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Math Help - Algebra, Problems For Fun (23)

  1. #1
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    Algebra, Problems For Fun (23)

    [\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]=?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    -\frac12\,=\,\cos120^\circ\,=\,cos(3\times40^\circ)  \,=\,4\cos^340^\circ-3\cos40^\circ

    \therefore\quad8\cos^340^\circ-6\cos40^\circ+1\ =\ 0

    \implies\ \left(2\cos40^\circ\right)^3-3(2\cos40^\circ)+1\ =\ 0

    The polynomial x^3-3x+1=0 is irreducible over \mathbb Z and hence over \mathbb Q (by Gauß’s lemma); hence it is the minimum polynomial for 2\cos40^\circ over \mathbb Q.

    \therefore\ [\mathbb{Q}(2\cos 40^{\circ}):\mathbb{Q}]\,=\,[\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}]\,=\,3

    The significance of this result is in showing that the angle 120° cannot be trisected using only compass and straightedge (since such a procedure would require [\mathbb{Q}(\cos 40^{\circ}):\mathbb{Q}] to be a power of 2).
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