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: 12


$$P(\frac{2}{5}, \frac{\sqrt 21}{5})$$

Work Step by Step

Since $(\frac{2}{5}, [])$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(\frac{2}{5})^{2}+(y)^{2}=1$$ $$\frac{2\times2}{5\times5}+y^{2}=1$$ $$\frac{4}{25}+y^{2}=1$$ $$y^{2}=1-\frac{4}{25}$$ $$y^{2}=\frac{25}{25}-\frac{4}{25}$$ $$y^{2}=\frac{21}{25}$$ $$\sqrt (y^{2})=\sqrt (\frac{21}{25})$$ $$y=±\frac{\sqrt 21}{5}$$ Since the coordinate needs to be in the 1st quadrant, the y value needs to be positive. Therefore, $y=\frac{\sqrt 21}{5}$ and $P(\frac{2}{5}, \frac{\sqrt 21}{5})$
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.