Given that \( \sin \alpha=\frac{2}{3} \) and \( 90^{\circ}<\alpha<180^{\circ} \), find \( \sin 2 \alpha \) and \( \cos 2 \alpha \). \( \sin 2 \alpha=\square \quad \cos 2 \alpha=\square \)

To find \( \sin 2\alpha \) and \( \cos 2\alpha \), we can use the double angle formulas: \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \] \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \] We are given that \( \sin \alpha = \frac{2}{3} \) and that \( \alpha \) is in the second quadrant, where sine is positive and cosine is negative. First, we need to find \( \cos \alpha \). We can use the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Substituting \( \sin \alpha \): \[ \left(\frac{2}{3}\right)^2 + \cos^2 \alpha = 1 \] \[ \frac{4}{9} + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 1 - \frac{4}{9} = \frac{5}{9} \] Since \( \alpha \) is in the second quadrant, \( \cos \alpha \) is negative: \[ \cos \alpha = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3} \] Now we can calculate \( \sin 2\alpha \): \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \left(\frac{2}{3}\right) \left(-\frac{\sqrt{5}}{3}\right) \] \[ = -\frac{4\sqrt{5}}{9} \] Next, we calculate \( \cos 2\alpha \): \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \] \[ = \left(-\frac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 \] \[ = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \] Thus, we have: \[ \sin 2\alpha = -\frac{4\sqrt{5}}{9} \quad \text{and} \quad \cos 2\alpha = \frac{1}{9} \] Final answers: \[ \sin 2\alpha = -\frac{4\sqrt{5}}{9} \quad \cos 2\alpha = \frac{1}{9} \]