MCQ
If $\sin \alpha = \frac{{ - 3}}{5},$ where $\pi < \alpha < \frac{{3\pi }}{2},$ then $\cos \frac{1}{2}\alpha = $
- ✓$\frac{{ - 1}}{{\sqrt {10} }}$
- B$\frac{1}{{\sqrt {10} }}$
- C$\frac{3}{{\sqrt {10} }}$
- D$\frac{{ - 3}}{{\sqrt {10} }}$
$\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } $ [$\because \alpha$ lies in $III^{rd}$ Quadrant]
$ = - \sqrt {1 - \frac{9}{{25}}} = - \frac{4}{5}$
$\therefore \,\,\,\cos (\alpha /2) = - \sqrt {\frac{{1 - \frac{4}{5}}}{2}} = - \frac{1}{{\sqrt {10} }}$.
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