已知向量a=(cosx,sinx),b=(-siny,cosy),且x属于[π/4,π/2]1:若y=3x,求丨a+b丨2:若y=3x,且不等式a*b-根号2*丨a+b丨*K大于等于-3/2恒成立,求实数K的取值范围

问题描述:

已知向量a=(cosx,sinx),b=(-siny,cosy),且x属于[π/4,π/2]
1:若y=3x,求丨a+b丨
2:若y=3x,且不等式a*b-根号2*丨a+b丨*K大于等于-3/2恒成立,求实数K的取值范围

(1)
|a+b|=√(a^2+b^2+2*ab)=√(cosx^2+sinx^2+siny^2+cosy^2+2*sinx*cosy-2siny*cosx)=√(2+sin4x-sin2x-sin4x-sin2x)=√2-2sin2x=√2*√(sinx-cox)^2 {因为x属于[π/4,π/2]},所以=√2*(sinx-cosx)=2sin(x-π/4)
(2)
ab=-cosxsin3x+sinxcos3x=-sin2x
ab-√2|a+b|k≥-3/2
所以-sin2x-2k(sinx-cosx)≥-3/2
-sin2x-2k√(1-sin2x)≥-3/2
1-sin2x-2k√(1-sin2x)+1/2≥0
设√(1-sin2x)=t,所以0≤t≤1
t²-2kt+1/2≥0
因为k≤(t²+1/2)/2t要成立,则k要小于(t²+1/2)/2t的min
因为(t²+1/2)/2t=1/2(t+1/2t)≥√2/2
∴k≤√2/2

1. Ia+bI=√[(cosx-siny)^2+(sinx+cosy)^2]
=√[2+2sin(x-y)]
因y=3x
则原式=√[2(1-sin2x)]
=√2*√(sinx-cosx)^2
∵x属于[π/4,π/2],sinx>cosx
∴原式=√2(sinx-cosx)
2. y=3x
a*b=sinxcosy-sinycosx=sin(x-y)=-sin2x
a*b-根号2*丨a+b丨*K=-sin2x-2k(sinx-cosx)≥-3/2
即(sinx-cosx)^2-2k(sinx-cosx)+1/2≥0恒成立
则△=(-2k)^2-4*(1/2)k^2-√2/2

1)向量a+b=(cosx-siny,sinx+cosy)=(cosx-sin3x,sinx+cos3x)
|a+b|²=(cosx-sin3x)²+(sinx+cos3x)²
=cos²x-2cosxsin3x+sin²3x+sin²x+2sinxcos3x+cos²3x
=2-2(cosxsin3x-sinxcos3x)
=2-2sin2x
=2(sin²x+cos²x-2sinxcosx)
=2(sinx-cosx)²
因为x∈[π/4,π/2]
所以sinx>cosx
则|a+b|=√2(sinx-cosx)
2)向量ab=-cosxsin3x+sinxcos3x=-sin2x
ab-√2|a+b|k≥-3/2
即-sin2x-2k(sinx-cosx)≥-3/2
-sin2x-2k√(1-sin2x)≥-3/2
1-sin2x-2k√(1-sin2x)+1/2≥0
设√(1-sin2x)=t,则0≤t≤1
t²-2kt+1/2≥0
即k≤(t²+1/2)/2t要恒成立,那么k要小于(t²+1/2)/2t的最小值
又(t²+1/2)/2t=1/2(t+1/2t)≥√2/2
∴k≤√2/2

1.|a+b|^2=(cosx-sin3x)^2+(sinx+cos3x)^2
=2+2(sinxcos3x-cosxsin3x)
=2-2sin2x,
=2(sinx-cosx)^2,x属于[π/4,π/2] ,
∴|a+b|=(sinx-cosx)√2.
2.a*b=-cosxsin3x+sinxcos3x=-sin2x,
不等式a*b-根号2*丨a+b丨*K大于等于-3/2恒成立,
即-sin2x-2(sinx-cosx)k>=-3/2,
k设t=sinx-cosx,x∈[π/4,π/2],则t∈[0,1],sin2x=1-t^2,
w=(t^2+1/2)/(2t)>=(√2)/2,
∴k