Programming Challenge: a classic trigonometry problem

[Albert Chan]

Here is an interesting symmetric pair of q guesses, derived at the same time.

1/v = 1/c - 1/u

Remember v is positive, but u can be both (u has sign of q): edge(u) = ±a
Let z = c/a

edge(1/v) = 1/c ± 1/a = (1 ± c/a) / c
edge(v/c) = (1 ± z)^-1

u = ±√(a² - y) = ±√(a² - (b² - v²)) = c * (1 + q)

(1 + q) = ±√((a²-b²)/c² + (v/c)²)

edge(q) = -(1 ± √(k + (1±z)^-2))

We redo previous post example, solve for both Q real roots.
I named negative q guess as "hi", only because it is bigger in size.

lua> S = require'solver'
lua> a, b, c = 40, 30, 15
lua> k = (a*a-b*b)/(c*c)
lua> Q = fn'q: (1+q)^2*(1-q^-2) - k'
lua> eps = -1e-9 -- swap worse point as first

lua> z = c/a
lua> lo = -(1 - sqrt(k + (1-z)^-2)) -- positive q guess
lua> hi = -(1 + sqrt(k + (1+z)^-2)) -- negative q guess

lua> S.secant(Q, lo, lo*1.01, eps, true)
1.3814094799322336 -- = lo
1.395223574731556
1.4441708057247238
1.4443988616014456
1.4443995665095293
1.4443995665182763

lua> S.secant(Q, hi*0.99, hi, eps, true)
-2.907888028932993 -- = hi
-2.878809148643663
-2.8807700072753484
-2.8807853577712663
-2.8807853490168784
-2.880785349016917
-2.880785349016917