Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Given the mathematical problem: \[ \cos \theta = \frac{8}{9}, \quad \tan \theta < 0 \] **Objective:** Find \(\sin \theta\). **Explanation:** 1. **Identify the Quadrant:** - Since \(\cos \theta\) is positive and \(\tan \theta\) is negative, \(\theta\) must be in the fourth quadrant. 2. **Use the Pythagorean Identity:** \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute \(\cos \theta = \frac{8}{9}\): \[ \sin^2 \theta + \left(\frac{8}{9}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{64}{81} = 1 \] \[ \sin^2 \theta = 1 - \frac{64}{81} \] \[ \sin^2 \theta = \frac{81}{81} - \frac{64}{81} \] \[ \sin^2 \theta = \frac{17}{81} \] 3. **Calculate \(\sin \theta\):** Since \(\theta\) is in the fourth quadrant, \(\sin \theta\) is negative: \[ \sin \theta = -\sqrt{\frac{17}{81}} \] \[ \sin \theta = -\frac{\sqrt{17}}{9} \] Therefore, \(\sin \theta = -\frac{\sqrt{17}}{9}\).
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Transcribed Image Text:Given the mathematical problem: \[ \cos \theta = \frac{8}{9}, \quad \tan \theta < 0 \] **Objective:** Find \(\sin \theta\). **Explanation:** 1. **Identify the Quadrant:** - Since \(\cos \theta\) is positive and \(\tan \theta\) is negative, \(\theta\) must be in the fourth quadrant. 2. **Use the Pythagorean Identity:** \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substitute \(\cos \theta = \frac{8}{9}\): \[ \sin^2 \theta + \left(\frac{8}{9}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{64}{81} = 1 \] \[ \sin^2 \theta = 1 - \frac{64}{81} \] \[ \sin^2 \theta = \frac{81}{81} - \frac{64}{81} \] \[ \sin^2 \theta = \frac{17}{81} \] 3. **Calculate \(\sin \theta\):** Since \(\theta\) is in the fourth quadrant, \(\sin \theta\) is negative: \[ \sin \theta = -\sqrt{\frac{17}{81}} \] \[ \sin \theta = -\frac{\sqrt{17}}{9} \] Therefore, \(\sin \theta = -\frac{\sqrt{17}}{9}\).
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