设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求f(x)在[π\4,11π\24]上的最大值和最小值
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![设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求f(x)在[π\4,11π\24]上的最大值和最小值](/uploads/image/z/3779832-48-2.jpg?t=%E8%AE%BE%CE%B1%E2%88%88R%2Cf%EF%BC%88x%EF%BC%89%3Dcosx%28asinx-cosx%29%2Bcos%5E%EF%BC%88%CF%80%5C2-x%EF%BC%89%E6%BB%A1%E8%B6%B3f%EF%BC%88-%CF%80%5C3%EF%BC%89%3Df%EF%BC%880%EF%BC%89%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B%CF%80%5C4%2C11%CF%80%5C24%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求f(x)在[π\4,11π\24]上的最大值和最小值
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)
求f(x)在[π\4,11π\24]上的最大值和最小值
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求f(x)在[π\4,11π\24]上的最大值和最小值
因为f(-π\3)=f(0),所以cos(-π\3)[asin(-π\3)-cos(-π\3)]+cos²(π\2+π\3)=cos0(asin0-cos0)+cos²π\2,即-1/2[-(√3)a/2+1/2]+3/4=-1,解得a=-2(√3).
f(x)=-2(√3)sinxcosx-cos²x+sin²x=-(√3)sin2x-cos2x=-2{[(√3)/2]sin2x+(1/2)cos2x}=-2sin(x+π/6).
因为x∈[π\4,11π\24],所以(x+π/6)∈[5π/12 ,5π/8].然后就在区间上找最值了.不一定算得对,不过过程就是这样,你自己再算算吧.