Question

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

Answer 1

$\because \tan x=\frac{-4}{3},xisin{II}^{nd}quadrantgiventhatxisin{II}^{nd}quadrantso,{90}^o<x<{180}^o\Rightarrow \frac{{90}^0}{2}<\frac{x}{2}<\frac{{180}^o}{2}\Rightarrow {45}^o<x<{90}^o\therefore \frac{x}{2}liesin{I}^{st}quadrantIn{I}^{st}quadrantsin,cos,tanarepositive\Rightarrow \sin \frac{x}{2},\cos \frac{x}{2},\tan \frac{x}{2}arepositivegiven,\tan \frac{x}{2}=\frac{-4}{3}weknowthat,\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}\Rightarrow \tan \left(\frac{2x}{2}\right)=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\Rightarrow \tan x=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\Rightarrow \frac{-4}{3}=\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\Rightarrow \tan \frac{x}{2}=\frac{-1}{2}0r2\because \frac{x}{2}isin{I}^{st}quadrant\therefore \tan \frac{x}{2}=2also,\tan \theta =\frac{p}{b}comparingabovevaluesweget,p=2kandb=1k\therefore h=\sqrt{p^2+b^2}\Rightarrow h=\sqrt{{2k}^2+k^2}=\sqrt{5}k\therefore \sin \frac{x}{2}=\frac{p}{h}=\frac{2k}{\sqrt{5}k}=\frac{2}{\sqrt{5}}and,\cos \frac{x}{2}=\frac{b}{h}=\frac{k}{\sqrt{5}k}=\frac{1}{\sqrt{5}}$