Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 7


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

Work Step by Step

Here, we have $\theta=-840^{\circ}$ Here, $\theta'=-840^{\circ}+(3) 360^{\circ}=240^{\circ}$ and $\theta''=\theta'-180^{\circ}=240^{\circ}-180^{\circ}=60^{\circ}$ The function $\sin$ is negative in Quadrant $II$. Now, $\sin \theta=\sin \theta'= -\sin \theta''=-\sin 60^{\circ}=-\dfrac{\sqrt 3}{2}$
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