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

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(B) Given $\sin 	heta = -\frac{4}{5}$ and $	heta$ lies in the $III$ quadrant $(180^\circ < 	heta < 270^\circ)$.Since $	heta$ is in the $III$ quadr

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $\sin 	heta = -\frac{4}{5}$ and $	heta$ lies in the $III$ quadrant $(180^\circ Since $	heta$ is in the $III$ quadrant,$\cos 	heta$ is negative.$\cos 	heta = -\sqrt{1 - \sin^2 	heta} = -\sqrt{1 - (-\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.We need to find $\cos \frac{	heta}{2}$. Using the half-angle formula: $\cos \frac{	heta}{2} = \pm \sqrt{\frac{1 + \cos 	heta}{2}}$.Since $180^\circ This means $\frac{	heta}{2}$ lies in the $II$ quadrant,where cosine is negative.Therefore,$\cos \frac{	heta}{2} = -\sqrt{\frac{1 + (-3/5)}{2}} = -\sqrt{\frac{2/5}{2}} = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}}$.

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

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