If sin theta = frac{{ - 4}}{5} and theta lie | vedclass

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Question

(b) Given that $\sin 	heta = - \frac{4}{5}$ and $	heta $ lies in the $III$ quadrant.

$ \Rightarrow \cos 	heta = \sqrt {1 - \frac{{16}}{{25}}} =

Vedclass · Accepted Answer

$ - \frac{1}{{\sqrt 5 }}$

Vedclass · Answer

(b) Given that $\sin 	heta = - \frac{4}{5}$ and $	heta $ lies in the $III$ quadrant.

$ \Rightarrow \cos 	heta = \sqrt {1 - \frac{{16}}{{25}}} = \pm \frac{3}{5}$

$\cos \frac{	heta }{2} = \pm \sqrt {\frac{{1 + \cos 	heta }}{2}} $

$= \sqrt {\frac{{1 - 3/5}}{2}} = \pm \sqrt {\frac{1}{5}} $

But $\cos \frac{	heta }{2} = - \frac{1}{{\sqrt 5 }}.$

since $\frac{	heta }{2}$ will be in $II$ quadrant.

Hence $\cos \frac{	heta }{2} = - \frac{1}{{\sqrt 5 }}$.

If $\sin \theta = \frac{{ - 4}}{5}$ and $\theta $ lies in the third quadrant, then $\cos \frac{\theta }{2} = $

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