If \[ 0\le x\le \pi \] and \(x\) lies in the IInd quadrant such that \[ \sin x=\frac14 \] find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \tan\frac{x}{2} \]

Solution: Given, \[ \sin x=\frac14 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \cos^2x = 1-\left(\frac14\right)^2 \] \[ = 1-\frac1{16} \] \[ = \frac{15}{16} \] \[ \cos x=\pm\frac{\sqrt15}{4} \] Since \(x\) lies in the IInd quadrant, \[ \cos x<0 \] Therefore, \[ \cos x=-\frac{\sqrt15}{4} \] Now, \[ \cos\frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}} \] Substituting the value of \(\cos x\): \[ = \pm\sqrt{ \frac{1-\frac{\sqrt15}{4}}{2} } \] \[ = \pm\sqrt{ \frac{4-\sqrt15}{8} } \] Since \(x\) lies in the IInd quadrant, \[ \frac{x}{2} \] lies in the Ist quadrant, where cosine is positive. Therefore, \[ \boxed{ \cos\frac{x}{2} = \sqrt{\frac{4-\sqrt15}{8}} } \] Now, \[ \sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \] Substituting: \[ = \pm\sqrt{ \frac{1+\frac{\sqrt15}{4}}{2} } \] \[ = \pm\sqrt{ \frac{4+\sqrt15}{8} } \] Since \(\frac{x}{2}\) lies in the Ist quadrant, sine is positive. Hence, \[ \boxed{ \sin\frac{x}{2} = \sqrt{\frac{4+\sqrt15}{8}} } \] Now, \[ \tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \] \[ = \sqrt{ \frac{4+\sqrt15}{4-\sqrt15} } \] Rationalizing: \[ = \sqrt{ \frac{(4+\sqrt15)^2}{16-15} } \] \[ = 4+\sqrt15 \] Therefore, \[ \boxed{ \tan\frac{x}{2} = 4+\sqrt15 } \]