Where am I violating the rules?
A few approximations
When making approximations, there is no legal or illegal. There are things that work better and things that don't. When making approximations that are supposed to work over a large range of values, often the plain Taylor series is not the best way to go. Instead, a polynomial or rational function that matches the function at a number of points is better. $$ \frac{\pi(\pi-x)x}{\pi^2-\left(4-\pi\right)(\pi-x)x}\tag{1} $$ matches the values and slopes of $\sin(x)$ at $0$, $\frac\pi2$, and $\pi$. However, it is always low.
If instead, we match the values at $0$, $\frac\pi6$,$\frac\pi2$, $\frac{5\pi}6$, and $\pi$ we get Mahabhaskariya's approximation $$ \frac{16(\pi-x)x}{5\pi^2-4(\pi-x)x}\tag{2} $$ which is both high and low, and the maximal error is about $\frac13$ of the one-sided error.
A good quadratic polynomial approximation also matches the values at $0$, $\frac\pi6$,$\frac\pi2$, $\frac{5\pi}6$, and $\pi$
$$
\frac{31}{10\pi^2}(\pi-x)x+\frac{18}{5\pi^4}(\pi-x)^2x^2\tag{3}
$$
The maximal error is about $\frac23$ that of Mahabhaskariya's.
If we want to extend to a cubic polynomial, we can try to match values at $0$, $\frac\pi6$, $\frac\pi4$, $\frac\pi2$
$$
\tfrac{9711-6400\sqrt2}{210\pi^2}(\pi-x)x+\tfrac{-7194+5120\sqrt2}{15\pi^4}(\pi-x)^2x^2+\tfrac{43488-30720\sqrt2}{35\pi^6}(\pi-x)^3x^3\tag{4}
$$
The maximum error of approximation $(4)$ is about $\frac1{40}$ that of approximation $(3)$.
Analysis of the functions in the question
The function $$ \frac{\pi(\pi-x)x}{\pi^2-(\pi-x)x}\tag{5} $$ has a maximum error about $40\times$ as big as $(3)$
The function $$ \frac{(\pi-x)x}\pi+\frac{(\pi-x)^2x^2}{\pi^3}+\left(\frac2{\pi^5}-\frac1{6\pi^3}\right)(\pi-x)^3x^3\tag{6} $$ has $30\times$ the maximum error of $(4)$. However, the coefficients of $(6)$ are more appealing.
@Claude Leibivici use the following two point Taylor series in x=-Pi, Pi $$\frac{z (z-\pi )^3 (z+\pi )^3}{48 \pi ^4}-\frac{5 z (z-\pi )^3 (z+\pi )^3}{16 \pi ^6}+\frac{3 z (z-\pi )^2 (z+\pi )^2}{8 \pi ^4}-\frac{z (z-\pi ) (z+\pi )}{2 \pi ^2}$$ the cuadratic error is superior to any formula above at the same grade