Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 645: 64


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

Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ First, we reduce the angle: $495^{\circ}-360^{\circ}=135^{\circ}$ Since we are in quadrant 2, we subtract from $180^{\circ}$ $180^{\circ}-135^{\circ}=45^{\circ}$ Thus, we have $\sin 45^{\circ}=\dfrac{\sqrt 2}{2}$ The value is positive because sine is positive in the 2nd quadrant.
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