Question

For any angle $A,\sqrt{1+sinA}=|cos\frac{A}{2}+sin\frac{A}{2}|,\sqrt{1-sinA}=|cos\frac{A}{2}-sin\frac{A}{2}|$
Use this information to answer the following.
$\sqrt{1+sinA}-\sqrt{1-sinA}=-2cos\frac{A}{2}\Rightarrow$

• A. $\frac{\pi }{2}<A<\frac{3\pi }{2}$
  
• B. $\frac{3\pi }{2}<A<\frac{7\pi }{2}$
  
• C. $\frac{5\pi }{2}<A<\frac{7\pi }{2}$
  
• D. $\frac{7\pi }{2}<A<\frac{9\pi }{2}$

Answer 1

The correct option is C $\frac{5\pi }{2}<A<\frac{7\pi }{2}$


For any angle $A,\sqrt{1+sinA}=|cos\frac{A}{2}+sin\frac{A}{2}|,\sqrt{1-sinA}=|cos\frac{A}{2}-sin\frac{A}{2}|$
Use this information to answer the following.
Given $\sqrt{1+sinA}=|cos\frac{A}{2}+sin\frac{A}{2}|,\sqrt{1-sinA}=|cos\frac{A}{2}-sin\frac{A}{2}|$
$\sqrt{1+sinA}-\sqrt{1-sinA}=-2cos\frac{A}{2}\Rightarrow |cos\frac{A}{2}+sin\frac{A}{2}|-|cos\frac{A}{2}-sin\frac{A}{2}|=-2cos\frac{A}{2}\Rightarrow \frac{5\pi }{2}<A<\frac{7\pi }{2}$