How can I prove that $$\cos \left(\frac{2}{3}\right)>\frac{\pi }{4}$$
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4$|\cos \left(\frac{2}{3}\right)-\frac{\pi }{4}| < 0.0005$, wow! – Jihad Jan 07 '15 at 11:54
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6The solution that @1234 had posted (and AlexSilva incorrectly criticized) was indeed a proof. The Taylor of $\cos(x)$ is alternating, so it is enough to add terms until they are smaller than the precision needed, which is as Jihad is saying, the 4-th digit. The terms needed are not many, 6 are more than enough. – Pp.. Jan 07 '15 at 11:59
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1For the purposes of the proof, can we assume that the decimal expansion of $\pi$ is known to a reasonable accuracy? Because if it isn't, computing $\frac{\pi}{4}$ to $4$ significant digits is likely part of the challenge. – Jan 07 '15 at 12:15
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1@Pp.., I didn't criticize the Taylor approach posted by $@$1234, but the way he has computed the value of the expansion of $\text{cos}(2/3)$. Contrary to you have been saying here, he didn't take some terms of the expansion for solving the problem. Actually, he just substituted the value of the serie by the value of $\text{cos}(2/3)$. Did you really think that was a proof? – Alex Silva Jan 07 '15 at 13:22
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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 $\cos{2\over3}>{\pi\over4}$.
Christian Blatter
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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 $$\cos{\dfrac{2}{3}}>\dfrac{\pi}{4}$$
math110
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Carlson inequality proof: you can google :Carlson inequality with arccosx – math110 Jan 07 '15 at 12:08
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2How many digits of $\pi$ does one need to know in order for this solution to work? – Jan 07 '15 at 12:08
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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 $\pi = [3; 7, \color{red}{15}, 1, 292,\ldots ]$.
Jack D'Aurizio
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I think so,$\cos{(4x)}=T_{4}(\cos{x})$,but why $T_{4}(\cos{(1/6)})>T_{4}(1-1/72)?$ – math110 Jan 07 '15 at 12:54
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@math110: because in a neighbourhood of the origin $\cos z\geq 1-\frac{z^2}{2}$. – Jack D'Aurizio Jan 07 '15 at 13:05
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@math110: nope, but is it increasing in a neighbourhood of $1$ since we know its roots and the roots of the derivative. – Jack D'Aurizio Jan 07 '15 at 13:30
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3Good, though still not free from the critique of numerical approximation. Almost like saying $$\cos \frac23 > \frac \pi 4 \iff [0; 1, 3, 1, 2, \dots] > [0; 1, 3, 1, 1, \dots]$$ – Macavity Jan 07 '15 at 13:34