Question

f(x)=cosx(asinx-cosx)+cos^2(派/2 -x)满足f(-派/3)=f(0)

求函数f(x)在[派/4,11派/24]上的最大值和最小值

Answer 1

f(x)=cosx(asinx-cosx)+cos^2(π/2 -x)
∵f(-派/3)=f(0)
∴cos(-π/3)[a*sin(-π/3)-cos(-π/3)]+cos²[π/2-(-π/3)]=-1+cos²π/2
∴1/2[-√3/2a-1/2]+3/4=-1
∴a=2√3
∴f(x)=cosx(2√3sinx-cosx)+cos^2(π/2 -x)
=2√3sinxcosx-cos²x+sin²x
=√3sin2x-cos2x
=2(√3/2sin2x-1/2cos2x)
=2sin(2x-π/6)
∵x∈[π/4,11π/24]
∴2x∈[π/2,11π/12]
∴2x-π/6∈[π/3,3π/4]
∴2x-π/6=π/2时，sin(2x-π/6)=1,f(x)max=2
2x-π/6=3π/4时，sin(2x-π/6)=√2/2,f(x)min=√2

追问

-cos^2 x sin^2 x不等于cos 2x 啊

追答

∵cos2x=cos²x-sin²x

∴-cos²x+sin²x=-cos2x