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How to Derive cos 72

Date: 6/19/96 at 15:41:40
From: Anonymous
Subject: How to Derive cos 72

Can you explain to me how to derive a radical expression for cos 72 
using a 36-72-72 triangle?

       - Michael

Date: 6/20/96 at 6:39:13
From: Doctor Anthony
Subject: Re: cos 72

Draw the triangle ABC with 36 degrees at point B, and A and C both 
equal to 72 degrees.  Now from C draw the bisector CD to meet AB at D. 
Then triangle ABC is similar to triangle ACD. If we let AC = b and 
BC = a, then because of isosceles triangles we also have CD = b and 
DB = b.  This makes DA = a-b.

Now from similar triangles BC/CA = CA/AD

              or             a/b = b/(a-b)

                          a(a-b) = b^2

                   a^2 -ab - b^2 = 0

Treat this as a quadratic in a, and use the quadratic formula:

          a = {b +or-sqrt(b^2+4b^2)}/2   we ignore -ve sign since a>b

          a = {b + b*sqrt(5)}/2

            = b(1+sqrt(5))/2   ......(1)

Now if we look at original triangle ABC and draw a perpendicular from 
B to AC to bisect AC we can see that cos(72) = (b/2)/a  = b/(2a)

Now from equation (1) above we can see that:

          1/(1+sqrt(5)) = b/(2a)

And so cos(72) = 1/(1+sqrt(5)) 

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Trigonometry

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