32 Ө 12

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Question
**Find \(\sin(\theta)\) and \(\cos(\theta)\). Give exact answers.**

### Diagram Explanation:

The image shows a right triangle with:

- One angle labeled \(\theta\),
- The opposite side of \(\theta\) is labeled with a length of 32,
- The adjacent side to \(\theta\) is labeled with a length of 12.

### Solution:

To find \(\sin(\theta)\) and \(\cos(\theta)\), we first use the Pythagorean theorem to determine the hypotenuse. 

The Pythagorean theorem states:
\[
c^2 = a^2 + b^2
\]
where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides.

Here, \(a = 32\) and \(b = 12\):

\[
c^2 = 32^2 + 12^2 = 1024 + 144 = 1168
\]

Thus,
\[
c = \sqrt{1168} = \sqrt{4 \times 292} = 2\sqrt{292} = 2\sqrt{4 \times 73} = 4\sqrt{73}
\]

### Trigonometric Functions:

- **Sine Function**: \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\)

  \[
  \sin(\theta) = \frac{32}{4\sqrt{73}} = \frac{8}{\sqrt{73}}
  \]

- **Cosine Function**: \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\)

  \[
  \cos(\theta) = \frac{12}{4\sqrt{73}} = \frac{3}{\sqrt{73}}
  \]

For exact answers, \(\sin(\theta) = \frac{8}{\sqrt{73}}\) and \(\cos(\theta) = \frac{3}{\sqrt{73}}\). To rationalize, multiply numerator and denominator by \(\sqrt{73}\):

- \(\sin(\theta) = \frac{8\sqrt{73}}{73}\)
- \(\cos(\theta) = \frac{3\sqrt{73}}{73
Transcribed Image Text:**Find \(\sin(\theta)\) and \(\cos(\theta)\). Give exact answers.** ### Diagram Explanation: The image shows a right triangle with: - One angle labeled \(\theta\), - The opposite side of \(\theta\) is labeled with a length of 32, - The adjacent side to \(\theta\) is labeled with a length of 12. ### Solution: To find \(\sin(\theta)\) and \(\cos(\theta)\), we first use the Pythagorean theorem to determine the hypotenuse. The Pythagorean theorem states: \[ c^2 = a^2 + b^2 \] where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides. Here, \(a = 32\) and \(b = 12\): \[ c^2 = 32^2 + 12^2 = 1024 + 144 = 1168 \] Thus, \[ c = \sqrt{1168} = \sqrt{4 \times 292} = 2\sqrt{292} = 2\sqrt{4 \times 73} = 4\sqrt{73} \] ### Trigonometric Functions: - **Sine Function**: \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\) \[ \sin(\theta) = \frac{32}{4\sqrt{73}} = \frac{8}{\sqrt{73}} \] - **Cosine Function**: \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\) \[ \cos(\theta) = \frac{12}{4\sqrt{73}} = \frac{3}{\sqrt{73}} \] For exact answers, \(\sin(\theta) = \frac{8}{\sqrt{73}}\) and \(\cos(\theta) = \frac{3}{\sqrt{73}}\). To rationalize, multiply numerator and denominator by \(\sqrt{73}\): - \(\sin(\theta) = \frac{8\sqrt{73}}{73}\) - \(\cos(\theta) = \frac{3\sqrt{73}}{73
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