Problem 8 - Algebra and Trigonometry

Question

\(7-12\) . Find the missing coordinate of \(P\) , using the fact that \(P\) lies on the unit circle in the given quadrant. \(P\left(\quad,-\frac{7}{25}\right) \quad \mathrm{IV}\)

Step-by-Step Solution

Answer

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

Step 1: Recall the Equation of the Unit Circle

The equation of the unit circle is given by \(x^2 + y^2 = 1\), where \((x, y)\) is a point on the circle. This means the sum of the squares of the coordinates of any point on the circle is always 1.

Step 2: Insert Known Values into the Equation

We're given that \(y = -\frac{7}{25}\) and need to find \(x\). We substitute \(y\) into the equation: \(x^2 + \left(-\frac{7}{25}\right)^2 = 1\).

Step 3: Simplify the Equation

Calculate \(\left(-\frac{7}{25}\right)^2 = \frac{49}{625}\). Thus, the equation becomes \(x^2 + \frac{49}{625} = 1\).

Step 4: Isolate \(x^2\)

Subtract \(\frac{49}{625}\) from both sides to get \(x^2 = 1 - \frac{49}{625}\).

Step 5: Simplify the Right Side

Convert 1 to \(\frac{625}{625}\) to have a common denominator: \(x^2 = \frac{625}{625} - \frac{49}{625} = \frac{576}{625}\).

Step 6: Solve for \(x\)

Take the square root of both sides: \(x = \pm \sqrt{\frac{576}{625}} = \pm \frac{24}{25}\).

Step 7: Determine the Quadrant's Impact on \(x\)

Since point \(P\) lies in the fourth quadrant (IV), where \(x > 0\) and \(y < 0\), \(x\) must be \(\frac{24}{25}\).

Key Concepts

Quadrants

When studying the unit circle, understanding the concept of quadrants is crucial. A unit circle is divided into four sections called quadrants, labeled as I, II, III, and IV.
These quadrants help determine the sign of the coordinates of any point on the circle.- **Quadrant I**: Here, both the x and y coordinates are positive, often written as \(x > 0, y > 0\).- **Quadrant II**: In this quadrant, x is negative, and y is positive, \(x < 0, y > 0\).- **Quadrant III**: Both coordinates are negative, \(x < 0, y < 0\).- **Quadrant IV**: x is positive while y is negative, \(x > 0, y < 0\).The importance of knowing which quadrant a point lies in helps in deciding the signs of the coordinates. For instance, in the original problem, knowing that the point lies in Quadrant IV tells us that the x-coordinate must be positive.

Finding Coordinates

Finding coordinates on the unit circle involves using the circle's equation to determine unknown values. Any point \( (x, y) \) on the unit circle must satisfy the equation: \ x^2 + y^2 = 1 \.Here's how you find a missing coordinate:

• **Step 1**: Identify the known coordinate and plug it into the equation. For instance, we knew \( y = -\frac{7}{25}\).
• **Step 2**: Simplify and solve. Plug in the values: \(x^2 + ( -\frac{7}{25})^2 = 1 \) and compute, leading to \ x^2 + \frac{49}{625} = 1 \.
• **Step 3**: Rearrange to find x. Subtract \frac{49}{625}\ from both sides: \ x^2 = \frac{576}{625}\.
• **Step 4**: Calculate the square root: \ x = \pm\frac{24}{25} \.

Only the positive root is selected here, as the point is in the fourth quadrant, affirming that \( x > 0 \). Understanding how to manipulate the equation is essential for accurately finding both x and y coordinates.

Equation of Circle

The equation of the circle is a foundational concept in trigonometry and algebra. For a unit circle, this is expressed as \[ x^2 + y^2 = 1 \]. Here, every point \( (x, y) \) on the circle satisfies this equation because each radius is of length 1.- **Understanding**: This equation essentially arises from the Pythagorean Theorem. It explains that in a right-angled triangle formed within the circle, the radius (hypotenuse) is always 1 when squared, equalling the squares of the x and y coordinates combined.- **Application**: With given coordinates, you plug them into this equation to test or find missing values.Knowing the principles behind the equation enables solving for unknowns easily while understanding the circle's fundamental properties. It boils down to stating how a circle's radius constrains its points, defined in Cartesian coordinates.