Question 7/9 (4 $v<>$ NEXT 7 Given $\cos \theta=\frac{4}{9}$ in Quadrant IV, determine $\sin \theta$. Because the solution is in Quadrant IV, $\sin \theta=$

Step 1 ：Given that \(\cos \theta = \frac{4}{9}\) in Quadrant IV, we need to find \(\sin \theta\).

Step 2 ：Using the Pythagorean identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\), we can solve for \(\sin \theta\).

Step 3 ：Since \(\cos \theta = \frac{4}{9}\), we have \(\sin^2(\theta) = 1 - \left(\frac{4}{9}\right)^2 = \frac{65}{81}\).

Step 4 ：As the angle is in Quadrant IV, \(\sin \theta\) is negative. Therefore, \(\sin \theta = -\sqrt{\frac{65}{81}}\).

Step 5 ：\boxed{-\frac{\sqrt{65}}{9}} is the final answer for \(\sin \theta\) in Quadrant IV.