漂亮的恒等公式

王守恩

\(1,\sin(\arccos(\frac{\sin\theta}{2}))≡\frac{\sqrt{3(\sin\theta)^2+4(\cos\theta)^2}}{2}\)

\(2,\sin(\arccos(\frac{\cos\theta}{2}))≡\frac{\sqrt{4(\sin\theta)^2+3(\cos\theta)^2}}{2}\)

\(3,\cos(\arcsin(\frac{\sin\theta}{2}))≡\frac{\sqrt{3(\sin\theta)^2+4(\cos\theta)^2}}{2}\)

\(4,\cos(\arcsin(\frac{\cos\theta}{2}))≡\frac{\sqrt{4(\sin\theta)^2+3(\cos\theta)^2}}{2}\)

王守恩

\(\bigg(\sin\big(\arccos(\frac{\sin\theta}{3})\big)-\frac{\cos\theta}{3}\bigg)\bigg(\sin\big(\arccos(\frac{\sin\theta}{3})\big)+\frac{\cos\theta}{3}\bigg)≡\frac{8}{9}\)

王守恩

\(1,\bigg(\sin\big(\arccos(\frac{\sin\theta}{n})\big)\bigg)^2+\bigg(\frac{\sin\theta}{n})\bigg)^2≡1\)

\(2,\bigg(\cos\big(\arcsin(\frac{\cos\theta}{n})\big)\bigg)^2+\bigg(\frac{\cos\theta}{n})\bigg)^2≡1\)

王守恩

\(\frac{\cos\theta}{3}=S(1)\)

\(\arcsin(\frac{\cos\theta}{3})=S(2)\)

\(\arccos(\arcsin(\frac{\cos\theta}{3}))=S(3)\)

\(\arcsin(\arccos(\arcsin(\frac{\cos\theta}{3})))=S(4)\)

\(\arccos(\arcsin(\arccos(\arcsin(\frac{\cos\theta}{3}))))=S(5)\)

\(\cdots\cdots\)

\(问: S(2026)=?\)

王守恩

有40个自然数:1,2,3,…,39,40,至少从这40个数中

取出40个不同数,才能确保其中有四个不同数的和等于10-13,151-154,

取出39个不同数,才能确保其中有四个不同数的和等于14-17,147-150.

取出38个不同数,才能确保其中有四个不同数的和等于18-21,143-146.

取出37个不同数,才能确保其中有四个不同数的和等于22-25,139-142.

取出36个不同数,才能确保其中有四个不同数的和等于26-29,135-138.

取出35个不同数,才能确保其中有四个不同数的和等于30-33,131-134.

取出34个不同数,才能确保其中有四个不同数的和等于34-37,127-130.

取出33个不同数,才能确保其中有四个不同数的和等于38-41,123-126.

取出32个不同数,才能确保其中有四个不同数的和等于42-45,119-122.

取出31个不同数,才能确保其中有四个不同数的和等于46-49,115-118.

取出30个不同数,才能确保其中有四个不同数的和等于50-53,111-114.

取出29个不同数,才能确保其中有四个不同数的和等于54-57,107-110.

取出28个不同数,才能确保其中有四个不同数的和等于58-61,103-106.

取出27个不同数,才能确保其中有四个不同数的和等于62-65,99-102.

取出26个不同数,才能确保其中有四个不同数的和等于66-69,95-98.

取出25个不同数,才能确保其中有四个不同数的和等于70-73,91-94.

取出24个不同数,才能确保其中有四个不同数的和等于74-77,87-90.

取出23个不同数,才能确保其中有四个不同数的和等于78-81,83-86.

取出22个不同数,才能确保其中有四个不同数的和等于82.

王守恩

1=  1
1=s(2 - 1]
1=s(2-s(3 - 2]
1=s(2-s(3-s(2)s(2]
1=s(2-s(3-s(2)s(5 - 3]
1=s(2-s(3-s(2)s(5-s(3)s(3]
1=s(2-s(3-s(2)s(5-s(3)s(8 - 5]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(5]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13 - 8]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(8]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21 - 13]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(13]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34 - 21]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(21]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55 - 34]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(34]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89 - 55]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(55]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(144 - 89]
1=s(2-s(3-s(2)s(5-s(3)s(8-s(5)s(13-s(8)s(21-s(13)s(34-s(21)s(55-s(34)s(89-s(55)s(144-s(89)s(89]

s=sqrt，]=多重()

王守恩

\(\displaystyle\lim_{n \to \infty}\bigg(1 - \sum_{k=1}^n\frac{n^{a - 1}}{n^a + k^b}\bigg)*n^{a - b}=\frac{1}{b+1}\)     a > b > 0。——凭的是"手感"！——软件出来是错误的！

Table[Limit[(1 - Sum[n^(a - 1)/(n^a + k^b)n^(a - b), {k, n}]), n -> \[Infinity]], {b, 8}, {a, b + 1, 9}]——软件出来是错误的！

王守恩

\(\displaystyle\lim_{n \to \infty}\bigg(\tan^{-1}A - \sum_{k=1}^{A*n}\frac{n}{n^2 + k^2}\bigg)*n=\frac{A^2}{2A^2+2}\)     A > 0。

王守恩

\(\displaystyle B=1, 2, 3, 4, 5, 6,\cdots\ \lim_{n \to \infty}\sum_{k=1}^{B*n}\frac{n}{n^2 + k}=1, 2, 3, 4, 5, 6, 7, 8, 9, \cdots\cdots\)