## Algebra and Trigonometry 10th Edition

$arccos[cos(-\frac{3\pi}{2})]=\frac{\pi}{2}$
$-\frac{3\pi}{2}$ does not lie in the range of $arccos~x$: ($0\leq x\leq \pi$). But, we know that the period of $cos~x=2\pi$ and that $-\frac{3\pi}{2}=\frac{\pi}{2}-2\pi$. So: $cos(-\frac{3\pi}{2})=cos(\frac{\pi}{2})$. Now, $\frac{\pi}{2}$ lies in the range of $arccos~x$. $arccos[cos(-\frac{3\pi}{2})]=\frac{\pi}{2}$