设α∈R,f(x)=cosx(asinx-cosx)+cos^(π\2-x)满足f(-π\3)=f( 0分
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/03 06:12:02
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)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
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
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)
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
设α∈R,f(x)=cosx(asinx-cosx)+cos2( π 2 -x)满足f(- π 3 )=f(0),求函数
(2011•江西模拟)设a∈R,f(x)=cosx(asinx−cosx)+cos2(π2−x)满足f(−π3)=f(0
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
1.化简f(x)=cosx(asinx-cosx)+cos(π/2-x)cos(π/2-x).2.化简f(x)=4cos
设函数f(x)=cos(x+2π/3)+2(cosx/2)^2,x∈R
设函数f(x)=sinxcosx-3^(1/2)cos(x+π)cosx (x∈R)求f(x)的最小正周期
设函数f(x)=cos^4x-2asinx×cosx-sin^4x的图像的一条对称轴的方程x=-π/8
已知函数f(x)=cos(x-π/3)cosx,x∈R.
已知向量m=(asinx,cosx),n=(sinx,bcosx),其中a,b,x∈R,设函数f(x)=m*n满足f(π