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-

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