Question

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

 

 

Answer 1

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

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

Here, x lies in the second quadrant.

\[\therefore \text{ sin } x = \frac{4}{5}\]

We know,

sin2x = 2sinx cosx

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

Now,

\[\text{ cos } x = 1 - 2 \sin^2 \frac{x}{2}\]
\[ \Rightarrow 2 \sin^2 \frac{x}{2} = 1 - \left( - \frac{3}{5} \right) = \frac{8}{5}\]
\[ \Rightarrow \sin^2 \frac{x}{2} = \frac{4}{5}\]
\[ \Rightarrow \sin\frac{x}{2} = \pm \frac{2}{\sqrt{5}}\]

Since x lies in the second quadrant,

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

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