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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![## Trigonometric Identities and Solutions
Given the following information:
\[ \csc(x) = 4 \]
\[ \tan(x) < 0 \]
We are to find \(\sin(2x)\), \(\cos(2x)\), and \(\tan(2x)\).
### Solution
First, use the given information:
Since \(\csc(x) = \frac{1}{\sin(x)}\), we can find \(\sin(x)\):
\[ \sin(x) = \frac{1}{\csc(x)} = \frac{1}{4} \]
Next, we need \(\cos(x)\). We use the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Substitute \(\sin(x) = \frac{1}{4}\):
\[ \left(\frac{1}{4}\right)^2 + \cos^2(x) = 1 \]
\[ \frac{1}{16} + \cos^2(x) = 1 \]
\[ \cos^2(x) = 1 - \frac{1}{16} = \frac{15}{16} \]
Since \(\tan(x) < 0\), \(\cos(x)\) must be negative:
\[ \cos(x) = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \]
Now, we find the values of the trigonometric identities for \(2x\):
### \(\sin(2x)\)
We use the double-angle formula:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
\[ \sin(2x) = 2 \left(\frac{1}{4}\right) \left(-\frac{\sqrt{15}}{4}\right) = 2 \left(-\frac{\sqrt{15}}{16}\right) = -\frac{\sqrt{15}}{8} \]
### \(\cos(2x)\)
We use the double-angle formula:
\[ \cos(2x) = \cos^2(x) - \sin^2(x) \]
\[ \cos(2x) = \left(-\frac{\sqrt{15}}{4}\right)^2 - \left(\frac{1}{4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2F379879d8-5083-4b52-b6fa-f1219f91f7a4%2Fvroy5ba_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Trigonometric Identities and Solutions
Given the following information:
\[ \csc(x) = 4 \]
\[ \tan(x) < 0 \]
We are to find \(\sin(2x)\), \(\cos(2x)\), and \(\tan(2x)\).
### Solution
First, use the given information:
Since \(\csc(x) = \frac{1}{\sin(x)}\), we can find \(\sin(x)\):
\[ \sin(x) = \frac{1}{\csc(x)} = \frac{1}{4} \]
Next, we need \(\cos(x)\). We use the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Substitute \(\sin(x) = \frac{1}{4}\):
\[ \left(\frac{1}{4}\right)^2 + \cos^2(x) = 1 \]
\[ \frac{1}{16} + \cos^2(x) = 1 \]
\[ \cos^2(x) = 1 - \frac{1}{16} = \frac{15}{16} \]
Since \(\tan(x) < 0\), \(\cos(x)\) must be negative:
\[ \cos(x) = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \]
Now, we find the values of the trigonometric identities for \(2x\):
### \(\sin(2x)\)
We use the double-angle formula:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
\[ \sin(2x) = 2 \left(\frac{1}{4}\right) \left(-\frac{\sqrt{15}}{4}\right) = 2 \left(-\frac{\sqrt{15}}{16}\right) = -\frac{\sqrt{15}}{8} \]
### \(\cos(2x)\)
We use the double-angle formula:
\[ \cos(2x) = \cos^2(x) - \sin^2(x) \]
\[ \cos(2x) = \left(-\frac{\sqrt{15}}{4}\right)^2 - \left(\frac{1}{4
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