1.7. Algebraic reformulation 25 Figure 1.7.2 B = (cos θ, sin θ). Note that ∠AOB is constructible and has θ radians. Hence the original angle has been trisected. The converse is left to the reader. The main result of this section is the following. 1.7.3. Theorem. Let ∠AOB be a constructible angle that has 3θ radians with 0 3θ π 2 let a = cos 3θ. The angle ∠AOB can be trisected using straightedge and compass alone if and only if there is a positive constructible number x satisfying 1.7.4 x3 − 3x − 2a = 0. Proof. Using (and iterating) the formula for the cosine of the sum of two angles, the reader can derive the following: cos 3θ = 4 cos3 θ − 3 cos θ. So if x 2 = cos θ, this trig identity becomes a = 4 x 2 3 − 3 x 2 = x3 2 − 3x 2 , which is equivalent to (1.7.4). So if ∠AOB can be trisected, cos θ is a constructible number, and therefore so is x = 1 2 cos θ. This value of x is positive and satisfies (1.7.4). For the converse, assume that a positive constructible number x exists that is a root of (1.7.4). Since 0 3θ π 2 , we have that 0 a 1. The polynomial p(t) = t3 − 3t − 2a can be graphed, and we see that it has exactly one positive real root. As pointed out before, p( 1 2 cos θ) = 0. Since this polynomial has only one positive root, it must be that 1 2 cos θ = x, a constructible number. Therefore cos θ is constructible, and, by Lemma 1.7.2, an angle having 3θ radians can be trisected.

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