Im and Difference Formulas Question 29, 5.2.61 Patacts HW Score: \( 90.3 \%, 34,67 \) of 30 points Points: 0.67 of 1 Find the ecact value of the expressions \( \cos (\alpha+\beta) . \sin (\alpha+\beta) \) and \( \tan (\alpha+\beta) \) under the following conditions: \( \cos (\alpha)=\frac{24}{25}, a \) les in quadiant \( V \), and \( \sin (\beta)=\frac{-2}{5}, \beta \) lecs in quadiant III. \( a \cdot \cos (\alpha+\beta)=\frac{-24 \sqrt{21}-14}{125} \) (Simplity your answer Type an exad answer, using radicals as needed. Rationalke all denominators Use integers or fractions for any numbers in the eppession) \( b \sin (a+\beta)=\frac{7 \sqrt{21}-48}{125} \) (Simplity your answer. Type an exad answer, using radicals as needed. Rationalize all denominators. Use integers or flactions for any numbers in the expression) c. \( \tan (\alpha+\beta)= \) \( \square \) (Simplify your answer. Type an evad answer, using radicals as needed. Ratonalke all denominators. Use integers or flactions for any numbers in the expression)

To solve the problem, we need to find the exact values of \( \cos(\alpha + \beta) \), \( \sin(\alpha + \beta) \), and \( \tan(\alpha + \beta) \) using the given conditions. ### Step 1: Determine \( \sin(\alpha) \) and \( \cos(\beta) \) 1. **For \( \alpha \)**: - We know \( \cos(\alpha) = \frac{24}{25} \). - Since \( \alpha \) is in quadrant V, \( \sin(\alpha) \) will be negative. - Using the Pythagorean identity: \[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \] \[ \sin^2(\alpha) + \left(\frac{24}{25}\right)^2 = 1 \] \[ \sin^2(\alpha) + \frac{576}{625} = 1 \] \[ \sin^2(\alpha) = 1 - \frac{576}{625} = \frac{49}{625} \] \[ \sin(\alpha) = -\frac{7}{25} \quad (\text{since } \alpha \text{ is in quadrant V}) \] 2. **For \( \beta \)**: - We know \( \sin(\beta) = -\frac{2}{5} \). - Since \( \beta \) is in quadrant III, \( \cos(\beta) \) will also be negative. - Using the Pythagorean identity: \[ \sin^2(\beta) + \cos^2(\beta) = 1 \] \[ \left(-\frac{2}{5}\right)^2 + \cos^2(\beta) = 1 \] \[ \frac{4}{25} + \cos^2(\beta) = 1 \] \[ \cos^2(\beta) = 1 - \frac{4}{25} = \frac{21}{25} \] \[ \cos(\beta) = -\frac{\sqrt{21}}{5} \quad (\text{since } \beta \text{ is in quadrant III}) \] ### Step 2: Use the Sum Formulas Now we can use the sum formulas for cosine and sine: 1. **Cosine of the sum**: \[ \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) \] Substituting the values: \[ \cos(\alpha + \beta) = \left(\frac{24}{25}\right)\left(-\frac{\sqrt{21}}{5}\right) - \left(-\frac{7}{25}\right)\left(-\frac{2}{5}\right) \] \[ = -\frac{24\sqrt{21}}{125} - \frac{14}{125} \] \[ = \frac{-24\sqrt{21} - 14}{125} \] 2. **Sine of the sum**: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \] Substituting the values: \[ \sin(\alpha + \beta) = \left(-\frac{7}{25}\right)\left(-\frac{\sqrt{21}}{5}\right) + \left(\frac{24}{25}\right)\left(-\frac{2}{5}\right) \] \[ = \frac{7\sqrt{21}}{125} - \frac{48}{125} \] \[ = \frac{7\sqrt{21} - 48}{125} \] ### Step 3: Calculate Tangent of the Sum Finally, we can find \( \tan(\alpha + \beta) \): \[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \] Substituting the values: \[ \tan(\alpha + \beta) = \frac{\frac{7\sqrt{21} - 48}{125}}{\frac{-24\sqrt{21} - 14}{125}} = \frac{7\sqrt{21} - 48}{-24\sqrt{21} - 14} \] ### Final Answers - \( a \cdot \cos(\alpha + \beta) = \frac{-24\sqrt{21} - 14}{125} \) - \( b \cdot \sin(\alpha + \beta) = \frac{7\sqrt{21} - 48}{125} \) - \( \tan(\alpha + \beta) = \frac{7\sqrt{21} - 48}{-24\sqrt{21} - 14} \) This is the simplified form of the tangent.