Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.1 The Inverse Sine, Cosine, and Tangent Functions - 6.1 Assess Your Understanding - Page 474: 42


$\dfrac{ \pi}{3}$

Work Step by Step

Value of certain composite functions formula: $f^{-1}(f(x)= \cos^{-1}{(\cos{x})} = x \hspace{25pt} 0 \leq x \leq \pi$ Note that $-\dfrac{5 \pi}{3} \text{ isn't in the interval } \left[0,\pi \right]$. Recal that: $\cos{\theta} = \cos{(\theta+2 \pi)}$ Thus, $\cos{\left(-\dfrac{5 \pi}{3} \right)} = \cos{\left(- \dfrac{5 \pi}{3} +2\pi\right) }= \cos{\left(\dfrac{\pi}{3} \right)}$ Since $\dfrac{\pi}{3} \text{ is in the interval } \left[0,\pi \right]$, then $\cos^{-1} {\left[\cos{\left(\dfrac{ \pi}{3} \right)} \right]} = \dfrac{ \pi}{3}$ Therefore, $\cos^{-1} {\left[\cos{\left(-\dfrac{ 5 \pi}{3} \right)} \right]} = \boxed{\dfrac{ \pi}{3}}$
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