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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![### Problem Statement
Find the exact values of \(\cos \theta\) if \(\sin \theta = -\frac{5}{13}\) and \(\sec \theta > 0\).
### Explanation
To solve this problem, we first need to consider the trigonometric identity:
\[ \sin^2 \theta + \cos^2 \theta = 1. \]
Given that \(\sin \theta = -\frac{5}{13}\), we can substitute this value into the identity:
\[ \left(-\frac{5}{13}\right)^2 + \cos^2 \theta = 1. \]
Calculating \(\left(-\frac{5}{13}\right)^2\):
\[ \frac{25}{169} + \cos^2 \theta = 1. \]
Subtracting \(\frac{25}{169}\) from both sides:
\[ \cos^2 \theta = 1 - \frac{25}{169}. \]
Convert 1 to a fraction with denominator 169:
\[ \cos^2 \theta = \frac{169}{169} - \frac{25}{169}. \]
\[ \cos^2 \theta = \frac{144}{169}. \]
Taking the square root of both sides, considering the positive value (since \(\sec \theta = \frac{1}{\cos \theta} > 0\)):
\[ \cos \theta = \frac{12}{13}. \]
Thus, the exact value of \(\cos \theta\) is \(\frac{12}{13}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99f990aa-ade1-4201-bad9-acb9a7debe3d%2Fa9ce7f78-4c85-41e0-a911-c095e091bba3%2Fruw27yj_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the exact values of \(\cos \theta\) if \(\sin \theta = -\frac{5}{13}\) and \(\sec \theta > 0\).
### Explanation
To solve this problem, we first need to consider the trigonometric identity:
\[ \sin^2 \theta + \cos^2 \theta = 1. \]
Given that \(\sin \theta = -\frac{5}{13}\), we can substitute this value into the identity:
\[ \left(-\frac{5}{13}\right)^2 + \cos^2 \theta = 1. \]
Calculating \(\left(-\frac{5}{13}\right)^2\):
\[ \frac{25}{169} + \cos^2 \theta = 1. \]
Subtracting \(\frac{25}{169}\) from both sides:
\[ \cos^2 \theta = 1 - \frac{25}{169}. \]
Convert 1 to a fraction with denominator 169:
\[ \cos^2 \theta = \frac{169}{169} - \frac{25}{169}. \]
\[ \cos^2 \theta = \frac{144}{169}. \]
Taking the square root of both sides, considering the positive value (since \(\sec \theta = \frac{1}{\cos \theta} > 0\)):
\[ \cos \theta = \frac{12}{13}. \]
Thus, the exact value of \(\cos \theta\) is \(\frac{12}{13}\).
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