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=√(1-\cos^2x)=√(1-(-0.2)^2)\approx1.02`

Answer 2

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