For a long time I've been fascinated by the way you can tile the 80 80 20 isosceles with smaller isosceles/equilateral triangles, like this
I also really like the geometric determination of sin 18 using recursive triangles, which you've epxlored on video.
Have you looked at combining the two? I realised today that the 80 80 20 tiling very quickly and elegantly shows that y=2sin(10) has to satisfy y=1/(3-y^2) ==> y^3-3y^2+1. After adjusting for the factor of 2 this is the same cubic as you talk about on your sin 10 video.
To see this, start with the isosceles tiling from cut-the-knot above, setting the base of the big 80 80 20 triangle to 1; this length then propagates through all the isosceles triangles in the diagram; and set y=2sin(10) as shown.
Because ABC and BCD are similar we know their sides are in proportion so y/1 = 1/AB = 1/(AF+1).
Now focussing on EFA we can add points G and H along FA so that FEH and AEG are congruent to each other and to BCD, so EG = EH =y; and because EGH is similar we know that GH = y^2, so AF = AG +FH - GH = 1+1-y^2 = 2-y^2. Substituting into y=1/(AF+1) gives y=1/(3-y^2) ==> y^3-3y+1=0, QED.
This diagram also gives you the connection with cos40=sin50: the two sides of the big isosceles triangle ABC, namely AB and AC have to be equal. We know AB=3-y^2 as just shown and can see that AC=1+y+2sin50. Equating these gives sin50=1-y/2-(y^2)/2.
Well done Josh. It's quite astonishing how pretty much all the relationships we take for granted (Pythagorous, sine law, cosine law, differentials in calculus, dot product etc etc .......) all come down to one basic property of triangles. If you increase the sides proportionally in any triangle, then the angles remain the same. A basic property of Eulidean Space that all trig, geometry and calculus depend directly upon.