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: 95

Answer

$\dfrac{\sqrt 3}{2}$

Work Step by Step

We are given that $f(\theta)=\sin \theta$ and $\theta=60^{\circ}$ We know form the unit circle that $\sin (60^{\circ})=\sin (\dfrac{\pi}{3})=\dfrac{\sqrt 3}{2}$ Thus, we have: $f(\theta)=\sin (60^{\circ}) \\=\sin (\dfrac{\pi}{3}) \\=\dfrac{\sqrt 3}{2}$
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