Can a rational Bézier curve take exactly the same shape as a part of the sine function?

In a Bézier curve, $x$ and $y$ are polynomials in the parameter $t$. Note that you can't just have "a part of the sine function": if $y(t) = \sin(x(t))$ for $t$ in some interval, since both sides of that equation are analytic functions on the complex plane the equation would be true for all complex numbers $t$. Since $y(t)$ is a polynomial, for any given value of $y$ (unless $y$ is constant) there are only finitely many $t$ and thus finitely many $x$. But this is not the case for the sine function: $\sin(n \pi) = 0$ for all integers $n$. So the sine curve can't be given exactly by a Bézier curve of any degree.

The answer to the question posted is that given a high enough order of rational bezier you can approximate $\sin(x)$ to within a given error, but I think the question that the author meant to ask was more along the lines of "how can a parametric quadratic bezier exactly describe a conic section but I can't get one to give me an exact $\sin$ approximation?"

The keyword here is "parametric". You can easily generate a parametric rational quadratic curve which exactly passes through all the points on a quarter of a circle: $$ arc(t) = \left \{(1-t^2)/(1+t^2),\, 2\,t\,/(1+t^2)\right \},\, t \in \left [0, 1 \right ] $$ The problem is that the variable t does not have to transcribe the arc at a constant angular rate for it to still make an exact arc, so the following is unfortunately true: $$ arc(t) \neq \left \{ \cos(t\: \pi/2),\, \sin(t\: \pi/2) \right \},\, t \in \left [0, 1 \right ] $$ As an example $arc(1/2) = \left \{ 0.6,\, 0.8 \right \}$ and indeed $\sqrt{1-0.6^2}=0.8$ so the point is on the unit circle, but clearly $\left \{ 0.6,\, 0.8 \right \} \neq \left \{ 1/\sqrt{2},\, 1/\sqrt{2} \right \}$