Prove $1-\frac{1}{2} x^{2} \leq \cos x,\: \forall x \in \mathbb{R}$ Prove that

$$1-\frac{1}{2} x^{2} \leq \cos x,\: \forall x \in \mathbb{R}$$

I am trying to prove this using taylor's theorem:
We have:
$$f(x)=f(0)+f'(0)x+\frac{x^2}{2!}f''(0)+...$$
Now approximating the function as a polynomial of degree $3$ we get:
$$\cos x=1-\frac{x^2}{2!}+\frac{\sin c}{6}x^3$$
where $c \in (0, x)$
Case $1.$ if $0 < x \le \pi$, then we have $\sin c \geq 0$ and $x^3 >0$, so we get
$$\cos x \geq 1-\frac{x^2}{2}$$
Case $2.$ If $\pi < x \leq 2\pi$, i am confused, because $\sin c<0$
Any help?

My answer's approach resembles that of MotylaNogaTomkaMazura but since his $f'[x]$ is wrong and I can't comment on his answer, I hope my answer will give you a better idea.
Let $$f[x] = \cos[x] - 1 + \frac{1}{2}x^2$$ for $x \ge 0$
$$f'[x] = x - \sin[x]$$
We can prove $f'[x] \ge 0$ by taking its derivative:
$$f''[x] = 1 - \cos[x] \ge 0$$ since $\cos[x] \le 1$.
Since $f'[0] = 0$ and $f''[x] \ge 0, f'[x] \ge 0.$
Then again, $f[0] = 1 - 1 + 0 = 0$ and $f'[x] \ge 0$; therefore $f[x] \ge 0$ and thus $$\cos[x] \ge 1 - 1/2 x^2$$
Since $f[x]$ is an even function, $$\cos[x] \ge 1 - 1/2 x^2$$ for all $x$.
