已知x属于[-π 3,2π 3],求函数y=-3sin^2x-4cosx 4

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

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

tanx=1/2tan2x=2tanx/(1-tan²x)=4/3sinx/cosx=tanx=1/2cosx=2sinxsin²x+cos²x=1所以sinx=√5/5

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=

f(x)=sin(2x+7π/4)+cos(2x-3π/4)=2sin(2x-π/4)由-π/2+2kπ≤2x-π/4≤π/2+2kπ,k∈Z,得-π/8+kπ≤x≤3π/8+kπ,k∈Z,即f(x)

a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),(1)a*b=(cos3x/2,sin3x/2)*(cosx/2,-sinx/2)=cos(3x/2)*cos(x/2)-

f（x）=√3sin²x+sinxcosx=√3/2(1-cos2x)+1/2sin2x=√3/2+sin(2x-π/6)f（x）=0,sin(2x-π/6)=-√3/2,2x-π/6=π/

a=(cos3/2x,sin3/2x),b=(cosx/2,-sinx/2),a^2=1b^2=1,ab=cos3/2xcosx/2-sin3/2xsinx/2=cos(3/2x_x/2)=cos2x

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

cosx=-3/5,所以,sin(x-60°)=sinxcos60°-cosxsin60°=0.9196；补充问题;公式：tan（α+β）=（tanα+tanβ）／（1－tanαtanβ）带入已知条件

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

x第二象限则cosx

｛x属于Nl1

x∈[-2π/9,π/6]3x+π/3∈[-π/3,5π/6]sin(3x+π/3)∈[-√3/2,1]2sin(3x+π/3)∈[-√3,2]函数的最大值=2函数的最小值=-√3

(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/

函数f(x)=4(cosxcosπ/2+sinxsinπ/2)*(sin2π/3cosx-cos2π/3sinx)-1.f(x)=4sinx*[(√3/2)cosx-(-1/2)sinx]-1.=2√

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

x∈(π/2,π)sinx=3/5所以cosx=-√[1-(3/5)^2]=-4/5所以tanx=sinx/cosx=-3/4所以tan(x+π/4）=(-3/4+1)/[1-(-3/4)*1]=1/

当x∈[-π/3,π/4]时-√3