Given that \( \sin \alpha=\frac{4}{7} \) and \( 90^{\circ}<\alpha<180^{\circ} \), find \( \sin 2 \alpha \) and \( \cos 2 \alpha \). \( \sin 2 \alpha=\square \cos 2 \alpha=\square \)

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 \( \sin \alpha = \frac{4}{7} \). Since \( \alpha \) is in the second quadrant (where sine is positive and cosine is negative), we need to find \( \cos \alpha \). Using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Substituting \( \sin \alpha \): \[ \left(\frac{4}{7}\right)^2 + \cos^2 \alpha = 1 \] Calculating \( \left(\frac{4}{7}\right)^2 \): \[ \frac{16}{49} + \cos^2 \alpha = 1 \] Subtracting \( \frac{16}{49} \) from both sides: \[ \cos^2 \alpha = 1 - \frac{16}{49} = \frac{49}{49} - \frac{16}{49} = \frac{33}{49} \] Taking the square root, we find \( \cos \alpha \): \[ \cos \alpha = -\sqrt{\frac{33}{49}} = -\frac{\sqrt{33}}{7} \] Now we can calculate \( \sin 2\alpha \): \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \left(\frac{4}{7}\right) \left(-\frac{\sqrt{33}}{7}\right) \] Calculating this: \[ \sin 2\alpha = 2 \cdot \frac{4}{7} \cdot -\frac{\sqrt{33}}{7} = -\frac{8\sqrt{33}}{49} \] Next, we calculate \( \cos 2\alpha \): \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \] Substituting the values we found: \[ \cos 2\alpha = \frac{33}{49} - \left(\frac{4}{7}\right)^2 = \frac{33}{49} - \frac{16}{49} = \frac{33 - 16}{49} = \frac{17}{49} \] Thus, the final results are: \[ \sin 2\alpha = -\frac{8\sqrt{33}}{49}, \quad \cos 2\alpha = \frac{17}{49} \] So, we have: \[ \sin 2\alpha = -\frac{8\sqrt{33}}{49}, \quad \cos 2\alpha = \frac{17}{49} \]