## Precalculus (6th Edition)

Sine and cosine do not have a common interval on which BOTH are one-to-one, and BOTH have range $[-1,1]$.
Sine is not one-to-one on $[0, \pi]$ and the interval does not contain x-values with negative function values. So, the whole range is not covered on this interval. On the other hand, the interval $[0, \pi]$ satisfies the cosine function needs (values range from -1 to 1, and cosine is one-to-one here). The same problem will occur with cosine, when restricted to the interval $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$, which fits the sine function.