sinxcosx的周期

y=√2/2*sin2x+(1+cos2x)/2-1/2=√2/2*sin2x+1/2*cos2x=√3/2*sin(2x+z)其中tanz=(1/2)/(√2/2)=√2/2所以T=2π/2=π

y=√3cos2x+sin2x=√[1²+(√3)²]sin(2x+z)=2sin(2x+z)其中tanz=√3/1=√3所以最大=2,最小=-2T=2π/2=π

f(x)=√3/2(1+cos2x)+1/2sin2x=sin(2x+∏/3)+√3/2,∴值域为〔-1+√3/2,1+√3/2],周期为∏.

f(x)=√3*(1+cos2x)/2+sin2x/2=√3/2+√3/2*cos2x+sin2x/2=√3/2+sin(2x+π/3)值域[√3/2-1,√3/2+1],周期为5π/6

函数f（x）=2sinxcosx=sin2x，故最小正周期是T=2πω=π，故答案为π．

f(x)＝cos2x-sin2x+2√3sinxcosx＝cos2x-sin2x+√3sin2x＝cos2x+(-1+√3)sin2x＝√(5-2√3)sin(2x+φ)＝(√5-√3)si(2x+φ

y=√3cos²x+sinxcosx=(√3/2)[cos2x+1]+(1/2)sin2x=sin(2x+π/3)+√3/2所以值域为[-1+√3/2,1+√3/2]最小正周期为2π/2=π

y=sin2x+2sinxcosx-cos2x=sin2x+sin2x-cos2x=2sin2x-cos2x=(√5)sin(2x-θ）其中θ满足cosθ=2/√5,sinθ=1/√5所以周期为π,值

y=cos2x+sinxcosx=cos2x+1/2*sin2x=√(1/4+1)*sin(2x+arctan2)于是,T=2π/2=π

y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下

f(x)=(cosx)^4-(sinx)^4-2sinxcosx=[(sinx)^2+(cosx)^2][(cosx)^2-(sinx)^2]-2sinxcosx=1*[(cosx)^2-(sinx)

f(x)=cos2x-√(3)sin2x=2（1/2cos2x-√3/2sin2x)=2(cos2xsinπ/6-sin2xcosπ/6)=2sin(2x-π/6)所以最小正周期T=2π/2=π

g

f（x）=1+tanx1−tanx=tanπ4+tanx1−tanπ4tanx=tan（x+π4），∵ω=1，∴T=π1=π，则函数f（x）的最小正周期为π．故选D

f(x)=（1/2）2sinx*cosx+√3*cosx^2=1/2sin2x+√3/2（2cos^2-1)+√3/2=1/2sin2x+√3/2cos2x+√3/2=cos60sin2x+cos2x

∵y=sin2x+2sinxcosx=1−cos2x2+sin2x=sin2x-12cos2x+12=52sin（2x+φ）+12，（tanφ=-12）∴其周期T=2π2=π．故答案为：π．

函数y=√3cos^2x+sinxcosx=√3/2*(1+cos2x)+1/2*sin2x=√3/2+√3/2*cos2x+1/2*sin2x=√3/2+sin(2x+π/3)周期T=π最大值√3/

y=1/2*sin2x+(1+cos2x)/2=1/2*(sin2x+cos2x)+1/2=√2/2*sin(2x+π/4)+1/2所以T=2π/2=πsin(2x+π/4）递增则2kπ-π/2

f(x)=√3*cos^2x+sinxcosx=√3*(1+cos2x)/2+sinxcosx=√3/2(1+cos2x)+1/2sin2x=sin(2x+∏/3)+√3/2∴值域为〔-1+√3/2,