Question

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

Answer 1

Give x is in quadrant II

So, ${90}^o<x<{180}^o$

dividing with 2 all sides

$95<x<90$

So, $\frac{x}{2}$ lies in $1^{st}$ quadrant

i.e., $\sin ,\cos \&\tan$ are positive

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

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

$-4(1-ta{n}^2\frac{x}{2})=6\tan \frac{x}{2}$

$-4+4{\tan }^2\frac{x}{2}=6\tan \frac{x}{2}$ let $\tan \frac{x}{2}=a$

$4a^2-6a-4=0$

$4a-8a+2a-4=0$

$(4a+2)(a-2)=0$

So, $a=\frac{-1}{2}$ or $a=2$

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

Since $\frac{x}{2}$ is in first quadrant

$\tan \frac{x}{2}$ should be positive

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

$\cos \frac{x}{2}=\frac{1}{\sqrt{2^2+1}}=\frac{1}{\sqrt{5}}$

$\sin \frac{x}{2}=\frac{2}{\sqrt{2^2+1}}=\frac{2}{\sqrt{5}}$

$\cos \frac{x}{2}=\frac{1}{\sqrt{5}}$

$\tan \frac{x}{2}=2$

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