设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数的单调递增
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设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f(0)求函数的单调递增区间
f(x)的区间
f(x)的区间
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
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
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