能使y=sin(2x q) 根号3cos(2x q)
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y=根号3cos²x+sinxcosx-根号3/2=√3(1/2cos2x+1/2)+1/2sin2x-√3/2=√3/2cos2x+√3/2+1/2sin2x-√3/2=√3/2cos2x
定义域需满足:sinx-√3/2>=0,即sinx>=√3/2,得:2kπ+π/3=
y=sinα+√3cosα=2(1/sinα+√3/2cosα)=2(sinαcos(π/3)+cosαsin(π/3))=2sin(α+π/3)
因为是偶函数,所以,sin(y+x)+根号3cos(x-y)=sin(y-x)+根号3cos(-x-y)所以,可以展开:sinycosx+cosysinx+根号3cosxcosy+根号3sinxsin
答案是这个,再问:具体过程再答:久等了~公式难打~若满意请采纳~
这题只要判断sin根号2-sin根号3,cos根号2-cos根号3的正负;这里要知道π/2≈1.57,√2≈1.41,√3≈1.73;于是√2<π/2<√3<ππ/2-√3<π/2-√2,所以sin根
y=(√3/2)sin(x+π/2)+cos(π/6-x)=(√3/2)cosx+cos(π/6)cosx+sin(π/6)sinx=(√3/2)cosx+(√3/2)cosx+(1/2)sinx=√
把函数y=√[sin(π/3-2x)]看成y=√u(u≥0),u=sinv,v=π/3-2x的复合函数,√u是增函数,v=π/3-2x是减函数,∴y递增sinv递减,y的增区间由(2k+1/2)π
y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3
(1)原式=2[(1/2)sin(x/2)+(根号3/2)cos(x/2)]=2sin[(x/2)+pi/3]所以当[(x/2)+pi/3]=2kpi+pi/2时,y最大值为2解得x=4kpi+pi/
y=-(cos²x-sin²x)-√3sinxcosx=-cos2x-(√3/2)sin2x=-sin(2x+π/6)或者y=-(cos²x-sin²x)-√3
是求两个函数(1)y=√(sinx)(2)y=√(cosx)的定义域吧还是求(3)y=√sin(cosx)定义域(1)要使y=√(sinx)有意义,须令sinx≥0所以2kπ≤x≤π+2kπ,k∈z即
y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2+(√3/2)sin(2x)-1=(√3/2)sin(2x)-(1/2)cos(2x)-1/2=sin(2x-π/6)-1
y=sin(2x-π/3)+根号3cos2x=sin2xcosπ/3-cos2xsinπ/3+2sinπ/3cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.振幅=
y=cos²x+2√3sinxcosx-sin²x=cos²x-sin²x+2√3sinxcosx=cos2x+2√3sinxcosx=cos2x+√3sin2
y=2cosX*[sinXcospai/3+cosXsinpai/3]-根号3sin平方X+sinX·cosX=2cosX(1/2sinX+√3/2cosX)-根号3sin平方X+sinX·cosX=
3sinx-2sin^2x-1>=0令sinx=tt范围为[-1,1]3t-2t^2-1>=0所以(t-1/2)(t-1)
∵cos^2x-cos^2y=(cosx-cosy)(cosx+cosy=)=-2(sin(x/2-y/2))(sin(x/2+y/2))(cos(x/2-y/2))(cos(x/2+y/2))=-s
∵-1≤cosx≤1,所以2-cosx恒≥1∴原式可化为√3sinx=2Y-cosxY===>√3sinx+cosxY=2Y三角代换:sin(x+a)=2Y/√(3+y²)∴-1≤2Y/√(