Answer
Both functions have the same amplitudes and periods but the graph of function $g$ is a horizontal shift of $\pi$ units to the left of $f$.
Work Step by Step
Using the equation of sine $y=a\cos b(x-c)$:
For $f(x)=\cos x=1\cos 1(x-0)$,
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{1}=2\pi$
$shift=none$
For $f(x)=\cos (x-\pi)=1\cos 1(x-(-\pi))$,
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{1}=2\pi$
$shift=\pi~to~the~left$
Thus, both functions have the same amplitudes and periods but the graph of function $g$ is a horizontal shift of $\pi$ units to the left of $f$.
