數學:證明這個三角函數可以這個分數形式寫出?

4x³ + 2x² - 3x - 1 = (x + 1)(4x² - 2x - 1)　　　　　　　　　cos(3θ) = 4cos³θ - 3cosθ

cos(3π/5) ...... ①
= 4cos³(π/5) - 3cos(π/5) ...... ②
= [4cos³(π/5) + 2cos²(π/5) - 3cos(π/5) - 1] - [2cos²(π/5) - 1]
= [cos(π/5) + 1] [4cos²(π/5) - 2cos(π/5) - 1] - cos(2π/5)
= [cos(π/5) + 1] [4cos²(π/5) - 2cos(π/5) - 1] - cos(π - 3π/5)
= [cos(π/5) + 1] [4cos²(π/5) - 2cos(π/5) - 1] + cos(3π/5) ...... ③

∵ ① = ③
cos(3π/5) = [cos(π/5) + 1] [4cos²(π/5) - 2cos(π/5) - 1] + cos(3π/5)
[cos(π/5) + 1] [4cos²(π/5) - 2cos(π/5) - 1] = 0
4cos²(π/5) - 2cos(π/5) - 1 = 0
cos(π/5) = [2 + √(2² + 4×4)] / (2×4)　　　或　　　[2 - √(2² + 4)] / (2×4) < 0 ( 捨去 )
cos(π/5) = (1 + √5)/4

cos(3π/5)
= 4[(1 + √5)/4]³ - 3[(1 + √5)/4] ...... [ 代 cos(π/5) = (1 + √5)/4 入 ② ]
= (1 + 3√5 + 15 + 5√5)/16 - (3 + 3√5)/4
= (16 + 8√5 - 12 - 12√5)/16
= (1 - √5)/4

∴ cos(3π/5) = (1 - √5)/4