Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove 

$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} +  \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$

I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$. 
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
 A: For any $x>0$ we have $\sin(x)<x$, hence by applying $\int_{0}^{t}(\ldots)\,dx$ to both sides we get $1-\cos t < \frac{t^2}{2}$. By applying $\int_{0}^{x}(\ldots)\,dt$ to both sides we get $x-\sin x<\frac{x^3}{6}$, which can be rearranged as $\sin(x)>x-\frac{x^3}{6}$. By performing the same trick again we also get $\sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}$ and the wanted inequality is proved for any $x>0$.
An equivalent approach is noticing that
$$ \iint_{0\leq a \leq b \leq x}(a-\sin a)\,da\,db,\qquad \iiint_{0\leq a\leq b\leq c\leq x}(a-\sin a)\,da\,db\,dc $$
are clearly positive.
We are dealing with odd functions, hence the reversed inequality holds over $\mathbb{R}^-$.
A: hint
Prove that
$$1-\frac {x^2}{2}\le \cos (x)\le 1-\frac {x^2}{2}+\frac {x^4}{24} $$
and integrate.
to prove this, it is easier to show that
$$-x\le -\sin (x)\le -x+\frac {x^3}{6} $$
or the easiest
$$0\le 1-\cos (x)\le \frac {x^2}{2} $$
then integrate twice between $0$ and $x $.

Taylor expansion give only local information.

A: Define
$$f(x) = \sin x - x + \frac{1}{6}x^3$$
and
$$g(x) = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \sin x$$
We want to show $f(x) >0$ and $g(x) > 0$ for $x > 0$.
Notice that $f$ and its first four derivatives vanish at $x=0$.  If $f$ has a positive zero, say $f(a_0) = 0$, then by Rolle's Theorem $f'$ has a zero $a_1$ with $0 < a_1 < a_0$.  Repeating this argument, we have $0 < a_4 < a_3 < a_2 < a_1 < a_0$ with $f^{(n)}(a_n) = 0$.  Since $f^{(4)}(x) = \sin x$, we must have $\pi \le a_4$, so $\pi < a_0$.  This shows $f(x)$ does not change sign for $0<x<\pi$.  Since $-x + (1/6) x^3 > 3/2$ and is increasing for $x > 3$ while $\sin x \ge -1$, we see that $f(x) > 0$ for all $x > 0$.
To see that $g(x) > 0$ for $x>0$, observe that $g(0) = g'(0) = 0$, $g''(x) = f(x)$, and $f(x) > 0$ for $x>0$, as we just showed.
A: Hint: The Maclaurin expansion of $\sin$ is alternating, and for "small enough" values of $x$, the absolute value of the terms are monotonically decreasing. What can you say in general about bounding an alternating series who's terms monotonically decrease?
For "large enough" values of $x$, it is good enough to notice the lower and upper bounds are not in $[-1,1]$ and to then observe that the lower bound is monotonically decreasing while the upper bound is monotonically increasing.
A: All of them fit in this infinite sequence of inequalities:
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots \ge 0 \ \forall x \in \mathbb{R}  \\
\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} -\cdots > 0 \ \forall x>0 \\
\frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots > 0  \ \forall x \in \mathbb{R^*}\\
\cdots \cdots \\
$$
It's enough to show the first sum is $\ge 0$ and $>0$ for $0<|x|< \epsilon$, and notice that the derivative of any of those sums is the previous sum. The first inequality follows from the identity of power series 
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots= 2 \left( \sum_{k \ge 0}\,(-1)^k\frac{x^{2k+1}}{2^{2k+1}(2k+1)!}\right)^2$$ which one can check directly. 
