Problem 14 - Algebra and Trigonometry

Question

\(13-18=\) The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(y\) -coordinate of \(P\) is \(-\frac{1}{3}\) and the \(x\) -coordinate is positive.

Step-by-Step Solution

Answer

The coordinates of point \( P \) are \( \left( \frac{2\sqrt{2}}{3}, -\frac{1}{3} \right) \).

Step 1: Understanding the Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. Any point \( P(x, y) \) on this circle must satisfy the equation \( x^2 + y^2 = 1 \).

Step 2: Substitute Known Value

Given that the \( y \)-coordinate of \( P \) is \(-\frac{1}{3}\), substitute this value into the unit circle equation: \( x^2 + \left(-\frac{1}{3}\right)^2 = 1 \).

Step 3: Simplify the Equation

Calculate \( \left(-\frac{1}{3}\right)^2 \), which equals \( \frac{1}{9} \). Substitute to get: \( x^2 + \frac{1}{9} = 1 \).

Step 4: Solve for \( x^2 \)

Subtract \( \frac{1}{9} \) from both sides: \( x^2 = 1 - \frac{1}{9} \).

Step 5: Further Simplify

Convert \( 1 \) to a fraction with a denominator of 9: \( 1 = \frac{9}{9} \). So, the equation becomes \( x^2 = \frac{9}{9} - \frac{1}{9} \).

Step 6: Calculate \( x^2 \)

Perform the subtraction: \( x^2 = \frac{8}{9} \).

Step 7: Solve for \( x \)

Take the square root of both sides to find \( x \). So, \( x = \sqrt{\frac{8}{9}} \) or \( x = -\sqrt{\frac{8}{9}} \). Since \( x \) is positive, \( x = \sqrt{\frac{8}{9}} \).

Step 8: Simplify \( x \)

Simplify \( \sqrt{\frac{8}{9}} \) to get \( \frac{\sqrt{8}}{3} \). Further simplify \( \sqrt{8} \) to \( 2\sqrt{2} \), which gives \( x = \frac{2\sqrt{2}}{3} \).

Step 9: Write the Coordinates

Now we have both coordinates for point \( P \): \( P(x, y) = \left( \frac{2\sqrt{2}}{3}, -\frac{1}{3} \right) \).

Key Concepts

Coordinate System

The coordinate system serves as the foundation for understanding how points are plotted in a plane. In the exercise, we work with the unit circle, which is a crucial concept in the coordinate system.
The unit circle is centered at the origin, where both the x and y axes intersect. It has a radius of 1.
Any point on the unit circle is represented as \( P(x, y) \), where the coordinates satisfy the equation \( x^2 + y^2 = 1 \).

• The origin is at the coordinates (0,0), which is the center of the unit circle.
• Positive x-coordinates extend to the right and negative to the left.
• Positive y-coordinates go upwards, and negative go downwards - useful for understanding why the y-coordinate -\(\frac{1}{3}\) situates below the x-axis.

Understanding this system allows us to locate and solve for any point based on the given x or y values.

Equation Solving

Equation solving is an essential skill in mathematics, especially when dealing with circles and trigonometric properties. In this exercise, we need to solve for the x-coordinate given the y-coordinate on a unit circle.
The unit circle equation \( x^2 + y^2 = 1 \) is our starting point. By substituting \( y = -\frac{1}{3} \), the task is to find the corresponding x-coordinate.

• First substitute into the equation: \( x^2 + \left(-\frac{1}{3}\right)^2 = 1 \).
• Calculating \( \left(-\frac{1}{3}\right)^2 \), we get \( \frac{1}{9} \).
• Subtract \( \frac{1}{9} \) from 1 to isolate \( x^2 \): \( x^2 = \frac{8}{9} \).
• Finding x involves taking the square root, resulting in \( x = \sqrt{\frac{8}{9}} \) or \( -\sqrt{\frac{8}{9}} \).

Since the x-coordinate is positive, we conclude that \( x = \frac{2\sqrt{2}}{3} \). This series of operations illustrates how to carefully rearrange and solve equations related to geometrical shapes.

Trigonometric Identities

Trigonometric identities offer a bridge between algebra and geometry, especially in relation to circular paths. In the unit circle, every point \( P(x, y) \) also has trigonometric significance. Each coordinate corresponds to a trigonometric identity:

• The x-coordinate represents \( \cos(\theta) \).
• Similarly, the y-coordinate stands for \( \sin(\theta) \).

In this example, since \( y = -\frac{1}{3} \), it suggests an angle \( \theta \) where the sine value corresponds to \(-\frac{1}{3} \), often involving specific or inverse trigonometric functions.
Additionally, leveraging the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \), which is the standard unit circle equation, re-affirms how the coordinates relate directly to cosine and sine.
Such identities simplify the process of understanding a point's relation to the circle, underscoring the harmonious blend of trigonometry and geometric placements.