已知sinθ,cosθ是关于x的方程x^2-ax+a=0(a∈R)的两个根，求cos(π/2-θ)^3+sin(π/2-θ)^3的值，

显示全部楼层 · 2012-6-14 12:07:36

sinθ+cosθ=a (1)sinθcosθ=a
(2)(1)两边平方:1+2sinθc0sθ=a^22sinθc0sθ=a^2-1 (3)(3)/(2):2=a-1/aa1=1+√2(舍去), a2=1-√2cos3(π/2-θ)+sin3(π/2-θ)=sin3θ+cos3θ=(sinθ+cosθ)(sin2θ-sinθcosθ+cos2θ)=a(1-a)=a-a2=1-√2-(1-√2)2=√2-3 (1)/(2):tanθ+1/tanθ=1tan(π-θ)-1/tanθ=-tanθ-1/t...

千问 · 2012-6-14 12:07:36

sinθ+cosθ=a (1)sinθcosθ=a
(2)(1)两边平方:1+2sinθc0sθ=a^22sinθc0sθ=a^2-1 (3)(3)/(2):2=a-1/aa1=1+√2(舍去), a2=1-√2cos3(π/2-θ)+sin3(π/2-θ)=sin3θ+cos...