Given cos 0 and angle 0 is in Quadrant IV, what is the exact value of sin 0 in 6. simplest form? Simplify all radicals if needed. Answer: Submit Answer rch a 50 近

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°.
Topic Video
Question
### Problem Statement
Given \( \cos \theta = \frac{\sqrt{11}}{6} \) and angle \( \theta \) is in Quadrant IV, what is the exact value of \( \sin \theta \) in simplest form? Simplify all radicals if needed.

#### Answer Submission
- **Answer:**
  - **Input Box:** [   ]
  - **Submit Button:** "Submit Answer"
  
#### Instructions
1. Enter your answer in the provided input box.
2. Click on "Submit Answer" to check your solution.

### Explanation of the Problem
In this problem, we are given the cosine value of an angle \(\theta\) in the fourth quadrant. Using trigonometric identities and properties of the unit circle, we need to determine the sine value of the same angle.

#### Steps to Solve:
1. Recall the Pythagorean identity:
   \[
   \sin^2 \theta + \cos^2 \theta = 1
   \]
2. Substitute the given \(\cos \theta\):
   \[
   \sin^2 \theta + \left(\frac{\sqrt{11}}{6}\right)^2 = 1
   \]
3. Simplify and solve for \(\sin \theta\):
   \[
   \sin^2 \theta + \frac{11}{36} = 1
   \]
   \[
   \sin^2 \theta = 1 - \frac{11}{36}
   \]
   \[
   \sin^2 \theta = \frac{36}{36} - \frac{11}{36}
   \]
   \[
   \sin^2 \theta = \frac{25}{36}
   \]
   \[
   \sin \theta = \pm \sqrt{\frac{25}{36}}
   \]
   \[
   \sin \theta = \pm \frac{\sqrt{25}}{\sqrt{36}}
   \]
   \[
   \sin \theta = \pm \frac{5}{6}
   \]
4. Since \(\theta\) is in Quadrant IV, \( \sin \theta \) must be negative:
   \[
   \sin \theta = -\frac{5}{6}
   \]

So the exact value of \( \sin \theta \) in simplest form is:
\[ \
Transcribed Image Text:### Problem Statement Given \( \cos \theta = \frac{\sqrt{11}}{6} \) and angle \( \theta \) is in Quadrant IV, what is the exact value of \( \sin \theta \) in simplest form? Simplify all radicals if needed. #### Answer Submission - **Answer:** - **Input Box:** [ ] - **Submit Button:** "Submit Answer" #### Instructions 1. Enter your answer in the provided input box. 2. Click on "Submit Answer" to check your solution. ### Explanation of the Problem In this problem, we are given the cosine value of an angle \(\theta\) in the fourth quadrant. Using trigonometric identities and properties of the unit circle, we need to determine the sine value of the same angle. #### Steps to Solve: 1. Recall the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] 2. Substitute the given \(\cos \theta\): \[ \sin^2 \theta + \left(\frac{\sqrt{11}}{6}\right)^2 = 1 \] 3. Simplify and solve for \(\sin \theta\): \[ \sin^2 \theta + \frac{11}{36} = 1 \] \[ \sin^2 \theta = 1 - \frac{11}{36} \] \[ \sin^2 \theta = \frac{36}{36} - \frac{11}{36} \] \[ \sin^2 \theta = \frac{25}{36} \] \[ \sin \theta = \pm \sqrt{\frac{25}{36}} \] \[ \sin \theta = \pm \frac{\sqrt{25}}{\sqrt{36}} \] \[ \sin \theta = \pm \frac{5}{6} \] 4. Since \(\theta\) is in Quadrant IV, \( \sin \theta \) must be negative: \[ \sin \theta = -\frac{5}{6} \] So the exact value of \( \sin \theta \) in simplest form is: \[ \
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