Find sinx2 cosx2 and tanx2 in cos x is equal to 13 x in quadrant three

Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in

Q (9)

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

Answers (1)

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

We know that
cos x = $2\cos^{2}\frac{x}{2} - 1$
   $2\cos^{2}\frac{x}{2} =$ cos x + 1
=   $\left ( -\frac{1}{3} \right )$ + 1 =   $\left ( \frac{-1+3}{3} \right )$ = $\frac{2}{3}$

$\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$

     $\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$
Now,
we know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
   $2\sin^{2}\frac{x}{2} = 1 - \cos x$
= 1 - $\left ( -\frac{1}{3} \right )$ =   $\frac{3+1}{3}$ = $\frac{4}{3}$

$2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because   $\sin\frac{x}{2}$ is +ve in given quadrant

$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$