设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数f(x)的单调递增区间
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![设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数f(x)的单调递增区间](/uploads/image/z/5265772-52-2.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%88+0%E5%88%86%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%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数f(x)的单调递增区间
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数f(x)的单调递增区间
设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数f(x)的单调递增区间
因为α∈R,f(x)=cosx(asinx-cosx)+cos^2(π/2-x)满足f(-π/3)=f(0)
所以有:cos(-π/3)[asin(-π/3)-cos(-π/3)]+cos^2[π/2-(-π/3)]=cos0(asin0-cos0)+cos^(π\2-0)=-1
解得:a=2√3
代入函数中得:
f(x)=cosx(2√3sinx-cosx)+cos^2(π\2-x)
=cosx(2√3sinx-cosx)+sin^2x
=2√3sinx*cosx-cos^2x+sin^2x
=√3*sin2x-cos2x
=2(√3/2*sin2x-1/2*cos2x)
=2(cosπ/6*sin2x-sinπ/6*cos2x)
=2*sin(2x-π/6)
对于正弦函数的单调递增区间为:
2π-π/2