Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Review Exercises - Page 285: 24

Answer

$-\frac{\sqrt3}{2}$

Work Step by Step

RECALL: Since sine and arcsine are inverse functions of each other, then, for all valid values of $x$, $\sin{(\arcsin(x))}=x$ Thus, $\sin{(\arcsin{(-\frac{\sqrt3}{2})})}=-\frac{\sqrt3}{2}$
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