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)/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)/2 (by the Quadratic Formula, discarding the negative root as extraneous), 2*(y/x)^2 = 3 + sqrt. 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)/(3+sqrt), = (1+sqrt)/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-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/
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