已知向量a=(m,sin(x π 6) ) b=(cosx,1)

a·b=cos(3x/2)*cos(x/2)-sin(3x/2)*)sin(x/2)=cos(3x/2+x/2)=cos2x,向量a+b=[cos(3x/2)+cos(x/2),sin(3x/2)-s

(1)a.b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=cos(2x)(2)|a+b|=√{[cos(3x/2)+cos(x/2)]^2+[sin(3x/2)-sin(x

再问：第三问怎么做..思路..

解m(-2sinx,cosx),n=(√3cox,2cosx)f(x)=1-mn=1-(-2√3sinxcosx+2cosxcosx)=2√3sinxcosx-2cosxcosx+1=√3sin2x-

(1)、设b=(xcosθ,xsinθ)=(√3,-1）xsinθ/xcosθ=tanθ=-1/√3θ=150(2)、(cosθ,sinθ)(√3,-1）=02sin(θ-π/3)=0x=60(3)、

(1)a*b=cos3x/2cosx/2-sin3x/2sinx/2=cos(3x/2+x/2)=cos2x=2cos²-1|a+b|=√(cos3x/2+cosx/2)²+(si

1,f(x)=2√3sin(x/4)xcos(x/4)+2cos^2(x/4）=√3sin(x/2）+cos(x/2）+1=2sin(x/2+π/6)+1f(x)的最小正周期T=2π/w=4π,2,f

向量B的坐标表示写的不大清楚再问：向量b=(1,2sin(x+π/4)),再答：(1)函数f(x)=向量a*向量b=cos(2x-π/3）+sin(x-π/4)*2sin(xπ/4）=1/2*cos2

答案如下,记得平面向量运算是没有 "×"乘运算的哦!

f(x)=1-2[cos(x+π/8)*sin(x+π/8)+sin^2(x+π/8)]=1-sin(2x+π/4)-2sin^2(x+π/8)=cos(2x+π/4)-sin(2x+π/4)=√2c

∵向量a∥向量b,∴sin(πx/2)cos(π/3)-sin(π/3)cos(πx/2)=0.sin(πx/2-π/3)=0.πx/2-π/3=0.x=2/3.令f(x)=sin(πx/2-π/3)

（1）f(x)的最小正周期为π（2）f(x)的值域为[-2,2] 过程如下图： 

（1）∵f（x）=a·b∴f（x）=cos（2x-π/3）×1+sin(x-π/4)×2sin(x+π/4)=1/2cos2x+√3/2sin2x+2[﹙√2/2sinx﹚²-﹙√2/2co

[1].cosx+1/3=1,cosx=2/3,x=arxcos2/3.[2].cos^2x=4/9,sin^2x=5/9,sinx=√5/3.cos(2x+π/3)=cos2xcosπ/3-sin2

（1）f(x)=m(a•b+3sin2x)=m(sin(x+π2)cosx-sin 2x+3)sin2x]=m(cos2x-sin 2x+3sin2x)=2msin(2x+π6)…

请问楼主第二问给的条件等式是不是多了一个c?应该是“(2a-c)cosB=bcosC”吧?否则没法做!1.m={√3sin(x/4),1},n={cos(x/4),cos^(x/4)}m*n=√3si

（1）f(x)=2nm+b=2（cosx/2的平方+cosx/2*sinx/2）+b=cosx+1+sinx+b(运用到正弦余弦的二倍角公式)=（根号2）*sin(x+45°)+1+b正弦的增区间在[

f(x)=sin(π/2+x)cosx+cos（π-x)sinx=(cosx)^2-cosxsinx=((1+cos2x)/2)-(1/2)sin2x=(1/2)(cos2x-sin2x)+1/2=(

letP(x,y)MP=cosθMA+sinθMBOP-OM=cosθ(OA-OM)+sinθ(OB-OM)(x,y-1)=cosθ(1,0)+sinθ(0,1)(x,y-1)=(cosθ,sinθ)