<s>[INST] Proving $\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$ I wish to prove the following inequality for $x\ne 0$:
$$\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$$
Using the fact that I already prove:
$$\cos x > 1 - \frac{x^2}{2}$$
My try:
$\cos x = 1 - \frac{x^2}{2} +\frac{x^4}{24} + R_2(x)$ where $R_2(x)$ is the remainder with the order of $n=2$.
By Lagrange's form of the remainder:  
$$R_2(x) = \frac{cos^{(3)}(c)}{(3)!}x^3 = \frac{\sin c}{6}x^3$$
WLOG we assume $x>0$ and $0<c<x$
Now, what's left to prove is that: $$R_2(x) < \frac{x^4}{24}$$
But, I got the strange result: $x<4$.  
What did I do wrong / what should I do instead? 
I wish to rely on this inequality: $\cos x > 1 - \frac{x^2}{2}$
 [/INST] By the symmetry of $\cos$ about $x=0$, we only need to prove it for positive $x$.  Since $\cos t\gt 1-\frac{t^2}{2!}$, we have for positive $x$ that 
$$\sin x=\int_0^x \cos t\,dt\gt \int_0^x \left(1-\frac{t^2}{2!}\right)\,dt=x-\frac{x^3}{3!}.$$
Do it again. We have 
$$1-\cos x=\int_0^x \sin t\,dt\gt \int_0^x \left(t-\frac{t^3}{3!}\right)\,dt,$$
which gives us what we want. And if we want to continue, there is nothing to stop us. 
Remark: One can do it with differentiation.  We first prove that for positive $x$ we have $\sin x\gt x-\frac{x^3}{3!}$. 
Let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$. We have $f(0)=0$. To show that $f(x)\gt 0$ for $x\gt 0$, it is enough to show that $f(x)$ is increasing.
Note that $f'(x)=\cos x-\left(1-\frac{x^2}{2!}\right)$. By the result you want to use, we have $f'(x)\gt 0$ if $x\gt 0$. It follows that $f(x)$ is increasing. 
For your question about $1-\frac{x^2}{2!}+\frac{x^4}{4!}$, do it again. Let $g(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cos x$. We have $g(0)=0$, and want to prove $g(x)$ is increasing for $x\gt 0$. 
For this, we will use the fact that $g'(x)\gt 0$. Note that $g'(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and we have just shown that this is positive. 
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