Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.2 - The Rectangular Coordinate System - 1.2 Problem Set - Page 25: 48


$(\frac{\sqrt5}{3}, -\frac{2}{3})$) is found on the graph of the unit circle.

Work Step by Step

A unit circle is a circle with a radius of one, with a center on the origin (0,0). Therefore the point $(\frac{\sqrt5}{3}, -\frac{2}{3})$), is found on the unit circle. To check for this, substitute the x and y values of the point into the area of a circle as well as the radius. $x^2+y^2=r$ $(\frac{\sqrt5}{3})^2+(-\frac{2}{3})^2=1$ Substitute. $\frac{5}{9}+\frac{4}{9}=1$ $\frac{9}{9}=1$ $1=1$
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