Question

Prove that

$2\sin \frac{A}{2}=\pm \sqrt{1-\sin A}\pm \sqrt{1+\sin A},$ when $\frac{A}{2}=\frac{19\pi }{11}$

Answer 1

$\frac{A}{2}=\frac{19\pi }{11}\Rightarrow \frac{A}{2}=\pi +\frac{8\pi }{11}=\frac{3\pi }{2}+\frac{5\pi }{22}$

$\therefore \frac{A}{2}$ is in 4th quadrant.

$\therefore \sin \frac{A}{2}$ is negative and $\cos \frac{A}{2}$ is positive.

$\mid \sin \frac{A}{2}\mid =\mid \cos \frac{A}{2}\mid$

$\therefore \cos \frac{A}{2}+\sin \frac{A}{2}<0$ and $\cos \frac{A}{2}-\sin \frac{A}{2}>0$

$2\sin \frac{A}{2}=\left(\cos \frac{A}{2}\right)-(\cos \frac{A}{2}-\sin \frac{A}{2})$

$=-\sqrt{{(\cos \frac{A}{2}+\sin \frac{A}{2})}^2}-\sqrt{{(\cos \frac{A}{2}-\sin \frac{A}{2})}^2}$

$\therefore 2\sin \frac{A}{2}=-\sqrt{{(\cos \frac{A}{2}+\sin \frac{A}{2})}^2}-\sqrt{{(\cos \frac{A}{2}-\sin \frac{A}{2})}^2}$

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