Question

If $\tan x=3/4,\pi <x<3\pi /2$. Find the value of $\sin x/2,\cos x/8$ and $\tan x/2$.

Answer 1

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

$180<x<\frac{3}{2}\times 180$

$=180<x<270$

$=90<\frac{x}{2}<\frac{270}{2}$

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

$\Rightarrow \frac{x}{2}$ lies in $2^{nd}$ quadrant.

$\sin \frac{x}{2}$ is positive.

$\cos \frac{x}{2}$ is negative.

$\tan \frac{x}{2}$ is negative.

$\tan x=\frac{3}{4}$

$\tan x=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$

$=\frac{3}{4}=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}$

$\Rightarrow 3-3{\tan }^2\frac{x}{2}=8\tan \frac{x}{2}$

$\Rightarrow 3{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}-3=0$

$\Rightarrow 3{\tan }^2\frac{x}{2}+9\tan \frac{x}{2}-\tan \frac{x}{2}-3=0$

$\Rightarrow 3\tan \frac{x}{2}(\tan \frac{x}{2}+3)-1(\tan \frac{x}{2}+3)=0$

$3\tan \frac{x}{2}-1=0$ & $\tan \frac{x}{2}+3=0$

$\Rightarrow 3\tan \frac{x}{2}=1$ $\tan \frac{x}{2}=-3$

$\Rightarrow \tan \frac{x}{2}=\frac{1}{3}$ (not possible as $\tan \frac{x}{2}$ is negative}

$\therefore \tan \frac{x}{2}=-3$

Consider, $\cos \frac{x}{2}=\frac{1}{\sec \frac{x}{2}}=\frac{-1}{\sqrt{1+{\tan }^2\frac{x}{2}}}$

$=\frac{-1}{\sqrt{1+(-3{)}^2}}$=$\frac{-1}{\sqrt{10}}$

$\therefore \cos \frac{x}{2}=\frac{-1}{\sqrt{10}}$

Consider,

$\sin \frac{x}{2}=\tan \frac{x}{2}\times \cos \frac{x}{2}$

$=(-3)(-1/\sqrt{10})$

$=\frac{3}{\sqrt{10}}$.