If sin alpha = -frac{3}{5}, where pi

Question

(A) Given $\sin \alpha = -\frac{3}{5}$ and $\pi < \alpha < \frac{3\pi}{2}$ (i.e.,$\alpha$ is in the $III^{rd}$ quadrant).In the $III^{rd}$ quadrant,$\

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{10}}$

Vedclass · Answer

(A) Given $\sin \alpha = -\frac{3}{5}$ and $\pi In the $III^{rd}$ quadrant,$\cos \alpha$ is negative.$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - (-\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.Since $\pi This implies $\frac{\alpha}{2}$ lies in the $II^{nd}$ quadrant,where $\cos$ is negative.The half-angle formula is $\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + \cos \alpha}{2}}$.Substituting the value of $\cos \alpha$:$\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-4/5)}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

If $\sin \alpha = -\frac{3}{5},$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

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