Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant II

solution:

Here, x is in quadrant II.



$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive.

It is given that $\sin x=\frac{1}{4}$.

$\cos ^{2} x=1-\sin ^{2} x=1-\left(\frac{1}{4}\right)^{2}=1-\frac{1}{16}=\frac{15}{16}$

$\Rightarrow \cos x=-\frac{\sqrt{15}}{4}[\cos x$ is negative in quadrant II]

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

$\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}} \quad\left[\because \sin \frac{x}{2}\right.$ is positive $]$

$=\sqrt{\frac{4+\sqrt{15}}{8}} \times \frac{2}{2}$

$=\sqrt{\frac{8+2 \sqrt{15}}{16}}$

$=\frac{\sqrt{8+2 \sqrt{15}}}{4}$

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

$\Rightarrow \cos \frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}} \quad\left[\because \cos \frac{x}{2}\right.$ is positive $]$

$=\sqrt{\frac{4-\sqrt{15}}{8} \times \frac{2}{2}}$

$=\sqrt{\frac{8-2 \sqrt{15}}{16}}$

$=\frac{\sqrt{8-2 \sqrt{15}}}{4}$

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

$=\sqrt{\frac{8+2 \sqrt{15}}{8-2 \sqrt{15}} \times \frac{8+2 \sqrt{15}}{8+2 \sqrt{15}}}$

$=\sqrt{\frac{(8+2 \sqrt{15})^{2}}{64-60}}=\frac{8+2 \sqrt{15}}{2}=4+\sqrt{15}$

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{\sqrt{8+2 \sqrt{15}}}{4}, \frac{\sqrt{8-2 \sqrt{15}}}{4}$,

and $4+\sqrt{15}$