Question

Find sin $\frac{x}{2}$, cos $\frac{x}{2}$ and tan $\frac{x}{2}$ in the following:

cos x = $\frac{-1}{3}$, x in quadrant III.

Answer 1

Here cos x = $\frac{-1}{3}$, x in quadrant III.

Now $\pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$

So $\frac{x}{2}$ lies in second quadrant.

$\therefore$ sin $\frac{x}{2}$ is positive and cos $\frac{x}{2}$, tan $\frac{x}{2}$ are negative.

Now cos $\frac{x}{2}=-\sqrt{\frac{\text{1+ cos x}}{2}}=-\frac{\sqrt{1-\frac{1}{3}}}{2}$

=- $\frac{1}{\sqrt{3}}\times \frac{\sqrt{5}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$

sin $\frac{x}{2}=\sqrt{\frac{\text{1- cos x}}{2}}=\sqrt{\frac{1+\frac{1}{3}}{2}}$

= $\sqrt{\frac{2}{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

tan $\frac{x}{2}=\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}=\frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}}=-\sqrt{2}$.