设f(x)=2根号3sin(兀-x)sinx-(sinx-

f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3)；y轴右侧的第一个最高点的横坐标为π/6,π/3ω

17/4根号3/2-321/64a/2带入后,平方,化简

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

f(x)=√2/2×cos[2x+(π/4)]+sin²x.=√2/2cos2x*√2/2-√2/2*sin2x*√2/2+sin²x=1/2cos2x-1/2sin2x+sin&

题目有错误,没办法做,请更正!

由题意可得:f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2=2sin(2x+π/3

f(x)=sin(2x＋π3)＋sin(2x－π/3)f(x)=sin(2x)cos(π/3)＋co(2x)sin(π/3)＋sin(2x)cos(π/3)－cos(2x)sin(π/3)f(x)=2

f（x）=2cosx*sin（x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+

f(x)=-根号3cos2X-sin2x=-2(根号3/2cos2x+1/2sin2x)=-2sin(2x+π/3)(1).T=2π/w=π（2）.由X属于[-π/3,π/3]得2x+π/3∈【-π/

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

题目是f(x=sin(x+π/3)+2sin(x+π/3)-√3cos(2π/3-x)!再问：设函数f(x)=sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x)再答：f(x)=

1、f(sin2)+f(sin(-2))=√(1-sin2)+√[1-sin(-2)]=√(1-sin2)+√(1+sin2)1-sin2=(sin1)^2+(cos1)^2-2sin1cos1=(s

f'(x)=(sinθ)x^2+((根号3)cosθ)x+tanθf'(π/4)=(sinθ)(π^2)/16+[(根号3)*(π/4)*cosθ]+tanθ题目很阴险啊,想让别人把θ和x弄混.

a·b=(2cosx,1)·(cosx,sqrt(3)sin2x)=2cosx^2+sqrt(3)sin2x=sqrt(3)sin2x+cos2x+1=2sin(2x+π/6)+1,故：f(x)=2s

函数f(x)=负根号3sin^2x+sinxcosx应该没^这个符号的吧?如果是没有的话f(x)=负根号3sin2x+sinxcosx=负根号3sin2x+sin2x=(1/2-3^(1/2))sin

1）因为f(x)=2cosx^2+根号3sin2x=1+2sin(2x+pi/6),所以最小周期T=pi.2)f（A)=2,且A大于0小于pi,所以A=pi/3,也就是60度,有A的余弦定理得b^2+

根据诱导公式进行化简这个要利用和角和倍角公式逆用

(sqrt是开方)f(x)=5sqrt(3)[cos(2x)]^2+sqrt(3)[sin(2x)]^2-4sin(x)cos(x),由于[cos(2x)]^2+[sin(2x)]^2=1,且2sin

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx=2c

f(x)=-根号3sin^2x+sinxcosx=-√3/2(1-cos2x)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=sin(2x+π/3)-√3/2(1)f((23π)/