你这里是不是少打个条件x∈[0,π/2]啊
f(x)=cos^4x-2sinxcosx-sin^4x
=(cosx)^2-(sinx)^2-2sinxcosx
=cos2x-sin2x
=(√2)cos(2x+π/4),
(1)f(x)的最小正周期是π.
(2)x∈[0,π/2],
∴2x+π/4∈[π/4,5π/4],
∴f(x)的最小值是-√2,这时2x+π/4=π,x=3π/8.
你这里是不是少打个条件x∈[0,π/2]啊
f(x)=cos^4x-2sinxcosx-sin^4x
=(cosx)^2-(sinx)^2-2sinxcosx
=cos2x-sin2x
=(√2)cos(2x+π/4),
(1)f(x)的最小正周期是π.
(2)x∈[0,π/2],
∴2x+π/4∈[π/4,5π/4],
∴f(x)的最小值是-√2,这时2x+π/4=π,x=3π/8.