[VA] SRC #009 - Pi Day 2021 Special

[Albert Chan]

(03-16-2021 09:09 PM)robve Wrote:  $$ \arctan u - \arctan v = \arctan\frac{u-v}{1+uv} $$

This is not quite right, LHS has possible range of ±pi, RHS is limited to ±pi/2
Correct identity is more complicated: see Sum of ArcTangents

We could use this instead: atan(u) ± atan(v) = atan2(u±v , 1∓uv)       (*)

y = atan(x) - atan((x-1)/(x+1))                       // y undefined when x = -1
= atan2(x - (x-1)/(x+1), 1 + x*(x-1)/(x+1))
= atan2((x²+1)/(x+1), (x²+1)/(x+1))             // numerator always positive.

If x > -1, 4*y = 4*atan(1) = pi
If x < -1, 4*y = 4*(atan(1) - pi) = -3*pi

(*) Proof is trivial: (1+u*i) * (1±v*i) = (1∓u*v) + (u±v)*i

Phase angle of two sides matched, and we have above atan2 identity