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
![**Problem 2:**
Find the exact value of \( \sin^{-1}\left(\sin\left(\frac{9\pi}{8}\right)\right) \). You must show all your work.
---
**Solution Explanation:**
To solve \( \sin^{-1}(\sin(x)) \), it is important to remember that \( \sin^{-1} \) (also written as arcsin) represents the inverse sine function, whose principal value is restricted between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
**Step-by-Step Process:**
1. **Identify the angle**: We are given \( x = \frac{9\pi}{8} \).
2. **Convert the angle**: Since \(\frac{9\pi}{8}\) is greater than \(\pi\), it falls outside the interval \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
3. **Reduce the angle**: Find a coterminal angle in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
- Calculate \( x - \pi \):
\[
\frac{9\pi}{8} - \pi = \frac{9\pi}{8} - \frac{8\pi}{8} = \frac{\pi}{8}
\]
4. **Evaluate the expression**: The reduction of the angle leads us to see \( \sin^{-1}(\sin(x)) = \sin^{-1}\left(\sin\left(\frac{\pi}{8}\right)\right) \).
5. **Final Answer**:
\[
\sin^{-1}(\sin(\frac{9\pi}{8})) = \frac{\pi}{8}
\]
Since \(\frac{\pi}{8}\) is within the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), this is the exact value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe708b20e-246d-43f9-aec0-b076b54d250e%2F78f74e84-6c61-4fa2-9584-a1efc715f6d2%2Fjp94qrp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 2:**
Find the exact value of \( \sin^{-1}\left(\sin\left(\frac{9\pi}{8}\right)\right) \). You must show all your work.
---
**Solution Explanation:**
To solve \( \sin^{-1}(\sin(x)) \), it is important to remember that \( \sin^{-1} \) (also written as arcsin) represents the inverse sine function, whose principal value is restricted between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
**Step-by-Step Process:**
1. **Identify the angle**: We are given \( x = \frac{9\pi}{8} \).
2. **Convert the angle**: Since \(\frac{9\pi}{8}\) is greater than \(\pi\), it falls outside the interval \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
3. **Reduce the angle**: Find a coterminal angle in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
- Calculate \( x - \pi \):
\[
\frac{9\pi}{8} - \pi = \frac{9\pi}{8} - \frac{8\pi}{8} = \frac{\pi}{8}
\]
4. **Evaluate the expression**: The reduction of the angle leads us to see \( \sin^{-1}(\sin(x)) = \sin^{-1}\left(\sin\left(\frac{\pi}{8}\right)\right) \).
5. **Final Answer**:
\[
\sin^{-1}(\sin(\frac{9\pi}{8})) = \frac{\pi}{8}
\]
Since \(\frac{\pi}{8}\) is within the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), this is the exact value.
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