Verify that the point lies on the graph of the unit circle. V3
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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![### Verifying a Point on the Unit Circle
#### Objective:
Verify that the point \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\) lies on the graph of the unit circle.
#### Methodology:
We check the point by showing that the coordinates satisfy the equation of the unit circle, which is given by:
\[ x^2 + y^2 = 1 \]
1. **Substitute Coordinates into the Equation:**
\[ x = \frac{1}{2} \]
\[ y = \frac{\sqrt{3}}{2} \]
2. **Calculate \( x^2 \) and \( y^2 \):**
\[ x^2 + y^2 = \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 \]
3. **Simplify each term:**
\[ \left( \frac{1}{2} \right)^2 = \frac{1}{4} \]
\[ \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \]
4. **Sum the terms:**
\[ x^2 + y^2 = \frac{1}{4} + \frac{3}{4} \]
5. **Combine the fractions:**
\[ x^2 + y^2 = \frac{1 + 3}{4} = \frac{4}{4} = 1 \]
#### Conclusion:
Thus, the point \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\) does lie on the graph of the unit circle.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3e9f930-7347-4a0a-9949-50203265dda3%2Fccdf492b-1b67-4bdb-84fe-721c41abef8a%2Fe62o0u.jpeg&w=3840&q=75)
Transcribed Image Text:### Verifying a Point on the Unit Circle
#### Objective:
Verify that the point \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\) lies on the graph of the unit circle.
#### Methodology:
We check the point by showing that the coordinates satisfy the equation of the unit circle, which is given by:
\[ x^2 + y^2 = 1 \]
1. **Substitute Coordinates into the Equation:**
\[ x = \frac{1}{2} \]
\[ y = \frac{\sqrt{3}}{2} \]
2. **Calculate \( x^2 \) and \( y^2 \):**
\[ x^2 + y^2 = \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 \]
3. **Simplify each term:**
\[ \left( \frac{1}{2} \right)^2 = \frac{1}{4} \]
\[ \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \]
4. **Sum the terms:**
\[ x^2 + y^2 = \frac{1}{4} + \frac{3}{4} \]
5. **Combine the fractions:**
\[ x^2 + y^2 = \frac{1 + 3}{4} = \frac{4}{4} = 1 \]
#### Conclusion:
Thus, the point \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\) does lie on the graph of the unit circle.
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