Respuesta :

[tex]\sin x>0[/tex] when [tex]x[/tex] is in quadrant II, so

[tex]\cos^2x+\sin^2x=1\implies\sin x=\sqrt{1-\cos^2x}=\sqrt{1-(-0.2)^2}\approx1.02[/tex]

dhiab

Answer:

Step-by-step explanation:

note : cos²x + sin²x = 1

         (- 0.2)² + sin²x =1

          0.04 + sin²x = 1

sin²x = 1 - 0.04

sin²x = 0.96

sinx = √0.96  or sinx = - √0.96

in quadrant II : sinx > 0    ...so : sinx = √0.96