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3.Trigonometrical Ratios, Functions and Identities
easy
If $\sin \theta = \frac{{ - 4}}{5}$ and $\theta $ lies in the third quadrant, then $\cos \frac{\theta }{2} = $
A
$\frac{1}{{\sqrt 5 }}$
B
$ - \frac{1}{{\sqrt 5 }}$
C
$\sqrt {\frac{2}{5}} $
D
$ - \sqrt {\frac{2}{5}} $
Solution
(b) Given that $\sin \theta = – \frac{4}{5}$ and $\theta $ lies in the $III$ quadrant.
$ \Rightarrow \cos \theta = \sqrt {1 – \frac{{16}}{{25}}} = \pm \frac{3}{5}$
$\cos \frac{\theta }{2} = \pm \sqrt {\frac{{1 + \cos \theta }}{2}} $
$= \sqrt {\frac{{1 – 3/5}}{2}} = \pm \sqrt {\frac{1}{5}} $
But $\cos \frac{\theta }{2} = – \frac{1}{{\sqrt 5 }}.$
since $\frac{\theta }{2}$ will be in $II$ quadrant.
Hence $\cos \frac{\theta }{2} = – \frac{1}{{\sqrt 5 }}$.
Standard 11
Mathematics
