User:Case2009

iteration
$$a_{n+1}=\frac{a_{n}+\frac{\sqrt[\frac{k}{2}]b}{a_n}}{\sqrt[\frac{k}{2}]b}=\sqrt[k]b$$

Ramanujan' formula
$$\frac{2}{\pi}-\frac{5}{8}=\sum_{n=1}^{\infty}{2n-1 \choose n}^3\frac{42n+5}{2^{12n}}$$

Approximation

$$\frac{1}{2^{12}\left(\frac{2}{\pi}-\frac{5}{8}\right)-47}\approx{2^{\frac{3}{4}}}$$

Catalan's Number
$$C_n=\frac{1}{n+1}{2n\choose n}=\frac{2{2n-1 \choose n}}{n+1}$$

n ≥ 0

1, 1, 2, 5, 14, 42, 132, 429, 1430,...

Relation

$$C_0=1$$

$$C_1=1{C_0}^2$$

$$C_2=2C_1$$

$$C_3=2C_2+{C_1}^2$$

$$C_4=2C_3+2C_1C_2$$

$$C_5=2C_4+2C_1C_3+{C_2}^2$$

$$C_6=2C_5+2C_1C_4+2C_2C_3$$

$$C_n=2C_{n-1}+2C_{1}C_{n-2}+2C_{2}C_{n-3}+...$$

Even

$$C_0=1$$

$$C_2=2C_1$$

$$C_4=2C_3+2C_1C_2$$

$$C_6=2C_5+2C_1C_4+2C_2C_3$$

$$C_n=2C_{n-1}+2C_{1}C_{n-2}+2C_{2}C_{n-3}+...$$

$$2=\frac{C_{2n}}{\sum_{j=0}^{n-1}C_jC_{2n-1-j}}$$

Odd

$$C_1=1{C_0}^2$$

$$C_3=2C_2+{C_1}^2$$

$$C_5=2C_4+2C_1C_3+{C_2}^2$$

$$C_n=2C_{n-1}+2C_{1}C_{n-2}+2C_{2}C_{n-3}+...$$

$$\frac{C_{2n+1}-{C_n}^2}{2}=\sum_{j=0}^{n-1}C_jC_{2n-j}$$

Formulae

$$\frac{2n+1}=\sum_{j=0}^{n-1}C_jC_{2n-1-j}$$

$$\frac{{2n-1 \choose n}{4n+1 \choose 2n+1}-{2n-1 \choose n}^3}{(n+1)^2}= C_n\sum_{j=0}^{n-1}C_jC_{2n-j}$$

Or

$$\frac{{4n+1 \choose 2n+1}-{2n-1 \choose n}^2}{n+1}= 2\sum_{j=0}^{n-1}C_jC_{2n-j}$$

Binomial
$${n \choose 1}^2=\sum_{j=1}^{2n-1}(-1)^{1+j}{2+j \choose 3}$$

$$2{2n+2 \choose 4}+{2n+1 \choose 2}^2={2n+5 \choose 5}+2\sum_{j=1}^{2n}(-1)^{1+j} {4+j \choose 5}$$

$$\sum_{j=1}^{2n+1}(-1)^{1+j}{2+j \choose 3}={2n+4 \choose 4}+2\sum_{j=1}^{2n}(-1)^{1+j}{3+j \choose 4}$$

$$\frac{k}{k-1}{kn-1 \choose n}={kn \choose n}$$

$${2n \choose 2n-2}=n{2n-1 \choose 2n-2}$$

$${2n+2 \choose 2n-2}={2n-1 \choose 2n-2}\sum_{j=1}^{n}j^2$$

or

$$\frac= \sum_{j=1}^{n}\left[\frac\right]^2$$

$${2n \choose 2}=n\left[{2n \choose 1}-1\right]$$

$${2n+1 \choose 2}=n{2n+1 \choose n}$$

Patterns

$$2{n+0 \choose n}{n+1 \choose n}={2n+2 \choose 2n+1}$$

$$2{n+0 \choose n}{n+2 \choose n}+{n+1 \choose n}^2= {2n+3 \choose 2n+1}$$

$$2{n+0 \choose n}{n+3 \choose n}+2{n+1 \choose n}{n+2 \choose n} ={2n+4 \choose 2n+1}$$

...

General formulae

$$2\sum_{j=0}^{k-1}{n+j\choose n}{n+2k-1-j\choose n}={2n+2k\choose 2n+1}$$

$${n+k\choose n }^2+2\sum_{j=0}^{k-1}{n+j\choose n}{n+2k-j\choose n}={2n+2k+1\choose 2n+1}$$

$$\sum_{i=1}^{n}(2i-1)^2=n^2+2\sum_{j=1}^{n-1}j(2n-j)={2n+1 \choose 2n-2}$$

$$\sum_{i=1}^{n}(3i-2)^2=n^3+3\sum_{j=1}^{n-1}j(2n-j)$$

Approximation
$$\sinh\left(\frac{3}{5}\right)\approx{\frac{2}{\pi}}$$

$$\left(5^2+2^2\right)^2-\left(3^2+4^2\right)^2=6^3$$

Where a, b and c are arithmetic terms

$$c^2=4b^2-4ab+a^2$$

$$a^2=c^2-4bc+4b^2$$

$$16^0=16^1\sum_{n=1}^{\infty}\frac{n^3}{e^{n\pi}-1}-16^2\sum_{n=1}^{\infty} \frac{n^3}{e^{4n\pi}-1}$$

Arithmetic quadratic equation
$$ax^2+bx+c=0$$

$$x=\frac{-b\pm\sqrt{\left(3b\right)^2-4abc\left(\frac{1}{a}+\frac{1}{b}+ \frac{1}{c}\right)}}{2a}$$

$$b=\frac{a+c}{2}$$

The root will always be complex number.

Problem
$$2n^2-1=k^2$$

''Solution n = 5. Is there any more?''

Alteration of Cosine Rule
$$a^2=\left(b-c\right)^2+4bc\sin^2\left(\frac{A}{2}\right)$$

$$a^2=\left(b-c\right)^2+4(s-b)(s-c)$$

Arctan
Where $$A = \frac{(n-1)(2n-1)}{n}$$

$$\frac{1}{2}\tan^{-1}\left(\frac{2n-1}{2n^2-2n}\right)=\tan^{-1}\left[\frac{-(A+1)\pm\sqrt{A^2+6A+1}} {2A}\right]$$

$$\frac{1}{2n-1}=\frac{-\left(A+1\right)\pm\sqrt{A^2+6A+1}}{2A}$$

pi
$$\frac{1}{2}\left(1-\frac{\pi}{4}\right)= \sum_{n=1}^{\infty}\frac{n}{n^2-(n+1)^2+(2n+1)^2}$$

$$\frac{\pi}{4}=2\tan^{-1}\left(\frac{1}{3}\right)+ 2\sin^{-1}\left(\frac{1}{5\sqrt{2}}\right)-\tan^{-1}\left(\frac{1}{7}\right)$$

Sine and Cosine
General

$$c^2=a^2+b^2$$

$$\sin^{-1}\left(\frac{b}{c}\right)-\sin^{-1}\left(\frac{a}{c}\right)= 2\sin^{-1}\left(\frac{b-a}{c\sqrt{2}}\right)$$

$$\cos^{-1}\left(\frac{a}{c}\right)-\cos^{-1}\left(\frac{b}{c}\right)= 4\cos^{-1}\left(\frac{a+\sqrt{a+c}}{\sqrt{c}}\cdot{\frac{\sqrt{2-\sqrt{2}}}{2}}\right)$$

Exact trigonometry constant
$$\cos\left(\frac{15}{2}\right)=\frac{\sqrt{2-\sqrt{2}}}{2} \times{\frac{\sqrt{3}+\sqrt{2+\sqrt{3}}}{\sqrt{2}}}$$

$$\sin{15}=\frac{\sqrt{2-\sqrt{3}}}{2}$$

$$\sin9=\sqrt{\frac{1}{2}-\frac{\sqrt{5+2\sqrt{5}}}{2(1+\sqrt{5})}}$$

$$\sin27=\sqrt{\frac{1}{2}+\frac{\sqrt{5-2\sqrt{5}}}{2(1-\sqrt{5})}}$$

$$\sin63=\sqrt{\frac{1}{2}-\frac{\sqrt{5-2\sqrt{5}}}{2(1-\sqrt{5})}}$$

$$\sin^2{9}-\sin^2{27}=\frac{(1-\sqrt{5})\sqrt{5+2\sqrt{5}}+(1-\sqrt{5})\sqrt{5+2\sqrt{5}}} {8}$$

$$\sin^2{9}-\sin^2{27}=\frac{\left(\sqrt{5}-1\right)\tan{9}-\left(\sqrt{5}+1\right)\tan{27}}{8}$$

$$\sin^2{27}-\cos^2{27}=\frac{\sqrt{5-2\sqrt{5}}}{1-\sqrt5}$$

$$\tan27=\sqrt{\frac{1-\sqrt5+\sqrt{5-2\sqrt5}}{1-\sqrt5-\sqrt{5-2\sqrt5}}}$$

Sin and Cos
$$\sin^2{80}+\sin^2{160}+\sin^2{320}=\cos^2{80}+\cos^2{160}+\cos^2{320}=\frac{3}{2}$$

$$\cos^2\left({2^{n+1}\times{5}}\right)+\cos^2\left({2^{n+2}\times{5}}\right)+ \cos^2\left({2^{n+3}\times{5}}\right)=\frac{3}{2}$$

n = 1

Sin and Cos
$$\cos^2{20}-\cos{40}\cos{80}=\frac{3}{4}$$

$$\cos^2{20}-\sin{40}\sin{80}=\frac{1}{4}$$

$$\sin{40}\sin{80}-\cos{40}\cos{80}=\frac{1}{2}$$

$$\sin{20}\sin{40}-\cos{20}\cos{40}=-\frac{1}{2}$$

Tangent
$$\tan{40}=\frac{\cos{20}+\cos{80}}{\sin{20}+\sin{80}}$$

$$\tan{C}=\frac{\cos{A}+\cos{B}}{\sin{A}+\sin{B}}$$

$$\tan{A_x}=\frac{\cos{A_1}+\cos{A_2}+...+\cos{A_n}}{\sin{A_1}+\sin{A_2}+...+\sin{A_n}}$$

Reciprocal
$$-\frac{1}{\cos{20}}+\frac{\sqrt{3}}{\sin{20}}=4$$

$$+\frac{1}{\cos{40}}+\frac{\sqrt{3}}{\sin{40}}=4$$

$$+\frac{1}{\cos{80}}-\frac{\sqrt{3}}{\sin{80}}=4$$

$$+\tan{40}\times\tan{80}-1=4\cos20$$

$$+\tan{20}\times\tan{80}+1=4\cos40$$

$$-\tan{20}\times\tan{40}+1=4\cos80$$

$$4=\tan40\times\frac{\tan80-\tan20}{\cos80+\cos20}$$

$$4=\frac{\tan80-\tan20}{\sin80+sin20}$$

$$\frac{4\cos20+1}{4\cos40-1}=\frac{\tan40}{\tan20}$$

Cubic equation
$$A^3-\frac{3A}{4}-\frac{1}{8}=0$$

$$A=\cos{20}$$

General

$$8Y^3-6Y\pm{X}=0$$

$$4Y^3-3Y\pm{X}=0$$

$$Y=\cos{\theta}$$

e.g

$$8\cos^3{25}-6\cos{25}-\frac{1}{\sqrt{2+\sqrt{3}}}=0$$

Cos
$$\frac{\cos{40}-\cos{80}}{\cos^3{40}-\cos^3{80}}=\frac{4}{3}$$

$$\frac{\cos{20}+\cos{40}}{\cos^3{20}+\cos^3{40}}=\frac{4}{3}$$

$$\frac{\cos{20}+\cos{80}}{\cos^3{20}+\cos^3{80}}=\frac{4}{3}$$

$$\frac{\cos{40}-\cos{80}}{\cos^2{40}-\cos^2{80}}=\cos{20}$$

sin(pi/a^n)
$$\sin\frac{45}{2^0}=\frac{\sqrt2}{2}$$

$$\sin\frac{45}{2^1}=\frac{\sqrt{2-\sqrt{2}}}{2}$$

$$\sin\frac{45}{2^2}=\frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2}$$

$$\sin\frac{45}{2^3}=\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}$$

$$\frac{2}{\pi}=\frac{\sqrt{2}}{2}{\cdot}\frac{\sqrt{2+\sqrt{2}}}{2}{\cdot}\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} {\cdot}\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\cdots$$

$$\frac{\sqrt{2}}{\pi}=\prod_{n=4}^{\infty}\left(1-2\sin\frac{\pi}{2^n}\right)$$

$$\sin\frac{15}{2^0}=\frac{\sqrt{2-\sqrt{3}}}{2}$$

$$\sin\frac{15}{2^1}=\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$

$$\sin\frac{15}{2^2}=\frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}}{2}$$

$$\frac{\pi}{3}\approx{2^1\sqrt{2-\sqrt{3_1}}}$$

$$\frac{\pi}{3}\approx{2^2\sqrt{2-\sqrt{2+\sqrt{3_2}}}}$$

$$\frac{\pi}{3}\approx{2^3\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3_3}}}}}$$

$$\frac{\pi}{3}=\lim_{n\to\infty}{2^n\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}_n}}}}$$

$$\frac{3}{\pi}=\frac{\sqrt{2+\sqrt{3}}}{2}{\cdot}\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}{2} {\cdot}\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}{2}\cdots$$

The Inverse Square Trig
$$\frac{1}{\cos^2{20}}+\frac{1}{\cos^2{40}}+\frac{1}{\cos^2{80}}=36$$

$$\frac{1}{\sin^2{20}}+\frac{1}{\sin^2{40}}+\frac{1}{\sin^2{80}}=12$$

$$\frac{1}{\tan^2{20}}+\frac{1}{\tan^2{40}}+\frac{1}{\tan^2{80}}=9$$

Others

$$\frac{1}{\sin20}+\frac{1}{\sin40}-\frac{1}{\sin80}=2\sqrt{3}$$

$$\frac{1}{\cos20}-\frac{1}{\cos40}-\frac{1}{\cos80}=-6$$

$$\frac{1}{\tan20}-\frac{1}{\tan40}+\frac{1}{\tan80}=\sqrt{3}$$

$$\left(\tan{20}\tan{40}\right)^2+\left(\tan{40}\tan{80}\right)^2+ \left(\tan{20}\tan{80}\right)^2=27$$

$$\tan{40}\tan{80}-\tan{20}\tan{80}+\tan{20}\tan{40}=3\cdot$$

$$\sin^2{20}+\sin^2{40}+\sin^2{80}=\frac{3}{2}$$

$$\cos^2{20}+\cos^2{40}+\cos^2{80}=\frac{3}{2}$$

$$\tan^2{20}+\tan^2{40}+\tan^2{80}=33\cdot$$

$$\sin{20}+\sin{40}-\sin{80}=0\cdots(1)$$

$$\cos{20}-\cos{40}-\cos{80}=0\cdots(2)$$

$$\tan{20}-\tan{40}+\tan{80}=3\sqrt{3}\cdots(3)$$

$$\frac{1}{1+\cos{40}}+\frac{1}{1+\cos{80}}+\frac{1}{1+\cos{160}}=18$$

$$\frac{1}{\left(1+\cos{40}\right)^2}+\frac{1}{\left(1+\cos{80}\right)^2}+ \frac{1}{\left(1+\cos{160}\right)^2}=276$$

$$\frac{1}{\left(1+\cos{40}\right)^3}+\frac{1}{\left(1+\cos{80}\right)^3}+ \frac{1}{\left(1+\cos{160}\right)^3}=4560$$

$$\frac{1}{1-\cos{40}}+\frac{1}{1-\cos{80}}+\frac{1}{1-\cos{160}}=6$$

$$\frac{1}{\left(1-\cos{40}\right)^2}+\frac{1}{\left(1-\cos{80}\right)^2}+ \frac{1}{\left(1-\cos{160}\right)^2}=20$$

$$\frac{1}{\left(1-\cos{40}\right)^3}+\frac{1}{\left(1-\cos{80}\right)^3}+ \frac{1}{\left(1-\cos{160}\right)^3}=80$$

Cube
$$\frac{1}{\cos^3{80}}+\frac{1}{\cos^3{40}}-\frac{1}{\cos^3{20}}=192$$

Series
$$\sin^2(10)-\sum_{j=1}^{3k-1}\sin^2\left(2^j\times{10}\right)=-\frac{3k-2}{2}$$

Where k = 1, 2 , 3 , ...

Others
$$\frac{\sin10\cos10-\sin20\cos20}{\sin^2{10}-\sin^2{20}}=\sqrt{3}$$

$$\frac{\tan20}{\tan40}+\frac{\tan40}{\tan80}+\frac{\tan80}{\tan160}=-15$$

$$\tan{20}\tan{80}\tan{160}+\tan{160}\tan^2{40}+\tan{40}\tan^2{80}=15\sqrt{3}$$

$$\frac{\sin{20}-\frac{1}{2}\sin{80}}{\sin^2{10}-\sin^2{20}}=\sqrt{3}$$

$$\sin20=\sin30\sin80-\sin60\sin10\cdots(1)$$

$$\sin40=\sin30\sin80+\sin60\sin10\cdots(2)$$

$$\frac{\sin40-\sin20}{\sin10}=\sqrt{3}$$

$$\frac{\sin20}{\sin10\sin80}=2$$

$$2\sin10+\frac{\sin40}{\sin80}=1$$

$$\frac{1}{4\sin^2{20}}-\frac{2\sin80}{\sqrt{3}}=1$$

$$\frac{\sin10\sin20\sin40\sin80}{\sin40-\sin20}=\frac{1}{8}$$

$$\sin10=\cos20-\cos40\cdots(1)$$

$$\frac{\sin40-\sin20}{\cos40-\cos20}=-\sqrt{3}$$

$$\frac{\sin40+\sin20}{\cos40+\cos20}=\frac{1}{\sqrt{3}}$$

Others and pi approximations
$$\sqrt{5}\approx{e-\frac{\sqrt{e-1}}{e}}$$

$$\pi\approx{\ln\ln\left(3\times{2^{\frac{159}{5}}}\right)}$$

$$\frac{e^{\pi}-\ln\left(3\times{2^n}\right)}{\ln2}+n\approx{31.8}$$

$$\frac{e^{\pi}-\ln\left(3\times{2^{\frac{n-1}{2}}}\right)}{\ln2}+\frac{n+1}{2}\approx{32.8}$$

$$\frac{\sqrt{34}}{16}\approx{\frac{\pi}}$$

$$\frac{e^{\pi}-\ln{pi}}{\ln5}\approx{\frac{41}{3}}$$

$$e^{\pi}-\ln{\pi}\approx{21.995}$$

$$\left(\frac{83}{7}\right)^{\frac{4}{9}}\approx{3.0014}$$

$$\frac{1}{\left(\sin20\sin40\sin160\right)^8}\approx{\frac{1}{\phi^2}+1}$$