Rational trig identities?

[Albert Chan]

(05-31-2021 09:51 PM)Albert Chan Wrote:  \(\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{1+x^2}+1} \right)\)

This covered n = powers-of-2:

With Pythagorean triples formula: (m^2-n^2)^2 + (2mn)^2 = (m^2+n^2)^2
This made above square root goes away.

atan((2*m*n)/(m^2-n^2)) = 2*atan( (2*m*n)/((m^2+n^2)+(m^2-n^2)) = 2*atan(n/m)

Let 2^p = m/n, above simplified to: 2*atan(1/2^p) = atan(2^(p+1)/(4^p-1))

Example:

2*atan(1/2) = atan(4/3)

2*atan(1/4) = atan(8/15)
4*atan(1/4) = atan((2*15*8)/(15^2-8^2)) = atan(240/161)

2*atan(1/8) = atan(16/63)
4*atan(1/8) = atan((2*63*16)/(63^2-16^2)) = atan(2016/3713)
8*atan(1/8) = atan((2*3713*2016)/(3713^2-2016^2)) = atan(14970816/9722113)