已知向量A=(sin(a π 6),1),b=(4,4cosa-
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/25 14:52:55
λ≥1或λ≤ -2详解点击下图
若a⊥b则ab=0sinω+cosω=0sinω=-cosω-兀/2
f(x)=a•b=sin(x2+π12)•cos(x2+π12)−cosx2•cosx2=12sin(x+π6)−1+cosx2=12(sinx•32−cosx•12)−12因为cosx=−35,且x
sin(α+π/6)=sinα·cos(π/6)+cosα·sin(π/6)=(√3/2)sinα+(1/2)cosα;所以sin(α+π/6)+cosα=(√3/2)sinα+(3/2)cosα;即
1:1/4(sinθ/1=(cosθ-2sinθ)/2,则2sinθ=cosθ-2sinθ,4sinθ=cosθ,tanθ=1/4)2:π/2(sinθ=1且cosθ-2sinθ=2,又0
向量a⊥向量b,所以cosa*cos²a+sina*sin²a=0tan³a=-1所以tana=-1即sina+cosa=0又sin²a+cos²a=
1.a=(cos(θ),-sin(θ)),b=(sin(θ),cos(θ)),a*b=0,故a⊥b2.x*y=k*a^2+t*(t^2+3)b^2=k+t*(t^2+3)=0(k+t^2)/t=-t^
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
f(x)=sin(x+π/6)cosx+cosxcos(x-π/3)-1/2cos(x-π/3)=cos(π/3-x)=cos[π/2-(x+π/6)]=sin(x+π/6)所以,f(x)=sin(x
|a+2b|^2=|a|^2+4a.b+|b|^2|a|^2=1,|b|^2=1,4a.b=4*(cosαcosβ+sinαsinβ)=4cosα-β=2所以|a+2b|^2=4,|a+2b|=2
(1),前面我未计算,如果f(x)=cos(2x-θ)是对的,则点(π/6,1)代人f(x)=cos(2x-θ)中,得:1=cos(2*π/6-θ).即,cos(π/3-θ)=cos0°.π/3-θ=
因为向量a和b垂直,所以,a*b=0,整理,得到4sin(a+π/6)+4cosa-根号3=0进一步整理,得到sin(a+π/6)+cosa=根号3/4,然后把sin(a+π/6)展开,得到√3/2s
|a+b|2=(sinθ+1)2+(cosθ+3)2=5+4sin(θ+π3),∴当θ=π6时,|a+b|2的最大值为5+4=9,故|a+b|的最大值为3.故答案为3
(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
老师说:数学题要自己做,给你找了公式http://zhidao.baidu.com/question/106090196.html自己学去,赶紧把分给我,不然期末考试扣你分
a*b=sinacosa+2=17/8sinacosa=1/8(sina-cosa)^2=1-2sinacosa=1-1/4=3/4由于a属于(0,Pai/4),故有sina
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=(
(1)f(x)=a*b=2sin(x+π/6)+2cosx=√3sinx+3cosx化成同名函数f(x)=2√3sin(x+π/6)(提取系数平方和)则2kπ-π/2≤x+π/6≤2kπ+π/2解得f
1A选择题要会小题小做,比如令φ=5π/6,可知a=(-√3,1)所以cos夹角=-1/2则夹角为120度经检验只有A符合则选A(想算可以,打字麻烦啊^-^)2D∵向量PA点成向量PB=向量PB点成向
cos(β-α)=4/5,tan²(α+β)={1/cos²[(β-α)+2α]}-1cos[(β-α)+2α]=(1/√2)(4/5±3/5)=1/5√2或者7/5√2tan&s