已知向量a[cos2分之3x,sin2分之3x],b=[cos2分之x]
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(1)f(x)=a•b+1=23sinxcosx−2cos2x+1=3sin2x−cos2x=2sin(2x−π6),∴函数f(x)的最小正周期T=2π2=π,由−π2+2kπ≤2x−π6≤2kπ+π
a=(√3sinx/2,cosx/2),b=(cosx/2,-cosx/2)f(x)=a·b=√3sin(x/2)cos(x/2)-(cosx/2)^2=(√3/2)sinx-(1/2)cosx-1/
首先,求俩括号里的两个向量的点击a·b=2+cos2θ;c·d=2-cos2θ;再者,看求什么,f(x).最终得根号1+cos2θ<根号1-cos2θ;得cos2θ<0;θ取值范围:π/4+kπ<θ<
sin(α+β)=-3/5,cos(α-β)=12/13,π/2
f(sin(x/2))=cosx+1=1-2(sin(x/2))^2+1=2-2(sin(x/2))^2令y=sin(x/2)则f(y)=2-2y^2令y=cos(x/2)f(cos(x/2))=2-
(1)∵a=(1−2cos2ωx2, 1),b=(-1,cos(ωx+π3)),∴f(x)=a•b=2cos2ωx2−1+cos(ωx+π3)=cosωx+12cosωx−32s
(1)a·b=(cos2/3x,sin2/3x)*(cos2/x,-sin2/x)=cos2/3x*cos2/x-sin2/3x*sin2/x=cos(2/3x+2/x)=cos8/3x|a+b|=√
a^2=1,b^2=1,a*b=cos(α/2)cos(β/2)+sin(α/2)sin(β/2)=cos[(α-β)/2],由|a-b|=2√5/5两边平方得a^2-2a*b+b^2=4/5,所以c
向量a乘以向量b=cos(3x/2)乘以cos(x/2)-sin(3x/2)乘以sin(x/2)=cos(3x/2+x/2)(余弦函数两角和公式)=cos2x因为x属于(0,π/2),则2x属于(0,
1f(x)=a·b+2λ|a+b|a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))=cos(2x)|a+b|^2=|a|^2+|b|^2+2a·b=2+2c
1)aXb=cos(2/3x)cos(2/x)-sin(2/3x)(sin(2/x)=cos(2/3x+2/x)=cos(8/3x)=0所以8/3x∈π/2+kπ,k∈Z即x∈3π/16+3kπ/8,
由直角得:OA·OB=(a-b)(a+b)=a²-b²=0∴‖a‖=‖b‖由等腰得:‖OA‖=‖OB‖即‖a-b‖=‖a+b‖∴√(a-b)²=√(a+b)²∴
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)-
(Ⅰ)∵向量a=(cos2ωx−sin2ωx,sinωx),b=(3,2cosωx),∴f(x)=a•b=(cos2ωx-sin2ωx,sinωx)•(3,2cosωx)=3cos2ωx+sin2ωx
(本小题满分12分)(Ⅰ)因为函数f(x)=a•b=(cos2ωx-sin2ωx,sinωx)•(3,2cosωx)=3(cos2ωx-sin2ωx)+2sinωxcosωx=3cos2ωx+sin2
已知xa向量+3b向量=c向量,a向量=(2x,3分之2),b向量=(x,y),c向量=(-1,3分之8),求实数x,y的值.XA+3B=C2x^2+3x=-1①2/3x+3y=8/3②=>2x^2+
f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2=1/2*sinx+√3/2*(1-cosx)+√3/2=1/2*sinx-√3/2*cosx+√3=sin(x
1.已知向量a=(sin3分之x,cos3分之x),b=(cos3分之x,根号3cos3分之x),函数f(x)=向量a·向量b则有:f(x)=sin(3分之x)cos(3分之x)+cos(3分之x)*