sin a =- 5 sas 2n cos B=- TSBS 17' Given: 2 Sin'a-Cosa Sin? %3 -Cos a=! cos'a sin B= SinB -8-64 1. COS a = 2. %3D 2. b82 Sin B CosB' 17 Sin 25Cos a tan a = Sina 7osa 3. 4. tan B = -3/5 - 4/5 sin(a+B)= Sin ACasB+ SinBCesA -3.5.15 15 17 4 20 5. 15 36 85 24 85 85 cos 2ß = 2C0s B5in-B 6.
sin a =- 5 sas 2n cos B=- TSBS 17' Given: 2 Sin'a-Cosa Sin? %3 -Cos a=! cos'a sin B= SinB -8-64 1. COS a = 2. %3D 2. b82 Sin B CosB' 17 Sin 25Cos a tan a = Sina 7osa 3. 4. tan B = -3/5 - 4/5 sin(a+B)= Sin ACasB+ SinBCesA -3.5.15 15 17 4 20 5. 15 36 85 24 85 85 cos 2ß = 2C0s B5in-B 6.
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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Did I do 5 and 6 correctly? More importantly number 6?
![### Trigonometric Identity Solutions
**Given:**
- \( \sin \alpha = -\frac{3}{5} \), \( \frac{3\pi}{2} \leq \alpha \leq 2\pi \)
- \( \cos \beta = -\frac{8}{17} \), \( \pi \leq \beta \leq \frac{3\pi}{2} \)
---
#### Problem 1:
\[ \cos \alpha = \sin^2 \alpha \cdot \cos^2 \alpha \]
Calculation:
- \(\sin^2 \alpha = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}\)
- \(\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25}\)
- \(\cos \alpha = \frac{4}{5}\)
---
#### Problem 2:
\[ \sin \beta = \sin^2 \beta - \cos^2 \beta \]
Calculation:
- \(\sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(-\frac{8}{17}\right)^2 = \frac{64}{289}\)
- \(\sin \beta = -\frac{15}{17}\)
---
#### Problem 3:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Calculation:
- Substitute values: \(\tan \alpha = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}\)
---
#### Problem 4:
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} \]
- Substitute values: \(\tan \beta = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8}\)
---
#### Problem 5:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \]
Calculation:
- Substitute values:
\[
\sin(\alpha + \beta) = \left(-\frac{3}{5}\right) \left(-\frac{8}{17}\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2367310-44c7-43e5-b110-21237a7a5b46%2F4a148dfa-4137-45fa-bddc-35200114054d%2Fhjs6gm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometric Identity Solutions
**Given:**
- \( \sin \alpha = -\frac{3}{5} \), \( \frac{3\pi}{2} \leq \alpha \leq 2\pi \)
- \( \cos \beta = -\frac{8}{17} \), \( \pi \leq \beta \leq \frac{3\pi}{2} \)
---
#### Problem 1:
\[ \cos \alpha = \sin^2 \alpha \cdot \cos^2 \alpha \]
Calculation:
- \(\sin^2 \alpha = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}\)
- \(\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25}\)
- \(\cos \alpha = \frac{4}{5}\)
---
#### Problem 2:
\[ \sin \beta = \sin^2 \beta - \cos^2 \beta \]
Calculation:
- \(\sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(-\frac{8}{17}\right)^2 = \frac{64}{289}\)
- \(\sin \beta = -\frac{15}{17}\)
---
#### Problem 3:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Calculation:
- Substitute values: \(\tan \alpha = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}\)
---
#### Problem 4:
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} \]
- Substitute values: \(\tan \beta = \frac{-\frac{15}{17}}{-\frac{8}{17}} = \frac{15}{8}\)
---
#### Problem 5:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \]
Calculation:
- Substitute values:
\[
\sin(\alpha + \beta) = \left(-\frac{3}{5}\right) \left(-\frac{8}{17}\
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