Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Secton 5.1 - The Unit Circle - 5.1 Exercises: 11


$$P(-\frac{2\sqrt 2}{3}, \frac{1}{3})$$

Work Step by Step

Since $([], \frac{1}{3})$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(x)^{2}+(\frac{1}{3})^{2}=1$$ $$x^{2}+\frac{1\times1}{3\times3}=1$$ $$x^{2}+\frac{1}{9}=1$$ $$x^{2}=1-\frac{1}{9}$$ $$x^{2}=\frac{9}{9}-\frac{1}{9}$$ $$x^{2}=\frac{8}{9}$$ $$\sqrt (x^{2})=\sqrt (\frac{8}{9})$$ $$x=±\frac{2\sqrt 2}{3}$$ Since the coordinate needs to be in the 2nd quadrant, the x value needs to be negative. Therefore, $x=-\frac{2\sqrt 2}{3}$ and $P(-\frac{2\sqrt 2}{3}, \frac{1}{3})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.