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=b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)
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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=b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)](/uploads/image/z/5166896-32-6.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A+B+C%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9a+b+c+%2C%E5%90%91%E9%87%8FM%3D%EF%BC%882cos+c%2F2%2C-sin%28A%2BB%29%EF%BC%89%2CN%3D%EF%BC%88cos+c%2F2%2C2sin%28A%2BB%29%EF%BC%89%2CM%E2%8A%A5N+%E8%8B%A5a2%3Db%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%EF%BC%8C%E8%A7%92A+B+C%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9a+b+c+%2C%E5%90%91%E9%87%8FM%3D%EF%BC%882cos+c%2F2%2C-sin%28A%2BB%29%EF%BC%89%EF%BC%8CN%3D%EF%BC%88cos+c%2F2%2C2sin%28A%2BB%29%EF%BC%89%2CM%E2%8A%A5N%E8%8B%A5a2%3Db2%2B1%2F2C2%2C%E8%AF%95%E6%B1%82sin%28A-B%29)
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=b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)
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=b
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
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=b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)
由于:M⊥N
则:M*N=0
即:(2cos(C/2),-sin(A+B))*(cos(C/2),2sin(A+B))=0
2cos^2(C/2)-2sin^2(A+B)=0
cos^2(C/2)=sin^2(A+B)
由于:A+B+C=π,
则:sin(A+B)=sin(π-C)=sinC
又:cos^2(C/2)=(1+cosC)/2
则:(1+cosC)/2=sin^2(C)=1-cos^2(C)
2cos^2(C)+cosC-1=0
则cosC=-1(舍)或1/2
则:C=π/3
因为a^2=b^2+(1/2)c^2
则由正弦定理:
sin^2(A)=sin^2(B)+(1/2)sin^2(C)
sin^2(A)-sin^2(B)=(1/2)sin^2(C)
[sin(A)+sin(B)][sin(A)-sin(B)]=(1/2)sin^2(A+B)
利用和差化积公式:
{2sin[(A+B)/2]cos[(A-B)/2]}*{2cos[(A+B)/2]sin[(A-B)/2]}=(1/2)sin^2(A+B)
逆用二倍角公式得:
sin(A+B)sin(A-B)=(1/2)sin^2(A+B)
因为sin(A+B)不等于0
则:sin(A-B)=(1/2)sin(A+B)=(1/2)sinC=√3/4
向量M=(2cos c/2,-sin(A+B)),N=(cos c/2, 2sin(A+B)),M⊥N
所以m点乘n=0,解得cosC=1/2
又sin(A-B)=sinAcosB-cosAsinB=a/2R乘(a2+c2-b2)/2ac-(b2+c2-a2)/2bc乘b/2R
经化简=(a2-b2)/2cR=1/2c2/2cR=1/2sinC=根号3/4