?
Favorite Answer
π/96 = (π/6)/16 so one idea is to cascade tan(θ/2) results
Let T = tan(2θ) and t = tan(θ)
T = 2t/(1 – t^2) and the lesser root for t is
t = [√(T^2 + 1) – 1]/T or if more convenient
t = √(1 + 1/T^2) – 1/T ..........................(1)
Start with T = tan(π/6) = √(3)/3 so that 1/T = √(3)
t = tan(π/12) = √(1 + 3) – √(3) = 2 – √(3)
Repeat that, but now T = tan(π/12) and 1/T = 2 + √(3), 1/T^2 = 7 + 4√(3)
tan(π/24) = t = √[1 + 1/T^2] – 1/T = √[8 + 4√(3)] – 2 - √(3)
Now you can see a plan to get to tan(π/48) and tan(π/96)
Repeating again, but with T = tan(π/24) = √[8 + 4√(3)] – 2 - √(3)
Expand T^2 and then note √(T^2) to keep to keep results in surd form.
But now it is your turn
Anonymous
23