Answer:
2nd option
Step-by-step explanation:
Using
sin²A = [tex]\frac{1-cos2A}{2}[/tex] ( take the square root of both sides )
sinA = ± [tex]\sqrt{\frac{1-cos2A}{2} }[/tex]
Replace 2A with θ and A with [tex]\frac{0}{2}[/tex] , then
sin ([tex]\frac{0}{2}[/tex] ) = ± [tex]\sqrt{\frac{1-cos20}{2} }[/tex]
