I have to prove that $$x-\frac{x^{3}}{6} < \sin(x) \quad\text{ for }\quad 0<x<\pi $$
I tried to define $f(x) = \sin(x) - (x-\frac{x^{3}}{6})$, and to differentiate it, but $f'(x) = \cos(x) -1 +\frac{1}{2}x^{2}$. I have no idea how to continue.
Thanks.
In fact $x-x^3/6< \sin x$ for all $x > 0.$ Proof: As you noted, we can define $f(x) = \sin x -x + x^3/6.$ Is $f(x)>0$ for $x > 0?$ Because $f(0)=0,$ this is true if $f'(x) = \cos x -1 + x^2/2 > 0$ for $x>0.$ Is this true? Because $f'(0)=0,$ this will hold if $f''(x) = -\sin x + x > 0$ for $x> 0.$ Well, hopefully the last inequality looks familiar.
Consider that $$ \sin(x)=\sum_{n\ge 0} (-1)^n\frac{x^{2n+1}}{(2n+1)!}. $$ Moreover, for all $x \in [0,\pi]$ and positive $n$ it holds $ (2n+3)(2n+2)\ge 20>\pi^2\ge x^2$, so that $$ \left|\frac{x^{2n+1}}{(2n+1)!}\right|> \left|\frac{x^{2n+3}}{(2n+3)!}\right|. $$ Therefore for each integer $m\ge 0$ and $x \in [0,\pi]$ we obtain $$ \sum_{n=0}^{2m+1} (-1)^n\frac{x^{2n+1}}{(2n+1)!} <\sin(x)<\sum_{n=0}^{2m} (-1)^n\frac{x^{2n+1}}{(2n+1)!}. $$ In particular, we have $$ x-\frac{x^3}{6}<\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}. $$
Using the approach you've started with, let $f(x)$ be the difference $$ f(x) = \sin(x) - (x-\frac{x^3}{6})$$ then we just want to show that this difference is positive for all values $0 < x < \pi$. This can be done by showing that $f(0) = 0$ and $f'(x)>0$ on the interval. $$ f'(x) = \cos(x) +\frac{x^2}{2} - 1 $$ The only critical point (where $f'(x)=0$) is at $x=0$, and so you just have to show that $f(x)>0$ for some value on the interval, i.e., $f(1)=.0403$.
Use the MacLaurin expansion with Lagrange type remainder term. $$\sin(x) = x - \frac{x^3}{6} + \frac{x^4}{4!}\sin(\xi)$$ where $0<\xi<x<\pi$. Now $\sin(\xi)$ is positive on this intervall showing that the remainder must be positive and proving the inequality.
Consider $$g(x)=\sin x-(x-\frac{x^3}{6})=\frac{x^3}{6}-x+\sin x$$. One has
$$g'(x)=\frac{x^2}{2}-1+\cos x$$ $$g''(x)=x-\sin x$$ $$g'''(x)=1+\cos x$$ It is clear that $g'''(x)\ge 0$ so $g''(x)$ is increasing so positive because $g''(0)=0$ which implies $g'(x)$ is increasing so positive since $g'(0)=0$. But this implies that $g(x)$ is increasing and positive because $g(0)=0$
Thus $$\sin x>x-\frac{x^3}{6}$$ (all on $0<x<\pi$)
