If sin alpha = frac{{336}}{{625}} and 450^cir | vedclass

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Question

(c) $\sin \alpha = \frac{{336}}{{625}}$

==> $\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - {{\left( {\frac{{336}}{{625}}} \right

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(c) $\sin \alpha = \frac{{336}}{{625}}$

==> $\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - {{\left( {\frac{{336}}{{625}}} ight)}^2}} $, 

                                                  [$\because  \alpha $ is in $II$ Quadrant] 

Now, $\cos \left( {\frac{\alpha }{2}} ight) = - \sqrt {\frac{{1 + \cos \alpha }}{2}} = - \frac{7}{{25}}$,

                                                    [ $\because \frac{\alpha }{2}$ is in $III$ Quadrant]

$	herefore \,\,\,\sin \left( {\frac{\alpha }{4}} ight) = + \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]

If $\sin \alpha = \frac{{336}}{{625}}$ and $450^\circ < \alpha < 540^\circ ,$ then $\sin \left( {\frac{\alpha }{4}} \right) = $

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