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?
 A: 
I am trying to prove this using Taylor's theorem ....

That is possible, just use the remainder for the term with exponent $2$, i.e.
$$
f(x)=f(0)+f'(0)x+f''(c) \frac{x^2}{2!}
$$
for some $c$ between $0$ and $x$. This gives
$$
 \cos(x) = 1 + 0 \cdot x +  \underbrace{(-\cos(c))}_{\ge -1} \cdot \frac{x^2}{2!}\ge 1 - \frac{x^2}{2} 
$$
for all $x \in \Bbb R$.
A: Consider a function $$f(x) =\cos x -1  + 2^{-1} x^2 $$ for $x\geq 0.$
Since $f'(x) =\sin x +x \geq 0$  thus the function is monotone increasing and therefore  $$f(x) \geq f(0)=0$$  for $x\geq 0,$
but $f(-x) =f(x) $ and hence $$f(x) \geq f(0)=0$$ for all $x\in \mathbb{R}.$
A: If $x=0$, then the inequality is correct.
Let $x≠0$, then we have
$$\begin{align}&1-\frac{1}{2} x^{2}≤ \cos x\\
\iff &\frac{1-\cos x}{2}≤\frac 14 x^2\\
\iff
&\sin^2\left(\frac x2\right)≤\frac {x^2}{4} \\\iff 
&\left|\sin\left(\frac x2\right)\right|≤\left|\frac x2\right|\\
\iff &|\sin t|≤|t|\\
\iff&\left|\frac{\sin t}{t}\right|≤1,\thinspace t≠0.\end{align}$$
This is the well-known fact that,
$$-1≤\cos x ≤ \frac{\sin x}{x} ≤ 1
$$
This completes the proof.
A: 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 $x$.
A: This proof uses that $|\sin y|\leq |y|.$
Consider the right triangle $A=(\cos x,0),$ $B=(\cos x,\sin x)$ and $C=(1,0).$
Then $$|BC|^2=(1-\cos x)^2+\sin^2 x=2-2\cos x.$$
But we can also show that $|BC|=2|\sin(x/2)|,$ by considering it as the base of an isosceles triangle, $OBC,$ with angle $x$ and equal sides $1.$
We also know that $|\sin y|\leq |y|.$
So $$2-2\cos x=4\sin^2\frac x2<x^2.$$
Rearranging gives your inequality.

This is almost entirely a geometric proof, except for the sine inequality we use.
We can show $|\sin y|\leq |y|$ geometrically, too, if we accept that the shortest path between two points is a line.
The line segment between $(\cos y,\sin y)$ and $(\cos y,-\sin y)$ is of length $2|\sin y|,$ while the path along the circle between the two points is length $2|y|.$ The circle path has to be at least as long as the shortest path, and we get our sine inequality.
