f(x)=根三sin(wx fai)

函数f(x)=sin(2x+φ)+根号三cos(2x+θ)=2sin（2x+θ+π/3）的定义域为：R则f(0)=0所以：0=sin（θ+π/3）,=>θ+π/3=2kπ,k∈Z即θ=2kπ-π/3所

f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3）cosx+sinx]cosx-(√3）（sinx)^2=2sinxcosx+(√3）[(cosx)^2-(

公式cos2x=1-2sin²x,可以知道sin²x=(1-cos2x)/2后面的√3sinxcosx=√3sin2x/2所以原式=-(cos2x)/2+(√3sin2x)/2+1

后面的sin应该有平方(1)f(x)=√3sin（2X－π/6）+2sin²(x-π/12)=√3sin（2X－π/6）+1-cos(2x-π/6)=3sin（2X－π/6）-cos(2x-

1、f(x)=1/2*sin(2x/3)+√3*(1+cos2x/3)/2=1/2*sin(2x/3)+√3/2*cos(2x/3)+√3/2=sin(2x/3)cosπ/3+cos(2x/3)sin

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)=5√3cos^x+√3sin^x-4sinxcosx=√3+2√3（1+cos2x)-2sin2x=3√3+4cos(2x+π/6),π/4

f(1)=3sin(2+π/3),f(2)=3sin(4+π/3),f(3)=3sin(6+π/3)2+π/3是第二象限角；4+π/3是第四象限角；6+π/3是第一象限角所以f(1)>0,f(2)0f

(1)f(x)=2sin(x+θ/2).cos(x+θ/2)+2倍根号三cos²(x+θ/2)-根号三=sin(2x+θ)+√3[cos(2x+θ)+1]-√3=sin(2x+θ)+√3[c

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x=[sinx+√3cosx+sinx]cosx-√3sin^2x=2sinxcosx+√3cos^2x-√3sin^2x=sin

f(x)=cos(x/2)=√3sin(x/2)cos(x/2)+cos^2(x/2)=√3/2sinx+1/2(cosx+1)=√3/2sinx+1/2cosx+1=sinxcos(π/6)+cos

f(x)=根号3cos^x+sinxcosx-根号3/2=根号3*(1+cos2x)/2+sin2x/2-根号3/2所以f(派/8)=根号3*(1+cos派/4)/2+sin(派/4)/2-根号3/2

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

（根号2＋根号6）÷4再问：如何做的？？？？？？？？？？谢谢

∵f(cosx)=sin3x∴f(sinx)=f[cos(π/2-x)]=sin[3(π/2-x)]=sin(3π/2-3x)=-cos3x选D再问：呃，不太明白怎么变的。。。。。。再答：f()括号内

答：手机提问无法在电脑中显示平方f(x)=√3/2-√3(sinwx)^2-sinwxcoswxf(x)=√3/2*[1-2(sinωx)^2]-(1/2)*2sinωxcosωxf(x)=(√3/2

f(x)=√3/2-√3sin²ωx-sinωxcosωx=√3/2(1-2sin²ωx)-1/2*2sinωxcosωx=√3/2*cos2wx-1/2sin2wx=cos2wx

2-x2再问：写下过程吧？？！！

f(x)=(sin²x+cos²x)+2cos²x+2√3sinxcosx-2=1+1+cos2x+√3sin2x-2=√3sin2x+cos2x=2sin(2x+π/6