Answer:
sinΘ = [tex]\frac{2}{3}[/tex]
Step-by-step explanation:
using the identity
sin²x + cos²x = 1 ( subtract cos²x from both sides )
sin²x = 1 - cos²x ( take square root of both sides )
sinx = ± [tex]\sqrt{1-cos^2x}[/tex]
given
cosΘ = - [tex]\frac{\sqrt{5} }{3}[/tex] , then
sinΘ = ± [tex]\sqrt{1-(-\frac{\sqrt{5} }{3})^2 }[/tex]
= ± [tex]\sqrt{1-\frac{5}{9} }[/tex]
= ± [tex]\sqrt{\frac{4}{9} }[/tex]
= ± [tex]\frac{2}{3}[/tex]
since Θ is in quadrant II where sinΘ > 0 , then
sinΘ = [tex]\frac{2}{3}[/tex]
