Question

If $\sin x=\frac{\sqrt{5}}{3}$ and x lies in IInd quadrant, find the values of $\cos \frac{x}{2},\sin \frac{x}{2}\mathrm{and}\tan \frac{x}{2}$

Answer 1

Given:
$\sin x=\frac{\sqrt{5}}{3}$
Using the identity $\cos x=\sqrt{1-{\sin }^{2}x}$, we get

$\cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-{\left(\frac{\sqrt{5}}{3}\right)}^2}=\pm \frac{2}{3}$

Since x lies in the 2nd quadrant, cosx is negative.

$\therefore \cos x=-\frac{2}{3}$

Now,

$$\begin{gathered} \cos x=1-2{\sin }^2\frac{x}{2} \\ \Rightarrow -\frac{2}{3}=1-2{\sin }^2\frac{x}{2} \\ \Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{5}{6}} \end{gathered}$$

Since x lies in the 2nd quadrant, $\frac{x}{2}$ lies in the 1st quadrant.

$\therefore \sin \frac{x}{2}=\sqrt{\frac{5}{6}}$

Again,

$$\begin{gathered} \cos x=2{\cos }^2\frac{x}{2}-1 \\ \Rightarrow -\frac{2}{3}=2{\cos }^2\frac{x}{2}-1 \\ \Rightarrow \cos \frac{x}{2}=\pm \frac{1}{\sqrt{6}} \\ \Rightarrow \cos \frac{x}{2}=\frac{1}{\sqrt{6}}\left(\because \frac{x}{2}<\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$
$\mathrm{Now},\tan \frac{x}{2}=\frac{\sin {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}}=\frac{{\displaystyle \frac{\sqrt{5}}{\sqrt{6}}}}{{\displaystyle \frac{1}{\sqrt{6}}}}=\sqrt{5}$