已知sin(x-2π)

(1+cotx)sin^2x=sin^2x+sinxcosx2sin(x+π/4)sin(x-π/4)=根号2(sinx+cosx)*根号2/2(sinx-cosx)=sin^2x-cos^2xf(x

1、由于函数g(x)=sin(2(x－a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3－2a)=±1,sin(2a－π/

f(x)=cos^2x+sinxcosx=(1+cos2x)/2+1/2*sin2x=1/2+1/2(cos2x+sin2x)=√2/2*(√2/2*cos2x+√2/2sin2x)+1/2=√2/2

sin(PI/6+2x)=cos(PI/2-PI/6-2x)=cos(PI/3-2x)=cos(2*(PI/6-x))=1-2*sin(PI/6-x)^2=1-2*(1/4)^2=7/8tan70*c

∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间：π/2+2

f(x)=cosx＋sinxf(x)=√2sin(x＋π/4)(1)递增区间：2kπ－π/2≤x＋π/4≤2kπ＋π/2得：2kπ－3/4π≤x≤2kπ＋π/4递增区间是：[2kπ－3π/4,2kπ＋

∵f（x）=[-sinx(-sinx)cos(π+x)]／[-2cosxsin(π-x)]=[sin²x(-cosx)]／(-2cosxsinx)=1/2sinx∴最小正周期T=2π∴函数图

f(x)=2sin(π-x)sin(π/2-x)=2sinxcosx=sin2x1)最小正周期=2π/2=π2）在区间[-派/6,派/2]上x=π/4时,有最大值=sinπ/2=1x=-π/6时,有最

π2x+π/6属于[π/2+2kπ,3π/2+2kπ]时为减区间,所以x属于[π/6+kπ,2π/3+kπ],k属于Z列表：三行2x+π/60π/2π3π/22πx(根据上面一行的值求出x对应的值)f

f(x)=2sinx*sin(π/2+x)-2sin^2x+1＝2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)

因为f(x)=sinx+cosx=√2sin（x+π/4）第一题T=2π/1=2π第二题当sin（x+π/4）=1时,为最大值,即f(x)=√2sin（x+π/4）=-1时,为最小值,即f(x)=-√

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

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

∵x∈[0，π3]，∴π3≤x+π3≤2π3，根据正弦函数的性质得，32≤sin(x+π3)≤1，则3≤2sin(x+π3)≤2，∴f（x）的值域是[3，2]．故答案为：[3，2]．

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3（cos*2x-sin*2x）=sin2x+根号3cos2x=2sin（2x+派

（Ⅰ）f(x)＝sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f（x）的最小正周期T＝2π2＝π（Ⅱ）当x∈[−π8，3π8]时，

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

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)