Find `sin ""(x)/(2), cos "" (x)/(2) and tan "" (x)/(2)` in each of the case:
`sin x = (1)/(4), x ` in II quadrant.

Answer

To find the values of \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \), and \( \tan \frac{x}{2} \) given that \( \sin x = \frac{1}{4} \) and \( x \) is in the second quadrant, we can follow these steps:

### Step 1: Find \( \cos x \)
We know the identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Given \( \sin x = \frac{1}{4} \), we can substitute this into the identity:
\[ \left(\frac{1}{4}\right)^2 + \cos^2 x = 1 \]
\[ \frac{1}{16} + \cos^2 x = 1 \]
\[ \cos^2 x = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \]
Now, taking the square root:
\[ \cos x = \pm \sqrt{\frac{15}{16}} = \pm \frac{\sqrt{15}}{4} \]
Since \( x \) is in the second quadrant, \( \cos x \) is negative:
\[ \cos x = -\frac{\sqrt{15}}{4} \]

### Step 2: Find \( \cos \frac{x}{2} \)
Using the half-angle formula:
\[ \cos x = 2 \cos^2 \frac{x}{2} - 1 \]
Substituting the value of \( \cos x \):
\[ -\frac{\sqrt{15}}{4} = 2 \cos^2 \frac{x}{2} - 1 \]
Rearranging gives:
\[ 2 \cos^2 \frac{x}{2} = 1 - \frac{\sqrt{15}}{4} \]
\[ 2 \cos^2 \frac{x}{2} = \frac{4}{4} - \frac{\sqrt{15}}{4} = \frac{4 - \sqrt{15}}{4} \]
\[ \cos^2 \frac{x}{2} = \frac{4 - \sqrt{15}}{8} \]
Taking the square root:
\[ \cos \frac{x}{2} = \sqrt{\frac{4 - \sqrt{15}}{8}} \]
Since \( \frac{x}{2} \) is in the first quadrant (as \( x \) is in the second quadrant), \( \cos \frac{x}{2} \) is positive:
\[ \cos \frac{x}{2} = \frac{\sqrt{4 - \sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}} = \frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}} \]

### Step 3: Find \( \sin \frac{x}{2} \)
Using the identity:
\[ \sin^2 \frac{x}{2} = 1 - \cos^2 \frac{x}{2} \]
Substituting the value of \( \cos^2 \frac{x}{2} \):
\[ \sin^2 \frac{x}{2} = 1 - \frac{4 - \sqrt{15}}{8} \]
\[ \sin^2 \frac{x}{2} = \frac{8}{8} - \frac{4 - \sqrt{15}}{8} = \frac{8 - (4 - \sqrt{15})}{8} = \frac{4 + \sqrt{15}}{8} \]
Taking the square root:
\[ \sin \frac{x}{2} = \sqrt{\frac{4 + \sqrt{15}}{8}} \]
Since \( \frac{x}{2} \) is in the first quadrant, \( \sin \frac{x}{2} \) is positive:
\[ \sin \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{\sqrt{8}} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}} \]

### Step 4: Find \( \tan \frac{x}{2} \)
Using the identity:
\[ \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \]
Substituting the values:
\[ \tan \frac{x}{2} = \frac{\frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}}{\frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}}} = \frac{\sqrt{4 + \sqrt{15}}}{\sqrt{4 - \sqrt{15}}} \]

### Final Answers
\[ \sin \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}, \quad \cos \frac{x}{2} = \frac{\sqrt{4 - \sqrt{15}}}{2\sqrt{2}}, \quad \tan \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{\sqrt{4 - \sqrt{15}}} \]