Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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, \(\sin \theta = -\frac{\sqrt{17}}{9}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63d16113-ff39-4f92-89f8-61ce82bfb495%2Fe9ff5f2a-c1de-422f-a460-55fffae8951d%2Fu6a28q6_processed.jpeg&w=3840&q=75)
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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