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Friday, August 29, 2014

Cosine of 72 degrees (and constructing a regular pentagon)

By using (say) DeMoivre's theorem, we have that (cos2π5+isin2π5)5=1
Expanding the left side as the fifth power of a binomial, equating the imaginary parts on both sides of the equation, and then replacing  sin272 with  1cos272
 isin72(5cos472+10cos272i2sin272+i4sin472)=0i
 16cos47212cos272+1=0
Solving this quadratic in cos272, we get
 cos272=12±8032
so
 cos272=6±2516=(5±1)242
 cos72=±5±14
where we can choose the correct value of the four possible values by noting that, because 72° is between 45° and 90°,  cos72 must lie between  1/2 and 0. Because  cos72 is positive, we choose the "+" before the fraction, and because  cos72 is less than 1/2, which in turn is less than 5+14, we choose the "-" in the numerator:
cos72=514

Constructing a regular pentagon

So we can construct cos72. For example, the diagonal of a 1-by-2 rectangle is 5. We could cut off one unit from a segment of length 5, then divide the segment of length 51 into four pieces of length 514. (Or we could construct the appropriate solution to the equation 4x2+2x1=0. See my post on Solving quadratic equations via geometric construction.)

Construct a unit circle centered at O, and construct a radius ¯OA.  Construct the point B on ¯OA so that ¯OB has length  cos72. If C is a point on the circle so that ¯BC is perpendicular to ¯OA, then COA is a 72° angle, and both A and C are vertices of a regular pentagon inscribed in the circle.

1 comment:

Bruce Yoshiwara said...

This method of evaluating cos(2pi/5) was shown to me by my then-UCLA office mate Larry Miller in my first year of grad school.