已知向量a (sinx,2√3sinx),
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a.b=(sinx-cosx)(sinx+cosx)+2cosxsinx=sin2x-cos2x=3/5=>(sin2x-cos2x)^2=9/251-2sin2xcos2x=9/25sin4x=16
(1)f(x)=a·b=2sin²x+2√3sinxcosx=1-cos2x+√3sin2x=2(√3/2*sin2x-1/2*cos2x)+1=2sin(2x-π/6)+1由2kπ-π/2
f(x)=a·b=-2sin²x+2√3sinxcosx+2=cos2x-1+√3sin2x+2=2[sin2x*(√3/2)+cos2x*(1/2)]+1=2[sin2x*cos(π/6)
f(x)=(2sinx)×(√3cosx)+(cosx+sinx)×(sinx-cosx)f(x)=2√3sinxcosx-(cos²x-sin²x)f(x)=√3sin2x-co
第一题:(1):f(x)=2倍sinx的平方+2倍根号3cosxsinx-1化简为:f(x)=-2cos(2x+π/3)显然f(x)在x=0处去最小为-1;在x=π/3处取最大为2(2):f(x)=-
f(x)=a*b-√3=2sinxcosx+2√3(sinx)^2-√3=sin2x-√3[1-2(sinx)^2]=sin2x-√3cos2x=2sin(2x-∏/3)1,2k∏
f(x)=2(cosx)^2+2根号3sinxcosx=cos2x+1+根号3sin2x=2sin(2x+Pai/6)+1单调增区间是:-Pai/2+2kPai
f(x)=2sinxcosx+√3sinxcosx=(1+√3/2)sin2xx∈R时,单调减区间为:-π/2+2kπ再问:那如果是已知向量a=(sinx,√3),b=(2cosx,cosx),定义f
1、f(x)=2sin²x+2√3sinxcosx=1-cos2x+√3sin2x=2sin(2x-π/6)+1.当x∈[0,π/2]时,f(x)∈[2,3];若f(x)关于直线x=a对称,
f(x)=a*b=2sinxcosx+2√3(sinx)^2=sin(2x)+√3[1-cos(2x)]=2sin(2x-π/3)+√3,因为y=f(x+φ)=2sin(2x+2φ-π/3)+√3为偶
f(x)=向量a×向量b=(sinx,√3cosx)*(cosx,cosx)=sinxcosx+√3cosxcosx=1/2(2sinxcosx+2√3cosxcosx)=1/2(sin2x+√3co
f(x)=mn+a=2cosx+2√3sinxcosx+a=√3sin2x+cos2x+1+a=2sin(2x+π/6)+1+a当a=-1时f(x)=2sin(2x+π/6)因为x∈(0,π)===>
已知向量a=(2cosx,sinx),向量b=(0,√3cosx),f(x)=|向量a+向量b|.(1)求f(π/6)的值.(2)当x∈(0,π/3)时,求f(x)的值域.(1)解析:∵向量a=(2c
(1)a+b=(2cosx,sinx+√3cosx)得到f(x)=|向量a+向量b|=√(4cosxcosx+sinxsinx+3cosxcosx+2√3sinxcosx)=√(6cosxcosx+2
函数f(x)=向量a×向量b=(2cosx,2sinx)×(cosx,√3cosx)=2√3(cosx)^2-2sinxcosx=√3(cos(2x)-1)-sin2x=sin(pi/3-2x)-√3
f(x)的表达式是什么……再答:再问:cos(x-π/3)怎么化成1/2cosx+√3/2sinx再答:cos(a-b)=cosacosb+sinasinb那么这里cos(x-π/3)=cosx·co
|向量2a-向量b|²=(2cosx-√3)²+(2sinx+1)²=4cos²x+4sin²x+3+1-4√3cosx+4sinx=8+4(sinx
f(x)=2√3cosx^2+2sinxcosx=sin2x+√3(cos2x+1)=sin2x+√3cos2x+√3=2sin(2x+π/3)+√3后面应该会解吧?
f(x)=a·b=sin²x-√3sinxcosx²=1/2-(cos2x+√3sin2x)/2=1/2-sin(2x+π/6)单调递增区间2x+π/6∈[(2n+1/2)π,(2
(1)a⊥b则:f(x)=sinxcosx+√3cosxcosx=sin2x/2+√3(1+cos2x)/2=sin(2x+π/3)+√3/2=0∴2x+π/3=2kπ+3π/2±π/6∴x=kπ+7