqrt{4-\sqrt{15}}}{\sqrt{4-\sqrt{15}}}$ = $\frac{\sqrt{16-15}}{4-\sqrt{15}}$ = $\frac{1}{4-\sqrt{15}}$ × $\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4+\sqrt{15}}}$ = $\frac{\sqrt{4+\sqrt{15}}}{1}$ = 4 + $\sqrt{15}$ Hence , sin $\frac{x}{2}$ = $\frac{\sqrt{8+2\sqrt{15}}}{4}$ , cos $\frac{x}{2}$ = $\frac{\sqrt{8-2\sqrt{15}}}{4}$ , tan $\frac{x}{2}$ = 4 + $\sqrt{15}$