Algebra 2 (1st Edition)

Published by McDougal Littell

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 29

Answer

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

Work Step by Step

When $P$ is the period of $f(x)$ then we have $f(x+p)=f(x)$ Here, we have $\cot (-\dfrac{8 \pi}{3})=-\cot (-\dfrac{8 \pi}{3})$ But $\dfrac{8 \pi}{3}- 2\pi=\dfrac{8 \pi}{3}-\dfrac{6 \pi}{3}=\dfrac{2 \pi}{3}$ Thus, we have $-\cot(\dfrac{2 \pi}{3})=-\tan [-\dfrac{\pi}{6}]$ or, $=\dfrac{\sqrt 3}{3}$

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