2a=5π/3 - 2b
sin2a=sin(5π/3 - 2b) = sin(2π-2b-π/3)
=sin(2b+π/3)= sin2bcosπ/3 +cos2bsinπ/3= sin2b /2 + √3cos2b/2
y=2-sin2a-cos2b
= 2-sin2b /2 -√3cos2b/2 -cos2b
=2-[sin2b+(√3+2)cos2b]/2
sin2b+(√3+2)cos2b= (√6+√2)sin(2b+θ)
tanθ=√3+2 > √3 θ∈(π/3,π/2)
2b+θ∈(π/3,3π/2)
sin(2b+θ) 能够取到最大值 1,不能取到最小值 -1
y=2-[sin2b+(√3+2)cos2b]/2
=2-(√6+√2)sin(2b+θ) /2
能取到最小值 （4-√6-√2）/2
没有最大值