We know that : Sin²θ + Cos²θ = 1
Given : Sinθ [tex]\bf{= \frac{4}{7}}[/tex]
[tex]\bf{\implies (\frac{4}{7})^2 + Cos^2(\theta) = 1}[/tex]
[tex]\bf{\implies Cos^2(\theta) = 1 - \frac{16}{49}}[/tex]
[tex]\bf{\implies Cos^2(\theta) =(\frac{49 - 16}{49})}[/tex]
[tex]\bf{\implies Cos^2(\theta) =(\frac{33}{49})}[/tex]
[tex]\bf{\implies Cos(\theta) = (\pm)(\frac{\sqrt{33}}{7})}[/tex]
Given : θ lies in Quadrant II
We know that : Cosθ is Negative in Quadrant II
[tex]\bf{\implies Cos(\theta) = (-)(\frac{\sqrt{33}}{7})}[/tex]
Option 3 is the Answer
