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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

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