Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$ Using Taylor's theorem show that 
$1-\frac{x^2}{2} < \cos x < 1-\frac{x^2}{2} + \frac{x^4}{24} \, \text{ for all } x \in \mathbb{R}$
I know this is true because of the Alternating series truncation error that alternates in sign, but is there a way to prove it?
 A: As said in comments these are not strict inequalities, but I can show you how to obtain the second inequality :
With Taylor theorem (Lagrange version) we have :
$$\forall x \in \mathbb R, \exists c\in (0,1),  \cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}\cos''''(cx)=1-\frac{x^2}2+\frac{x^4c^4}{24}\cos(cx)$$
$$\cos(cx)\leq 1$$
$$\cos(x)\leq 1 -\frac{x^2}2+\frac{x^4c^4}{24}\leq 1 -\frac{x^2}2+\frac{x^4}{24}$$
You can apply the same kind of reasonning for the other inequality.
A: We keep integrating the inequality $\cos x\le x$, for $x\ge 0$, in the interval $[0,x]$, and we recursively obtain
$$
\cos x \le 1\quad\Longrightarrow\quad
\sin x \le x\quad\Longrightarrow\quad
1-\cos x \le \frac{x^2}{2!}\quad\Longrightarrow\quad
1- \frac{x^2}{2!}\le\cos x \quad\Longrightarrow\quad
x-\frac{x^3}{3!}\le \sin x\quad\Longrightarrow\quad \frac{x^2}{2!}-\frac{x^4}{4!}\le 1-\cos x\quad\Longrightarrow\quad
\cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}\quad\Longrightarrow\quad\cdots
$$
In general, continuing this we obtain
$$
\sum_{k=0}^{2n-1}\frac{(-1)^kx^{2k}}{(2k)!}\le\cos x\le \sum_{k=0}^{2n}\frac{(-1)^kx^{2k}}{(2k)!}
$$
Note that it holds, not only for $x\ge 0$, but for $x<0$, since all the functions above are even. 
