已知x属于-π3,2π3求函数y=cos

f（x）=1-2sin^2x-sinx+3=4-2（sin^2x+1/2sinx+1/16）+1/16=-2（sinx+1/4）^2+65/16当x属于【π/6,π/2】,sinx属于【1/2,1】所

y=-2cos^2x+2sinx+3/2=-2+2sin^2x+2sinx+3/2=-2+2(sin^2x+sinx+1/4)+1=2(sinx+1/2)^2-1当sinx=-1/2即x=-π/6时,

y=3sin(π/6-2x)=--3sin（2x-π/6）(与y=3sin（2x-π/6）的单调区间相反)令-π/2+2kπ

遇到绝对值一定想分类讨论所以x>=0时,f(x)=x^2-2xx

1,已知函数f(x)=3sin(2x+π/6),若x属于[-π/6,π/6],求f(x)值域2x+π/6属于[-π/6,π/2]那么sin(2x+π/6)属于【-1/2,1】那么值域是[-3/2,3]

f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx

f(x)=-1+√3sin2x+2cos2x=-1+√7sin(2x+t)想想,你题目cos2x前面的系数2是不是多余的?

∵-1==1∴f(x)=sin(2x+π/6)+3/2>0∵sinx的周期是2π,∴sin2x的周期为π∵f(x)>0∴f(x)的最小正周期为π.f'(x)=2con(2x+π/6)当2nπ-π/2

f(x)=cos(2x-2π/3)-cos2x=-2sin（2x-π/3）sin（-2π/3）=√3sin（2x-π/3）最小正周期π,单调递增区间2kπ-π/2

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

sin^2x-cosx+3=-cos^2x-cosx+4令cosx=y则原式为-y^2-y+4配方-(y+1/2)^2+17/4由x∈R则-1

(cosx)^2-sinx+3=1-(sinx)^2-sinx+3=-(sinx)^2-sinx+4=-(sinx)^2-sinx-1/4+1/4+4=-[(sinx)^2+sinx+1/4]+17/

令t=2x+π/4则7π/12

F(x)=根号3(sin^2X-cos^2X)+2sinXcosX=√3cos2x+sin2x=2sin(2x+π/3)(1)X属于[0,2π/3]2x+π/3属于[π/3,5π/3]2sin(2x+

解(1)由f（x）=sin（π/3-2x）=-sin(2x-π/3)当2kπ-π/2≤2x-π/3≤2kπ+π/2,k属于Z时,y是减函数,即kπ-π/12≤x≤kπ+5π/12,k属于Z时,y是减函

f(1)=0x1--->f(x)=1/((x^21)/(x-1))=1/(x12/(x-1))=1/((x-1)2/(x-1)2)x>1--->f(x)=xf(x)>=1/(-2sqrt(2)2),x

f(x)=-根号3sin^2x+sinxcosx=-√3/2(1-cos2x)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=sin(2x+π/3)-√3/2(1)f((23π)/

（1）π（2）7/9再问：可以写出过程吗再答：fx=sinx-sin(x-π/3)=sinx-(1/2sinx-根号3/2cosx)=1/2sinx+根号3/2cosx=sin(x+π/3)所以T=2

当x∈[-π/3,π/4]时-√3

f(x)=sinxcosx-根号3sinx平方=(sin2x)/2+(根号3cos2x)-根号3/2=sin(2x+60°)-根号3/2所以最小正周期T=2π/2=πx属于〔0,π／2〕2x+60°属