If 630^{circ}

Question

(A) Given $630^{\circ} < 	heta < 810^{\circ}$,dividing by $4$ gives $157.5^{\circ} < \frac{	heta}{4} < 202.5^{\circ}$.Since $	an 	heta = -\frac{7}

Vedclass · Accepted Answer

$-\sqrt{\frac{7+5 \sqrt{2}}{10 \sqrt{2}}}$

Vedclass · Answer

(A) Given $630^{\circ} Since $	an 	heta = -\frac{7}{24}$ and $	heta$ is in the $4$th quadrant $(630^{\circ} Then $\sec^2 	heta = 1 + 	an^2 	heta = 1 + \frac{49}{576} = \frac{625}{576}$,so $\sec 	heta = \frac{25}{24}$ (since $\cos 	heta > 0$ in the $4$th quadrant).Thus $\cos 	heta = \frac{24}{25}$.Using the half-angle formula $\cos^2 \frac{	heta}{2} = \frac{1 + \cos 	heta}{2} = \frac{1 + 24/25}{2} = \frac{49}{50}$,so $\cos \frac{	heta}{2} = \pm \frac{7}{5 \sqrt{2}}$.Since $315^{\circ}  0$,so $\cos \frac{	heta}{2} = \frac{7}{5 \sqrt{2}}$.Then $\cos^2 \frac{	heta}{4} = \frac{1 + \cos \frac{	heta}{2}}{2} = \frac{1 + \frac{7}{5 \sqrt{2}}}{2} = \frac{5 \sqrt{2} + 7}{10 \sqrt{2}}$.Since $157.5^{\circ} < \frac{	heta}{4} < 202.5^{\circ}$,$\cos \frac{	heta}{4}$ is negative,so $\cos \frac{	heta}{4} = -\sqrt{\frac{7 + 5 \sqrt{2}}{10 \sqrt{2}}}$.

If $630^{\circ} < \theta < 810^{\circ}$ and $\tan \theta = -\frac{7}{24}$,then $\cos \left(\frac{\theta}{4}\right) = $

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