設一方程: 16x4-8x3-12x2+8x-1=0
試求sin-1x =??
galaxylee 寫到:16x^4-8x^3-12x^2+8x-1=0(2x-1)(8x^3-6x+1)=0x=1/2是一根8x^3-6x+1=0 => 2(3x-4x^3)=1令x=sinθ 上式變成 2sin3θ=1 => sin3θ=1/23θ=(π/6)+2kπ,或(5π/6)+2kπ,其中k為整數θ=(π/18)+(2kπ)/3,或(5π/18)+(2kπ)/3可歸納出sinθ=sin(π/18) 或 sin(5π/18),sin(-7π/18)所以方程式的四根為 x = 1/2(=sin(π/6))、sin(π/18)、sin(5π/18)、sin(-7π/18)故 sin^(-1)x = π/6、π/18、5π/18、-7π/18