If A = 580^circ,then which one of the followi | vedclass

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(C) Given $A = 580^\circ$. We know that $\sqrt{1 + \sin A} = \sqrt{(\sin(A/2) + \cos(A/2))^2} = |\sin(A/2) + \cos(A/2)|$ and $\sqrt{1 - \sin A} = \sqr

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$2 \sin \left( \frac{A}{2} \right) = -\sqrt{1 + \sin A} - \sqrt{1 - \sin A}$

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(C) Given $A = 580^\circ$. We know that $\sqrt{1 + \sin A} = \sqrt{(\sin(A/2) + \cos(A/2))^2} = |\sin(A/2) + \cos(A/2)|$ and $\sqrt{1 - \sin A} = \sqrt{(\cos(A/2) - \sin(A/2))^2} = |\cos(A/2) - \sin(A/2)|$.Since $A = 580^\circ$,$A/2 = 290^\circ$. In the fourth quadrant $(270^\circ  0$.Also,$|\sin(290^\circ)| > |\cos(290^\circ)|$ because $290^\circ$ is closer to $270^\circ$ than $360^\circ$.Thus,$\sin(A/2) + \cos(A/2) And $\cos(A/2) - \sin(A/2) > 0$,so $\sqrt{1 - \sin A} = \cos(A/2) - \sin(A/2)$.Adding these: $\sqrt{1 + \sin A} + \sqrt{1 - \sin A} = -\sin(A/2) - \cos(A/2) + \cos(A/2) - \sin(A/2) = -2 \sin(A/2)$.Therefore,$2 \sin(A/2) = -\sqrt{1 + \sin A} - \sqrt{1 - \sin A}$.

If $A = 580^\circ$,then which one of the following is true?

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