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 ?
 A: We'll prove that your inequality is true for all $0<x<\frac{\pi}{2}$.
Indeed, $$f'(x)=\cos{x}-1+\frac{x^2}{2};$$
$$f''(x)=x-\sin{x}$$ and
$$f'''(x)=1-\cos{x}>0.$$
Thus, $$f''(x)>f''(0)=0,$$
which gives
$$f'(x)>f'(0)=0$$ and
$$f(x)>f(0)=0.$$
Now, $$g'(x)=\cos{x}-1+\frac{x^2}{2}-\frac{x^4}{24};$$
$$g''(x)=-\sin{x}+x-\frac{x^3}{3}=-f(x)<0.$$
Thus, $$g'(x)<g'(0)=0,$$
which gives
$$g(x)<g(0)=0$$ and we are done!
By the way, for $x<0$ your inequality is wrong. 
A: 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 $(0,\varepsilon)$.
A: From $$3^k\,\sin\frac{x}{3^k}-3^{k-1}\,\sin\frac{x}{3^{k-1}}=4\cdot3^{k-1}\,\sin^3\frac{x}{3^k},$$ by summing the telescoping series,
$$x-\sin x=4\,\sum^\infty_{k=1}3^{k-1}\,\sin^3\frac{x}{3^k}.$$ Since $\sin x<x$ for $x>0$, the RHS is $\le x^3/6$.
BTW, the inequality for $\sin x$ can be shown in a similar way, because
$$\frac{\sin x}{x}=\prod^\infty_{k=1}\cos\frac{x}{2^k}<1$$ for $x\neq0$.
A: Well, $f(0)=0$ and
$$f'(x)=\cos x-1+\frac{x^2}2.$$
If we could show that $f'(x)>0$ when $x>0$, then it would follow
that $f'(x)>0$ when $x>0$.
A: Taking my answer in
Is this really equal to sin x?
one more step:
If you start with
$\sin'(x) = \cos(x),
\cos'(x) = -\sin(x),
\sin(0) = 0,
\cos(0) = 1,
\sin^2(x)+\cos^2(x) = 1$,
you can proceed like this
(not original with me):
$$\sin(x)
=\int_0^x \cos(t)dt
\le\int_0^x dt
=x
$$
$$\cos(x)-\cos(0)
=\int_0^x -\sin(t) dt
=-\int_0^x \sin(t) dt
\ge-\int_0^x t dt
=-\frac{x^2}{2}\\
\text{ so }
\cos(x)
\ge 1-\frac{x^2}{2}
$$
$$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$$
$$\cos(x)-\cos(0)
=\int_0^x -\sin(t) dt
=-\int_0^x \sin(t) dt
\ge-\int_0^x (t-\frac{t^3}{6}) dt
=-\frac{x^2}{2}+\frac{x^4}{24}\\
\text{ so }
\cos(x)
\le 1-\frac{x^2}{2}+\frac{x^4}{24}
$$
$$\sin(x)
=\int_0^x \cos(t)dt
\le\int_0^x (1-\frac{x^2}{2}+\frac{x^4}{24})dt
=x-\frac{x^3}{6}+\frac{x^5}{120}
$$
By induction you can
derive the power series
for sin and cos
and show that they are
enveloping
(the sum is between any two
consecutive sums).
