已知向量a=(cosx/2,sinx/2),b=(cosx/2,-sinx/2),c=(1,-1),

已知向量a=(cosx/2,sinx/2),b=(cosx/2,-sinx/2),c=(1,-1),

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.======以下答案可供参考======供参考答案1:(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 =0x = 0 or πf''(x) = -4cosxf''(0) = -4 f''(π) = 4 >0 (min)max f(x)=f(0) = 4minf(x) = f(π) =-4供参考答案2:1、(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

我好好复习下