Proof of Cosine of 36 Degrees

Date: 11/26/2001 at 21:21:03
From: John Kuli
Subject: The proof of Cosine of 36 degrees.

Hello! I'm in Calculus 1 at Wright State University. As a bonus 
question my teacher asked us to solve the following:

Prove: cosine {(pi)/5} = {1+5^(1/2)}/4

I was wondering if you could help me out.

Thanks.
John

Date: 11/26/2001 at 22:37:27
From: Doctor Paul
Subject: Re: The proof of Cosine of 36 degrees.

To compute cos(36) notice that cos(36) = cos(72/2)

Now apply the half angle formula:

   cos(36) = +- sqrt[(1+cos(72))/2]

Now we need to know how to compute cos(72).

This has already been done in our archives, where a person asked how 
to compute sin(18) but sin(18) = cos(72):

   Finding Values by Hand
   http://mathforum.org/dr.math/problems/victor8.26.98.html   

Following the math derived in the link above, 

   cos(72) = -0.25 + 0.25*SQRT(5)

Plugging this in above (and taking the positive square root) gives:

   cos(36) = sqrt[(1 + (-0.25 + 0.25*SQRT(5)))/2]

   = sqrt[(3+sqrt(5))/8]     (verify this!)

How do we show this is the same as (1 + sqrt(5))/4 ?

Start with something we know to be true:

   8 + 16*sqrt(5) + 40 = 48 + 16*sqrt(5)

then rewrite as:

   1 + 2*sqrt(5) + 5        3 + sqrt(5)
   -----------------   =    -----------
           16                    8

now take the square root of both sides:

   1+sqrt(5)
   -------- =  sqrt[(3+sqrt(5))/8]
       4


Thus cos(36) = sqrt[(3+sqrt(5))/8] = (1+sqrt(5))/4

You may ask, how did I know to write

   8 + 16*sqrt(5) + 40 = 48 + 16*sqrt(5)  ?

The answer is I didn't. But I knew the two things had to be equal so I 
set them equal to each other and then showed they were the same (I did 
this on some scratch paper). Then I just copied my work in reverse 
order to establish the desired result.

I hope this helps. Please write back if you'd like to talk about this 
some more - especially if you have trouble following anything I've 
written or if you have trouble following the math in the link I 
referred to.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/