Determining the constructability of angles

[trap101]

Are 6 degrees, 5 degrees, and 7.5 degrees constructable?

So based on the theorems that I know, these angles are constructable iff the cos[itex]\vartheta[/itex] is constructible. So all that is left for me to do is show if cos[itex]\vartheta[/itex] is constructable.

For 6 degrees, I tried to get a relatonship with the trig identity of cos(3[itex]\vartheta[/itex]) = 4cos³([itex]\vartheta[/itex]) - 3cos([itex]\vartheta[/itex]) because from there I could use the fact that if I can obtain a solution of the 3rd degree polynomial, then based on the rational roots theorem I could determine if the angle is constructible. My issue is, I couldn't find a simple relationship for 6degrees. I tried some higher level trig identities and "attempted" to simplify, but it started to appear futile.

So now I'm at a cross roads. Same with the other two.

 

[Dick]

Are 6 degrees, 5 degrees, and 7.5 degrees constructable?

So based on the theorems that I know, these angles are constructable iff the cos[itex]\vartheta[/itex] is constructible. So all that is left for me to do is show if cos[itex]\vartheta[/itex] is constructable.

For 6 degrees, I tried to get a relatonship with the trig identity of cos(3[itex]\vartheta[/itex]) = 4cos³([itex]\vartheta[/itex]) - 3cos([itex]\vartheta[/itex]) because from there I could use the fact that if I can obtain a solution of the 3rd degree polynomial, then based on the rational roots theorem I could determine if the angle is constructible. My issue is, I couldn't find a simple relationship for 6degrees. I tried some higher level trig identities and "attempted" to simplify, but it started to appear futile.

So now I'm at a cross roads. Same with the other two.

Think about polygons that have constructible angles. A pentagon is constructible. That means 72 degrees is constructible. If that's constructible then 36 degrees is also constructible. Why? 30 degrees is also constructible. Why? That would mean 36-30 is also constructible. Why?