Question

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

 

 

Answer 1

\[\cos x = - \frac{3}{5}\] Using the identity

\[\cos2\theta = \cos^2 \theta - \sin^2 \theta\] , we get

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

It is given that x lies in the third quadrant. This means that

\[\frac{x}{2}\]  lies in the second quadrant.

\[\therefore \cos\frac{x}{2} = - \frac{1}{\sqrt{5}}\]

Again,

\[\text{ cos } x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{3}{5} = \left( - \frac{1}{\sqrt{5}} \right)^2 - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{3}{5} = \frac{1}{5} - \sin^2 \frac{x}{2}\]
\[ \Rightarrow - \frac{1}{5} - \frac{3}{5} = - \sin^2 \frac{x}{2}\]
\[ \Rightarrow \frac{4}{5} = \sin^2 \frac{x}{2}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \frac{2}{\sqrt{5}}\]

It is given that x lies in the third quadrant. This means that

\[\frac{x}{2}\]  lies in the second quadrant.

\[\therefore \sin\frac{x}{2} = \frac{2}{\sqrt{5}}\]

\[Now, \]
\[\text{ sin } x = \sqrt{1 - \cos^2 x}\]
\[ \Rightarrow \text{ sin } x = \sqrt{1 - \left( - \frac{3}{5} \right)}^2 \]
\[\text{ sin } x = \sqrt{1 - \frac{9}{25}} = \pm \frac{4}{5}\]

Since x lies in the third quadrant, sinx is negative.

\[\therefore \text{ sin } x = - \frac{4}{5}\]
\[ \Rightarrow \sin2x = 2\text{ sin } x\text{ cos } x\]
\[ \Rightarrow \sin2x = 2 \times \left( - \frac{4}{5} \right) \times \left( - \frac{3}{5} \right)\]
\[ \Rightarrow \sin2x = \frac{24}{25}\]