How to Derive cos 72Date: 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 Check out our web site! http://mathforum.org/dr.math/ |
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