设函数f(x)=2根号3sinx-cosx)cosx (

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&

你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏／8代入原式子,即sin(∏／4+φ)=1或者-1,再用(-π

题目有错误,没办法做,请更正!

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/

1）由三角函数和差化积公式：f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6

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)=

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

f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知：y=-3

1：(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2

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+

1）f(x)=sin(2x+φ)一条对称轴是X=π/8则kπ+π/2=2*π/8+φ===>φ=kπ+π/4因为-π

2x+φ＝kπ＋π/2,x＝(kπ＋π/2－φ)/2(kπ＋π/2－φ)/2＝π/8当k＝0时,φ＝π/4

1.由f(x)=sin(2x+φ)一条对称轴是直线x=π/2可得：在x=π/2时,函数取极值.则2*π/2+φ=kπ+π/2(k∈Z)φ=kπ-π/2又-π

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)/

f(x)=sin(2x+a)是R上的偶函数有f(x)=f(-x);sin(2x+a)=sin(-2x+a)=cos(π/2-(-2x+a))=cos(π/2+2x-a)余弦函数为R上的偶函数,a=π/