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: 6

Answer

Yes, ($-\frac{5}{7},-\frac{2\sqrt 6}{7}$) is a point on the unit circle.

Work Step by Step

Since ($-\frac{5}{7},-\frac{2\sqrt 6}{7}$) is an (x, y) coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(-\frac{5}{7})^{2}+(-\frac{2\sqrt 6}{7})^{2}=1$$ $$\frac{-5\times-5}{7\times7}+\frac{-2\sqrt 6\times-2\sqrt 6}{7\times7}=1$$ $$\frac{25}{49}+\frac{4\times6}{49}=1$$ $$\frac{25}{49}+\frac{24}{49}=1$$ $$\frac{49}{49}=1$$ We end up with 1=1, which is true. Therefore, ($-\frac{5}{7},-\frac{2\sqrt 6}{7}$) is a point on the unit circle.
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