Use the information given below to find \( \sin (\alpha+\beta) \). \( \sin \alpha=\frac{4}{5} \), with \( \alpha \) in quadrant II \( \sin \beta=-\frac{4}{5} \), with \( \beta \) in quadrant IV

Step 1: Use the identity for \(\sin(\alpha + \beta)\): \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] Step 2: Find \(\cos \alpha\) using \(\sin^2 \alpha + \cos^2 \alpha = 1\): \[ \cos^2 \alpha = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \] Since \(\alpha\) is in quadrant II, \(\cos \alpha = -\frac{3}{5}\). Step 3: Find \(\cos \beta\) using \(\sin^2 \beta + \cos^2 \beta = 1\): \[ \cos^2 \beta = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \] Since \(\beta\) is in quadrant IV, \(\cos \beta = \frac{3}{5}\). Step 4: Substitute the values into the identity: \[ \sin(\alpha + \beta) = \left(\frac{4}{5}\right)\left(\frac{3}{5}\right) + \left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right) \] Step 5: Simplify the expression: \[ \sin(\alpha + \beta) = \frac{12}{25} + \frac{12}{25} = \frac{24}{25} \] Step 6: Since both angles are in quadrants where sine is negative, adjust the sign: \[ \sin(\alpha + \beta) = -\frac{24}{25} \]