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.

i.e.,  \({\pi}<x<\frac{3\pi}{2}\)

\(⇒\frac{\pi}{2}<\frac{x}{2}<\frac{3\pi}{4}\)

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

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

cos x=1-2 sin² \(\frac{x}{2}\)

\(⇒sin^2\frac{x}{2}=\frac{1-cosx}{2}\)

\(⇒sin^2\frac{x}{2}=\frac{1-(-\frac{1}{3})}{2}=\frac{(1+\frac{1}{3})}{2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}\)

\(⇒sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}\,\,\,\,\,\,\,[sin\frac{x}{2}\,is \,positive]\)

\(∴\,sin\frac{x}{2}=\frac{\sqrt2}{\sqrt3}×\frac{\sqrt3}{\sqrt3}=\frac{\sqrt6}{\sqrt3}\)

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  \(sin\frac{x}{2},cos\frac{x}{2}\,and\,tan\frac{x}{2}\,are\,\frac{\sqrt6}{3},\frac{-\sqrt3}{3},\,and-{\sqrt2}\)