QPaperGen

\(\textbf{Step 1: Determine quadrant of \(\frac{x}{2}\)}\)

Since \(x\) lies in the second quadrant, i.e., \(\frac{\pi}{2} < x < \pi\), then:
\[
\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}
\]
So, \(\frac{x}{2}\) lies in the first quadrant. Therefore:
\[
\sin\left(\frac{x}{2}\right) > 0,\quad \cos\left(\frac{x}{2}\right) > 0,\quad \tan\left(\frac{x}{2}\right) > 0
\]

\(\textbf{Step 2: Use Pythagorean identity to find \(\cos x\)}\)

\[
\sin^2 x + \cos^2 x = 1 \Rightarrow \left(\frac{1}{4}\right)^2 + \cos^2 x = 1
\Rightarrow \cos^2 x = 1 - \frac{1}{16} = \frac{15}{16}
\Rightarrow \cos x = -\frac{\sqrt{15}}{4} \quad \text{(since \(x\) is in 2nd quadrant)}
\]

\(\textbf{Step 3: Use half-angle identity for sine}\)

\[
2\sin^2\left(\frac{x}{2}\right) = 1 - \cos x = 1 + \frac{\sqrt{15}}{4}
\Rightarrow \sin^2\left(\frac{x}{2}\right) = \frac{4 + \sqrt{15}}{8}
\Rightarrow \sin\left(\frac{x}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}}
\]

\(\textbf{Step 4: Use Pythagorean identity to find cosine}\)

\[
\cos^2\left(\frac{x}{2}\right) = 1 - \sin^2\left(\frac{x}{2}\right) = 1 - \frac{4 + \sqrt{15}}{8} = \frac{4 - \sqrt{15}}{8}
\Rightarrow \cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}}
\]

\(\textbf{Step 5: Compute tangent}\)

\[
\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} =
\frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} =
\sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}
\]

Now rationalize:
\[
\tan\left(\frac{x}{2}\right) = \sqrt{\frac{(4 + \sqrt{15})^2}{(4 - \sqrt{15})(4 + \sqrt{15})}} =
\sqrt{\frac{16 + 8\sqrt{15} + 15}{16 - 15}} =
\sqrt{\frac{31 + 8\sqrt{15}}{1}} = \sqrt{31 + 8\sqrt{15}}
\]

\(\textbf{Final Answer:}\)
\[
\sin\left(\frac{x}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}}, \quad
\cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}}, \quad
\tan\left(\frac{x}{2}\right) = \sqrt{31 + 8\sqrt{15}}
\]