Question

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

 

 

Answer 1

Given:

\[\text{ sin } x = \frac{\sqrt{5}}{3}\] Using the identity \[\text{ cos } x = \sqrt{1 - \sin^2 x}\] , we get

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

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

\[\therefore \text{ cos } x = - \frac{2}{3}\]

Now,

\[\text{ cos } x = 1 - 2 \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{2}{3} = 1 - 2 \sin^2 \frac{x}{2}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \sqrt{\frac{5}{6}}\]

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,

\[\text{ cos } x = 2 \cos^2 \frac{x}{2} - 1\]
\[ \Rightarrow - \frac{2}{3} = 2 \cos^2 \frac{x}{2} - 1\]
\[ \Rightarrow \cos\frac{x}{2} = \pm \frac{1}{\sqrt{6}}\]
\[ \Rightarrow \cos\frac{x}{2} = \frac{1}{\sqrt{6}} \left( \because \frac{x}{2} < \frac{\pi}{2} \right)\]

\[\text{ Now,}  \tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{5}}{\sqrt{6}}}{\frac{1}{\sqrt{6}}} = \sqrt{5}\]