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 trick is to get rid of the transcendental function.
As the sine is monotonic in $(0,\pi/2)$,
$$x>\arcsin\left(x-\frac{x^3}3\right),$$ and, as the two members coincide for $x=0$, by differentiation
$$1>\frac{1-x^2}{\sqrt{1-\left(x-\dfrac{x^3}3\right)^2}}.$$
Then it suffices that for $x<1$,
$$1-\left(x-\dfrac{x^3}3\right)^2>(1-x^2)^2,$$
$$-\frac{x^6}9-\frac{x^4}3+x^2>0,$$ which is true.
