Question

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

Answer 1

Here, x is in quadrant III. ${\pi}<x<\frac{3\pi}{2}$

i.e., $=\frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}$

Therefore, $cos\frac{x}{2}\,and\,,tan\frac{x}{2}\,and\,tan\frac{x}{2}\,are\,negative\,,whereas\,sin\frac{x}{2}\,is\,positive.$

it is given that $cos x=-\frac{1}{3}.$

$sin^2\frac{x}{2}=\frac{1-cosx}{2}$

$sin2\frac{x}{2}=\frac{1-(-\frac{1}{3})}{2}=\frac{(1+\frac{1}{3})}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}$

$sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}\,[sin\frac{x}{2}\,is \,positive]$

$sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}×\frac{\sqrt3}{\sqrt3}=\frac{\sqrt6}{\sqrt3}$

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

$=cos^2\frac{x}{2}=\frac{1+cosx}{2}=\frac{1+(-\frac{1}{3})}{2}=\frac{(\frac{3-1}{2})}{2}=\frac{(\frac{2}{3})}{2}=\frac{1}{3}$

$=cos\frac{x}{2}=-\frac{1}{\sqrt3}\,[cos\frac{x}{2}\,is\,negative]$

$=cos\frac{x}{2}=-\frac{1}{\sqrt3}×\frac{\sqrt3}{\sqrt3}=\frac{\sqrt3}{3}$

$tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{(\frac{\sqrt2}{\sqrt3})}{(\frac{-1}{\sqrt3})}=-{\sqrt2}$

Thus, the respective values of $sin\frac{x}{2},cos\frac{x}{2}and\,tan\frac{x}{2}\,are\,\frac{\sqrt6}{3},\frac{-\sqrt3}{3},\,and-{\sqrt2}$