3.Trigonometrical Ratios, Functions and Identities
easy

यदि $\sin \theta  = \frac{{ - 4}}{5}$ तथा $\theta $ तीसरे चतुर्थांश में हो, तो $\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

दिया है $\sin \theta  =  – \frac{4}{5}$ एवं $\theta $ तृतीय चतुर्थांष में है

$ \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}} $

लेकिन $\cos \frac{\theta }{2} =  – \frac{1}{{\sqrt 5 }}$

चूँकि $\frac{\theta }{2}$ द्वितीय चतुर्थांष  में है।

अत: $\cos \frac{\theta }{2} =  – \frac{1}{{\sqrt 5 }}$.

Standard 11
Mathematics

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