Proving of $\cos (\frac{2}{3})>\frac{\pi }{4}$

Solution 1:

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 $\cos{2\over3}>{\pi\over4}$.

Solution 2:

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 $$\cos{\dfrac{2}{3}}>\dfrac{\pi}{4}$$

Solution 3:

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 $\pi = [3; 7, \color{red}{15}, 1, 292,\ldots ]$.