Problem 9 - Algebra and Trigonometry

Question

\(9-14\) . Find the missing coordinate of \(P\) , using the fact that \(P\) lies on the unit circle in the given quadrant. $$ \frac{\text { Coordinates }}{P\left(-\frac{3}{5},\right.} ) \frac{\text { Quadrant }}{\text { III }} $$

Step-by-Step Solution

Answer

The missing coordinate of P is \(y = -\frac{4}{5}\).

Step 1: Understanding the Unit Circle

Points on the unit circle satisfy the equation \(x^2 + y^2 = 1\). Given that \(P\) is on the unit circle, this equation holds for its coordinates.

Step 2: Substitute the Given x-Coordinate

Since we know \(x = -\frac{3}{5}\), substitute this value into the unit circle equation: \((-\frac{3}{5})^2 + y^2 = 1\).

Step 3: Simplify the Equation

Calculate \((-\frac{3}{5})^2 = \frac{9}{25}\). Substitute this result into the equation to get \(\frac{9}{25} + y^2 = 1\).

Step 4: Solve for y^2

To isolate \(y^2\), subtract \(\frac{9}{25}\) from both sides: \(y^2 = 1 - \frac{9}{25}\).

Step 5: Simplify Further

Convert \(1\) into \(\frac{25}{25}\) and perform the subtraction: \(y^2 = \frac{25}{25} - \frac{9}{25}\). This results in \(y^2 = \frac{16}{25}\).

Step 6: Find y

Take the square root of both sides to find \(y\). Since point \(P\) is in Quadrant III, where \(y\) is negative: \(y = -\frac{4}{5}\).

Key Concepts

Understanding Quadrants

While observing the unit circle, it's essential to understand the concept of quadrants. The coordinate plane is divided into four quadrants, each representing a distinct section:

• Quadrant I: Both x and y coordinates are positive.
• Quadrant II: x is negative, while y is positive.
• Quadrant III: Both x and y coordinates are negative.
• Quadrant IV: x is positive, and y is negative.

Points on the unit circle situated within each of these quadrants exhibit sign patterns that aid in determining specific coordinates. For instance, since point \( P \left(-\frac{3}{5}, \right)\) is located in Quadrant III, this indicates that both x and y coordinates must be negative. This information is fundamental when solving for unknown coordinates on the unit circle.

Cartesian Coordinates

The term Cartesian coordinates refers to the system used to define every point in a plane through two numbers: the x-coordinate and the y-coordinate. In the Cartesian plane:

• The x-coordinate signifies horizontal position (left or right from the origin).
• The y-coordinate signifies vertical position (upwards or downwards from the origin).

When dealing with the unit circle, understanding these coordinates is vital because any point \( P(x, y) \) on the circle satisfies the equation \( x^2 + y^2 = 1 \). This equation is derived from the circle's radius of 1. Thus, given one coordinate, you can solve for the other, as illustrated in the exercise where \( x = -\frac{3}{5} \), allowing us to find \( y = -\frac{4}{5} \).

Exploring Trigonometric Identities

Trigonometric identities pave a thorough understanding of relationships between angles and sides in right triangles or unit circles. Critical identities include sin, cos, and tan functions that relate an angle to ratios of sides. On the unit circle:

• The cosine of an angle \( \theta \) corresponds to the x-coordinate of the point on the circle.
• The sine of an angle \( \theta \) corresponds to the y-coordinate of that point.
• Tangent \( (\tan \theta) \) results from the ratio \( \frac{\sin \theta}{\cos \theta} \).

These identities let us work beyond just finding coordinates; they allow exploration into angles and distances, critical when considering trigonometric functions in different quadrants. For example, knowing \( x = -\frac{3}{5} \) and \( y = -\frac{4}{5} \) in Quadrant III helps calculate tan \( \theta = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \). Recognizing and applying these identities effectively assists in robust understanding and problem-solving.