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 Sine recurrence (Posted on 2019-01-08)
If a0 = sin2 (π/45) and an+1 = 4an(1 - an) for n >= 0, find the smallest positive integer n such that an = a0

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Full solution | Comment 5 of 6 |
Apply the procedure to sin2 (a)

an+1 = 4sin2 (a)(1 - sin2 (a))
4sin2 (a)cos2 (a)
= (2sin(a)cos(a))^2
= (sin(2a))^2
sin2 (2a)

Each term just doubles the angle from the last.

Now, in the interval [0,2π) we know
sin2 (1π/45) = sin2 (44π/45) = sin2 (46π/45) = sin2 (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

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 2019-01-08 11:36:32

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