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Homework Help: Determining the constructability of angles

  1. Apr 1, 2013 #1
    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.
  2. jcsd
  3. Apr 1, 2013 #2


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    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?
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