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If $\sin \alpha = \frac{{336}}{{625}}$ and $450^\circ < \alpha < 540^\circ ,$ then $\sin \left( {\frac{\alpha }{4}} \right) = $
$\frac{1}{{5\sqrt 2 }}$
$\frac{7}{25}$
$\frac{4}{5}$
$\frac{3}{5}$
Solution
(c) $\sin \alpha = \frac{{336}}{{625}}$
==> $\cos \alpha = – \sqrt {1 – {{\sin }^2}\alpha } = – \sqrt {1 – {{\left( {\frac{{336}}{{625}}} \right)}^2}} $,
[$\because \alpha $ is in $II$ Quadrant]
Now, $\cos \left( {\frac{\alpha }{2}} \right) = – \sqrt {\frac{{1 + \cos \alpha }}{2}} = – \frac{7}{{25}}$,
[ $\because \frac{\alpha }{2}$ is in $III$ Quadrant]
$\therefore \,\,\,\sin \left( {\frac{\alpha }{4}} \right) = + \sqrt {\frac{{1 – \cos (\alpha /2)}}{2}} = \sqrt {\frac{{1 + \frac{7}{{25}}}}{2}} = \frac{4}{5}$,
[ $\because \,\frac{\alpha }{4}$ is in $II$ Quadrant]
