Apply the procedure to
sin^{2} (a)
a_{n+1} = 4sin^{2} (a)(1  sin^{2} (a))
= 4sin^{2} (a)cos^{2} (a)
= (2sin(a)cos(a))^2
= (sin(2a))^2
= sin^{2} (2a)
Each term just doubles the angle from the last.
Now, in the interval [0,2π) we know
sin^{2} (1π/45) = sin^{2} (44π/45) = sin^{2} (46π/45) = sin^{2} (89π/45)
So the question becomes how many times do we double the original angle to get one of these?
Here's a table of n, multiple of π/45 mod 90
0 1
1 2
2 4
3 8
4 16
5 32
6 64
7 38
8 76
9 62
10 34
11 68
12 46 < we have a hit. The answer is 12.

Posted by Jer
on 20190108 11:36:32 