Problem 2 - Algebra and Trigonometry

Question

\(1-6=\) Show that the point is on the unit circle. $$ \left(-\frac{5}{13}, \frac{12}{13}\right) $$

Step-by-Step Solution

Answer

Yes, the point \((-\frac{5}{13}, \frac{12}{13})\) is on the unit circle.

Step 1: Understand the Unit Circle

The unit circle is a circle centered at the origin (0,0) with a radius of 1. A point \((x, y)\) is on the unit circle if \(x^2 + y^2 = 1\).

Step 2: Identify the Coordinates

The given point is \((-\frac{5}{13}, \frac{12}{13})\). We will use these coordinates to verify if the point lies on the unit circle.

Step 3: Calculate \(x^2\)

Using the x-coordinate \(-\frac{5}{13}\), calculate \(x^2\). \[x^2 = \left(-\frac{5}{13}\right)^2 = \frac{25}{169}\]

Step 4: Calculate \(y^2\)

Using the y-coordinate \(\frac{12}{13}\), calculate \(y^2\). \[y^2 = \left(\frac{12}{13}\right)^2 = \frac{144}{169}\]

Step 5: Add \(x^2\) and \(y^2\)

Now, add \(x^2\) and \(y^2\) to check if their sum equals 1. \[x^2 + y^2 = \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1\]

Step 6: Conclusion

Since the sum \(x^2 + y^2\) equals 1, the point \(\left(-\frac{5}{13}, \frac{12}{13}\right)\) is on the unit circle.

Key Concepts

Cartesian Coordinates

In the world of mathematics, points on a plane are most often described using Cartesian coordinates. A pair of numbers, \(x, y\), precisely pinpoints a location when graphed on a two-dimensional grid. The grid is made up of a horizontal axis (typically labeled the x-axis) and a vertical axis (known as the y-axis). Each coordinate \(x, y\) tells you how far along the x and y axes a point is located.

When you plot a point such as \(-\frac{5}{13}, \frac{12}{13}\), you're essentially saying: "Move \(-\frac{5}{13}\) units along the x-axis, and \(\frac{12}{13}\) units up along the y-axis." This system is incredibly useful because it allows for a consistent framework to refer to any point and apply mathematical principles like distance, midpoint, and shortly exploring the nature of circles.

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry and geometry that relates to the sides of a right triangle. In terms of the unit circle context, it states that for any angle, \(\theta\), the square of the sine of \(\theta\) and the cosine of \(\theta\) add up to 1: \(\sin^2(\theta) + \cos^2(\theta) = 1\). This identity is derived from the Pythagorean Theorem, which is more widely known as \(a^2 + b^2 = c^2\), applicable to right-angled triangles.

But here's how it comes into play with the unit circle:

• The unit circle is the set of all points \(x, y\) that satisfy the equation \(x^2 + y^2 = 1\).
• On the unit circle, "sine" is analogous to \(y\) and "cosine" to \(x\).

Hence, any point \((x, y)\) on the unit circle follows the Pythagorean Identity, reaffirming it as being on the circle if \(x^2 + y^2 = 1\). This aligns directly with the original problem of showing how the point \(-\frac{5}{13}, \frac{12}{13}\) conforms.

Circle Equation

A circle, especially the unit circle, can be described easily with an equation. The equation \(x^2 + y^2 = r^2\) represents a circle with radius \(r\) centered at the origin \((0,0)\). When \(r = 1\), the equation simplifies to \(x^2 + y^2 = 1\), describing the unit circle.

Here's why this is significant:

• Every point \(x, y\) on this circle will satisfy the equation \(x^2 + y^2 = 1\).
• To determine if a specific point is on the unit circle, simply substitute its coordinates into this equation.

For example, substituting the point \(-\frac{5}{13}, \frac{12}{13}\) into \(x^2 + y^2 = 1\), we previously calculated \(x^2 = \frac{25}{169}\) and \(y^2 = \frac{144}{169}\). Adding these results yields 1, confirming the point’s presence on the unit circle.

This logical sequence shows how circle equations tie back into verifying point placement on geometric constructs like circles.