已知x属于[﹣π 6,π 2},求函数y

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=（sinx+1)(cosx+1）=sinx+cosx+sinxcosx+1=√2*sin(x+π/4)+1/2*sin2x+1=√2*sin(x+π/4)-1/2*cos(2x+π/2)+1=√2

f(x)=1/2cos(2x+π/6)+sinxcosx=1/2cos(2x+π/6)+1/2sinxcosx=根下2乘以sin（2x+π/6+π/4)=根下2乘以sin（2x+5π/12)2x+5π

f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosx(√3/2cosx+1/2sinx)-√3i

值域：【0.5-b,0.5+b】,b=(根号2)/2,sin2a=1/3+(1/6)*根号(14）f(x)=(1-cos(2x))/2+sin(2x)/2=1/2+c*(sin(2x-pai/4))/

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

已知tanx=根号2,x属于(π,3π/2),第三象限.sinx/cosx=根号2sinx=cosx根号2因为：sin²x+cos²x=1所以,3cos²x=1cosx=

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]

∵-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

[-0.5,0.5]

cos2x=1-2sin^2x=7/8∵x∈（π/2,π）∴sinx=1/4cosx=-√15/4sin(x+π/6)-sin2x=sinxcosπ/6+cosxsinπ/6-2sinxcosx=1/

平方sin²x+2sinxcosx+cos²x=4/91+2sinxcosx=4/92sinxcosx=-5/90则cosx0(sinx-cosx)²=sin²

(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

tanx=-2,x属于[0,π],x=π-arctan2再问：可以说明一下怎么得到答案吗再答：arctanA∈（-π/2，π/2）arctan2∈（0，π/2）tanx=-2，x∈[0,π],，可知x

两角和公式cosxcosy-sinxsiny=17√2/26,cosx已知sinx根据sin^2x+cos^2x=1得到设cosy为a然后把siny表示出来解方程ok

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)/

当x∈[-π/3,π/4]时-√3

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cosx^2+1,=-sin2x-cos2x+3sin2x-cos2x,=2sin2x-2cos2x=2√2sin(2x-π/4)最小正周