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]
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