Prove that $1-\frac{x^2}{2}<\cos{x}<1-\frac{x^2}{2}+\frac{x^4}{24}$ and $x \in (0,\pi/2]$
Proof:
for $1-\frac{x^2}{2}<\cos{x}$
$\implies 1<\frac{x^2}{2}+\cos{x}$
Then, by mean value theorem
$\cos{x} +x^2/2 - 1 = (-\sin{c} +c)(x)$
$\implies \cos{x} -x^2/2 - 1 >0$ ( I put the lower bound on this)
and hence the inequality follows.
for $\cos{x}<1-\frac{x^2}{2}+\frac{x^4}{24}$
$\implies \cos{x}+\frac{x^2}{2} -\frac{x^4}{24}<1$
Again, by MVT, we have that:
$\cos{x}+\frac{x^2}{2} -\frac{x^4}{24} -1 = (-\sin{c} +c -\frac{c^3}{6})(x)$
I don't know how to proceed any further. Can anyone please help and also verify the first part of the inequality?
Thank you