Sin 2B Cos2B 17. tan 2B = 289 289 %3D cos(a – B) = CosACosB+ SinA 18. %3D Sin2B=2Sin BCesB 17
Sin 2B Cos2B 17. tan 2B = 289 289 %3D cos(a – B) = CosACosB+ SinA 18. %3D Sin2B=2Sin BCesB 17
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°.
Related questions
Question
Did I do 17 and 18 right?
![Certainly! Here is the transcription and detailed explanation of the mathematical content in the image:
---
### Transcription of Mathematical Content
#### Problem 17
- Expression: \(\tan 2\beta = \frac{\sin 2B}{\cos 2B}\)
- Equation: \(\sin 2B = 2 \sin B \cos B\)
1. Simplification:
\[
2\left(\frac{\sqrt{5}}{3}\right) \left(\frac{2}{3}\right) = 2 \cdot \frac{2\sqrt{5}}{9}
\]
2. Result:
\[
\frac{4\sqrt{5}}{9}
\]
#### Problem 18
- Expression: \(\cos(\alpha - \beta) = \cos A \cos B + \sin A \sin B\)
1. Calculation:
\[
\left(\frac{-15}{17}\right)\left(\frac{2}{3}\right) + \left(\frac{-8}{17}\right)\left(\frac{\sqrt{5}}{3}\right)
\]
\[
= \frac{-30}{51} + \frac{-8\sqrt{5}}{51}
\]
2. Combined Result:
\[
\frac{-30 - 8\sqrt{5}}{51}
\]
3. Simplified Final Result:
\[
- \frac{30 + 8\sqrt{5}}{51}
\]
### Explanation of Content
The content involves trigonometric identities and involved calculations:
1. **Trigonometric Identities**:
- The double angle identity for sine, \(\sin 2B = 2 \sin B \cos B\), is used to simplify expressions.
- A formula for cosine of the difference of angles, \(\cos(\alpha - \beta)\), is used.
2. **Algebraic Manipulations**:
- Fractional multiplication and simplification are prominent in the problem-solving process.
- Radical expressions (\(\sqrt{5}\)) are involved, emphasizing the need for careful handling of irrational numbers.
This transcription should assist students in understanding the application of trigonometric identities and algebraic techniques in complex expressions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2367310-44c7-43e5-b110-21237a7a5b46%2F7a3b229f-baef-472c-b5fa-0dc2e76b9d85%2F7gwewh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Here is the transcription and detailed explanation of the mathematical content in the image:
---
### Transcription of Mathematical Content
#### Problem 17
- Expression: \(\tan 2\beta = \frac{\sin 2B}{\cos 2B}\)
- Equation: \(\sin 2B = 2 \sin B \cos B\)
1. Simplification:
\[
2\left(\frac{\sqrt{5}}{3}\right) \left(\frac{2}{3}\right) = 2 \cdot \frac{2\sqrt{5}}{9}
\]
2. Result:
\[
\frac{4\sqrt{5}}{9}
\]
#### Problem 18
- Expression: \(\cos(\alpha - \beta) = \cos A \cos B + \sin A \sin B\)
1. Calculation:
\[
\left(\frac{-15}{17}\right)\left(\frac{2}{3}\right) + \left(\frac{-8}{17}\right)\left(\frac{\sqrt{5}}{3}\right)
\]
\[
= \frac{-30}{51} + \frac{-8\sqrt{5}}{51}
\]
2. Combined Result:
\[
\frac{-30 - 8\sqrt{5}}{51}
\]
3. Simplified Final Result:
\[
- \frac{30 + 8\sqrt{5}}{51}
\]
### Explanation of Content
The content involves trigonometric identities and involved calculations:
1. **Trigonometric Identities**:
- The double angle identity for sine, \(\sin 2B = 2 \sin B \cos B\), is used to simplify expressions.
- A formula for cosine of the difference of angles, \(\cos(\alpha - \beta)\), is used.
2. **Algebraic Manipulations**:
- Fractional multiplication and simplification are prominent in the problem-solving process.
- Radical expressions (\(\sqrt{5}\)) are involved, emphasizing the need for careful handling of irrational numbers.
This transcription should assist students in understanding the application of trigonometric identities and algebraic techniques in complex expressions.
![The image contains a mathematical problem involving trigonometric identities and calculations. Below is a transcription and explanation suitable for an educational context:
### Given:
- \(\sin \alpha = -\frac{8}{17}\), with \(\pi \leq \alpha \leq \frac{3\pi}{2}\)
- \(\cos \beta = \frac{2}{3}\), with \(\frac{3\pi}{2} \leq \beta \leq 2\pi\)
### Problem:
1. **Calculate \(\cos \alpha\):**
Using the identity:
\[
\cos^2 \alpha = 1 - \sin^2 \alpha
\]
Substituting the given \(\sin \alpha\):
\[
\cos^2 \alpha + \left(-\frac{8}{17}\right)^2 = 1
\]
\[
\cos^2 \alpha = 1 - \frac{64}{289}
\]
\[
\cos^2 \alpha = \frac{225}{289}
\]
Taking the square root:
\[
\cos \alpha = \sqrt{\frac{225}{289}} = \frac{15}{17}
\]
Since \(\pi \leq \alpha \leq \frac{3\pi}{2}\), \(\cos \alpha\) is negative:
\[
\cos \alpha = -\frac{15}{17}
\]
2. **Calculate \(\sin \beta\):**
Using the identity:
\[
\sin \beta = \sqrt{1 - \cos^2 \beta}
\]
Substituting the given \(\cos \beta\):
\[
\sin^2 \beta = 1 - \left(\frac{2}{3}\right)^2
\]
\[
\sin^2 \beta = 1 - \frac{4}{9}
\]
\[
\sin^2 \beta = \frac{5}{9}
\]
Taking the square root:
\[
\sin \beta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
\]
Since \(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2367310-44c7-43e5-b110-21237a7a5b46%2F7a3b229f-baef-472c-b5fa-0dc2e76b9d85%2Fzkyjbym_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical problem involving trigonometric identities and calculations. Below is a transcription and explanation suitable for an educational context:
### Given:
- \(\sin \alpha = -\frac{8}{17}\), with \(\pi \leq \alpha \leq \frac{3\pi}{2}\)
- \(\cos \beta = \frac{2}{3}\), with \(\frac{3\pi}{2} \leq \beta \leq 2\pi\)
### Problem:
1. **Calculate \(\cos \alpha\):**
Using the identity:
\[
\cos^2 \alpha = 1 - \sin^2 \alpha
\]
Substituting the given \(\sin \alpha\):
\[
\cos^2 \alpha + \left(-\frac{8}{17}\right)^2 = 1
\]
\[
\cos^2 \alpha = 1 - \frac{64}{289}
\]
\[
\cos^2 \alpha = \frac{225}{289}
\]
Taking the square root:
\[
\cos \alpha = \sqrt{\frac{225}{289}} = \frac{15}{17}
\]
Since \(\pi \leq \alpha \leq \frac{3\pi}{2}\), \(\cos \alpha\) is negative:
\[
\cos \alpha = -\frac{15}{17}
\]
2. **Calculate \(\sin \beta\):**
Using the identity:
\[
\sin \beta = \sqrt{1 - \cos^2 \beta}
\]
Substituting the given \(\cos \beta\):
\[
\sin^2 \beta = 1 - \left(\frac{2}{3}\right)^2
\]
\[
\sin^2 \beta = 1 - \frac{4}{9}
\]
\[
\sin^2 \beta = \frac{5}{9}
\]
Taking the square root:
\[
\sin \beta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
\]
Since \(\
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