求函数f(x)=cosx+cos(x+π/3)的最大值,并求函数f(x)取最大值时所对应的x的集合
来源:百度知道 编辑:UC知道 时间:2024/07/05 15:15:05
应用和差化积公式:
cosx+cos(x+π/3)
=2cos[(2x+π/3)/2]*cos(-π/6)
=(根号3)*cos[(2x+π/3)/2]
=(根号3)*cos(x+π/6)
所以最大值是: 根号3
最大值点为 x+π/6=2kπ
所以x =2kπ - π/6 (k属于整数)
f(x)=cosx+cos(x+π/3)
=cosx+1/2cosx-√3/2sinx
=√3(√3/2cosx-1/2sinx)
=√3cos(x+π/6)
cos(x+π/6)=1,
f(x)max=√3
此时x的集合x=2kπ-π/6
f(x)=cosx+cos(x+π/3)
=2cos(x+π/6)cos(-π/6)
= (3开根号) cos(x+π/6)
-1<=cos(x+π/6)<=1
f(x)max=(3开根号)
此时:x+π/6=2kπ+π/2
x=2kπ+π/3 ,k为正整数。
f(x)
=cosx+cosxcosπ/3-sinxsinπ/3
=3/2cosx-√3/2sinx
=√3(√3/2cosx-1/2sinx)
=√3(sinπ/3cosx-cosπ/3sinx)
=√3sin(π/3-x)
当x=-π/6+2kπ(k∈Z)时,f(x)有最大值√3
原式展开:
f(x)=cosx+cosxcosπ/3-sinxsinπ/3
=3/2cosx-√3/2sinx
=√3(√3/2cosx-1/2sinx)
=√3cos(30°+x)
当x=60°+n*360°时 f(x)max=√3
f(x)=cosx+cos(x+π/3)
=cosx+1/2cosx-√3/2sinx