# Prove that $1-\frac{x^2}{2}<\cos{x}<1-\frac{x^2}{2}+\frac{x^4}{24}$

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

• – Lord Shark the Unknown May 17 '18 at 5:20
• Duplicate: see math.stackexchange.com/questions/2783301/… – Kavi Rama Murthy May 17 '18 at 5:56
• I saw the post. However, I'm interested in knowing what i've done is right or wrong and doing the second inequality using just the MVT instead of employing the notion of limits, series or power series etc. Can you help? – A.Asad May 17 '18 at 6:13