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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
Transcribed Image Text:**(b) Compute \( \sin 2\theta \). Hint:** \( \sin 2\theta = \sin(\theta + \theta) \)
This hint suggests using the angle addition formula for sine.
![**Problem Statement:**
Suppose that \(\sin \theta = \frac{1}{7}\) and \(0 < \theta < \pi/2\).
(a) Compute \(\cos \theta\).
---
**Explanation:**
Given that \(\sin \theta = \frac{1}{7}\) and the angle \(\theta\) is in the first quadrant \((0 < \theta < \pi/2)\), we need to find \(\cos \theta\).
To find \(\cos \theta\), we use the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Plug in the value of \(\sin \theta\):
\[
\left(\frac{1}{7}\right)^2 + \cos^2 \theta = 1
\]
\[
\frac{1}{49} + \cos^2 \theta = 1
\]
Subtract \(\frac{1}{49}\) from both sides:
\[
\cos^2 \theta = 1 - \frac{1}{49} = \frac{49}{49} - \frac{1}{49} = \frac{48}{49}
\]
Take the square root of both sides:
\[
\cos \theta = \sqrt{\frac{48}{49}} = \frac{\sqrt{48}}{7}
\]
Since \(\theta\) is in the first quadrant, \(\cos \theta\) is positive. Thus:
\[
\cos \theta = \frac{\sqrt{48}}{7}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e29160-8b58-485e-8a5b-b6eaf19fb950%2F27c79476-c922-4419-85ce-7769fc71dc0b%2Fvva8ddn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose that \(\sin \theta = \frac{1}{7}\) and \(0 < \theta < \pi/2\).
(a) Compute \(\cos \theta\).
---
**Explanation:**
Given that \(\sin \theta = \frac{1}{7}\) and the angle \(\theta\) is in the first quadrant \((0 < \theta < \pi/2)\), we need to find \(\cos \theta\).
To find \(\cos \theta\), we use the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Plug in the value of \(\sin \theta\):
\[
\left(\frac{1}{7}\right)^2 + \cos^2 \theta = 1
\]
\[
\frac{1}{49} + \cos^2 \theta = 1
\]
Subtract \(\frac{1}{49}\) from both sides:
\[
\cos^2 \theta = 1 - \frac{1}{49} = \frac{49}{49} - \frac{1}{49} = \frac{48}{49}
\]
Take the square root of both sides:
\[
\cos \theta = \sqrt{\frac{48}{49}} = \frac{\sqrt{48}}{7}
\]
Since \(\theta\) is in the first quadrant, \(\cos \theta\) is positive. Thus:
\[
\cos \theta = \frac{\sqrt{48}}{7}
\]
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