Numerical analysis , using Taylor Theorem

Question

On interval [$-\pi ,\pi$ ] , show that $1-\frac{{x}^{2}}{2}\le \cos\left(x\right)$ by using Taylor theorem.

I thought that choose ${x}_{0}=0$ and $f(x)=\cos\left(x\right)$. Then,

$f\text{'}(x)=-\sin\left(x\right)$ , $f\text{'}\text{'}(x)=-cos\left(x\right)$ and $f\text{'}\text{'}\text{'}(x)=sin\left(x\right)$ then by Taylor theorem, expansion of function is ,

$\cos(x)=1-\frac{{x}^{2}}{2}+\sin\left(\xi (x)\right)\frac{{x}^{3}}{6}$

But then I stuck here. Any help will be appreciated.

Answer 1

Case 1: $x \in [0, \pi]$, then $ \xi(x) \in [0,\pi]$, hence $\sin\left(\xi (x)\right)\frac{{x}^{3}}{6} \ge 0.$ Thus $1-\frac{{x}^{2}}{2}\le \cos\left(x\right)$

Case 2: $ x \in [- \pi,0]$. Then $ -x \in [0,\pi]$. Case 1 now shows:

$1-\frac{{x}^{2}}{2}= 1-\frac{{(-x)}^{2}}{2} \le \cos(-x) = \cos(x).$