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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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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

$\begin{array}{1 1} \sin \frac{x}{2} =\frac{-\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{- \sqrt 3}{3}; \tan \frac{x}{2} =\sqrt 2 \\ \sin \frac{x}{2} =\frac{\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{\sqrt 3}{3}; \tan \frac{x}{2} =\sqrt 2 \\ \sin \frac{x}{2} =\frac{\sqrt 6}{3} ; \cos \frac{x}{2} =\frac{- \sqrt 3}{3}; \tan \frac{x}{2} = - \sqrt 2 \\ \sin \frac{x}{2} =\frac{- \sqrt 6}{3} ; \cos \frac{x}{2} =\frac{ \sqrt 3}{3}; \tan \frac{x}{2} = - \sqrt 2 \end{array}$

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  • $\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$
answered Apr 18, 2014 by sreemathi.v
 

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