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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![### Question 2
Find all solutions of the equation sec^2(x) - 2 = 0.
The answer is \( A + Bk\pi \) where \( k \) is any integer and \( 0 < A < \frac{\pi}{2} \),
\[
A = \quad , \quad B =
\]
This problem requires you to find the values of \( A \) and \( B \) that satisfy the given equation. The solution should have a specific format where \( A \) is a constant within the specified range, and \( B \) involves \( k \), where \( k \) is any integer.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1c81f19-6b44-4c48-9f6b-8f6cecd1e5f5%2Fb590f11b-a9fc-4d51-851b-69ebfc2ba251%2Fu87gakdm_processed.png&w=3840&q=75)
Transcribed Image Text:### Question 2
Find all solutions of the equation sec^2(x) - 2 = 0.
The answer is \( A + Bk\pi \) where \( k \) is any integer and \( 0 < A < \frac{\pi}{2} \),
\[
A = \quad , \quad B =
\]
This problem requires you to find the values of \( A \) and \( B \) that satisfy the given equation. The solution should have a specific format where \( A \) is a constant within the specified range, and \( B \) involves \( k \), where \( k \) is any integer.
---
![### Trigonometric Identities and Calculations
#### Problem Statement:
If \( \sin x = -\frac{3}{5} \), \( x \) in quadrant III, then:
- \( \sin 2x = \underline{\hspace{50pt}} ; \)
- \( \cos 2x = \underline{\hspace{50pt}} ; \)
- \( \tan 2x = \underline{\hspace{50pt}} .\)
#### Explanation:
This problem involves applying trigonometric identities to find the values of \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \), given that \( \sin x = -\frac{3}{5} \) and \( x \) is in the third quadrant. In the third quadrant, sine is negative and cosine is also negative.
**1. Find \( \cos x \):**
Using the Pythagorean identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Given:
\[ \sin x = -\frac{3}{5} \]
We square it:
\[ \left(-\frac{3}{5}\right)^2 + \cos^2 x = 1 \]
\[ \frac{9}{25} + \cos^2 x = 1 \]
\[ \cos^2 x = 1 - \frac{9}{25} \]
\[ \cos^2 x = \frac{25}{25} - \frac{9}{25} \]
\[ \cos^2 x = \frac{16}{25} \]
Since \( x \) is in the third quadrant, \( \cos x \) is negative:
\[ \cos x = -\frac{4}{5} \]
**2. Calculate \( \sin 2x \):**
Using the double-angle identity for sine:
\[ \sin 2x = 2 \sin x \cos x \]
\[ \sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \]
\[ \sin 2x = 2 \cdot \frac{12}{25} \]
\[ \sin 2x = \frac{24}{25} \]
**3. Calculate \( \cos 2x \):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1c81f19-6b44-4c48-9f6b-8f6cecd1e5f5%2Fb590f11b-a9fc-4d51-851b-69ebfc2ba251%2Fz481ag7_processed.png&w=3840&q=75)
Transcribed Image Text:### Trigonometric Identities and Calculations
#### Problem Statement:
If \( \sin x = -\frac{3}{5} \), \( x \) in quadrant III, then:
- \( \sin 2x = \underline{\hspace{50pt}} ; \)
- \( \cos 2x = \underline{\hspace{50pt}} ; \)
- \( \tan 2x = \underline{\hspace{50pt}} .\)
#### Explanation:
This problem involves applying trigonometric identities to find the values of \( \sin 2x \), \( \cos 2x \), and \( \tan 2x \), given that \( \sin x = -\frac{3}{5} \) and \( x \) is in the third quadrant. In the third quadrant, sine is negative and cosine is also negative.
**1. Find \( \cos x \):**
Using the Pythagorean identity:
\[ \sin^2 x + \cos^2 x = 1 \]
Given:
\[ \sin x = -\frac{3}{5} \]
We square it:
\[ \left(-\frac{3}{5}\right)^2 + \cos^2 x = 1 \]
\[ \frac{9}{25} + \cos^2 x = 1 \]
\[ \cos^2 x = 1 - \frac{9}{25} \]
\[ \cos^2 x = \frac{25}{25} - \frac{9}{25} \]
\[ \cos^2 x = \frac{16}{25} \]
Since \( x \) is in the third quadrant, \( \cos x \) is negative:
\[ \cos x = -\frac{4}{5} \]
**2. Calculate \( \sin 2x \):**
Using the double-angle identity for sine:
\[ \sin 2x = 2 \sin x \cos x \]
\[ \sin 2x = 2 \left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \]
\[ \sin 2x = 2 \cdot \frac{12}{25} \]
\[ \sin 2x = \frac{24}{25} \]
**3. Calculate \( \cos 2x \):
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