11/07/15 09:25:44.09 mnstTaxgP
>>593
cosA - cos2A = 2sin[(A+2A)/2] sin[(2A-A)/2]
cosA - cos2A = 2sin(3A/2) sinA/2
A = 36
cos36 - cos72 = 2sin(54) sin(18)
cos36 - cos72 = 2 cos36 cos 72
Let 36 = x
cosx - cos2x = 2cosx cos2x
cosx - 2cos^2x + 1 = 2cosx [ 2cos^2x - 1]
cosx - 2cos^2 x + 1 = 4cos^3 x - 2cosx
4cos^3x + 2cos^2x - 3cosx - 1 = 0
solve the equation putting cosx = y
4y^3 + 2y^2 - 3y - 1 = 0
U'll get
cosx = -1
cosx = (1/4) [1 - sqrt5]
cosx = (1/4) [1 + sqrt5] <= Valid answer
cos36 = (1/4) [ 1 + sqrt5]
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