Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 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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