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 380: 108

Answer

$\dfrac{\sqrt3-1}{2}$

Work Step by Step

RECALL: $(f-g)(x)=f(x)-g(x)$ Use the rule above to obtain: $(f-g)(60^\circ)=\sin{60^\circ}-\cos{60^\circ}=\dfrac{\sqrt3}{2}-\dfrac{1}{2}=\dfrac{\sqrt3-1}{2}$
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