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Chapter 5: Problem 7

Show that the point is on the unit circle. $$\left(-\frac{\sqrt{5}}{3}, \frac{2}{3}\right)$$

Short Answer

Expert verified
The point \(-\frac{\sqrt{5}}{3}, \frac{2}{3}\) is on the unit circle since its coordinates satisfy the equation \(x^2 + y^2 = 1\).

Step by step solution

01

Recall the Unit Circle Equation

The equation of the unit circle is given by \(x^2 + y^2 = 1\). To verify if a point \((x, y)\) is on the unit circle, substituting the x and y values into this equation should result in a true statement.
02

Substitute the Coordinates into the Equation

Substitute \(x = -\frac{\sqrt{5}}{3}\) and \(y = \frac{2}{3}\) into the unit circle equation:\[\left(-\frac{\sqrt{5}}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = 1\]
03

Simplify the Squared Terms

Calculate \(\left(-\frac{\sqrt{5}}{3}\right)^2\) and \(\left(\frac{2}{3}\right)^2\):- \(\left(-\frac{\sqrt{5}}{3}\right)^2 = \frac{5}{9}\)- \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)
04

Add the Results

Add the two results obtained from squaring the coordinates:\[\frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1\]
05

Verify the Result

Since the sum of the squares equals \(1\), the equation \(x^2 + y^2 = 1\) holds true, confirming that the point \(-\frac{\sqrt{5}}{3}, \frac{2}{3}\) is indeed on the unit circle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of a Circle
An equation of a circle is a mathematical expression that represents any circle on the coordinate plane. The general equation of a circle centered at the origin (0,0) with radius \( r \) is \( x^2 + y^2 = r^2 \). For a unit circle, specifically, the radius is equal to 1.

This simplifies our equation to \( x^2 + y^2 = 1 \). The concept is simple yet powerful because it allows us to determine if a point lies on the unit circle or not. By substituting the coordinates of any point \((x, y)\) into the equation, we can check if the sum of the squares is equal to 1.
  • If true, the point is on the circle.
  • If false, the point is outside or inside the circle, depending on whether the sum is greater or lesser than 1, respectively.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. It allows us to describe geometric shapes in a numerical way, using algebraic expressions.

The unit circle is a perfect example of how we use coordinate geometry to verify points on a curve. With coordinates like \((x, y)\), you can easily check their positions relative to the unit circle by substituting these values into its equation.
  • This practical approach makes it easy to manipulate points and curves numerically.
  • It connects algebra with geometry, offering a concrete way to solve geometric problems.
Trigonometric Identity
The unit circle is closely connected to trigonometry. In fact, it's central to understanding trigonometric identities, which are equations involving trigonometric functions that are true for all values of the variables.

For example, in the unit circle, the coordinate \(x\) is the cosine of the angle \(\theta\), and \(y\) is the sine of \(\theta\). This leads to the well-known identity \(\sin^2\theta + \cos^2\theta = 1\).
  • This identity mirrors the unit circle equation, \(x^2 + y^2 = 1\), but expressed in trigonometric terms.
  • Understanding this connection is a key part of learning how angles and their respective sine and cosine values interplay on the circle.

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Most popular questions from this chapter

An initial amplitude \(k\), damping constant \(c,\) and frequency \(f\) or period \(p\) are given. (Recall that frequency and period are related by the equation \(f=1 / p .\) ) (a) Find a function that models the damped harmonic motion. Use a function of the form \(y=k e^{-c t} \cos \omega t\) in Exercises 21-24 and of the form \(y=k e^{-c t} \sin \omega t\) in Exercises 25-28 (b) Graph the function. $$k=1, \quad c=1, \quad p=1$$

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