Question

Is each point on the unit circle? How do you know? a) \(\left(-\frac{3}{4}, \frac{1}{4}\right)\) b) \(\left(\frac{\sqrt{5}}{8}, \frac{7}{8}\right)\) c) \(\left(-\frac{5}{13}, \frac{12}{13}\right)\) d) \(\left(\frac{4}{5},-\frac{3}{5}\right)\) e) \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\) f) \(\left(\frac{\sqrt{7}}{4}, \frac{3}{4}\right)\)

Short answer

Only points (c), (d), (e), and (f) are on the unit circle.

Step-by-step solution

Understand the Unit Circle Definition

A point \( (x, y) \) is on the unit circle if it satisfies the equation \( x^2 + y^2 = 1 \).

Check Point (a)

Calculate \( x^2 + y^2 \) for \( \bigg(-\frac{3}{4}, \frac{1}{4} \bigg) \): \(-\frac{3}{4}^2 + \frac{1}{4}^2 = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8} \). Since \( \frac{5}{8} \eq 1 \), the point is not on the unit circle.

Check Point (b)

Calculate \( x^2 + y^2 \) for \( \bigg(\frac{\sqrt{5}}{8}, \frac{7}{8} \bigg) \): \( \bigg(\frac{\sqrt{5}}{8}\bigg)^2 + \bigg(\frac{7}{8}\bigg)^2 = \frac{5}{64} + \frac{49}{64} = \frac{54}{64} = \frac{27}{32} \). Since \( \frac{27}{32} \eq 1 \), the point is not on the unit circle.

Check Point (c)

Calculate \( x^2 + y^2 \) for \( \bigg(-\frac{5}{13}, \frac{12}{13} \bigg) \): \(-\frac{5}{13}^2 + \frac{12}{13}^2 = \frac{25}{169} + \frac{144}{169} = \frac{169}{169} = 1 \). Since \( 1 = 1 \), the point is on the unit circle.

Check Point (d)

Calculate \( x^2 + y^2 \) for \( \bigg(\frac{4}{5}, -\frac{3}{5} \bigg) \): \( \frac{4}{5}^2 + -\frac{3}{5}^2 = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \). Since \( 1 = 1 \), the point is on the unit circle.

Check Point (e)

Calculate \( x^2 + y^2 \) for \( \bigg(-\frac{\sqrt{3}}{2}, -\frac{1}{2} \bigg) \): \( \bigg(-\frac{\sqrt{3}}{2}\bigg)^2 + \bigg(-\frac{1}{2}\bigg)^2 = \frac{3}{4} + \frac{1}{4} = 1 \). Since \( 1 = 1 \), the point is on the unit circle.

Check Point (f)

Calculate \( x^2 + y^2 \) for \( \bigg(\frac{\sqrt{7}}{4}, \frac{3}{4} \bigg) \): \( \bigg(\frac{\sqrt{7}}{4}\bigg)^2 + \bigg(\frac{3}{4}\bigg)^2 = \frac{7}{16} + \frac{9}{16} = \frac{16}{16} = 1 \). Since \( 1 = 1 \), the point is on the unit circle.

Key concepts

precalculus

Precalculus serves as a bridge between algebra and calculus. It prepares students for more advanced mathematical concepts. One crucial aspect of precalculus is understanding the unit circle. The unit circle is a perfect introduction to trigonometry and its fundamentals.
The unit circle can help students learn about angles, radians, and trigonometric functions such as sine and cosine.
Mastering these concepts is essential for those who aim to dive deeper into calculus and beyond.

trigonometry

Trigonometry is the study of relationships between the angles and sides of triangles. A significant part of trigonometry revolves around the unit circle, which has a radius of one unit.
The coordinates of points on this circle provide the values of trigonometric functions for various angles.
For example, the coordinate \((x, y)\) corresponds to \((cos(\theta), sin(\theta))\). This relationship allows us to solve many trigonometric problems by using simple but powerful formulas.

coordinate geometry

Coordinate geometry, also known as analytic geometry, involves representing geometric figures using coordinates and algebra.
The unit circle's equation is \[ x^2 + y^2 = 1 \]. By using the Pythagorean theorem, we check if points lie on the unit circle.
For instance, for point \((\frac{4}{5}, -\frac{3}{5})\), we evaluate \((\frac{4}{5})^2 + (-\frac{3}{5})^2 = 1\), confirming it is on the unit circle.
This technique applies to any point we want to verify.

circle equation

The equation of a circle in coordinate geometry is usually written as \[ (x-h)^2 + (y-k)^2 = r^2 \], where \(h, k\) is the circle's center and \(r\) is its radius. For the unit circle, the center is at \(0, 0\) and the radius is 1, simplifying the equation to \[ x^2 + y^2 = 1 \].
This form is essential for identifying points on the unit circle. By squaring and adding the \(x\) and \(y\) coordinates of a given point, if the sum equals 1, the point lies on the unit circle.
This fundamental principle is applied to solve many precalculus and trigonometry problems, providing a strong foundation for more advanced studies.