10. Show that there is a root of cos(x): = in the interval (0,7). TT Do NOT find the root. Explain all of your steps and reasoning, and name any theorems used.
10. Show that there is a root of cos(x): = in the interval (0,7). TT Do NOT find the root. Explain all of your steps and reasoning, and name any theorems used.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.4: Multiple-angle Formulas
Problem 40E
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**Mathematics Problem Solving**
**Problem: Root of the Cosine Function**
**Task**:
Show that there is a root of \( \cos(x) = \frac{x}{\pi} \) in the interval \( \left(0, \frac{\pi}{2}\right) \).
**Instructions**:
Do NOT find the root. Explain all of your steps and reasoning, and name any theorems used.
---
**Solution Guide**:
To show that there is a root of \( \cos(x) = \frac{x}{\pi} \) in the interval \( \left(0, \frac{\pi}{2}\right) \), we can use the **Intermediate Value Theorem** (IVT).
**Step-by-Step Reasoning**:
1. **Define the Function**:
Let \( f(x) = \cos(x) - \frac{x}{\pi} \).
2. **Analyze the Endpoints of the Interval**:
- At \( x = 0 \):
\[ f(0) = \cos(0) - \frac{0}{\pi} = 1 - 0 = 1. \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) - \frac{\frac{\pi}{2}}{\pi} = 0 - \frac{1}{2} = -\frac{1}{2}. \]
3. **Apply the Intermediate Value Theorem (IVT)**:
- **IVT Statement**:
If \( f \) is continuous on the interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \).
- **Check Continuity**:
The function \( f(x) = \cos(x) - \frac{x}{\pi} \) is continuous on \( \left[0, \frac{\pi}{2}\right] \) because both \( \cos(x) \) and \( \frac{x}{\pi} \) are continuous functions.
- **Evaluate Signs**:
At \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d01a9fa-2d22-41b6-ab89-d6b153d8821e%2F7ef7380a-9436-444c-8a82-61a58f6d13a9%2Fbf0jozf_processed.png&w=3840&q=75)
Transcribed Image Text:---
**Mathematics Problem Solving**
**Problem: Root of the Cosine Function**
**Task**:
Show that there is a root of \( \cos(x) = \frac{x}{\pi} \) in the interval \( \left(0, \frac{\pi}{2}\right) \).
**Instructions**:
Do NOT find the root. Explain all of your steps and reasoning, and name any theorems used.
---
**Solution Guide**:
To show that there is a root of \( \cos(x) = \frac{x}{\pi} \) in the interval \( \left(0, \frac{\pi}{2}\right) \), we can use the **Intermediate Value Theorem** (IVT).
**Step-by-Step Reasoning**:
1. **Define the Function**:
Let \( f(x) = \cos(x) - \frac{x}{\pi} \).
2. **Analyze the Endpoints of the Interval**:
- At \( x = 0 \):
\[ f(0) = \cos(0) - \frac{0}{\pi} = 1 - 0 = 1. \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) - \frac{\frac{\pi}{2}}{\pi} = 0 - \frac{1}{2} = -\frac{1}{2}. \]
3. **Apply the Intermediate Value Theorem (IVT)**:
- **IVT Statement**:
If \( f \) is continuous on the interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \).
- **Check Continuity**:
The function \( f(x) = \cos(x) - \frac{x}{\pi} \) is continuous on \( \left[0, \frac{\pi}{2}\right] \) because both \( \cos(x) \) and \( \frac{x}{\pi} \) are continuous functions.
- **Evaluate Signs**:
At \(
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