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: 61


$-\sqrt 3$.

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 make the angle positive: $-210^{\circ}+360^{\circ}=150^{\circ}$ We are in quadrant 2, so the reference angle is: $180^{\circ}-150^{\circ}=30^{\circ}$ Thus, we have $ \cot 30^{\circ}=-\sqrt 3$ The answer is negative because cotangent is negative in the second quadrant.
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