Question

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

Answer 1

As $\tan x=-\frac{4}{3},\frac{\pi }{2}<x<\pi$

i.e $x$ lies in $2nd$ quadrant

Hence $\tan x=-\frac{4}{3}\Rightarrow \sin x=\frac{4}{\sqrt{4^2+3^2}}=\frac{4}{5}$

And $\cos x=-\frac{5}{\sqrt{4^2+3^2}}=-\frac{3}{5}$

Now using $1-\cos x=2{\sin }^2\frac{x}{2}\Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}$, we get

$\sin \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{3}{5})}{2}}=\pm \sqrt{\frac{8}{10}}$

As $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $sine$ is positive in $1st$ quadrant

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

Using $1+\cos x=2{\cos }^2\frac{x}{2}\Rightarrow \cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}$

We get, $\cos \frac{x}{2}=\pm \sqrt{\frac{1+(-\frac{3}{5})}{2}}=\pm \sqrt{\frac{2}{10}}$

$\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $cos$ is positive $1st$ quadrant

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

Using $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\Rightarrow \tan \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}$

We get, $\tan \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{3}{5})}{1+(-\frac{3}{5})}}=\pm \sqrt{4}$

As $\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$ and $tan$ is positive in $1st$ quadrant

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