Solution

We have seen before how it is possible to cancel the inner inverse trig with the outer trig. We could look at the semi-circle for the Cosine Inverse to see that no matter what \(x\) coordinate we provide on the Unit Semi-Circle, there will always only be one angle that corresponds with it:

The difficulty that this problem poses, is that the input to the Inverse Trig function is \(5x\). Now, the Domain of the Inverse Trig is \([-1,1]\), (the left and right most \(x\) value of the Unit Circle). In order to determine the domain of \(\cos^{-1}(5x)\), we need to undestand that \(5x\) represents a horizontal compression by a factor of 1/5. Thus, the valid domain for \(\cos^{-1}(5x)\) is \(\left[-\frac{{1}}{{5}},\frac{{1}}{{5}}\right]\).

Finally, the answer to our question is then \(\cos(\cos^{-1}(5x)) = 5x\) with a domain of \(\left[-\frac{{1}}{{5}},\frac{{1}}{{5}}\right]\). You can actually visually verify this by graphing \(y=\cos(\cos^{-1}(5x))\) on a graphing utility (note that on some graphing utilities \(\cos^{-1}(x) = \arccos(x)\)