Question

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

2:-$\cos x=-\frac{1}{3}$, x in quadrant $III$

Answer 1

Part 1^st

given,

$\tan x=-\frac{4}{3}$ ,x in 2^nd quadrant

We have to find $\tan \frac{x}{2}=?,\cos \frac{x}{2}=?,\sin \frac{x}{2}=?$

Consider ,

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

$\tan \left(2.\frac{x}{2}\right)=\frac{-4}{3}$

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

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

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

$4{\tan }^2\frac{x}{2}-6\tan \frac{x}{2}-4=0$

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

By solving this equation, we get

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

We take $\tan \frac{x}{2}=-\frac{1}{2}$ because in 2^nd quadrant tanx is negative

Hence ,

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

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

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

${\tan }^2\frac{x}{2}+1=\frac{5}{4}$

${\sec }^2\frac{x}{2}=\frac{5}{4}$

$\sec \frac{x}{2}=\pm \frac{\sqrt{5}}{2}$

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

Because in 2^nd quadrant cosx is negative

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

Now ,

${\cos }^2\frac{x}{2}={(-\frac{\sqrt{5}}{2})}^2=\frac{5}{4}$

$1-{\cos }^2\frac{x}{2}=1-\frac{5}{4}=\frac{-1}{4}$

$\sin \frac{x}{2}=\pm \frac{\sqrt{-1}}{2}$

Because in 2^nd quadrant cosx is negative

$\sin \frac{x}{2}=\frac{\sqrt{-1}}{2}$

Part 2nd .

Given,

$\cos x=-\frac{1}{3}$ and $x$ is in III quarant.

Now,

$\because \cos A=2{\cos }^2\frac{A}{2}-1$

$\therefore {\cos }^2\frac{A}{2}=\sqrt{\frac{\cos A+1}{2}}$

$\therefore {\cos }^2\frac{x}{2}=\sqrt{\frac{\cos x+1}{2}}=\sqrt{\frac{-\frac{1}{3}+1}{2}}=\pm \sqrt{\frac{1}{3}}$

But cosx is negative in III quadrant

$\cos \frac{x}{2}=-\sqrt{\frac{1}{3}}$

$\because \sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}=\sqrt{\frac{1-(-\frac{1}{3})}{2}}$

But in II quadrant sinx is negative

$\mathrm{Sin}\frac{x}{2}=\pm \sqrt{\frac{2}{3}}$

$=-\sqrt{\frac{2}{3}}$

$\mathrm{Tan}\frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\sqrt{2}$