Use the function value to find the indicated trigonometric value in the specified quadrant. 4 function value sin (0) = 5 quadrant III trigonometric value cos(8) cos(0) =
Use the function value to find the indicated trigonometric value in the specified quadrant. 4 function value sin (0) = 5 quadrant III trigonometric value cos(8) cos(0) =
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
![### Finding the Cosine Value in the Specified Quadrant
Use the function value to find the indicated trigonometric value in the specified quadrant.
#### Given:
- **Function value:** \( \sin(\theta) = -\frac{4}{5} \)
- **Quadrant:** III
- **Trigonometric value to find:** \( \cos(\theta) \)
#### Problem:
Determine the value of \( \cos(\theta) \).
\[ \cos(\theta) = \large{\boxed{}} \]
Since the sine function is negative and we are in the third quadrant, where both sine and cosine are negative, use the Pythagorean identity:
\[ \sin^{2}(\theta) + \cos^{2}(\theta) = 1 \]
Substitute the given sine value:
\[ \left(-\frac{4}{5}\right)^{2} + \cos^{2}(\theta) = 1 \]
\[ \frac{16}{25} + \cos^{2}(\theta) = 1 \]
Solve for \( \cos^{2}(\theta) \):
\[ \cos^{2}(\theta) = 1 - \frac{16}{25} \]
\[ \cos^{2}(\theta) = \frac{25}{25} - \frac{16}{25} \]
\[ \cos^{2}(\theta) = \frac{9}{25} \]
Taking the square root of both sides and considering that cosine is negative in the third quadrant:
\[ \cos(\theta) = -\sqrt{\frac{9}{25}} \]
\[ \cos(\theta) = -\frac{3}{5} \]
Thus, the value of \( \cos(\theta) \) is:
\[ \cos(\theta) = \large{\boxed{-\frac{3}{5}}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa0e38307-1ade-44bc-b712-aaeda4c58098%2Fa56173f3-b581-4de1-ac2e-9aa02ca96157%2Fmbk978_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Cosine Value in the Specified Quadrant
Use the function value to find the indicated trigonometric value in the specified quadrant.
#### Given:
- **Function value:** \( \sin(\theta) = -\frac{4}{5} \)
- **Quadrant:** III
- **Trigonometric value to find:** \( \cos(\theta) \)
#### Problem:
Determine the value of \( \cos(\theta) \).
\[ \cos(\theta) = \large{\boxed{}} \]
Since the sine function is negative and we are in the third quadrant, where both sine and cosine are negative, use the Pythagorean identity:
\[ \sin^{2}(\theta) + \cos^{2}(\theta) = 1 \]
Substitute the given sine value:
\[ \left(-\frac{4}{5}\right)^{2} + \cos^{2}(\theta) = 1 \]
\[ \frac{16}{25} + \cos^{2}(\theta) = 1 \]
Solve for \( \cos^{2}(\theta) \):
\[ \cos^{2}(\theta) = 1 - \frac{16}{25} \]
\[ \cos^{2}(\theta) = \frac{25}{25} - \frac{16}{25} \]
\[ \cos^{2}(\theta) = \frac{9}{25} \]
Taking the square root of both sides and considering that cosine is negative in the third quadrant:
\[ \cos(\theta) = -\sqrt{\frac{9}{25}} \]
\[ \cos(\theta) = -\frac{3}{5} \]
Thus, the value of \( \cos(\theta) \) is:
\[ \cos(\theta) = \large{\boxed{-\frac{3}{5}}} \]
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