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,...
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Question
#### Diagram Explanation
In the diagram, a circle is centered at the origin of an \(xy\)-coordinate system. The circle is denoted as the "unit circle", meaning it has a radius of 1. The points marked on the diagram are:
- \( P(x, y) \) located on the circumference of the circle.
- \( (1, 0) \) showing a point on the circle where the circle intersects the x-axis.
Note: The figure is not drawn to scale.
#### Solution Approach
1. **Equation of the Unit Circle:**
The general equation for a unit circle centered at the origin (0,0) is:
\[
x^2 + y^2 = 1
\]
2. **Substitute the Provided Coordinates:**
We are given that the y-coordinate of point P is \(\frac{1}{3}\). Substituting \(y = \frac{1}{3}\) into the unit circle equation:
\[
x^2 + \left(\frac{1}{3}\right)^2 = 1
\]
3. **Simplify the Equation:**
\[
x^2 + \frac{1}{9} = 1
\]
4. **Isolate \(x^2\):**
Subtract \(\frac{1}{9}\) from both sides:
\[
x^2 = 1 - \frac{1}{9}
\]
5. **Find a Common Denominator:**
Convert 1 into a fraction with the same denominator:
\[
1 = \frac{9}{9}
\]
6. **Subtract the Fractions:**
\[
x^2 = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}
\]
7. **Solve for \(x\):**
\[
x = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0fac6bec-f0b2-4056-963f-cf73547656d5%2F582683f9-2802-4725-b8c0-d9ea31242011%2F53mjan_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Website Content:**
---
### Understanding Coordinates on the Unit Circle
#### Problem Statement
The point \( P = \left(x, \frac{1}{3}\right) \) lies on the unit circle shown below. What is the value of \( x \) in simplest form?
#### Diagram Explanation
In the diagram, a circle is centered at the origin of an \(xy\)-coordinate system. The circle is denoted as the "unit circle", meaning it has a radius of 1. The points marked on the diagram are:
- \( P(x, y) \) located on the circumference of the circle.
- \( (1, 0) \) showing a point on the circle where the circle intersects the x-axis.
Note: The figure is not drawn to scale.
#### Solution Approach
1. **Equation of the Unit Circle:**
The general equation for a unit circle centered at the origin (0,0) is:
\[
x^2 + y^2 = 1
\]
2. **Substitute the Provided Coordinates:**
We are given that the y-coordinate of point P is \(\frac{1}{3}\). Substituting \(y = \frac{1}{3}\) into the unit circle equation:
\[
x^2 + \left(\frac{1}{3}\right)^2 = 1
\]
3. **Simplify the Equation:**
\[
x^2 + \frac{1}{9} = 1
\]
4. **Isolate \(x^2\):**
Subtract \(\frac{1}{9}\) from both sides:
\[
x^2 = 1 - \frac{1}{9}
\]
5. **Find a Common Denominator:**
Convert 1 into a fraction with the same denominator:
\[
1 = \frac{9}{9}
\]
6. **Subtract the Fractions:**
\[
x^2 = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}
\]
7. **Solve for \(x\):**
\[
x = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8
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