Problem 10 - 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. $$ \frac{\text { Coordinates }}{P\left(-\frac{7}{25},\right.} ) \frac{\text { Quadrant }}{\text { IV }} $$

Step-by-Step Solution

Answer

The missing coordinate is \( -\frac{24}{25} \).

Step 1: Identify the Properties of a Unit Circle

A unit circle is a circle with a radius of 1, centered at the origin (0,0) in the coordinate plane. If a point \( P \left( x,y \right) \) lies on the unit circle, then it satisfies the equation \( x^2 + y^2 = 1 \). The given point \( P \left( -\frac{7}{25}, y \right) \) must also satisfy this equation.

Step 2: Substitute the Known Value into the Unit Circle Equation

Substitute \( x = -\frac{7}{25} \) into the unit circle equation: \[ \left( -\frac{7}{25} \right)^2 + y^2 = 1 \] Calculate the square of \( x \) to continue.

Step 3: Calculate the Square of \( x \) and Simplify

Calculate \( \left( -\frac{7}{25} \right)^2 \): \[ \frac{49}{625} \]Now, substitute back in: \[ \frac{49}{625} + y^2 = 1 \]Subtract \( \frac{49}{625} \) from both sides to solve for \( y^2 \).

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

Subtract \( \frac{49}{625} \) from 1: \[ y^2 = 1 - \frac{49}{625} \]\[ y^2 = \frac{625}{625} - \frac{49}{625} \]\[ y^2 = \frac{576}{625} \]To find \( y \), take the square root of both sides.

Step 5: Determine the Correct Sign for \( y \) in Quadrant IV

Take the square root: \[ y = \pm \frac{24}{25} \]In Quadrant IV, the \( y \)-coordinate is negative, so \( y = -\frac{24}{25} \).

Step 6: State the Complete Coordinate of \( P \)

Thus, the complete coordinates for the point \( P \) are \( \left( -\frac{7}{25}, -\frac{24}{25} \right) \).

Key Concepts

Understanding Quadrants

In the coordinate plane, quadrants divide the plane into four sections. These sections are numbered counterclockwise starting from the positive x-axis. Each quadrant has unique properties based on the signs of its coordinates.

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

In this exercise, we know the point lies in Quadrant IV. This tells us the y-coordinate is negative, which will be an important fact when determining the missing coordinate.

Exploring the Coordinate Plane

The coordinate plane is a two-dimensional surface where every point is defined by a pair of numbers, known as coordinates. This pair usually consists of an x-coordinate (horizontal placement) and a y-coordinate (vertical placement).
The intersecting lines in the plane are called axes:

• The x-axis is the horizontal line.
• The y-axis is the vertical line.

The point where these axes intersect is known as the origin, denoted as (0,0). The coordinates of a point tell how far it is from the origin in both directions.

Using Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides of the equality are defined.
One of the fundamental identities is the Pythagorean identity, which relates the trigonometric functions sine and cosine:\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]For any point on the unit circle:

• x coordinate is the value of the cosine function: \( \cos(\theta) \)
• y coordinate is the value of the sine function: \( \sin(\theta) \)

Using these identities, we can find unknown sides or angles in right triangles and solve equations involving trigonometric components.

Techniques for Solving Equations

Solving equations is about finding all possible values of the variable that make the equation true. In the context of the unit circle and trigonometric equations, this often involves making use of known identities and positions of angles.
To determine the unknown coordinate for a point on the unit circle:

• Start with the equation \( x^2 + y^2 = 1 \) which must be satisfied because the radius of a unit circle is 1.
• Substitute any known value of x or y to find the missing value by isolating the missing term on one side of the equation.
• Solve for the unknown by using algebraic manipulation, which may include adding, subtracting, multiplying, dividing, and taking square roots.

It’s essential to consider the quadrant in determining the correct sign for your solution.