Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 505: 5



Work Step by Step

We know that if $\cos{(y)}=x$, then $\cos^{-1}{x}=y.$ Hence, if $\cos{(\frac{5\pi}{6})}=-\frac{\sqrt3}{2}$, then by the definition of the inverse function: $\cos^{-1}(-\frac{\sqrt3}{2})=\frac{5\pi}{6}$. (because it is in the right domain, in $[0,\pi$)
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