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 identities: \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \] \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \] We are given \( \sin \alpha = \frac{2}{3} \) and since \( 90^{\circ} < \alpha < 180^{\circ} \), \( \alpha \) is in the second quadrant where cosine is negative. First, we can find \( \cos \alpha \) using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] \[ \left(\frac{2}{3}\right)^2 + \cos^2 \alpha = 1 \] \[ \frac{4}{9} + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 1 - \frac{4}{9} = \frac{5}{9} \] \[ \cos \alpha = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3} \] Now, substitute \( \sin \alpha \) and \( \cos \alpha \) into the double angle formulas: \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \left(\frac{2}{3}\right) \left(-\frac{\sqrt{5}}{3}\right) = -\frac{4\sqrt{5}}{9} \] \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \left(-\frac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \] Thus, the final results are: \[ \sin 2 \alpha = -\frac{4\sqrt{5}}{9} \quad \cos 2 \alpha = \frac{1}{9} \]