一題難數一題難數數數數

Part C.
cos 36 = (sin 18 + 1)/(2 sin 18 + 1)............... (Proved in part b)
Let sin 18 = x
so cos 36 = (x + 1)/(2x + 1)...................(1)
cos 72 = sin (90 - 72) = sin 18 = x
Also, cos 72 = 2 cos 36^2 - 1
that is x = 2 cos 36^2 - 1
2 cos 36^2 = 1 + x................(2)
From (1) and (2)
(1 + x)/2 = [(x + 1)/(2x + 1)]^2
(1 + x)(2x + 1)^2 = 2(x + 1)^2
(1 + x)(4x^2 + 4x + 1) = 2x^2 + 4x + 2
4x^2 + 4x + 1 + 4x^3 + 4x^2 + x = 2x^2 + 4x + 2
4x^3 + 8x^2 + 5x + 1 = 2x^2 + 4x + 2
4x^3 + 6x^2 + x - 1 = 0
since x = sin 18, so sin 18 is the root of this equation.