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.
If $A={280}^{\circ }$ then $\sqrt{1+sinA}+\sqrt{1-sinA}=$

• A. $2sin{50}^{\circ }$
  
• B. $2cos{50}^{\circ }$
  
• C. $2sin{10}^{\circ }$
  
• D. $2cos{10}^{\circ }$

Answer 1

The correct option is A $2sin{50}^{\circ }$


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 $\sqrt{1+sinA}=|cos\frac{A}{2}+sin\frac{A}{2}|,\sqrt{1-sinA}=|cos\frac{A}{2}-sin\frac{A}{2}|$
If A =280 then
$\sqrt{1+sinA}+\sqrt{1-sinA}=|cos\frac{A}{2}+sin\frac{A}{2}|+|cos\frac{A}{2}-sin\frac{A}{2}|$
$=|cos{140}^{\circ }+sin{140}^{\circ }|+|cos{140}^{\circ }-sin{140}^{\circ }|=|-sin{50}^{\circ }+cos{50}^{\circ }|+|-sin{50}^{\circ }-cos{50}^{\circ }|=sin{50}^{\circ }-cos{50}^{\circ }+sin{50}^{\circ }+cos{50}^{\circ }=2sin{50}^{\circ }$