Question

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，

，求tan(π-θ)-1/tanθ的值

Answer 1

sinθ+cosθ=a (1)
sinθcosθ=a (2)
(1)两边平方:
1+2sinθc0sθ=a^2
2sinθc0sθ=a^2-1 (3)
(3)/(2):2=a-1/a
a1=1+√2(舍去), a2=1-√2
cos³(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=1
tan(π-θ)-1/tanθ=-tanθ-1/tanθ
=-(tanθ+1/tanθ)=-1

Answer 2

sinθ+cosθ=a (1)
sinθcosθ=a (2)
(1)两边平方:
1+2sinθc0sθ=a^2
2sinθc0sθ=a^2-1 (3)
(3)/(2):2=a-1/a
a1=1+√2(舍去), a2=1-√2
cos³(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=1
tan(π-θ)-1/tanθ=-tanθ-1/tanθ
=-(tanθ+1/tanθ)=-1