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
解 原题应为f(x)=cosx(asinx-cosx)+cos^2(π/2-x)=acosxsinx-cos^2（x）+cos^2(π/2-x)
=(a/2)sin2x-（1+cos2x）/2+（1+cos2(π/2-x）)/2=(a/2)sin2x-cos2x
所以f(-π/3)=(a/2)sin2(-π/3)-cos2(-π/3)=(a/2)*(-√3/2)+1/2
f(0)==(a/2)sin2*0-cos2*0=1
即(a/2)*(-√3/2)+1/2=1 a=-2/√3

周期函数啊

答案是 2 和 根号2 这是重庆的高考题