Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 403: 113

Answer

$\dfrac{\sqrt2}{4}$

Work Step by Step

Recall the rule: $(p \circ g)(x)=p \ [(g(x)] $ Let us consider: $p(x)=\frac{x}{2}$ and $g(x)=\cos x$ Thus, we have: $(p\circ g)(315^\circ)=p [g(315^\circ)] \\ =p(\cos{315^\circ}) \\ =p (\dfrac{\sqrt2}{2}) \\ \\ =\dfrac{(\frac{\sqrt 2}{2}) }{2} \\ =\dfrac{\sqrt2}{4}$
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