已知向量a=(cosx/2,sinx/2),b=(cosx/2,-sinx/2),c=(1,-1),求证（a+b）垂直于（a-b）
问题描述：
已知向量a=(cosx/2,sinx/2),b=(cosx/2,-sinx/2),c=(1,-1),求证（a+b）垂直于（a-b）
设函数f(x)=(|a+c|平方-3）（|b+c|平方-3）（-π/2
数学人气：606 ℃时间：2019-10-23 07:08:57
优质解答
向量a+向量b=(cosx/2+cosx/2,sinx/2-sinx2)=(2cosx/2,0)
向量a-向量b=(cosx/2-cosx/2,sinx/2-(-sinx/2)=(0,2sinx/2)
(a+b).(a-b)=2cosx/2*0+0*2sinx/2=0.
∴(a+b)⊥(a-b).
|a+c|^2=(cosx/2+1)^2+(sinx/2-1)^2=cos^2(x/2)+2cos(x/2)*1+1+sin^2(x/2)-2sixnx/2+1.
=2+1-2(simx/2-cosx/2).
=3-2√2si(x/2-π/4)
|b+c|^2 =(cosx/2+1)^2+(-sinx/2-1)^2
=cos^2x/2+2cosx/2*1+1+sin^2x/2+2sinx/2+1.
=3+2√2sin(x/2+π/4).
|a+c|^2-3=2√2(sin(x/2-π/4).
|b+c|^2-3=2√2(sin(x/2+π/4).
(|a+c|^2-3)(|b+c|^3-3)=(2√2)^2(sinx/2-π/4)[sin(x/2+π/4].
=4cosx.
∴f(x)=4cosx.cos
∵x∈[-π/2,π/2],当x=-π/2,或π/2时,f(x)min=0,当x=0时,f (x)max=4cos0=4
∴x∈【-π/2,π/2],f(x)max=4,f(x)min=0.
向量a-向量b=(cosx/2-cosx/2,sinx/2-(-sinx/2)=(0,2sinx/2)
(a+b).(a-b)=2cosx/2*0+0*2sinx/2=0.
∴(a+b)⊥(a-b).
|a+c|^2=(cosx/2+1)^2+(sinx/2-1)^2=cos^2(x/2)+2cos(x/2)*1+1+sin^2(x/2)-2sixnx/2+1.
=2+1-2(simx/2-cosx/2).
=3-2√2si(x/2-π/4)
|b+c|^2 =(cosx/2+1)^2+(-sinx/2-1)^2
=cos^2x/2+2cosx/2*1+1+sin^2x/2+2sinx/2+1.
=3+2√2sin(x/2+π/4).
|a+c|^2-3=2√2(sin(x/2-π/4).
|b+c|^2-3=2√2(sin(x/2+π/4).
(|a+c|^2-3)(|b+c|^3-3)=(2√2)^2(sinx/2-π/4)[sin(x/2+π/4].
=4cosx.
∴f(x)=4cosx.cos
∵x∈[-π/2,π/2],当x=-π/2,或π/2时,f(x)min=0,当x=0时,f (x)max=4cos0=4
∴x∈【-π/2,π/2],f(x)max=4,f(x)min=0.
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答
向量a+向量b=(cosx/2+cosx/2,sinx/2-sinx2)=(2cosx/2,0)
向量a-向量b=(cosx/2-cosx/2,sinx/2-(-sinx/2)=(0,2sinx/2)
(a+b).(a-b)=2cosx/2*0+0*2sinx/2=0.
∴(a+b)⊥(a-b).
|a+c|^2=(cosx/2+1)^2+(sinx/2-1)^2=cos^2(x/2)+2cos(x/2)*1+1+sin^2(x/2)-2sixnx/2+1.
=2+1-2(simx/2-cosx/2).
=3-2√2si(x/2-π/4)
|b+c|^2 =(cosx/2+1)^2+(-sinx/2-1)^2
=cos^2x/2+2cosx/2*1+1+sin^2x/2+2sinx/2+1.
=3+2√2sin(x/2+π/4).
|a+c|^2-3=2√2(sin(x/2-π/4).
|b+c|^2-3=2√2(sin(x/2+π/4).
(|a+c|^2-3)(|b+c|^3-3)=(2√2)^2(sinx/2-π/4)[sin(x/2+π/4].
=4cosx.
∴f(x)=4cosx.cos
∵x∈[-π/2,π/2],当x=-π/2,或π/2时,f(x)min=0,当x=0时,f (x)max=4cos0=4
∴x∈【-π/2,π/2],f(x)max=4,f(x)min=0.