Answer
$y=4\sin(2\pi x).$
Work Step by Step
If a function has form $y=A\sin(\omega x)$, then there $|A|$ stands for the amplitude and $T =\frac{2\pi}{\omega}$ stands for the period.
First we identify $\omega$ and $A$ and then plug them into the form of $y=A\sin(\omega x)$.
Hence $A=|4|=4$, $\omega=\frac{2\pi}{T}=\frac{2\pi}{1}=2\pi.$
Thus $y=A\sin(\omega x)=4\sin(2\pi x).$
