## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 70

#### Answer

$-1$

#### 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 IV so its reference angle is: $=2\pi - \dfrac{7\pi}{4} \\=\dfrac{\pi}{4}$ The value of cotangent in Quadrant IV is negative. Thus, $\cot{\frac{7\pi}{4}} = -\cot{\frac{\pi}{4}}$ RECALL: $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ Thus, $-\cot{\frac{\pi}{4}} = -\dfrac{1}{\tan{\frac{\pi}{4}}}$ $\dfrac{\pi}{4}$ is a special angle whose tangent value is $1$. Thus, $\cot{\dfrac{7\pi}{4}}=-\dfrac{1}{1}=-1$

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