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: 99



Work Step by Step

Let us consider: $f(\theta)=\sin \theta$ We are given that $\theta=60^{\circ}$ We know form the unit circle that $\sin (60^{\circ})=\sin (\dfrac{\pi}{3})=\dfrac{\sqrt 3}{2}$ Thus, we have: $[f(\theta)]^2=[\sin (60^{\circ})]^2 \\=[\sin\dfrac{\pi}{3}]^2 \\ =[\dfrac{\sqrt 3}{2}]^2 \\=\dfrac{3}{4}$
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