Precalculus (6th Edition) Blitzer

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

Chapter 4 - Test - Page 647: 11


$\dfrac{\sqrt 3}{3}$

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: $(2\pi - \theta)$ Thus, the reference angle of $\dfrac{ 6 \pi}{3}+\dfrac{3\pi}{3}-\dfrac{ 8\pi}{3}=\dfrac{\pi}{3}$ So, $ \cot \dfrac{ \pi}{3}=\dfrac{1}{\tan \dfrac{ \pi}{3}}=\dfrac{\sqrt 3}{3}$ Thus, we have $ \cot \dfrac{\pi}{3}=\dfrac{\sqrt 3}{3}$ because $\theta$ lies in Quadrant-III.
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