13. If 180° ≤ 0 ≤ 270° and cos 0 Cos os 블. 8 10' find
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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Question
![### Problem Statement:
**13.** If \( 180^\circ \leq \theta \leq 270^\circ \) and \( \cos \theta = -\frac{8}{10} \), find \( \cos \frac{\theta}{2} \).
---
### Analysis:
The problem requires calculating the cosine of half the angle given certain conditions about the original angle. Specifically, it provides a range for \(\theta\) and the cosine value of \(\theta\).
#### Steps to Solve:
1. **Range of \(\theta\)**: The angle \(\theta\) is in the third quadrant since \(180^\circ \leq \theta \leq 270^\circ\). In this quadrant, the cosine of \(\theta\) is negative.
2. **Calculate \(\cos \frac{\theta}{2}\)**: To find this value, we use the half-angle formula for cosine.
#### Half-Angle Formula for Cosine:
\[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \]
Since \(\theta\) is in the third quadrant, \(\frac{\theta}{2}\) will fall within the range \(90^\circ \leq \frac{\theta}{2} \leq 135^\circ\), which corresponds to the second quadrant where cosine is negative. Thus, we will take the negative root.
3. **Substitute the Given Value**:
\[ \cos \theta = -\frac{8}{10} = -0.8 \]
4. **Apply the Formula**:
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 - 0.8}{2}} \]
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{0.2}{2}} \]
\[ \cos \frac{\theta}{2} = -\sqrt{0.1} \]
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1}{10}} \]
\[ \cos \frac{\theta}{2} = -\frac{1}{\sqrt{10}} \]
\[ \cos \frac{\theta}{2} = -\frac{\sqrt{10}}{10} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F050b7896-96e1-4a53-ae28-da4cd34956eb%2F06cf02b5-138f-4209-ac26-16becbb4f3ad%2Fi0y8lk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**13.** If \( 180^\circ \leq \theta \leq 270^\circ \) and \( \cos \theta = -\frac{8}{10} \), find \( \cos \frac{\theta}{2} \).
---
### Analysis:
The problem requires calculating the cosine of half the angle given certain conditions about the original angle. Specifically, it provides a range for \(\theta\) and the cosine value of \(\theta\).
#### Steps to Solve:
1. **Range of \(\theta\)**: The angle \(\theta\) is in the third quadrant since \(180^\circ \leq \theta \leq 270^\circ\). In this quadrant, the cosine of \(\theta\) is negative.
2. **Calculate \(\cos \frac{\theta}{2}\)**: To find this value, we use the half-angle formula for cosine.
#### Half-Angle Formula for Cosine:
\[ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \]
Since \(\theta\) is in the third quadrant, \(\frac{\theta}{2}\) will fall within the range \(90^\circ \leq \frac{\theta}{2} \leq 135^\circ\), which corresponds to the second quadrant where cosine is negative. Thus, we will take the negative root.
3. **Substitute the Given Value**:
\[ \cos \theta = -\frac{8}{10} = -0.8 \]
4. **Apply the Formula**:
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1 - 0.8}{2}} \]
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{0.2}{2}} \]
\[ \cos \frac{\theta}{2} = -\sqrt{0.1} \]
\[ \cos \frac{\theta}{2} = -\sqrt{\frac{1}{10}} \]
\[ \cos \frac{\theta}{2} = -\frac{1}{\sqrt{10}} \]
\[ \cos \frac{\theta}{2} = -\frac{\sqrt{10}}{10} \]
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