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扫码下载作业帮拍照答疑一拍即得

1、(a＋b)*(a－b)=|a|²－|b|²，因|a|=1，|b|=1，则(a＋b)*(a－b)=0，即：(a＋)垂直(a－b)；
2、|a＋c|²=|a|²＋2a*c＋|c|²=1＋2[cos(x/2)－sin(x/2)]＋2=3＋2[cos(x/2)－sin(x/2)]，|b＋c|²=|b|²＋2b*c＋|c|²=1＋2[cos(x/2)＋sin(x/2)]＋2=3＋2[cos(x/2)＋sin(x/2)]，则f(x)=2[cos(x/2)－sin(x/2)]×2[cos(x/2)＋sin(x/2)]==4[cos²(x/2)－sin²(x/2)]=4cosx
因x∈[－π/2，π/2]，则cosx∈[0，1]，f(x)∈[0，4]，则f(x)最小值是0，最大值是4

(a+b).(a-b)
=(2cos(x/2),0).(0,2sin(x/2))
=0
=> (a+b)垂直于(a-b)
|a+c|^2
= (cos(x/2)+1)^2+(sin(x/2)-1)^2
= 3 + 2(cos(x/2)- sin(x/2))
=3+2√2cos(x/2+π/4)
|b+c|^2
= (cos(x/2)+1)^2+(-sin(x/2)-1)^2
=3+2(cos(x/2)+sin(x/2) )
= 3+2√2cos(x/2-π/4)
f(x)
=(|a+c|^2-3)(|b+c|^2-3)
= 8cos(x/2+π/4)cos(x/2-π/4)
f'(x) = -4[ cos(x/2-π/4)sin(x/2+π/4) + cos(x/2+π/4) sin(x/2-π/4) ]
=-4sinx =0
x = 0 or π
f''(x) = -4cosx
f''(0) = -4 f''(π) = 4 >0 (min)
max f(x)=f(0) = 4
minf(x) = f(π) =-4

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.