Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 5 - Section 5.1 - The Unit Circle - 5.1 Exercises - Page 408: 19

Answer

$P(\frac{-\sqrt{2}}{3},\frac{-\sqrt{7}}{3})$

Work Step by Step

We have the following equation for the unit circle: $x^{2}+y^{2}=1$, so we plug in $P(\frac{-\sqrt{2}}{3},y)$, $(\frac{-\sqrt{2}}{3})^{2}+y^{2}=1$ $ y^{2}=1 - \frac{2}{9}= \frac{7}{9}$ $y = \frac{\sqrt{7}}{3}$or $y = - \frac{\sqrt{7}}{3}$ but P lies below the x-axis so the y-coordinate is negative so $y = - \frac{\sqrt{7}}{3}$
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