Question

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

Answer 1

Given $\sin x=\frac{1}{4},x$ in Quadrant $II$.

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

We know that ${\cos }^{2}x=1-{\sin }^{2}x$

But $\sin x=\frac{1}{4}$ then

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

Since $\cos$ is negative in second quadrant , $\cos x=-\frac{\sqrt{15}}{4}$

Using $1-\cos x=2{\sin }^2\frac{x}{2}$ i.e,

$$\begin{gathered} 1-\cos x=2{\sin }^2\frac{x}{2} \\ \sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}} \\ \sin \frac{x}{2}=\pm \sqrt{\frac{1-\left(-{\displaystyle \frac{\sqrt{15}}{4}}\right)}{2}} \\ =\pm \sqrt{\frac{4+\sqrt{15}}{8}} \\ As\frac{\mathrm{\pi }}{2}<x<\mathrm{\pi }\Rightarrow \frac{\mathrm{\pi }}{4}<\frac{\mathrm{x}}{2}<\frac{\mathrm{\pi }}{2} \end{gathered}$$

Since $\sin$ is positive in first quadrant

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