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Find the exact value of the expressions \( \cos (\alpha+\beta), \sin (\alpha+\beta) \) and \( \tan (\alpha+\beta) \) under the following conditions: \[ \sin (\alpha)=\frac{15}{17}, \alpha \text { lies in quadrant I, and } \sin (\beta)=\frac{4}{5}, \beta \text { lies in quadrant II. } \]

Ask by Lowe Howell. in the United States
Jan 25,2025

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Answer

\[ \cos(\alpha + \beta) = -\frac{84}{85}, \quad \sin(\alpha + \beta) = -\frac{13}{85}, \quad \tan(\alpha + \beta) = \frac{13}{84} \]

Solution

Given: - \(\sin \alpha = \frac{15}{17}\) with \(\alpha\) in Quadrant I. - \(\sin \beta = \frac{4}{5}\) with \(\beta\) in Quadrant II. **Step 1: Determine \(\cos \alpha\) and \(\cos \beta\)** 1. For \(\alpha\) in Quadrant I: \[ \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17} \] 2. For \(\beta\) in Quadrant II: \[ \cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \] (Note: \(\cos \beta\) is negative in Quadrant II.) **Step 2: Use the Angle Addition Formulas** 1. **\(\cos(\alpha + \beta)\)** \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \left(\frac{8}{17}\right)\left(-\frac{3}{5}\right) - \left(\frac{15}{17}\right)\left(\frac{4}{5}\right) = -\frac{24}{85} - \frac{60}{85} = -\frac{84}{85} \] 2. **\(\sin(\alpha + \beta)\)** \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \left(\frac{15}{17}\right)\left(-\frac{3}{5}\right) + \left(\frac{8}{17}\right)\left(\frac{4}{5}\right) = -\frac{45}{85} + \frac{32}{85} = -\frac{13}{85} \] 3. **\(\tan(\alpha + \beta)\)** \[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{-\frac{13}{85}}{-\frac{84}{85}} = \frac{13}{84} \] **Final Answer:** \[ \cos(\alpha + \beta) = -\frac{84}{85}, \quad \sin(\alpha + \beta) = -\frac{13}{85}, \quad \tan(\alpha + \beta) = \frac{13}{84} \]

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Bonus Knowledge

To find the exact values of \( \cos(\alpha+\beta) \), \( \sin(\alpha+\beta) \), and \( \tan(\alpha+\beta) \), we first need to find \( \cos(\alpha) \) and \( \cos(\beta) \). Given \( \sin(\alpha) = \frac{15}{17} \) and \( \alpha \) is in quadrant I, we can use the Pythagorean identity: \[ \cos^2(\alpha) + \sin^2(\alpha) = 1 \implies \cos^2(\alpha) + \left(\frac{15}{17}\right)^2 = 1, \] \[ \cos^2(\alpha) + \frac{225}{289} = 1 \implies \cos^2(\alpha) = 1 - \frac{225}{289} = \frac{64}{289} \implies \cos(\alpha) = \frac{8}{17}. \] Next, for \( \beta \), we have \( \sin(\beta) = \frac{4}{5} \) and \( \beta \) is in quadrant II. Since cosine is negative in quadrant II, we find: \[ \cos^2(\beta) + \sin^2(\beta) = 1 \implies \cos^2(\beta) + \left(\frac{4}{5}\right)^2 = 1, \] \[ \cos^2(\beta) + \frac{16}{25} = 1 \implies \cos^2(\beta) = 1 - \frac{16}{25} = \frac{9}{25} \implies \cos(\beta) = -\frac{3}{5}. \] Now we can apply the angle sum formulas: 1. **Finding \( \sin(\alpha + \beta) \)**: \[ \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta). \] Substituting the values: \[ \sin(\alpha + \beta) = \left(\frac{15}{17}\right)\left(-\frac{3}{5}\right) + \left(\frac{8}{17}\right)\left(\frac{4}{5}\right), \] \[ = -\frac{45}{85} + \frac{32}{85} = -\frac{13}{85}. \] 2. **Finding \( \cos(\alpha + \beta) \)**: \[ \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta). \] Substituting the values: \[ \cos(\alpha + \beta) = \left(\frac{8}{17}\right)\left(-\frac{3}{5}\right) - \left(\frac{15}{17}\right)\left(\frac{4}{5}\right), \] \[ = -\frac{24}{85} - \frac{60}{85} = -\frac{84}{85}. \] 3. **Finding \( \tan(\alpha + \beta) \)**: \[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}. \] Substituting the values: \[ \tan(\alpha + \beta) = \frac{-\frac{13}{85}}{-\frac{84}{85}} = \frac{13}{84}. \] Thus, the exact values are: \[ \sin(\alpha + \beta) = -\frac{13}{85}, \quad \cos(\alpha + \beta) = -\frac{84}{85}, \quad \tan(\alpha + \beta) = \frac{13}{84}. \]

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