Question

Find $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\tan x=-\frac{4}{3},x$ in quadrant $II.$

Answer 1

Step $1:$ Check sign for $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$
Given that $x$ is in quadrant $II.$
$\Rightarrow {90}^{\circ }<x<{180}^{\circ }$
$\Rightarrow \frac{{90}^{\circ }}{2}<\frac{x}{2}<\frac{{180}^{\circ }}{2}$
$\Rightarrow {45}^{\circ }<\frac{x}{2}<{90}^{\circ }$
$\Rightarrow \frac{x}{2}$ lies in $1^{st}$ quadrant.
$\therefore \sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are positive.

Step $2:$
$\tan x=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$
$\Rightarrow -\frac{4}{3}=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$
$\Rightarrow -4+4{\tan }^2\frac{x}{2}=6\tan \frac{x}{2}$
$\Rightarrow 2{\tan }^2\frac{x}{2}-3\tan \frac{x}{2}-2=0$
$\Rightarrow 2\tan \frac{x}{2}(\tan \frac{x}{2}-2)+1(\tan \frac{x}{2}-2)=0$
$\Rightarrow (2\tan \frac{x}{2}+1)(\tan \frac{x}{2}-2)=0$
$\Rightarrow (2\tan \frac{x}{2}+1)=0,(\tan \frac{x}{2}-2)=0$
$\Rightarrow \tan \frac{x}{2}=-\frac{1}{2},\tan \frac{x}{2}=2$
But $\tan \frac{x}{2}$ is positive.
$\Rightarrow \tan \frac{x}{2}=2$

Step $3:$
$1+{\tan }^{2}x={\sec }^{2}x,$ Replacing $x$ by $\frac{x}{2}$
$1+{\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}$
$\Rightarrow 1+(2{)}^2={\sec }^2\frac{x}{2}$
$\Rightarrow {\sec }^2\frac{x}{2}=5\Rightarrow \sec \frac{x}{2}=\pm \sqrt{5}$
But $\frac{x}{2}$ lies in $1^{st}$quadrant, $\sec \frac{x}{2}$ is positive.
$\Rightarrow \sec \frac{x}{2}=\sqrt{5}$
$\therefore \cos \frac{x}{2}=\frac{1}{\sqrt{5}}$

Step $4:$
${\sin }^{2}x+{\cos }^{2}x=1$ Replacing $x$ by $\frac{x}{2}$
${\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1$
$\Rightarrow {\sin }^2\frac{x}{2}+\frac{1}{5}=1\Rightarrow {\sin }^2\frac{x}{2}=\frac{4}{5}$
$\Rightarrow \sin \frac{x}{2}=\pm \frac{2}{\sqrt{5}}$
But $\sin \frac{x}{2}$ is positive.
$\therefore \sin \frac{x}{2}=\frac{2}{\sqrt{5}}$
$\tan \frac{x}{2}=,\cos \frac{x}{2}=\frac{1}{\sqrt{5}}$ and $\sin \frac{x}{2}=\frac{2}{\sqrt{5}}$