Question

If $\tan x=\frac{3}{4},\pi <x<\frac{3\pi }{2}$, find the value of $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}.$

Answer 1

Step 1: check quadrant for $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$

Given that ${180}^{\circ }<x<{270}^{\circ }$

$\Rightarrow \frac{{180}^{\circ }}{2}<\frac{x}{2}<\frac{{270}^{\circ }}{2}$
$\Rightarrow \frac{x}{2}$ lies in $2^{nd}$ quadrant.
In $2^{nd}$ quadrant, $\sin$ is positive. $\cos$ and $\tan$ are negative
$\therefore \sin \frac{x}{2}$ is positive, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative.

Step $2:$ solve for the value of $\tan \frac{x}{2}$
$\tan x=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\Rightarrow \frac{3}{4}=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$
$\Rightarrow 3-3{\tan }^2\frac{x}{2}=8\tan \frac{x}{2}$
$\Rightarrow 3{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}-3=0$
$\Rightarrow 3{\tan }^2\frac{x}{2}+9\tan \frac{x}{2}-\tan \frac{x}{2}-3=0$
$\Rightarrow 3\tan \frac{x}{2}(\tan \frac{x}{2}+3)-1(\tan \frac{x}{2}+3)=0\Rightarrow (3\tan \frac{x}{2}-1)(\tan \frac{x}{2}+3)=0$
$\Rightarrow \tan \frac{x}{2}=\frac{1}{3}$ or $\tan \frac{x}{2}=-3$
Since $\frac{x}{2}$ lies in $2^{nd}$ quadrant.
$\therefore \tan \frac{x}{2}$ is negative.
$\Rightarrow \tan \frac{x}{2}=-3$

Step $3:$
$1+{\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}$
$\Rightarrow 1+(-3{)}^2={\sec }^2\frac{x}{2}\Rightarrow {\sec }^2\frac{x}{2}=1+9\Rightarrow \sec \frac{x}{2}=\pm \sqrt{10}$

But $\frac{x}{2}$ lies in $2^{nd}$ quadrant, therefore $\sec \frac{x}{2}$ is negative
$\Rightarrow \sec \frac{x}{2}=-\sqrt{10}$
$\therefore \cos \frac{x}{2}=\frac{-1}{\sqrt{10}}$

Step $4:$
Solve for the value of $\sin \frac{x}{2}$
${\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1$

$\Rightarrow {\sin }^2\frac{x}{2}=1-{(-\frac{1}{\sqrt{10}})}^2\Rightarrow {\sin }^2\frac{x}{2}=\frac{9}{10}$
$\Rightarrow \sin \frac{x}{2}=\pm \frac{3}{\sqrt{10}}$
But $\sin \frac{x}{2}$ is positive
$\therefore \sin \frac{x}{2}=\frac{3}{\sqrt{10}}$