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: 97



Work Step by Step

Let us consider: $f(\theta)=\sin \theta$ and $\theta=60^{\circ}$ We wish to find: $f(\dfrac{\theta}{2})$ When $\theta=60^{\circ}$, then $\dfrac{\theta}{2}=\dfrac{60^{\circ}}{2}=30^{\circ}$ We know from the unit circle that $\sin (30^{\circ})=\sin (\dfrac{\pi}{6})=\dfrac{1}{2}$ Thus, we have: $f(\dfrac{\theta}{2})=\sin (30^{\circ}) \\=\sin (\dfrac{\pi}{6}) \\ =\dfrac{1}{2}$
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