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# week19 - Proof that one can not Trisect an angle of 60...

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Proof that one can not Trisect an angle of 60 degrees with Straight Edge and Compass Original Notes adopted from February 12, 2002 (Week 19) © P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong F, r F, r>0 & if r F, F( r) = {a + b r: : a,b F} Tower of number fields is a collection F 0 F 1 F 2 ......... F n of number fields such that F 0 = Q to i, there exists r i F i such that r > 0, r i F i i+1 = F i ( r i ) Collection of surds is U {F n : F n at top of a tower} S, S number field ("the surd field") Set of Constructible Numbers: We showed: C is a number field Also, if r C, r > 0, then r C. We showed: Theorem: C = S. We'll show: Can't Trisect (with compass and straightedge) an angle of 60 degrees. Can construct angle of 60 degrees. If we could trisect that angle of 60 degrees, we could construct an angle of 20 degrees. We'll show that this is impossible. Lemma 1: If angle θ (acute) is constructible, then cos θ is a constructible number. cos θ = x/1 = x. Constructible x from θ . cos θ constructible. Suffices for: Theorem: That 60 degrees not trisectable to show cos20 degrees not constructible. I.e. Show cos20 degrees is not a surd.

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Recall: cos(A+B) = cosAcosB – sinAsinB cos2 θ = cos 2 θ - sin 2 θ = 2 cos 2 θ -1 cos3 θ = cos(2 θ + θ ) cos3 θ = cos2
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week19 - Proof that one can not Trisect an angle of 60...

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