设f(x)=2倍根号三sin(π-x)sinx-(sinx-cosx)²

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

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

1.2sin(30+2a)+4=5sin(30+2a)=1/230+2a=3030+2a=150a=0a=60tana=√32.f(x)=3sin²x+2√3sinxcosx+5cos

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!

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

第一题：（1）：f(x)=2倍sinx的平方+2倍根号3cosxsinx-1化简为：f(x)=-2cos（2x+π／3）显然f(x)在x=0处去最小为-1；在x=π／3处取最大为2（2）：f（x）=-

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

(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)=(1-cos4x)/2+√3/2sin4x=sin(4x-π/6)+1/2解不等式π/2+2kπ≤4x-π/6≤3π/2+2kπ(k∈Z)得π/6+kπ/2≤x≤5π/12+kπ/2(k∈Z

(1)因为sin²x=（1-cos2x）/2,sin(2x-π/3)=1/2sin2x-根号3/2cos2x所以函数f(x)=2倍根号3sin²x-sin(2x-π/3)=-sin

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

f(x)=6cos²x+√3sin2x=3+3cos2x+√3sin2x而f(A)=3-2√3化简得√3/2cos2A+1/2sin2A=-1--->sin(2A+60°)=-1解得A=60

原式=根号3倍的sin2x-cos2x化为2sin（2x-π/6）递增区间（kπ-π/6,kπ+π/3）递减（kπ+π/3,kπ+5π/6）最小值iffx=kπ-π/6时=sin（2kπ-π/2）=-

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)=1/(√x)²+2√xf(x)=1/x²+2x则1/x²+2x-x>2两边乘x²>0x³-2x²+1>0x³-x&su

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

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)/