这个数学问题sinθ=√3／3 .该怎么解决呢?若sinθ=√3／3 ,则：cos(π-θ) / cos[sin(3π/2 +θ)] +cos(2π-θ) / cos(π+θ)*sin(π/2+θ)-sin(3π/2+θ)的值?

这个数学问题sinθ=√3／3 .该怎么解决呢?若sinθ=√3／3 ,则：cos(π-θ) / cos[sin(3π/2 +θ)] +cos(2π-θ) / cos(π+θ)*sin(π/2+θ)-sin(3π/2+θ)的值?
这个数学问题sinθ=√3／3 .该怎么解决呢?
若sinθ=√3／3 ,则：cos(π-θ) / cos[sin(3π/2 +θ)] +cos(2π-θ) / cos(π+θ)*sin(π/2+θ)-sin(3π/2+θ)的值?

这个数学问题sinθ=√3／3 .该怎么解决呢?若sinθ=√3／3 ,则：cos(π-θ) / cos[sin(3π/2 +θ)] +cos(2π-θ) / cos(π+θ)*sin(π/2+θ)-sin(3π/2+θ)的值?
cosθ=土√[1-(sinθ)^2]=土（√6）/3,
cos(π-θ)=-cosθ,
sin(3π/2 +θ)=-cosθ,
cos(2π-θ)=cosθ,
cos(π+θ)=-cosθ,
sin(π/2+θ)=cosθ,
∴cos(π-θ) / sin(3π/2 +θ) +cos(2π-θ) / cos(π+θ)*sin(π/2+θ)-sin(3π/2+θ)
=1-1*cosθ+cosθ
=1.