Question 15.5.1: If tan x=-4/3 and π/2<x<π, find the values of (i) sin......
If \tan x=\frac{-4}{3} and \frac{\pi}{2}<x<\pi , find the values of
(i) \sin \frac{x}{2}, \quad (ii) \cos \frac{x}{2} , \quad (iii) \tan \frac{x}{2} .
Since x lies in Quadrant II, we have \cos x<0 .
Also, \frac{\pi}{2}<x<\pi \Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2} \\ \\ \Rightarrow \frac{x}{2} lies in Quadrant I
\Rightarrow \sin \frac{x}{2}>0 and \cos \frac{x}{2}>0 .
(i) 2 \sin ^{2} \frac{x}{2}=(1-\cos x)=\left(1+\frac{3}{5}\right)=\frac{8}{5} \\ \\\begin{array}{l}\Rightarrow \sin ^{2} \frac{x}{2}=\frac{8}{10}=\frac{4}{5} \\ \\\Rightarrow \sin \frac{x}{2}=+\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \left[\because \sin \frac{x}{2}>0\right] .\end{array}
(ii) 2 \cos ^{2} \frac{x}{2}=(1+\cos x)=\left(1-\frac{3}{5}\right)=\frac{2}{5} \\ \\\begin{array}{l}\Rightarrow \cos ^{2} \frac{x}{2}=\left(\frac{2}{5 \times 2}\right)=\frac{1}{5} \\ \\\Rightarrow \cos \frac{x}{2}=+\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}} \quad\left[\because \cos \frac{x}{2}>0\right] .\end{array}
(iii) \tan \frac{x}{2}=\frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}=\left(\frac{2}{\sqrt{5}} \times \sqrt{5}\right)=2 .