fx等于2sin×cosx+2分之根号3

fx=2sinxcosx=sin2x所以最小正周期2π/2=π[-π/6,π/2]上,2x∈[-π/3,π]最小-√3/2,最大1

f(x_=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=cos2x所以T=2π/2=πf(α/2)=cosα=1/3sin²α+cos²

f(x)=sin2x+2√3cosxcosx=sin2x+√3(1+cos2x)=sin2x+√3cos2x+√3=2sin(2x+π/3)+√3T=2π/2=π

(sin(cosx^2))'=cos(cosx^2)*(cosx^2)'=cos(cosx^2)*(-sinx^2)*2x=-[2xcos(cosx^2)*sinx^2]

f(x)=cosx-cos(x+π/2)=cosx+sinx=3/4sin^2x+cos^2x+2sinxcosx=9/162sinxcosx=sin2x=9/16-1=-7/16

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

f(0)=sin(0-π/6)+cos0=sin(-π/6)+cos0=-1/2+1=1/2如果想问的是化简后的结果,那么：f(x)=sin(x-π/6)+cosx=sinxcos(π/6)-cosx

f（x）=2（sinx+cosx）.cosx=2sinxcosx+2(cosx)^2=sin2x+2(cosx)^2-1+1=sin2x+cos2x+1所以f(x)的最小正周期为π

向量m=(2sinx/4,2sin^2x/4-1),n=(cosx/4,-√3)f(x)=mn=2sin(x/4)cos(x/4)-√3[2sin^2(x/4)-1]=sin(x/2)+√3cos(x

f（x）=2sinx*cosx=sin2x.可以求最小正周期、值域、单调递增区间.tana的平方=(1-cosa的平方)/cosa的平方=4,可以求cosa的平方=0.2.f（a）=2*tana*co

fx=4cos²x-2+1-cos²x-4cosx=3cos²x-4cosx-1令t=cosx则-1≤t≤1即求[3t²-4t-1]的最值

原式通分=[(sinx-cosx)²+(sinx+cosx)²]/(cosx+sinx)(cosx-sinx)=2(sin²x+cos²x)/(cos²

答：f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x

1-cosx=1-cos(x/2+x/2)=1-cos²x/2+sin²x/2=2sin²x/2

f(x)=2sin(x-π/6)cosx+2cos²x=(2sinxcosπ/6-2cosxsinπ/6)cosx+2cos²x=√3sinxcosx-cos²x+2co

第一题A.第二题B

（1）化简可得f(x)=(sin(x/2))^2+((√3)/2)sinx-0.5f'(x)=sin(x/2)cos(x/2)+((√3)/2)cosx=sinx+√3cosx=0√3cosx=-si

你的分析前一半是对的,一直到“那么2x的单调增区间是[-4分之π,4分之π]”.2x的单调递增区间是[-π/2,π/2],x的才是[-π/4,π/4].所以函数在x=-π/3处取得最小值为-2分之根号

等于+-根号下(1+cos)/2

f(x)=sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2sin(x+π/4)再问：为什么提根号2？用的半角公式还是什么？再答：√2是sinx与cosx系数平方和的平方根也就是√（