Use the triangle to find sin(23), cos(23), sin(), cos() 25 24 B
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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### Trigonometric Identities and Exact Values
#### 6. Use the triangle to find sin(2β), cos(2β), sin(β/2), cos(β/2)
In the given problem, we have a right triangle with the following dimensions:
- Hypotenuse (opposite the right angle) = 25 units
- Adjacent side to angle β = 24 units
- Opposite side to angle β = a units
Since it's a right triangle, we can use the Pythagorean theorem to find the value of \( a \):
\[ a^2 + 24^2 = 25^2 \]
\[ a^2 + 576 = 625 \]
\[ a^2 = 49 \]
\[ a = 7 \]
Using these values:
\[ \sin(\beta) = \frac{opposite}{hypotenuse} = \frac{7}{25} \]
\[ \cos(\beta) = \frac{adjacent}{hypotenuse} = \frac{24}{25} \]
Using double-angle formulas and half-angle formulas:
- **Double Angle Formulas**
\[ \sin(2\beta) = 2 \sin(\beta) \cos(\beta) \]
\[ \sin(2\beta) = 2 \left( \frac{7}{25} \right) \left( \frac{24}{25} \right) = \frac{2 \times 7 \times 24}{625} = \frac{336}{625} \]
\[ \cos(2\beta) = \cos^2(\beta) - \sin^2(\beta) \]
\[ \cos(2\beta) = \left( \frac{24}{25} \right)^2 - \left( \frac{7}{25} \right)^2 \]
\[ \cos(2\beta) = \frac{576}{625} - \frac{49}{625} = \frac{527}{625} \]
- **Half Angle Formulas**
\[ \sin\left(\frac{\beta}{2}\right) = \sqrt{\frac{1 - \cos(\beta)}{2}} \]
\[ \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{1 + \cos(\beta)}{2}} \]
#### 7. Write](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6841f1bf-ebcf-44b9-bde6-b1a56299f544%2F491ce828-1633-496a-a4f0-ec8867c9d5d3%2F8rtior9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Trigonometric Identities and Exact Values
#### 6. Use the triangle to find sin(2β), cos(2β), sin(β/2), cos(β/2)
In the given problem, we have a right triangle with the following dimensions:
- Hypotenuse (opposite the right angle) = 25 units
- Adjacent side to angle β = 24 units
- Opposite side to angle β = a units
Since it's a right triangle, we can use the Pythagorean theorem to find the value of \( a \):
\[ a^2 + 24^2 = 25^2 \]
\[ a^2 + 576 = 625 \]
\[ a^2 = 49 \]
\[ a = 7 \]
Using these values:
\[ \sin(\beta) = \frac{opposite}{hypotenuse} = \frac{7}{25} \]
\[ \cos(\beta) = \frac{adjacent}{hypotenuse} = \frac{24}{25} \]
Using double-angle formulas and half-angle formulas:
- **Double Angle Formulas**
\[ \sin(2\beta) = 2 \sin(\beta) \cos(\beta) \]
\[ \sin(2\beta) = 2 \left( \frac{7}{25} \right) \left( \frac{24}{25} \right) = \frac{2 \times 7 \times 24}{625} = \frac{336}{625} \]
\[ \cos(2\beta) = \cos^2(\beta) - \sin^2(\beta) \]
\[ \cos(2\beta) = \left( \frac{24}{25} \right)^2 - \left( \frac{7}{25} \right)^2 \]
\[ \cos(2\beta) = \frac{576}{625} - \frac{49}{625} = \frac{527}{625} \]
- **Half Angle Formulas**
\[ \sin\left(\frac{\beta}{2}\right) = \sqrt{\frac{1 - \cos(\beta)}{2}} \]
\[ \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{1 + \cos(\beta)}{2}} \]
#### 7. Write
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