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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![### Problem Statement
Evaluate the function \( f(x) = 7 \sin x - \cos \left(\frac{1}{2} x\right) \) at \( x = \frac{\pi}{3} \).
### Explanation
- **Function Definition**: The function \( f(x) \) is composed of two trigonometric components: \( 7 \sin x \) and \( -\cos \left(\frac{1}{2} x\right) \).
- \( 7 \sin x \): This part multiplies the sine of \( x \) by 7.
- \( -\cos \left(\frac{1}{2} x\right) \): This part finds the cosine of half of \( x \), then negates it.
- **Evaluation Point**: We need to evaluate this function specifically at \( x = \frac{\pi}{3} \).
### Steps to Solution
1. **Substitute** \( x = \frac{\pi}{3} \) into the equation:
\[ f\left(\frac{\pi}{3}\right) = 7 \sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{1}{2} \cdot \frac{\pi}{3}\right) \]
2. **Calculate Trigonometric Values**:
- \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
- \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
3. **Substitute the Trigonometric Values**:
\[ f\left(\frac{\pi}{3}\right) = 7 \left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} \]
4. **Simplify**:
\[ f\left(\frac{\pi}{3}\right) = \frac{7\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \]
\[ f\left(\frac{\pi}{3}\right) = \frac{6\sqrt{3}}{2} \]
\[ f\left(\frac{\pi}{3}\right) = 3\sqrt{3} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc760f331-1bd4-4b33-b6d2-ac26f1dcca1f%2Ff65c4a9f-da5a-48a1-8d7a-b2dbe7e11b2a%2F3p7lwyh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Evaluate the function \( f(x) = 7 \sin x - \cos \left(\frac{1}{2} x\right) \) at \( x = \frac{\pi}{3} \).
### Explanation
- **Function Definition**: The function \( f(x) \) is composed of two trigonometric components: \( 7 \sin x \) and \( -\cos \left(\frac{1}{2} x\right) \).
- \( 7 \sin x \): This part multiplies the sine of \( x \) by 7.
- \( -\cos \left(\frac{1}{2} x\right) \): This part finds the cosine of half of \( x \), then negates it.
- **Evaluation Point**: We need to evaluate this function specifically at \( x = \frac{\pi}{3} \).
### Steps to Solution
1. **Substitute** \( x = \frac{\pi}{3} \) into the equation:
\[ f\left(\frac{\pi}{3}\right) = 7 \sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{1}{2} \cdot \frac{\pi}{3}\right) \]
2. **Calculate Trigonometric Values**:
- \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
- \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
3. **Substitute the Trigonometric Values**:
\[ f\left(\frac{\pi}{3}\right) = 7 \left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} \]
4. **Simplify**:
\[ f\left(\frac{\pi}{3}\right) = \frac{7\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \]
\[ f\left(\frac{\pi}{3}\right) = \frac{6\sqrt{3}}{2} \]
\[ f\left(\frac{\pi}{3}\right) = 3\sqrt{3} \
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