f(x)=2sin²(π/4-x)-√3(sin²x-cos²x)
=2[√2/2sinx-√2/2cosx]²-√3cos2x
=(sinx-cosx)²-√3cos2x
=1-sin2x-√3cos2x
=1-2sin(2x+π/3)
所以最小周期T=π,最小值为-1..
f(x)=2sin²(π/4-x)-√3(sin²x-cos²x)
=2[√2/2sinx-√2/2cosx]²-√3cos2x
=(sinx-cosx)²-√3cos2x
=1-sin2x-√3cos2x
=1-2sin(2x+π/3)
所以最小周期T=π,最小值为-1..