Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding - Page 379: 31

Answer

$\dfrac {1+\sqrt {2}}{2}$

Work Step by Step

$\sin 45^o+\cos 60^o=\sin \dfrac {\pi }{4}+\cos \dfrac {\pi }{3}=\dfrac {\sqrt {2}}{2}+\dfrac {1}{2}=\dfrac {1+\sqrt {2}}{2}$
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