<s>[INST] Help with proving the following inequality I have to prove that $0\leq \sin(x)-(x-x^3/3!+x^5/5!-x^7/7!)\leq x^9/9!$, $x>0$
With the help of Taylor's theorem I got the following inequality,
$-x^9/9!\leq \sin(x)-(x-x^3/3!+x^5/5!-x^7/7!)\leq x^9/9!$. How should I proceed further? Thanks in advance!
EDIT: Applying Taylor's theorem to the function $f(x)=\ sinx$ in $[0,x]$ with remainder after nine terms, I got,
$sinx=x-x^3/3!+x^5/5!-x^7/7!+\sin(\theta x)x^9/9!$
Now,$-1 \leq \sin(\theta x)x^9/9! \leq 1$
Putting the value of $\sin(\theta x)x^9/9!$ I got the previous inequality.
 [/INST] The lower bound is obtained by applying Taylor's theorem with the remainder after seven terms. The seventh derivative of $\sin(x)$ is $-\cos(x)$. Therefore, if $ x > 0$ then we have with some $\theta \in (0, 1)$,
$$
 \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}(-\cos(\theta x))
\ge x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}
$$
since $\cos(\theta x)) \le 1$.
For the upper bound it should be
$$
 \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}(\cos(\theta' x))
\le x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+ \frac{x^9}{9!}
$$
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