已知a=(cosx,sinx),b=(cosy,siny)a与b之间有关系
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![已知a=(cosx,sinx),b=(cosy,siny)a与b之间有关系](/uploads/image/f/4219076-20-6.jpg?t=%E5%B7%B2%E7%9F%A5a%EF%BC%9D%28cosx%2Csinx%29%2Cb%EF%BC%9D%28cosy%2Csiny%29a%E4%B8%8Eb%E4%B9%8B%E9%97%B4%E6%9C%89%E5%85%B3%E7%B3%BB)
(1)∵a∥b,a=(sinx+cosx,sinx-cosx),b=(sinx,cosx)∴cosx(sinx+cosx)=sinx(sinx-cosx),整理得sin2x+cos2x=0,∴tan2
令t=sinx-cosx=2sin(x-π4),由已知,2kπ≤x≤2kπ+π2,k∈Z,∴t∈[-1,1],f(t)=1−t2−(a−4)t+a,∴g(t)=-t2-(a-4)t+a+1≥0在[-1
f(x)=a*b=2sinxcosx+(sinx+cosx)(cosx-sinx)=sin2x+cos2x=√2sin(2x+π/4)1)当x∈[0,π/2]时2x+π/4∈[π/4,π/4+π]当2
解f(x)=a*b=2sinxcosx+2cos²x=sin2x+2cos²x-1+1=sin2x+cos2x+1=√2sin(2x+π/4)+1∴最小正周期为:T=2π/2=π∵
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
sinx+3cosx+a=0在(0,π)内有两相异的解α,β∴sin(x+π3)=−a2在(0,π)内有两相异的解α,β令f(x)=sin(x+π3)的对称轴是x=π6∴α+β=π3故答案为:π3
(1)a•b=sin2x+cos2x−2sinxcosx(1分)=1-sin2x=15(3分)∴sin2x=45(4分)(2)∵x∈(0,π4),2x∈(0,π2)∴cos2x=35(5分)于是tan
1)a-b=(-2cosx,2sinx/2-2cosx/2)f(x)=2+sinx-(1/4)[4cos²x+4(sin²x/2+cos²x/2-2sinx/2cosx/
(Ⅰ)∵f(x)=a•b=(cosx+sinx)•(cosx-sinx)+sinx•2cosx…(2分)=cos2x-sin2x+2sinxcosx=cos2x+sin2x…(4分)=2(22•cos
f(x)=向量a.向量b.=(1+sin2x)*1+(sinx-cosx)*(sinx+cosx).=1+sin2x-(cos^2x-sin^2x).=1+sin2x-cos2x.=1+√2sin(2
(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k
∵.a•b=cos2x−sin2x+23sinxcosx=cos2x+3sin2x=2sin(2x+π6)∴sin(2x+π6)=513∵x∈[−π4,π6],∴x∈[−π3,π2]∴cos(2x+π
sinx+cosy=a两边平方得2sinxcosy=a^2-1cosx+siny=a两边平方得2sinycosx=a^2-1两个式子相减的sinxcosy-cosxsiny=0即sin(x-y)=0当
解析:∵a*b=(cosx+sinx,sinx)*(cosx-sinx,2cosx)=(cosx+sinx)(cosx-sinx)+2sinxcosx=[(cosx)^2-(sinx)^2]+2sin
①f(x)=a•b=cosx(sinx+3cosx)+(cosx−3sinx)sinx…(2分)⇒f(x)=2sinxcosx+3(cos2x−sin2x)=sin2x+3cos2x…(4分)⇒f(x
解(1)f(x)=a•b=(sinx,−2cosx)•(sinx+3cosx,−cosx)=sin2x+3sinxcosx+2cos2x=sin(2x+π6)+32(4分)∴f(x)的最小正周期是π(
(I)f(x)=a•b=2cos2x+23sinxcosx=2sin(2x+π6)+1,故函数的周期为π.令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,可得 kπ-π3≤x≤
f(x)=(cosx+sinx)(cosx-sinx)+2sinxcosx=cos2x+sin2x=根号2sin(2x+Pi/4)故最小正周期T=2Pi/2=Pi(2)如果a//b,则有(cosx+s
1.f(x)=a·b=sinxsinx+sinxcosx=(1-cos2x)/2+1/2sin2x=1/2(sin2x-cos2x)+1/2=√2/2sin(2x-π/4)+1/2-π/2+2kπ
(1)f(x)=sinxcosx+cos^2x=1/2sin2x+1/2(2cos^2x-1)+1/2=1/2(sin2x+cos2x)+1/2=(根号2)/2sin(2x+π/4)+1/2T=2π/