A) \[\frac{1}{5\sqrt{2}}\]
B) \[\frac{7}{25}\]
C) \[\frac{4}{5}\]
D) \[\frac{3}{5}\]
Correct Answer: C
Solution :
\[\sin \alpha =\frac{336}{625}\] Þ \[\cos \alpha =-\sqrt{1-{{\sin }^{2}}\alpha }=-\sqrt{1-{{\left( \frac{336}{625} \right)}^{2}}}\], [\[\because \]\[\alpha \]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]
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