f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值

f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值

f(x)=cos(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0),求函数f(x)在[π/4,11π/24]上最大值和最小值
f(x)=asinxcosx-cos²x+sin²x
=(a/2)sin2x-cos2x
f(0)=-1,f(-π/3)=-(√3)a/4+1/2
则：-(√3)a/4+1/2=-1
-(√3)a/4=-3/2
a=2√3
所以,f(x)=√3sin2x-cos2x=2sin(2x-π/6)
x∈[π/4,11π/24]
则2x-π/6∈[π/3,3π/4]
则sin(2x-π/6)∈[√2/2,1]
所以,f(x)∈[√2,2]
即f(x)在x∈[π/4,11π/24]上的最大值为2,最小值为√2