设函数f(x)=sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)

来源：学生作业帮编辑：作业帮分类：综合作业时间：2024/05/14 19:28:24

设函数f(x)=sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)
求f(π/6) f(π/3)的值..
求过程,求速度..

因为：f(π/6)=sinπ/2+2sin(-π/6)-根号3cos(π/2)=1-1-0=0
所以：f(π/6) f(π/3)=0
再问：原式=sin（x+π/3）+√3cos（x+π/3）+2sin（x-π/3） =2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3) =2sin(x+π/3+π/6)+2sin(x-π/3) =2[sin(x+π/2)+sin(x-π/3)] =2(cosx+√3/2sinx-1/2cosx) =2(√3/2sinx+1/2cosx) =2sin(x+π/6) 为什么百度上有这个答案。。和你的不一样
再答：你看求得是什么？ f(π/6)——表示当x=π/6时的函数值 f(π/3)——表示当x=π/3时的函数值所以，——当然我这样算是正确的！！亲

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