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?

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 $\mathbb{R}^-$.
