Question

If $A={340}^{\circ }$, prove that

$2\sin \frac{A}{2}=-\sqrt{1+\sin A}+\sqrt{1-\sin A}$

Answer 1

$2\sin \frac{A}{2}=-\sqrt{1+\sin A}+\sqrt{1-\sin A};A={340}^{\circ }$

LHS: $2\sin \frac{A}{2}=2\sin {170}^{\circ }=2\sin ({180}^{\circ }-{10}^{\circ })=2\sin {10}^{\circ }$

RHS: $-\sqrt{1+\sin A}+\sqrt{1-\sin A}$

$=-\sqrt{\cos {}^2\frac{A}{2}+\sin {}^2\frac{A}{2}+2.\sin \frac{A}{2}.\cos \frac{A}{2}}+\sqrt{\cos {}^2\frac{A}{2}+\sin {}^2\frac{A}{2}-2.\sin \frac{A}{2}.\cos \frac{A}{2}}=-|\cos \frac{A}{2}+\sin \frac{A}{2}|+|\cos \frac{A}{2}-\sin \frac{A}{2}|=-|\cos 110^{\circ }+\sin {10}^{\circ }|+|\cos {170}^{\circ }-\sin {170}^{\circ }|=-|\cos {10}^{\circ }+\sin {10}^{\circ }|+|-\cos {10}^{\circ }-\sin {10}^{\circ }|=-\cos {10}^{\circ }+\sin {10}^{\circ }+\cos {10}^{\circ }+\sin {10}^{\circ }(\sin {10}^{\circ }<\cos {10}^{\circ })=2\sin {10}^{\circ };\therefore LHS=RHS(\sin {10}^{\circ },\cos {10}^{\circ }>0)$

Hence proved.