在△ABC中,角A B C所对的边a b c ,向量M=（2cos c/2,-sin(A+B)）,N=（cos c/2,

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在△ABC中,角A B C所对的边a b c ,向量M=（2cos c/2,-sin(A+B)）,N=（cos c/2,2sin(A+B)）
,M⊥N
若a2=b2+1/2C2,试求sin(A-B)的值

A+B+C =π
A+B = π-C
M⊥N
=>M.N=0
(2cosC/2,-sin(A+B)).(cosC/2,2sin(A+B)) =0
(cosC/2)^2- (sin(A+B))^2 =0
(cosC/2)^2 - (sinC)^2 =0
(cosC+1)/2 - (sinC)^2 =0
2(cosC)^2 + cosC -1 =0
(2cosC-1)(cosC+1 ) =0
cosC = 1/2
C = π/3
a^2 = b^2 + (1/2)c^2
= b^2 + c^2 - (1/2) c^2
by cosine rule
(1/2) c^2 = 2bc cosA
cosA = c/(4b) (1)
Also
b^2 = a^2- (1/2)c^2
= a^2 + c^2 - (3/2)c^2
by cosine rule
(3/2)c^2 = 2ac cosB
cosB = 3c/(4a) (2)
b/sinB = c/sinC
c/4b = sinC/(4sinB) = cosA ( from (1) )
also
a/sinA =c/sinC
3c/(4a) = 3sinC/(4sinA) = cosB ( from (2) )
sin(A-B)
= sinAcosB - cosAsinB
= sinA(3sinC/(4sinA) ) - (sinC/(4sinB) ).sinB
= sinC( 3/4 - 1/4)
= (1/2) sinC
= √3/4

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