<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] 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.
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