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: 67

Answer

$\dfrac{\sqrt{3}}{2}$

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 I so its reference angle is: $=\pi - \dfrac{2\pi}{3} \\=\dfrac{\pi}{3}$ The value of sine in Quadrant II is positive. Thus, $\sin{\frac{2\pi}{3}} = \sin{\frac{\pi}{3}}$ $\dfrac{\pi}{3}$ is a special angle whose sine value is $\dfrac{\sqrt{3}}{2}$. Thus, $\sin{\dfrac{2\pi}{3}}=\sin{\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{2}$
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