Question

 If 0 ≤ x ≤ π and x lies in the IInd quadrant such that  \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and }  \tan\frac{x}{2}\]

 

 

Answer 1

\[\sin x = \frac{1}{4}\]

\[\therefore \text{ sin } x = \sqrt{1 - \cos^2 x}\]
\[ \Rightarrow \left( \frac{1}{4} \right)^2 = 1 - \cos^2 x\]
\[ \Rightarrow \frac{1}{16} - 1 = - \cos^2 x\]
\[ \Rightarrow \frac{15}{16} = \cos^2 x\]
\[ \Rightarrow \text{ cos } x = \pm \frac{\sqrt{15}}{4}\]
Since x lies in the 2nd quadrant, cos x is negative.

Thus,

\[\text{ cos } x = - \frac{\sqrt{15}}{4}\]

Now, using the identity

\[\text{ cos } x = 2 \cos^2 \frac{x}{2} - 1\] , we get 

\[- \frac{\sqrt{15}}{4} = 2 \cos^2 \frac{x}{2} - 1\]
\[ \Rightarrow - \frac{\sqrt{15}}{8} = \cos^2 \frac{x}{2} - \frac{1}{2}\]
\[ \Rightarrow \cos^2 \frac{x}{2} = \frac{4 - \sqrt{15}}{8}\]
\[ \Rightarrow \cos\frac{x}{2} = \pm \frac{4 - \sqrt{15}}{8}\]

Since x lies in the 2nd quadrant and \[\frac{x}{2}\]  lies in the 1st quadrant, \[\cos\frac{x}{2}\]  is positive.

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

Again,

\[\text { cos } x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{\sqrt{15}}{4} = $\left( \sqrt{\frac{4 - \sqrt{15}}{8}} \right)^2$ - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{\sqrt{15}}{4} = $\frac{4 - \sqrt{15}}{8}$ - \sin^2 \frac{x}{2}\]
\[ \Rightarrow \sin^2 \frac{x}{2} = \frac{4 + \sqrt{15}}{8}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \sqrt{\frac{4 + \sqrt{15}}{8}} = \sqrt{\frac{4 + \sqrt{15}}{8}} \]

Now,

\[\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\]
\[ = \frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} = \sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}\]
\[ = \sqrt{\frac{\left( 4 + \sqrt{15} \right)\left( 4 + \sqrt{15} \right)}{\left( 4 - \sqrt{15} \right)\left( 4 + \sqrt{15} \right)}}\]
\[ = \frac{4 + \sqrt{15}}{4^2 - \left( \sqrt{15} \right)^2} = \frac{4 + \sqrt{15}}{16 - 15} = 4+\sqrt{15}\]