<s>[INST] 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?
 [/INST] 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.

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