Question

If cosX = -0.2, and x is in quadrant II, then what it the value of sinX​

Answer 1

$\sin x>0$ when $x$ is in quadrant II, so

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

Answer 2

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