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# Find $\sin \large\frac{x}{2},$$\cos \large\frac{x}{2} and \tan \large\frac{x}{2} in each of the following \cos x=-\large\frac{1}{3}.x in quadrant III \begin{array}{1 1} \sin \frac{x}{2} =\frac{-\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{- \sqrt 3}{3}; \tan \frac{x}{2} =\sqrt 2 \\ \sin \frac{x}{2} =\frac{\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{\sqrt 3}{3}; \tan \frac{x}{2} =\sqrt 2 \\ \sin \frac{x}{2} =\frac{\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{- \sqrt 3}{3}; \tan \frac{x}{2} = - \sqrt 2 \\ \sin \frac{x}{2} =\frac{- \sqrt 6}{3} ; \cos \frac{x}{2} =\frac{ \sqrt 3}{3}; \tan \frac{x}{2} = - \sqrt 2 \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • \cos x=2\cos^2\large\frac{x}{2}$$-1$
• $\sin \large\frac{x}{2}$$=\pm\sqrt{1-\cos^2\large\frac{x}{2}} \cos x=2\cos^2\large\frac{x}{2}$$-1$
$-\large\frac{1}{3}$$=2\cos^2\large\frac{x}{2}$$-1$
$2\cos^2\large\frac{x}{2}$$=1-\large\frac{1}{3} 2\cos^2\large\frac{x}{2}=\frac{2}{3} \cos^2\large\frac{x}{2}=\frac{2}{6} \cos\large\frac{x}{2}=\pm\frac{1}{\sqrt 3} Since x lies in III quadrant \pi < x < \large\frac{3\pi}{2} \large\frac{\pi }{2}$$< \large\frac{x}{2}$$< \large\frac{3\pi}{4} \therefore \large\frac{x}{2} lies in II quadrant \therefore \cos\large\frac{x}{2}=\frac{-1}{\sqrt 3} \sin \large\frac{x}{2}$$=\pm\sqrt{1-\cos^2\large\frac{x}{2}}$
$\sin \large\frac{x}{2}$$=\pm\sqrt{1-\large\frac{1}{3}}=\pm\sqrt{\large\frac{2}{3}} \therefore \large\frac{x}{2} lies in II quadrant \therefore \tan \large\frac{x}{2}=\frac{\sin\Large\frac{x}{2}}{\cos\Large\frac{x}{2}}=\large\frac{\sqrt{\Large\frac{2}{3}}}{-\Large\frac{1}{\sqrt 3}} \tan \large\frac{x}{2}$$=-\sqrt 2$