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 474: 43

Answer

$-\dfrac{ \pi}{5}$

Work Step by Step

Value of certain composite functions formula: $f^{-1}(f(x)= \sin^{-1}{(\sin{x})} = x \hspace{25pt} -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$ Note that $\dfrac{4 \pi}{5} \text{ isn't in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$. Recall that: $\tan{\theta} = \tan{(\theta - \pi)}$ Thus, $\tan{\left(\dfrac{4 \pi}{5} \right)} = \tan{\left(\dfrac{4 \pi}{5} - \pi \right) }= \tan{\left(-\dfrac{\pi}{5} \right)}$ Since $-\dfrac{\pi}{5} \text{ is in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$, then $\tan^{-1} {\left[\tan{\left(-\dfrac{ \pi}{5} \right)} \right]} = -\dfrac{ \pi}{5}$ Therefore, $\tan^{-1} {\left[\tan{\left(\dfrac{ 4\pi}{5} \right)} \right]} = \boxed{-\dfrac{ \pi}{5}}$

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