Prove by Taylor's series method that $|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$ for $x\in [-1,1]$ Prove by Taylor's series method that $$|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|<\frac{1}{7!}$$ for $x\in [-1,1]$
Attempt:
Let $f(x)=\sin{x}$, then $f^n(x)=\sin{(\frac{n\pi}{2}+x)}, ~x\in \mathbb{R}$
Then $$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\frac{x^4}{4!}f^{iv}(0)+\frac{x^5}{5!}f^{v}(0)+\frac{x^6}{6!}f^{vi}(0)+\frac{x^7}{7!}f^{vii}(\theta x)\\=x-x^3/3!+x^5/5!+\frac{x^7}{7!}f^{vii}(\theta x)$$
then $$|\sin{x}-(x-\frac{x^3}{3!}+\frac{x^5}{5!})|=\frac{1}{7!}|f^{vii}(\theta x)|$$

By the Taylor expansion with Lagrange remainder,
$$
\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\cos\xi
$$
for some $\xi\in(-1,1)$ (actually $\xi$ can be chosen between $0$ and $x$, but it's not fully relevant; the only fact we need is that $\xi\ne0$ and $\xi\in[-1,1]$). Therefore
$$
\left|\sin x-x+\frac{x^3}{3!}-\frac{x^5}{5!}\right|
=\left|\frac{x^7}{7!}\cos\xi\right|
$$
Now $|x^7\cos\xi|<1$, for $x\in[-1,1]$, because $\lvert\cos\xi\rvert<1$.
