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: 80


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

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: $180^{\circ}-\theta $ c) Quadrant -III: $\theta - 180^o $ d) Quadrant -IV: $360^{\circ}-\theta $ Reference angle is equal to $ =\dfrac{13 \pi}{3}-4 \pi=\dfrac{\pi}{3}$ Since, $\cot (\dfrac{\pi}{3})=\dfrac{\sqrt 3}{3}$ So, $\cot \dfrac{13\pi}{3}=\dfrac{\sqrt 3}{3}$; Because $\theta $ lies in Quadrant-I.
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