<s>[INST] Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} =  0 \\ $
It seems that x = 0 is a critical point. 
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' =  -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
 [/INST] Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again over $\color{blue}{(0,x)}$ we get
$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz< \int_0^x\cos z dz=\sin x}$$
that is
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x}$$

continuing with this process you get,
$$\color{blue}{1-\frac{x^2}{2}< \cos x< 1-\frac{x^2}{2}+\frac{x^4}{24} }$$
$$\color{blue}{x-\frac{x^3}{6} <\sin x< x-\frac{x^3}{6} +\frac{x^5}{5!}}$$

More generally for  $n\geq1$, by induction we get
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}<\cos x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k}}{(2k)!}+\frac{x^{4n}}{(4n)!} }$$
$$\color{blue}{\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!} <\sin x<\sum_{k=0}^{2n-1}(-1)^k\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{4n+1}}{(4n+1)!}}$$
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