prove that $ x-\frac{1}{6}x^3<\sin(x)
prove that : 
  there exists a deleted neighborhood of $x=0$
  such that :
  $$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5
$$
MyTry:
let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$
And :
$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$
Now what ?

You have\begin{align*}\sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-\frac{x^9}{9!}+\frac{x^{11}}{11!}-\cdots\\&=x-\frac{x^3}6+\frac{x^5}{5!}\left(1-\frac{x^2}{6\times7}\right)+\frac{x^9}{9!}\left(1-\frac{x^2}{10\times11}\right)+\cdots\\&>x-\frac{x^3}6\end{align*}if $0<x<\sqrt{42}$. A similar argument proves that $\displaystyle\sin x<x-\frac{x^3}6+\frac{x^5}{120}$ in some interval $(0,\varepsilon)$.
