Trigonometry (11th Edition) Clone

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

Chapter 3 - Quiz (Sections 3.1-3.3) - Page 126: 10



Work Step by Step

$\sin s=\frac{\sqrt 3}{2}$ can be written as $s=\sin^{-1} (\frac{\sqrt 3}{2})$ Ensuring that the calculator is in radians, we type $\sin^{-1} (\frac{\sqrt 3}{2})$ into the calculator and solve: $s=\sin^{-1} (\frac{\sqrt 3}{2})=\frac{\pi}{3}$ Since the interval is $[\frac{\pi}{2},\pi]$, we subtract the answer from $\pi$: $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ Therefore, the exact value of $s$ in the interval $[\frac{\pi}{2},\pi]$ is $\frac{2\pi}{3}$.
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