将函数fx=sin(2x π÷3)的图像向右平移 个单位

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

(1)f(x)=2sin(2x+π/3)+2由2x+π/3=kπ+π/2,k∈Z得2x=kπ+π/6,k∈Z对称轴方程为x=kπ/2+π/12,k∈Z(2)g（x）=f（x）+m=2sin(2x+π/

设函数fx=sin(φ-2x)(0

fx=2sin(2x+pai/6)振幅A=2最小正周期T=2pai/2=paix∈【0,pai/]2xE[0,2pai]2x+pai/6E[pai/6,2pai+pai/6]很明显,设u=2x+pai

周期等于2派.g(x)=2sinx;基函数再问：有过程吗？？再答：这可以看出来，还要过程吗，，，，周期等于2派/x前的数1===2派；；g(x)=2sint(x+pi/3+p1/3)=2sinx;si

1、最小正周期T=2π/2=π;最大值=2×1+2=4；2、单调递增式时-π/2+2kπ≤2x+π/3≤π/2+2kπ(k∈Z)-5π/6+2kπ≤2x≤π/6+2kπ(k∈Z)-5π/12+kπ≤x

1.T=πfx=2cosxsin（x+π/3）-√3sin^2x+sinxcosx=cosxsinx+√3cos^2x-√3sin^2x+sinxcosx=2sinxcosx+√3cos2x=sin2

1,=1/2sinwxcoswx+（1+cos2wx）/2=1/2sin2wx+1/2cos2wx+1/2=根号2/2sin(2wx+π/4)+1/22π/2w=π解得w=12,根号2/2sin(2x

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/

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)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

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

T=2π/2=π[-1,1]最大值为1,最小值为-1

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

(1)fx=sin(2x+φ)经过点(π/12,1)sin(π/6+φ)=1∴π/6+φ=π/2+2kπ,k∈Z∴φ=π/3+2kπ,k∈Z∵0

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

f(x)=sin(2x+π/6)-cos2x+1所以为2π/2=πf(x)=根号3/2sin2x-(cos2x)/2+1=sin(2x-π/6)+1所以最大值为2,x=π/2+2kπ-π/6=π/3+

f(x)=√3sin2x-2sin²x=√3sin2x-(1-cos2x)=2sin(2x+π/6)-1∴当sin(2x+π/6)=1时f(x)max=2*1-1=1