已知sinθcosθ是关于x的方程x^2-ax+a=0的两个根 (1)求cos3(π/2-θ)+sin3(π/2-θ)的值 (2)求tan(π-θ)-1/tanθ的值
(1)解析：∵sinθ,cosθ是关于x的方程x^2-ax+a=0的两个根
∴sinθ+cosθ=a,sinθcosθ=a
(sinθ+cosθ)^2=1+2a=a^2==>a1=1-√2,a2=1+√2
x^2-(1-√2)*x+1-√2 =0
∴sinθ+cosθ=(1-√2),sinθcosθ=1-√2
x^2-(1+√2)*x+1+√2 =0无实解
cos3(π/2-θ)+sin3(π/2-θ)=-(cos3θ+sin3θ)=-[cosθ-4(sinθ)^2cosθ+4(cosθ)^2sinθ-sinθ]
=sinθ-cosθ+4sinθcosθ(sinθ-cosθ)
=(sinθ-cosθ)(1+4sinθcosθ)
=√[(sinθ+cosθ)^2-4sinθcosθ](1+4sinθcosθ)
=√(a^2-4a)*(1+4a)
=√(2√2-1)*(5-4√2)
(2)解析：∵sinθ+cosθ=a,sinθcosθ=a
∴tan(π-θ)-1/tanθ=-tanθ-1/tanθ=-[(tanθ)^2+1]/tanθ=-(secθ)^2/tanθ
=-1/(cosθsinθ)=-1/a=1+√2