TITLE: Proving of $\cos (\frac{2}{3})>\frac{\pi }{4}$
QUESTION [10 upvotes]: How can I prove that $$\cos \left(\frac{2}{3}\right)>\frac{\pi }{4}$$

REPLY [7 votes]: When $0< x\leq1$ then
$$\cos x>1-{x^2\over2}+{x^4\over24}-{x^6\over 720}\ .$$
Putting $x:={2\over3}$ gives
$$\cos{2\over3}>{25781\over 32805}>{11\over14}\ .\tag{1}$$
When $0<y<1$ then
$$\sin y>y-{y^3\over6}+{y^5\over120}-{y^7\over5040}\ .$$
Putting $y:={11\over21}$ gives
$$\sin{11\over21}>{4540399710451\over9077486246640} \ >{1\over2}\ .$$
From $(1)$ we therefore obtain
$$\sin\left({2\over3}\cos{2\over3}\right)>\sin\left({2\over3}\cdot{11\over14}\right)=\sin{11\over21}>{1\over2}\ .$$
This implies ${2\over3}\cos{2\over3}>{\pi\over6}$, which is the same as $\cos{2\over3}>{\pi\over4}$.