Arcos tangentes

tg(a-b)=tg(a)-tg(b)/1+tg(a)*tg(b) si llamamos x=tg(a) e y=tg(b) con lo que a=arctan(x) e b=arctan(y)  tendremos que:

tg(a-b)=x-y/1+xy => a-b=arctan(x-y/1+xy)  y con ello

arctan(x)-arctan(y)=arctan(x-y/1+xy)  y entonces:

arctan(1/n)=arctan(1/(n+1))+arctan(1/(n^2+n+1))

Puesto que lo unico que hemos hecho es x=1/n e y= 1/n+1 y observamos que

((1/n)-1/(n+1))/1+(1/n)*(1/(n+1))= 1/(n^2+n+1)

De la misma forma haciendo x=1/n  e y=1/(n+2) obten  

 emos con una pequeña manipulación y para cuando n sea impar:

arctan (1/n)=arctan(1/(n+2))+arctan(1/((n+1)^2/2))

Para n=n+s con s>2 no funciona por lo que no hay mas formulas de este tipo.Dando en ellas valores a n obtenemos:

arctan(1/2) = arctan(1/ 3) + arctan(1/ 7)
arctan(1/3) = arctan(1/ 4) + arctan(1/13)
arctan(1/4) = arctan(1/ 5) + arctan(1/21)
arctan(1/5) = arctan(1/ 6) + arctan(1/31)
arctan(1/5) = arctan(1/ 7) + arctan(1/18)
arctan(1/6) = arctan(1/ 7) + arctan(1/43)
arctan(1/7) = arctan(1/ + arctan(1/57)
arctan(1/7) = arctan(1/ 9) + arctan(1/32)
arctan(1/7) = arctan(1/12) + arctan(1/17)

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibpi.html
http://pi.pixelhack.se/?action=about

Arctan (1) = arctan (1 / 2) + arctan (1 / 5) + arctan (1 / 8)? 

Let me add a third way to work this problems.

Since complex multiplication adds arguments, multiply three complex
numbers with arguments of arctan(1/2), arctan(1/5), and arctan(1/8).

(2+i)(5+i)(8+i) = 65+65i

The product has an argument of arctan(65/65) = arctan(1).

~ por leonsotelo en Marzo 30, 2008.