Question

${\cos }^{-1}(\cos x)=xwhere\_\_\_\_\_.$

Short answer

${\cos }^{-1}(\cos x)=xwhere\mathbf{0}\mathbf{\le }\mathit{x}\mathbf{\le }\mathbf{\pi }$

Step-by-step solution

Step 1. Given Information:

Given expression: ${\cos }^{-1}(\cos x)=xwhere\mathrm{\_\_\_\_\_}$

We want to fill the gap in given expression.

Step 2. Solution:

The function of cosine and its inverse has the following property:

$f^{-1}\left(f\right(x\left)\right)={\cos }^{-1}(\cos x)=xwhere0\le x\le \mathrm{\pi }$

Therefore,${\cos }^{-1}(\cos x)=xwhere0\le x\le \mathrm{\pi }$