If sin alpha = frac{336}{625} and 450^circ

Question

(C) Given $\sin \alpha = \frac{336}{625}$ and $450^\circ < \alpha < 540^\circ$.Since $450^\circ < \alpha < 540^\circ$,$\alpha$ lies in the second quad

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(C) Given $\sin \alpha = \frac{336}{625}$ and $450^\circ Since $450^\circ In the second quadrant,$\cos \alpha$ is negative.$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left( \frac{336}{625} ight)^2} = -\sqrt{\frac{625^2 - 336^2}{625^2}} = -\sqrt{\frac{(625-336)(625+336)}{625^2}} = -\sqrt{\frac{289 	imes 961}{625^2}} = -\frac{17 	imes 31}{625} = -\frac{527}{625}$.Now,$225^\circ $\cos \left( \frac{\alpha}{2} ight) = -\sqrt{\frac{1 + \cos \alpha}{2}} = -\sqrt{\frac{1 - \frac{527}{625}}{2}} = -\sqrt{\frac{98}{1250}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.Finally,$112.5^\circ In the second quadrant,$\sin \left( \frac{\alpha}{4} ight)$ is positive.$\sin \left( \frac{\alpha}{4} ight) = \sqrt{\frac{1 - \cos(\alpha/2)}{2}} = \sqrt{\frac{1 - (-7/25)}{2}} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

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

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