Calculate : $\displaystyle \sin (6^\circ ),\cos (6^\circ )$
sin 6 degrees? hmnnnnn . . . . from this post http://www.mathhelpforum.com/math-he...tml#post350585
cos 72 = sin 18 = (sqrt(5) -1)/4, also sin 3A = 3(sin A) - 4(sin^3 A); then
sin 18 = 3(sin 6) - 4(sin^3 6); substituting
(sqrt 5 - 1)/4 = 3(sin 6) - 4(sin^3 6); let x = sin 6.
3x - 4x^3 = (sqrt 5 - 1)/4, rearranging
4x^3 - 3x - (1 - sqrt 5)/4 = 0, graphing it we have 3 roots and only one is valid,
that is . . . . x = sin 6 = (1/8)(-1 - sqrt 5 + sqrt(30 - 6 sqrt 5))
to get cos 6 will have the same pattern