Question

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

Answer 1

Step $1:$ Check sign for $\sin \frac{x}{2},\cos \frac{x}{2}$ and $\tan \frac{x}{2}$
Given that $x$ is in quadrant $III.$
$\Rightarrow {180}^{\circ }<x<{270}^{\circ }$
$\Rightarrow \frac{{180}^{\circ }}{2}<\frac{x}{2}<\frac{{270}^{\circ }}{2}$
$\Rightarrow {90}^{\circ }<\frac{x}{2}<{135}^{\circ }$
$\Rightarrow \frac{x}{2}$ lies in $2^{nd}$ quadrant.
$\therefore \sin$ is positive, $\cos$ and $\tan$ are negative.

Step $2:$
$\cos x=2{\cos }^2\frac{x}{2}-1$
$\Rightarrow -\frac{1}{3}=2{\cos }^2\left(\frac{x}{2}\right)-1$
$\Rightarrow 2{\cos }^2\left(\frac{x}{2}\right)=1-\frac{1}{3}$
$\Rightarrow 2{\cos }^2\left(\frac{x}{2}\right)=\frac{2}{3}$
$\Rightarrow \cos \left(\frac{x}{2}\right)=\pm \frac{1}{\sqrt{3}}$
But $\cos \frac{x}{2}$ is negative
$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{3}}$

Step $3:$
Solve for the value of $\sin \frac{x}{2}$
${\sin }^{2}x+{\cos }^{2}x=1$ Replacing $x$ by $\frac{x}{2}$
${\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1$
$\Rightarrow {\sin }^2\frac{x}{2}+\frac{1}{3}=1\Rightarrow {\sin }^2\frac{x}{2}=\frac{2}{3}$
$\Rightarrow \sin \frac{x}{2}=\pm \sqrt{\frac{2}{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\Rightarrow \sin \frac{x}{2}=\pm \frac{\sqrt{6}}{3}$
But $\sin \frac{x}{2}$ is positive
$\therefore \sin \frac{x}{2}=\frac{\sqrt{6}}{3}$

Step $4:$
$\tan x=\frac{\sin x}{\cos x}$ Replacing $x$ by $\frac{x}{2}$
$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$
$\Rightarrow \tan \frac{x}{2}=\frac{\frac{\sqrt{6}}{3}}{-\frac{1}{\sqrt{3}}}=\frac{\sqrt{6}}{3}\times -\frac{\sqrt{3}}{1}=\frac{\sqrt{3}\times \sqrt{2}}{3}\times -\frac{\sqrt{3}}{1}$
$\Rightarrow \tan \frac{x}{2}=-\sqrt{2}$
$\therefore \sin \frac{x}{2}=\frac{\sqrt{6}}{3},\cos \frac{x}{2}=-\frac{\sqrt{3}}{3}$ and $\tan \frac{x}{2}=-\sqrt{2}$