Proving of $\cos (\frac{2}{3})>\frac{\pi }{4}$ How can I prove that $$\cos \left(\frac{2}{3}\right)>\frac{\pi }{4}$$
 A: 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 $\cos{2\over3}>{\pi\over4}$.
A: Use Carlson inequality
$$\arccos{x}>\dfrac{6\sqrt{1-x}}{2\sqrt{2}+\sqrt{1+x}},0\le x<1$$
then let
$x=\frac{\pi}{4}$,then we have
$$\arccos{\dfrac{\pi}{4}}>\dfrac{6\sqrt{1-\dfrac{\pi}{4}}}{2\sqrt{2}+\sqrt{1+\frac{\pi}{4}}}\approx 0.66741060\cdots>\dfrac{2}{3}$$
see wolf
so
$$\cos{\dfrac{2}{3}}>\dfrac{\pi}{4}$$
A: The inequality is equivalent to:
$$ T_4\left(\cos\frac{1}{6}\right) > \frac{\pi}{4},\tag{1} $$
but since, using the Taylor series of the cosine function in a neighbourhood of zero:
$$ T_4\left(\cos\frac{1}{6}\right) > T_4\left(1-\frac{1}{72}\right),\tag{2} $$
it is sufficient to show that:
$$ 1-8\left(\frac{71}{72}\right)^2+8\left(\frac{71}{72}\right)^4 > \frac{\pi}{4}, \tag{3} $$
or:
$$ \pi < \frac{2638369}{839808} = [3; 7, 16, 1, 1, 6, 2, 2, 12, 1, 1, 1, 2]\tag{4}$$
that is true since $\pi =  [3; 7, \color{red}{15}, 1, 292,\ldots ]$.
