Given a square piece of paper, prove using paper folding alone this trigonometric identity:
arctan(1/3) + arctan(1) = arctan(2)
*** No geometry instrument aids like straightedge, compass etc are allowed.
Note that arctan(2)arctan(1/3) = arctan(1).
'Unfold' means unfold the last folding action, not earlier folds.
Fold the paper in half horizontally once.
Fold the paper in half vertically four times. Unfold.
Fold the paper along the last vertical crease.
Fold this rectangle diagonally. Unfold.
Count 8 vertical creases away from the larger  angled diagonal crease. Fold the paper backwards along the 8th crease.
This will create a square. Fold it diagonally so that the diagonal meets the larger  angled diagonal crease at a corner. Unfold.
The original diagonal crease is arctan(2) (precisely, arctan (15/8))
The diagonal of the square is arctan(1).
The angle between them is arctan(1/3) (precisely, arctan (7/23))
The governing formula is u^2v^2= uv+1; with {u=5,v=3} this gives {16,30,34} for which{8,15,17} is a solution. Larger solutions are more accurate but require a lot of folding; e.g. arctan(67104/33553) = 1.99994  arctan(1) =arctan(33551/100657) =0.33332... etc.
However, this doesn't affect the underlying principle.
Edited on June 18, 2014, 12:07 pm

Posted by broll
on 20140618 11:53:25 