## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 69

#### Answer

$-2$

#### Work Step by Step

RECALL: (1)The reference angle of a given angle is equal to the smallest acute angle that the terminal side makes with the x-axis. (2) Based on the location of the terminal side of an angle $\theta$, the reference angle can be found using the formula: (i) Quadrant I: $\theta$ (ii) Quadrant II: $\pi-\theta$ (iii) Quadrant III: $\theta-\pi$ (iv) Quadrant IV: $2\pi-\theta$ The given angle is in Quadrant III so its reference angle is: $=\dfrac{7\pi}{6}-\pi \\=\dfrac{\pi}{6}$ The value of cosecant in Quadrant III is negative. Thus, $\csc{\frac{7\pi}{6}} = -\csc{\frac{\pi}{6}}$ RECALL: $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ Thus, $-\csc{\frac{\pi}{6}} = -\dfrac{1}{\sin{\frac{\pi}{6}}}$ $\dfrac{\pi}{6}$ is a special angle whose sine value is $0.5$. Thus, $\csc{\dfrac{7\pi}{6}}=-\dfrac{1}{0.5}=-2$

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