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If \( \sin \alpha=0.504 \) and \( \cos \beta=0.679 \) with both angles' terminal rays in Quadrant-1, find the values of \( \tan (\alpha+\beta)=\square \) Your answers should be accurate to 4 decimal places. Submit Question

Ask by Reyes Zimmerman. in the United States
Jan 24,2025

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Answer

\( \tan(\alpha + \beta) \approx 4.5054 \)

Solution

To find \( \tan(\alpha + \beta) \) given \( \sin \alpha = 0.504 \) and \( \cos \beta = 0.679 \) with both angles in Quadrant I, follow these steps: 1. **Determine \( \cos \alpha \) and \( \sin \beta \):** - Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \): \[ \cos \alpha = \sqrt{1 - (0.504)^2} = \sqrt{1 - 0.254016} = \sqrt{0.745984} \approx 0.8637 \] - Similarly, since \( \sin^2 \beta + \cos^2 \beta = 1 \): \[ \sin \beta = \sqrt{1 - (0.679)^2} = \sqrt{1 - 0.461041} = \sqrt{0.538959} \approx 0.7341 \] 2. **Use the tangent addition formula:** \[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \] Plugging in the values: \[ \tan(\alpha + \beta) = \frac{0.504 \times 0.679 + 0.8637 \times 0.7341}{0.8637 \times 0.679 - 0.504 \times 0.7341} \] - **Calculate the numerator:** \[ 0.504 \times 0.679 = 0.342216 \\ 0.8637 \times 0.7341 \approx 0.63304 \\ \text{Numerator} = 0.342216 + 0.63304 \approx 0.975256 \] - **Calculate the denominator:** \[ 0.8637 \times 0.679 \approx 0.58645 \\ 0.504 \times 0.7341 \approx 0.3699864 \\ \text{Denominator} = 0.58645 - 0.3699864 \approx 0.2164636 \] - **Compute the tangent:** \[ \tan(\alpha + \beta) = \frac{0.975256}{0.2164636} \approx 4.5054 \] **Final Answer:** \[ \tan(\alpha + \beta) \approx 4.5054 \]

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The Deep Dive

To find \( \tan(\alpha + \beta) \), we can use the tangent addition formula: \[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \] First, we need to find \( \tan \alpha \) and \( \tan \beta \). From the given information: - \( \sin \alpha = 0.504 \): Using the Pythagorean identity, we can determine \( \cos \alpha \): \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (0.504)^2 \approx 1 - 0.254016 = 0.745984 \implies \cos \alpha \approx \sqrt{0.745984} \approx 0.8642 \] Thus, \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{0.504}{0.8642} \approx 0.5825 \] - \( \cos \beta = 0.679 \): Similarly, we find \( \sin \beta \) using the Pythagorean identity: \[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - (0.679)^2 \approx 1 - 0.460084 = 0.539916 \implies \sin \beta \approx \sqrt{0.539916} \approx 0.7342 \] Thus, \[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{0.7342}{0.679} \approx 1.0806 \] Now substituting these values into the tangent addition formula: \[ \tan(\alpha + \beta) \approx \frac{0.5825 + 1.0806}{1 - (0.5825)(1.0806)} \] Calculating: \[ \tan(\alpha + \beta) \approx \frac{1.6631}{1 - 0.6290} = \frac{1.6631}{0.3710} \approx 4.48 \] Thus, the value of \( \tan(\alpha + \beta) \) accurate to four decimal places is: \[ \tan(\alpha + \beta) \approx 4.48 \]

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