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?

A standard approach is to let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and to show that $f(x)\gt 0$ if $x\gt 0$. 
Note that $f(0)=0$. We will be finished if we can show that $f(x)$ is increasing in the interval $(0,\infty)$.
Note that $f'(x)=\cos x-1+\frac{x^2}{2!}$. We will be finished if we can show that $f'(x)\gt 0$ in the interval $(0,\infty)$.
Note that $f'(0)=0$. We will be finished if we can show that $f'(x)$ is increasing in $(0,\infty)$.
So we will be finished if we can prove that $f''(x)\gt 0$ in the interval $(0,\infty)$.
We have $f''(x)=-\sin x+x$. Since $f''(0)=0$, we will be finished if we can show that $f'''(x)\ge 0$ on $(0,\infty)$, with equality only at isolated points. This is true. 
Or else for the last step  we can use the geometrically evident fact that $\frac{\sin x}{x}\lt 1$ if $x\gt 0$. 
Remark: It is more attractive to integrate than to differentiate, but we used the above approach because differentiation comes before integration in most calculus courses. 
For the integration approach, let $x$ be positive. Since $\sin t\lt t$ on $(0,x)$, we have $\int_0^x (t-\sin t)\,dt\gt 0$. Integrate. We get $\cos x+\frac{x^2}{2}-1\gt 0$ (Mean Value Theorem for integrals), so $\cos t+\frac{t^2}{2}-1\gt 0$ if $t\gt 0$. 
Integrate from $0$ to $x$. We get $\sin x+\frac{x^3}{3!}-x\gt 0$, or equivalently $\sin x\gt x-\frac{x^3}{3!}$. Nicer, by a lot.
