Inequality involving Maclaurin series of $\sin x$ Question:
If $T_n$ is $\sin x$'s $n$th Maclaurin polynomial. 
Prove that $\forall 0<x<2, \forall m \in \Bbb N,~ T_{4m+3}(x)<\sin(x)<T_{4m+1}(x)$
Thoughts
I think I managed to prove the sides, proving that $T_3>T_1$ and adding $T_{4m}$ on both sides, but about the middle, I frankly have no clue...

Your thoughts are not quite right. Firstly, $T_{3} < T_{1}$ for $0 < x < 2$, and for that matter showing $T_{3} > T_{1}$ would be antithetical to what you are looking to prove. Secondly, truncated Taylor series are not additive in the way that you are claiming, i.e. it is NOT true that $T_{3} + T_{4m} = T_{4m + 3}$. 
However, consider the Maclaurin series for $\sin x$ and write out all sides of that inequality as follows: 
$$T_{4m+3} < \sin x < T_{4m+1}$$
is the same as writing:
$$x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdots + \frac{x^{4m+1}}{(4m+1)!} - \frac{x^{4m+3}}{(4m+3)!} < x - \frac{x^{3}}{3!} + \cdots < x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \cdots + \frac{x^{4m+1}}{(4m+1)!}$$
It is obvious from here that since $x> 0$, $T_{4m+3} < T_{4m+1}$. We must now complete the proof. All this amounts to proving is that for $0 < x < 2$, the following inequalities hold: 
$$\sum_{n = 2m+2}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} > 0$$
$$\sum_{n = 2m+1}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} < 0$$
Why this is what we need can be seen by looking at the terms of the Maclaurin series for $\sin x$ and comparing with each side of the inequality. I leave these details to you... feel free to comment if you need additional help. I imagine it well help to expand what these inequalities are out of series form.
