Question

If $2\sin \frac{A}{2}=\sqrt{1+\sin A}+\sqrt{1-\sin A}$, then $\frac{A}{2}$ lies between

• A. $2n\pi +\frac{\pi }{4}$ and $2n\pi +\frac{3\pi }{4},n\in Z$
• B. $2n\pi +\frac{\pi }{4}$ and $2n\pi +\frac{\pi }{4},n\in Z$
• C. $2n\pi -\frac{3\pi }{4}$ and $2n\pi -\frac{\pi }{4},n\in Z$
• D. $-\infty$ and $+\infty$

Answer 1

The correct option is A $2n\pi +\frac{\pi }{4}$ and $2n\pi +\frac{3\pi }{4},n\in Z$

$2sin\frac{A}{2}=\sqrt{1+sinA}+\sqrt{1-sinA}\Rightarrow 2sin\frac{A}{2}>0$
sq. both sides. i.e. $\frac{A}{2}\in [0,\pi ]$
$4si{n}^2\frac{A}{2}=1+sinA+1-sinA+2\sqrt{1-si{n}^{2}A}$ $A\in [0,2\pi ]$
$2(1-cosA)=2+2cosA$
$2-2cosA=2+2cosA$
$CosA=0$
So, $A\in [2n\pi \pm \frac{\pi }{2}]$ and $[2n\pi \pm \frac{3\pi }{2}]$
So, $\frac{A}{2}\in [2n\pi \pm \frac{\pi }{4}]$ and $[2n\pi \pm \frac{3\pi }{4}]$
but since A has to be +ve, other wise sin become -ve in ease if A is -ve
So $\frac{A}{2}\in [2n\pi +\frac{\pi }{4}]$ and $[2n\pi +\frac{3\pi }{4}]$