Problem 13 - 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. $$ \frac{\text { Coordinates }}{-P\left(\frac{2}{7},\right.} ) \frac{\text { Quadrant }}{\text { IV }} $$

Step-by-Step Solution

Answer

The missing coordinate is \(y = -\frac{3\sqrt{5}}{7}\).

Step 1: Know the Unit Circle Properties

A point \(P(x, y)\) that lies on the unit circle satisfies the equation \(x^2 + y^2 = 1\). Since the radius of the unit circle is 1, any point that satisfies this equation lies on this circle.

Step 2: Substitute the Known Coordinate

We are given that \(P\left(\frac{2}{7}, y\right)\) lies on the unit circle. Substitute \(x = \frac{2}{7}\) into the unit circle equation: \(\left(\frac{2}{7}\right)^2 + y^2 = 1\).

Step 3: Simplify the Equation

Calculate \(\left(\frac{2}{7}\right)^2 = \frac{4}{49}\). Substitute this into the equation: \(\frac{4}{49} + y^2 = 1\).

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

Subtract \(\frac{4}{49}\) from both sides of the equation to solve for \(y^2\): \(y^2 = 1 - \frac{4}{49}\).

Step 5: Calculate the Right-Hand Side

Find a common denominator and calculate \(1 - \frac{4}{49}\). This is \(\frac{49}{49} - \frac{4}{49} = \frac{45}{49}\).

Step 6: Solve for \(y\)

Take the square root of both sides: \(y = \pm \sqrt{\frac{45}{49}} = \pm \frac{\sqrt{45}}{7}\).

Step 7: Determine the Sign of \(y\) Based on Quadrant

In the fourth quadrant, the \(x\)-coordinate is positive and the \(y\)-coordinate is negative. Therefore, \(y = -\frac{\sqrt{45}}{7}\).

Step 8: Simplify the Expression for \(y\)

Since \(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\), the expression for \(y\) becomes \(y = -\frac{3\sqrt{5}}{7}\).

Key Concepts

Quadrants of the Coordinate Plane

The coordinate plane is divided into four sections known as quadrants. Each quadrant is defined by the positive and negative markings of the x and y coordinates.

Here's a simple guide to remember:

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

In Quadrant IV, where our point is located, the x-coordinate is positive and the y-coordinate is negative. This rule helps determine the sign of any unknown coordinates based on the quadrant they are in.

Coordinates on a Circle

The unit circle is a crucial concept in trigonometry. It is a circle with a radius of 1 centered at the origin of the coordinate plane. Every point on the unit circle has coordinates ewline

• that can be expressed as (\(cos(\theta)\),\(sin(\theta)\)), indicating the cosine and sine values of an angle \(\theta\) measured from the positive x-axis.
  
• These coordinates fit into the unit circle equation: \(x^2 + y^2 = 1\), ensuring that any point P(x, y) on the circle will maintain this relationship.

In this exercise, knowing one coordinate, \(\frac{2}{7}\), we use the unit circle equation to determine the y-coordinate, while considering the quadrant to find the correct sign of y.

Pythagorean Identity

The Pythagorean Identity is a famous equation that plays a foundational role in trigonometry:

• \[x^2 + y^2 = 1\]

This identity mirrors the Pythagorean theorem, where the equation reflects the idea of a right triangle whose hypotenuse (in this case, the radius of the unit circle) is 1.

Using the identity, we solve for the missing y-coordinate by substituting the known x-value, simplifying, and solving for y using the relationship: \(x^2 + y^2\) equals 1. The sign of y is determined based on which quadrant the point resides in.

Trigonometric Concepts

Trigonometry is the study of relationships between the angles and sides of triangles. On the unit circle, trigonometric functions such as sine, cosine, and tangent become very handy.

Every point P(x, y) on the unit circle can be associated directly with these functions:

• Cosine represents the x-coordinate: \(cos(\theta) = x\)
• Sine represents the y-coordinate: \(sin(\theta) = y\)

In this specific scenario, understanding these functions aids in deducing unknown values, like resolving for y given x. The quadrant criteria further ensure the correct trigonometric sign is applied. So, if you know where the point lies, you know the sign of these values.