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 ?

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).
