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