Respuesta :
Answer:
see explanation
Step-by-step explanation:
Using the identity
sin²x + cos²x = 1 then
sin²x = 1 - cos²x
sin x = ± [tex]\sqrt{1-cos^2x}[/tex]
Given cos θ = [tex]\frac{2}{3}[/tex] , then
sinθ = ± [tex]\sqrt{1-(\frac{2}{3})^2 }[/tex] = ± [tex]\sqrt{1-\frac{4}{9} }[/tex] = ± [tex]\sqrt{\frac{5}{9} }[/tex] = ± [tex]\frac{\sqrt{5} }{3}[/tex]
Also
tanθ = [tex]\frac{sin0}{cos0}[/tex]
= [tex]\frac{\frac{\sqrt{5} }{3} }{\frac{2}{3} }[/tex]
= [tex]\frac{\sqrt{5} }{3}[/tex] × [tex]\frac{3}{2}[/tex] = ± [tex]\frac{\sqrt{5} }{2}[/tex]
