## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 64

#### 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: $180^0-\theta$ (iii) Quadrant III: $\theta-180^o$ (iv) Quadrant IV: $360^o-\theta$ The given angle is in Quadrant III. Use the formula in (iii) above to find its reference angle: $=240^0-180^o \\=60^o$ Note that $240^o$ is in Quadrant III, where secant is negative. Thus, $\sec{240^o} = -\sec{60^o}$ RECALL: $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $60^o$ is a special angle whose cosine value is $0.5$. Thus, $\sec{240^o} \\= -\sec{60^o} \\=-\dfrac{1}{\cos{60^o}} \\=-\dfrac{1}{0.5} \\=-2$

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