x = Analytic form of cos x

I've been trying to write an article for a long time and talk to you:

\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{π^2}{6}But there has been no chance, this summer vacation just have free, so the hand itch, write this article, for everyone Entertainment (yǎ) (Shǎng).
This article assumes that the reader loves mathematics and has mastered the high mathematics knowledge.

1. First we need to review the trigonometric functions.

For any corner X, we have {\sin^2 x}+\cos^2x=1, which is one thing with the Pythagorean theorem.

Next is an important formula that is recommended for readers to understand through drawing

\sin (2x) = 2 \sin x \cos x

Then, by drawing a trigonometric image, we can easily verify the following two formulas

\begin{align*}\cos X & = \sin (x+\frac{π}{2}) \ \sin x & =\sin (π-x) \end{align*}

2. Now we can begin to prove it.

(This certificate was taken from the February 2002 109th issue of the American Journal of Mathematics, pp. 196-200 author Josef Hofbauer. )

2.1

Since \sin (2x) = 2 \sin x \cos x,

\sin x= 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})Take the countdown, square, get\frac{1}{\sin^2 X} = \frac{1}{4}\frac{1}{\sin^2 (X/2) \cos^2 (X/2)}And then according to{\sin^2 x}+\cos^2x=1, we have\frac{1}{\sin^2 X} = \frac{1}{4}\frac{\sin^2 (X/2) +\cos^2 (X/2)}{\sin^2 (X/2) \cos^2 (X/2)}=\frac{1}{4} (\frac{1}{\ Cos^2 (X/2)} + \frac{1}{\sin^2 (X/2)}) Next Use properties,\cos (X/2) = \sin ((x+π)/2), available in relational
\ \ \ \ \ \ \frac{1}{\sin^2 x} = \frac{1}{4} (\frac{1}{\sin^2\frac{x}{2}} + \frac{1}{\sin^2\frac{x+π}{2}}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)This is one of the most important steps in proving that we call this relationship "(*)".

By definition

\sin (Π/2) =\, \sin90°=1then squared, take the reciprocal, and reuse (*) The formula, we have\begin{align*} 1 & = \frac{1}{\sin^2 (Π/2)} \ \ & =\frac{1}{4} (\frac{1}{\sin^2 (Π/4)} + \frac{1}{\sin^2 (3Π/4)} \ \ & =\frac{1}{4^2} (\frac{1}{\sin^2 (Π/8)} + \frac{1}{\sin^2 (3Π/8)}+\frac{1}{\sin^2 (5Π/8)} + \frac{1}{\sin^2 (7π/ 8)}) \ \ & = \dots \end{align*}can do so all the time.

Now, using Identity \sin (π-x) =\sin x, you get

\begin{align*} 1 & =\frac{2}{4^2} (\frac{1}{\sin^2 (Π/8)} + \frac{1}{\sin^2 (3Π/8)}) \ \ & =\frac{2}{4^3} (\ Frac{1}{\sin^2 (Π/16)} + \frac{1}{\sin^2 (3Π/16)}+\frac{1}{\sin^2 (5Π/16)} + \frac{1}{\sin^2 (7Π/16)}) \ \ & ={\frac{2 }{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5Π/32)} +\dots + \frac{1}{\sin^2 (15Π/32)})} \ \ & = \dots \end{align*}we call this relationship "(* *)" type

2.2

Some readers may ask why, as was done just now, the reason is immediately clear, the purpose is only one: so that all \sin () values are acute.

Because for acute angle x, we have \sin x < x < \tan x

Take the countdown, square, get

\frac{1}{\sin^2 X} > \frac{1}{x^2} > \frac{1}{\tan^2 x} and we know
\frac{1}{\tan^2 x}=\frac{\cos^2 x}{\sin^2 x}=\frac{1-\sin^2 x}{\sin^2 x}=\frac{1}{\sin^2 x}-1 So
\begin{align}\frac{1}{\sin^2 x}-1<\frac{1}{x^2}<\frac{1}{\sin^2 X}\end{align} Now combine the previously deduced relationship (* *):
\begin{align}1= \frac{2}{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5Π/32)} +\dots + \frac{1}{\sin^2 (15Π/32)}) \end{align} We can get the following unequal relationship
\begin{align} 1-\frac{2}{4^4}*2^3 & <{\frac{2}{4^4} (\frac{1}{(Π/32) ^2} + \frac{1}{(3Π/32) ^2}+\frac{1}{( 5Π/32) ^2} +\dots + \frac{1}{(15Π/32) ^2})} < 1\\ 1-\frac{2}{4^4}*2^3 & < {\frac{2}{4^4}* 4^5 (\frac{1}{π^2} + \FR ac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\dots + \frac{1}{(15π) ^2})} <1\\ 1-\frac{1}{2^4}& < {\ 8\, (\frac{1}{π^2} + \ frac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\dots + \frac{1}{(15π) ^2})} <1 \end{align}

(Please note that these three unequal relations (3) (4) (5) may take time to read carefully. In particular (3), is the most difficult to understand the full text of a step, I hope readers can read patiently: How can be derived from the previous formula (1) (2) Out of (3)? ）

2.3

By observing, we can find that before in (* *), we only use the

1 = {\frac{2}{4^4} (\frac{1}{\sin^2 (Π/32)} + \frac{1}{\sin^2 (3Π/32)}+\frac{1}{\sin^2 (5Π/32)} +\dots + \frac{1}{\ Sin^2 (15Π/32)})}if before, make more use of (*) several times, so that the number of items \sin in (* *) is increased from 8=2^3 to 2^n

Then there are

1-\frac{1}{2^{n+1}} < {8 (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2} +\cdots + \frac{1}{((2^{n+1}-1) π) ^2 })}<1when n is large,\frac{1}{2^{n+1}} can be ignored, so we have {\,8\, (\frac{1}{π^2} + \frac{1}{(3π) ^2}+\frac{1}{(5π) ^2}+\frac{1}{(7π) ^2} +\cdots)}= 1that\frac{1}{1^2}+\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +\cdots =\frac{π^2}{8}

2.4

Now we are only one step away from the conclusion,

Make

\zeta (2) = \frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\so\frac{\zeta (2)}{4} = \frac{1}{2^2}+\frac{1}{4^2} +\frac{1}{6^2} +\frac{1}{8^2} +\frac{1}{10^2} +\cdotssubtract two, you can get\zeta (2)-\frac{\zeta (2)}{4} = \frac{1}{1^2}+\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +\cdots =\frac{\pi^2}{8} \ \ \ \ \ \ \ \ \ \ \ \so 3\zeta (2)/4 =Π^2/8, seek \zeta (2) =Π^2/6, that is, we want to testify the conclusion:

\frac{1}{1^2}+\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} +\frac{1}{5^2} +\cdots\ =\ \frac{π^2}{6}

Proof Complete!

How about, fun, math is always like this, with the most ingenious logic chain to construct the most beautiful proof.
As long as there is a little curiosity, and enough patience, everyone can enjoy the fun of maths.
I wish you a pleasant summer vacation.

Jabo Name
June 20, 2014 at Columbus, Ohio, USA
(Last updated on June 19, 2016, thank Sun Ho again for reading the first draft of this article, and pointed out a number of clerical errors, has been corrected.) )

x = Analytic form of cos x