求函数f(x)=(根号3 sinx sin(π 2

此题如果去化归的话,可能会比较复杂所以用分析法（1）在定义域内,sin平方x的周期显然是π,根号3sinxcosx=根号1.5*sin2x,所以周期也是π；综上,函数f(x)的最小周期为π（2）(si

f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).=√3sin(x+π/2)+sinx=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2sin(x

f(x)=2sinxcosx+2√3sin²x-√3=2sinxcosx+√3sin²x+√3(sin²x-1)=2sinxcosx+√3sin²x-√3cos

f(x)=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)-1≤sin(x-π/6)≤1-2≤f(x)≤2值域是[-2,2]

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(X)=负根号3sin平方x+sinxcosx=根号3/2-根号3/2cos2x+1/2sin2x=sin(2x-π/3)+根号3/2所以,周期为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

再问：十分感谢！第二题也帮我解答一下吧再答：

f(x)=3sin平方x+2根号3sinxcosx-3cos平方x=根号3sin2x-3cos2x=2根号3sin(2x-π/3）

f(x)=sin2(x+π)+根号3sin(x+π)sin(π-x)-1\2=sin2x－根号3sin²x-1/2=sin2x＋根号3/2cos2x－1=根号7/2sin（2x＋γ）－1co

f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx

f(x)=2cosx*(1/2*sinx+√3/2*cosx)-√3sin²x+sinxcosx=sinxcosx+√3*cos²x-√3sin²x+sinxcosx=2

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

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

f(25π/6)=f(π/6）3sin²x+sinxcosx=3sin²x+0.5sin2xf（x)=--根号下3sin²x+0.5sin2x然后根据函数的单调性就可求出

(1)已知函数f(x)=-(√3)sin²x+sinxcosx,求f(25π/6).f(x)=-(√3)sin²x+sinxcosx=(√3/2)(cos2x-1)+(1/2)si

f(x)=√3sin(2x-π／6)+2sin^2(x-π／12)=√3sin(2x-π／6)+1-cos(2x-π／6)=2(√3/2sin(2x-π／6)-1/2cos(2x-π／6))+1=2(

(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)=sin²x+√3sinx*sin(x+π/2)=sin²x+√3sinxcosx=(1/2)(1-cos2x)+(√3/2)sin2x=(√3/2)sin2x-(1/2)