已知F(X)=2sin(1 2x-π 6)图像

①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

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)=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-√

f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,

(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＋

周期是派最大值是2X=1/3派+K派（-1/6派+K派,1/3派+K派）

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)=sin2x＋cos2x－1=√2sin(2x＋π/4)－1.1、最小正周期是π,最大值时2x＋π/4=2kπ＋π/2,即x=kπ＋π/4,k是整数.再问：已知函数f(x)=2sin(∏-X)

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)=-√

（1）偶函数,则f(x)=f(-x)即：sin(2x+φ)=sin(-2x+φ),根据积化和差公式sin(2x)*cos(φ)+cos(2x)*sin(φ)=sin(-2x)*cos(φ)+cos(-

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

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

（根号2＋根号6）÷4再问：如何做的？？？？？？？？？？谢谢

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

f(x)=2sin(派-x)cosx=2sinxcosx=sin2x最小正周期=2pi/2=pi(pi就是“派”）f(-pi/6)=sin(-pi/3)=-(根号3)/2f(pi/2)=sin(pi)

f(sinx-1)=cos2x+2cos2x=1-2sin^2xf(sinx-1)=3-2sin^2x=-2(sinx-3/4)^2+7.5f(x)=-2(x+1/4)^2+7.5