Question

$\cos x=-\frac{1}{3}$, x in quadrant III. Find the value of $\sin \frac{x}{2},\cos \frac{x}{2},\tan \frac{x}{2}$

Answer 1

$\cos x=-\frac{1}{3},\pi <x<\frac{3\pi }{2}$

i.e. $x$ lies in $3rd$ quadrant

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

We get, $\sin \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{1}{3})}{2}}=\pm \sqrt{\frac{4}{6}}$

As $\pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$ and $sin$ is positive in $2nd$ quadrant

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

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

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

As $\pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$ and $cos$ is negative in $2nd$ quadrant

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

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 $\tan \frac{x}{2}=\pm \sqrt{\frac{1-(-\frac{1}{3})}{1+(-\frac{1}{3})}}=\pm \sqrt{2}$

As $\pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$ and $tan$ is negative in $2nd$ quadrant

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