昨晚的回答不知何故又被隱了, 簡單回答吧. 令 x = cos(3π/5) 則 cos(9π/5) = 4x^3 - 3x cos(6π/5) = 2x^2 - 1 cos(9π/5) + cos(6π/5) = 0 即 4x^3 + 2x^2 - 3x - 1 = (x+1)(4x^2 - 2x - 1 ) = 0. 因 π/2 < 3π/5 < π, 故 0 > cos(3π/5) > -1 故 cos(3π/5) = (1-√5)/4.
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昨晚的回答不知何故又被隱了, 簡單回答吧. 令 x = cos(3π/5) 則 cos(9π/5) = 4x^3 - 3x cos(6π/5) = 2x^2 - 1 cos(9π/5) + cos(6π/5) = 0 即 4x^3 + 2x^2 - 3x - 1 = (x+1)(4x^2 - 2x - 1 ) = 0. 因 π/2 < 3π/5 < π, 故 0 > cos(3π/5) > -1 故 cos(3π/5) = (1-√5)/4.
