## Precalculus (6th Edition)

Fill the blank with ... $\cos y$ ...
When we restrict the domain of $\cos x$ to $[0,\pi]$, it becomes one-to-one and has an inverse. For inverse functions, $f(f^{-1}(y))=y$ and $f^{-1}(f(x))=x.$ So, with $\cos^{-1}x$ being the inverse of $\cos x$ (on the restricted domain), $\cos y=\cos[\cos^{-1}(x)]=x.$ Fill the blank with ... $\cos y$ ...