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

Published by Pearson

# Chapter 6 - Analytic Trigonometry - Section 6.1 The Inverse Sine, Cosine, and Tangent Functions - 6.1 Assess Your Understanding - Page 473: 20

#### Answer

$\dfrac{\pi}{6}$

#### Work Step by Step

$y = \tan^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \tan{y}$ $\text{where } \hspace{15pt} -\infty \leq x \leq \infty \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$ Let $\theta = \tan^{-1} {\left(\dfrac{\sqrt{3}}{3} \right)} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$ Since $\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{3} \right)}$, then $\tan{\theta} =\dfrac{\sqrt{3}}{3}$ Referring to Table 3 on page 472: The only angle in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$ whose tangent is $\dfrac{\sqrt{3}}{3}$ is $\dfrac{\pi}{6}$. Therefore, $\tan^{-1}{\left(\dfrac{\sqrt{3}}{3} \right)} = \dfrac{\pi}{6}$.

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