Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 71

Answer

$1$

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: $180^{\circ}-\theta $ c) Quadrant -III: $\theta - 180^o $ d) Quadrant -IV: $360^{\circ}-\theta $ Reference angle is equal to $ =\dfrac{9 \pi}{4}-2 \pi=\dfrac{\pi}{4}$ Since, $\tan (\dfrac{\pi}{4})=1$ So, $\tan \dfrac{9\pi}{4}=1$; Because $\theta $ lies in Quadrant-I.
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