$ \cos x\geq 1-\frac{x^2}{2} $ Prove that for $x\in\mathbb{R}$
$$
\cos x\geq 1-\frac{x^2}{2}.
$$
My try:
Consider $g(x)=\cos(x)-1+\frac{x^2}{2}.$
If I differentiate $g(x)$ then we get $g'(0)>0$ so locally we get $g(x)>g(0)=0$ and then we can see that the function is increasing for any $x$ the function is increasing and hence we have $g(x)\geq 0$ for any $x \geq 0$. But I am getting that if $x<0$ then $g(x) \leq 0.$ So this inequality isn't true in general for all $x \in \Bbb R$.
But, if we use Taylor's theorem with Lagrange's remainder then also I am not sure what will be the point $\zeta\in [-x,0]$ where $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4}\cos(\zeta).$

$$\cos x-1+\frac{x^2}2\ge0$$ and equality holds at $x=0$.
Then differentiating,
$$-\sin x+x\ge 0$$ and equality holds at $x=0$.
Finally,
$$-\cos x+1\ge 0.$$
So $-\sin x+x$ grows from $0$ and is non negative, and $\cos x-1+\dfrac{x^2}2$ grows from $0$ and is non-negative.

This technique works for Taylor developments to arbitrary orders.
