## Calculus: Early Transcendentals 8th Edition

$(a) cos^{-1}(-1) = \pi$ $(b) sin^{-1}(0.5) = \frac{\pi}{6}$
$(a) cos^{-1}(-1) = \pi$ because $cos(\pi) = -1$ and $\pi$ lies between $0$ and $\pi$ $(b) sin^{-1}(0.5) = \frac{\pi}{6}$ because $sin(\frac{\pi}{6}) = 0.5$ and $\frac{\pi}{6}$ lies between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$