Question

$\sin x=\frac{1}{4}$, x in quadrant II. Find the value of $\sin \frac{x}{2}$

Answer 1

$\sin x=\frac{1}{4},x$ in $2nd$ quadrant

$\Rightarrow \frac{\pi }{2}<x<\pi$ and $cos$ is negative in $2nd$ quadrant

Using $1-\cos x=2{\sin }^2\frac{x}{2}\Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}$

We get, $\sin \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{\sqrt{15}}{4})}{2}}=\pm \sqrt{\frac{4+\sqrt{15}}{8}}$

As $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $sin$ is positive in $1st$ quadrant

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

Using $1+\cos x=2{\cos }^2\frac{x}{2}\Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}$

We get $\cos \frac{x}{2}=\pm \sqrt{\frac{1+(-\frac{\sqrt{15}}{4})}{2}}=\pm \sqrt{\frac{4-\sqrt{15}}{8}}=\pm \frac{\sqrt{8-2\sqrt{15}}}{4}$

As $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $cos$ is positive in first quadrant

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

Using $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\Rightarrow \tan \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}$

We get $\tan \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{\sqrt{15}}{4})}{1+(-\frac{\sqrt{15}}{4})}}=\pm \sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}$

As $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $tan$ is positive in first quadrant

$\therefore \tan \frac{x}{2}=4+\sqrt{15}$