Question

If $\tan =-\frac{4}{3},x$ lies in II quadrant, then find the value of $\sin \frac{x}{2}$.

Answer 1

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

$\sin x=\frac{4}{5},\cos x=\frac{-3}{5}$

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

$\Rightarrow \frac{-3}{5}=2\times {\cos }^2\frac{x}{2}-1$

$\therefore 1-\frac{3}{5}=2{\cos }^2\frac{x}{2}$

$\Rightarrow 2{\cos }^2\frac{x}{2}=\frac{2}{5}$

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

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

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

$\Rightarrow \frac{4^2}{5}=2\times \sin \frac{x}{2}\times \frac{1}{\sqrt{5}}$

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