Find sinx2 cosx2 and tanx2 in tan x is equal to 43 x in quadrant two

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

Q (8)

$\small \tan x = - \frac{4}{3}$ , x in quadrant II

Answers (1)

tan x = $-\frac{4}{3}$
We know that.
   $\sec^{2}x = 1 + \tan^{2}x$
   $= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$ =   $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant, which is why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now, $cos x = \frac{1}{\sec x}$ =   $-\frac{3}{5}$
We know that.
$cos x = 2\cos^{2}\frac{x}{2}- 1$ ( $\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$ )
   $-\frac{3}{5}+ 1 = 2$ $\cos^{2}\frac{x}{2}$

=   $\frac{-3+5}{5}$ = $2\cos^{2}\frac{x}{2}$

      $\frac{2}{5}$ = $2\cos^{2}\frac{x}{2}$
   $\cos^{2}\frac{x}{2}$ =   $\frac{1}{5}$
$\cos\frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of    $\cos\frac{x}{2}$ is +ve

$\cos\frac{x}{2}$ =   $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
   $cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$ = 1 -   $(-\frac{3}{5})$ =   $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

   $\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$