Rewrite sin 2 cos as an algebraic expression in v. 4 sin 2 cos 4 믐 =
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![**Rewrite the Expression**
**Topic**: Algebra and Trigonometry
**Objective**: Learn to rewrite \(\sin \left( 2 \cos^{-1} \frac{v}{4} \right)\) as an algebraic expression in \(v\).
### Problem Statement
Rewrite the following expression in terms of \(v\):
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) \]
### Transformation Steps
1. **Given Expression:**
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) = \sin \left( 2 \theta \right) \]
where \(\theta = \cos^{-1} \frac{v}{4}\).
2. **Use the Double-Angle Formula for Sine:**
\[ \sin (2 \theta) = 2 \sin \theta \cos \theta \]
3. **Identify \( \sin \theta \) and \( \cos \theta \):**
- By definition of \( \theta \),
\[ \cos \theta = \frac{v}{4} \]
- Use the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
\[ \sin^2 \theta + \left( \frac{v}{4} \right)^2 = 1 \]
\[ \sin^2 \theta + \frac{v^2}{16} = 1 \]
\[ \sin^2 \theta = 1 - \frac{v^2}{16} \]
\[ \sin \theta = \sqrt{1 - \frac{v^2}{16}} \]
4. **Combine the Results:**
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) = 2 \left( \sqrt{1 - \frac{v^2}{16}} \right) \left( \frac{v}{4} \right) \]
\[ = 2 \left( \frac{v}{4} \sqrt{1 - \frac{v^2}{16}} \right) \]
\[ = \frac{v}{2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0b581ba-098c-432d-a731-9ec346ce3593%2Fa117a606-0381-4f7f-9240-a605eb17d6c1%2F5l3qm1g_processed.png&w=3840&q=75)
Transcribed Image Text:**Rewrite the Expression**
**Topic**: Algebra and Trigonometry
**Objective**: Learn to rewrite \(\sin \left( 2 \cos^{-1} \frac{v}{4} \right)\) as an algebraic expression in \(v\).
### Problem Statement
Rewrite the following expression in terms of \(v\):
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) \]
### Transformation Steps
1. **Given Expression:**
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) = \sin \left( 2 \theta \right) \]
where \(\theta = \cos^{-1} \frac{v}{4}\).
2. **Use the Double-Angle Formula for Sine:**
\[ \sin (2 \theta) = 2 \sin \theta \cos \theta \]
3. **Identify \( \sin \theta \) and \( \cos \theta \):**
- By definition of \( \theta \),
\[ \cos \theta = \frac{v}{4} \]
- Use the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
\[ \sin^2 \theta + \left( \frac{v}{4} \right)^2 = 1 \]
\[ \sin^2 \theta + \frac{v^2}{16} = 1 \]
\[ \sin^2 \theta = 1 - \frac{v^2}{16} \]
\[ \sin \theta = \sqrt{1 - \frac{v^2}{16}} \]
4. **Combine the Results:**
\[ \sin \left( 2 \cos^{-1} \frac{v}{4} \right) = 2 \left( \sqrt{1 - \frac{v^2}{16}} \right) \left( \frac{v}{4} \right) \]
\[ = 2 \left( \frac{v}{4} \sqrt{1 - \frac{v^2}{16}} \right) \]
\[ = \frac{v}{2}
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