Question

$2\cos \frac{A}{2}=\pm \sqrt{1-\sin A}\pm \sqrt{1+\sin A}$, when $\frac{A}{2}=-{140}^{\circ }$

Answer 1

$2\cos \frac{A}{2}=\pm \sqrt{1-\sin A}\pm \sqrt{1+\sin A};\frac{A}{2}={140}^{\circ }$

LHS,

$2\cos \frac{A}{2}=2\cos {140}^{\circ }=2\cos ({180}^{\circ }-{40}^{\circ })=-2\cos {40}^{\circ }$

RHS,

$\pm \sqrt{1-\sin A}\pm \sqrt{1+\sin A}=\sqrt{\cos {}^2\frac{A}{2}+\sin {}^2\frac{A}{2}-2\sin \frac{A}{2}.\cos \frac{A}{2}}+\sqrt{\cos {}^2\frac{A}{2}+\sin {}^2\frac{A}{2}+2\sin \frac{A}{2}.\cos \frac{A}{2}}=\pm |\cos \frac{A}{2}-\sin \frac{A}{2}|\pm |\cos \frac{A}{2}+\sin \frac{A}{2}|=\pm |-\cos {40}^{\circ }-\sin {40}^{\circ }|\pm |-\cos {40}^{\circ }+\sin {40}^{\circ }|=+\cos {40}^{\circ }+\sin {40}^{\circ }+\cos {40}^{\circ }-\sin {40}^{\circ }=2\cos {40}^{\circ }(\sin {40}^{\circ }<\cos {40}^{\circ })\therefore |\sin {40}^{\circ }-\cos {40}^{\circ }|=-\sin {40}^{\circ }+\cos {40}^{\circ }$