Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set: 68

Answer

$-\dfrac{\sqrt{2}}{2}$

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 II so its reference angle is: $=\pi - \dfrac{3\pi}{4} \\=\dfrac{\pi}{4}$ The value of cosine in Quadrant II is negative. Thus, $\cos{\frac{3\pi}{4}} = -\cos{\frac{\pi}{4}}$ $\dfrac{\pi}{4}$ is a special angle whose cosine value is $\dfrac{\sqrt{2}}{2}$. Thus, $\cos{\dfrac{3\pi}{4}}=-\cos{\dfrac{\pi}{4}}=-\dfrac{\sqrt{2}}{2}$
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