函数Fx=cos(2x π 3) sinx^2 cos

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

f(x)=1+cos2x+cos[2x+π/3]-1=cos2x+cos[2x+π/3]=2cos(2x+π/6)cos(π/6)=√3cos(2x+π/6)1）最小正周期T=2π/2=π单调增区间：

若cosα=3/5.α属于(3π/2,2π),sinα=-4/5把f(2α+π/3)代入fx=√2cos(x-π/12),化简原式=cos2α-sin2αcos2α-sin2α怎么化简的就不用我说了吧

=(1/2)sin2x-(根号3/2)cos2x+(根号3/2)cos2x+(1/2)sin2x+1+cos2x-1=sin2x+cos2x=根号2sin(2x+pi/4)最小正周期为pi-pi/4再

f(x)=√3sin2x+cos2x=2(sin2x*√3/2+cos2x*1/2)=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6)所以f(π/6)=2sin(2×π/

先化简f(x)=2根号3sinxcosx+2cos^2x-1=根号3sin2x+cos2x=2(根号3/2sin2x+1/2cos2x)=2sin(2x+π/6）则T=2π/ω=2π/2=πy=sin

fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2（α）-1fx=cos2x+根号3s

fx=sin2x-根号3*(1+cos2x)+a+根号3=2sin(2x-60°)+aT=pi,增区间[k*pi-pi/6,k*pi+5pi/12],k属于Z 2.由题意得-5pi/6<

f(x)=cos(2x-4π/3)+2cos^2x=cos(2x-4π/3)+cos2x+1=2cos(2x-2π/3)cos2π/3+1=1-√3cos(2x-2π/3)1.当cos(2x-2π/3

 再答：这是高一的题目吧再答：不谢，复习加油

F（X）=cos(√3x+t）F'（X）=-√3sin(√3x+t）F（X）+F'（X）=cos(√3x+t）-√3sin(√3x+t）是奇函数所以F（0）+F'（0）=0即cost-√3sint=0

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/

f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0

设函数fx=2cos^2(π/4-x)+sin(2x+π/3)-1=cos(PI/2-2x)+sin(2x+PI/3)=sin(2x)+sin(2x)/2+cos(2x)*sqrt(3)/2=sqrt

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

①f(x)=cos﹙2x－4π／3﹚＋2cos²x=cos2xcos4π/3+sin2xsin4π/3+1+cos2x=1/2cos2x-√3/2sin2x+1=cos(2x+π/3)+1当

f(x)＝√3sin2x＋cos2x＝2sin(2x＋π/6)∴f(x0)＝2sin(2x0＋π/6)＝6/5∴sin(2x0＋π/6)＝3/5∵x0∈[π/4,π/2]∴2x0＋π/6∈[2π/3,

解f(x)=2cos^2x+2√3sinxcosx-1=√3sin2x+cos2x=2sin(2x+π/6)∴最小正周期为：2π/2=π再答：不懂追问再问：在三角形ABC中，角ABC所对的边分别是ab