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 402: 21



Work Step by Step

We need to compute $\sin(\dfrac{11\pi}{2})$ The trigonometric function $\sin$ has a period of $2\pi$. So, we will find a value where the argument is between $0$ and $2 \pi$ in terms of the special angles. Therefore, $\sin(\dfrac{11\pi}{2})=\sin (\dfrac{(3+8)\pi}{2})\\ =\sin(\dfrac{3\pi}{2}+4\pi)\\=\sin(\dfrac{3\pi}{2})\\=-1$
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