已知→a=(cosA,sinA),→b=(cosB,sinB),0小于B小于A小于派.(1)若|→a - →b|=根号2,求证:→a垂足→b.(2)设→c=(0,1),若→a + →b=→c,求A,B的值.

已知→a=(cosA,sinA),→b=(cosB,sinB),0小于B小于A小于派.
(1)若|→a - →b|=根号2,求证:→a垂足→b.(2)设→c=(0,1),若→a + →b=→c,求A,B的值.

已知→a=(cosA,sinA),→b=(cosB,sinB),0＜B＜A＜π
(1)若|→a - →b|=√2,求证:→a⊥→b.
(2)设→c=(0,1),若→a + →b=→c,求A,B的值.
(1)∵→a - →b=(cosA-cosB,sinA-sinB)
∴|→a - →b|=√[(cosA-cosB)²+(sinA-sinB)²]
=√(cos²A+cos²B-2cosAcosB+sin²A+sin²B-2sinAsinB)
=√[2-2(cosAcosB+sinAsinB)]
=√[2-2cos(A-B)]
∵|→a - →b|=√2
∴√[2-2cos(A-B)]=√2
∴cos(A-B)=0
∵0＜B＜A＜π
∴0＜A-B＜π
∴A-B=π/2即A=B+π/2
∴→a=(cosA,sinA)=(-sinB,cosB)
∴→a×→b=-sinBcosB+cosBsinB=0
∴→a⊥→b
(2)∵→c=→a + →b=(cosA+cosB,sinA+sinB),→c=(0,1)
∴cosA+cosB=0
sinA+sinB=1
∴cos²A+cos²B+2cosAcosB=0
sin²A+sin²B+2sinAsinB=1
两式相加,得
2+2cos(A-B)=1
∴cos(A-B)=-1/2
∵0＜A-B＜π
∴A-B=2π/3-----①
∵0＜B＜A＜π,cosA+cosB=0
∴A+B=π--------②
由①+②得,2A=5π/3即A=5π/6----③
由②+③得,5π/6+B=π即B=π/6
∴A=5π/6,B=π/6