Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set: 29

Answer

$\frac{\sqrt {13}}{7}$

Work Step by Step

$sin^{2}t$ + $cos^{2}t$ = 1 sin t = $\frac{6}{7}$ $(\frac{6}{7})^{2}$ + $cos^{2}t$ = 1 $\frac{36}{49}$ + $cos^{2}t$ = 1 $cos^{2}t$ = 1 -$\frac{36}{49}$ $cos^{2}t$ = $\frac{13}{49}$ cos t = $\sqrt \frac{13}{49}$ = $\frac{\sqrt {13}}{7}$
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