Trig Identities

Date: 04/16/99 at 03:08:31
From: Was
Subject: Trig Identities for "odd" angles

I have to prove that cos 36 (or sin 54) = (1+sqrt(5))/4. I can 
calculate it out and show that it does, which is nice, because it 
means I've done the first bit of the problem right, but I have no 
experience prooving the identity of "odd" angles like that. Is there a 
way? Are there identities that I don't know? Can I see some proofs for 
them?

Date: 04/16/99 at 13:54:50
From: Doctor Rob
Subject: Re: Trig Identities for "odd" angles

Thanks for writing to Ask Dr. Math.

Indeed, cos(36 degrees) = (1+sqrt[5])/4.  You can prove that by
considering the following diagram:

                                   R
                                 _,o
                              _,' / \
                           _,' 36/36 \
                   x+y _,-'     /     \
                    _,'        /       \
                _,-'         y/         \y
             _,'             /           \
          _,'               /             \
       _,'36     y     108 /72     x     72\
     o'-------------------o-----------------o
   P                     S                   Q

All three triangles are isosceles, because each has two angles equal, 
so making SQ = x and QR = y, the above labels are correct for the 
lengths of sides. Using the fact that triangles PQR and RSQ are 
similar (since they both have angles of 36, 72, and 72 degrees), you
get

   PQ/QR = RS/SQ,
   (x+y)/y = y/x,
   1 + (y/x) = (y/x)^2,
   y/x = (1+sqrt[5])/2     (by the Quadratic Formula, discarding the
                            negative root as extraneous),
   2*(y/x)^2 = 3 + sqrt[5].

Then applying the Law of Cosines to triangle RSQ,

   SQ^2 = RS^2 + QR^2 - 2*RS*QR*cos(<QRS),
   x^2 = y^2 + y^2 - 2*y^2*cos(36 degrees),
   cos(36 degrees) = (2*y^2-x^2)/(2*y^2),
                   = (2*[y/x]^2-1)/(2*[y/x]^2),
                   = (2+sqrt[5])/(3+sqrt[5]),
                   = (1+sqrt[5])/4,

by rationalizing the denominator.

You can avoid the Law of Cosines altogether by dropping a 
perpendicular from S to QR at T and using the Pythagorean Theorem 
on the two triangles QST and RST, then subtracting one equation 
from the other, and solving for RT in terms of x and y. Then 
RT/RS = cos(<QRS).

This only works for 36 degrees.  From this you can get

   sin(18 degrees) = (sqrt[5]-1)/4

with a half-angle formula. That will allow you to compute all the 
trigonometric functions of any multiple of 3 degrees in terms of 
sqrt(2), sqrt(3), sqrt(5), using only square roots and rational 
combinations.

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