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: 62

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: $180^0-\theta$ (iii) Quadrant III: $\theta-180^o$ (iv) Quadrant IV: $360^o-\theta$ The given angle is in Quadrant IV. Use the formula in (iv) above to find its reference angle: $=360^0-300^o \\=60^o$ Note, however, that $300^o$ is i Quadrant IV, where sine is negative. Thus, $\sin{300^o} = -\sin{60^o}$ $60^o$ is a special angle whose sine value is $\dfrac{\sqrt{3}}{2}$. Thus, $\sin{300^o} = -\dfrac{\sqrt{3}}{2}$
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