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 473: 23


$\dfrac{5 \pi}{6}$

Work Step by Step

$y = \cos^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \cos{y}$ $\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} 0 \leq y \leq \pi$ Let $\theta = \cos^{-1} {\left(-\dfrac{\sqrt{3}}{2} \right)} \hspace{20pt} 0 \leq \theta \leq \pi$ Since $\theta = \cos^{-1} {\left(-\dfrac{\sqrt{3}}{2} \right)}$, then $\cos{\theta} = -\dfrac{\sqrt{3}}{2}$ Referring to Table 2 on page 469: The only angle in the interval $\left[0,\pi \right]$ whose cosine is $-\dfrac{\sqrt{3}}{2}$ is $\dfrac{5 \pi}{6}$. Therefore, $\cos^{-1} {\left(-\dfrac{\sqrt{3}}{2} \right)} = \dfrac{5 \pi}{6}$.
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