题目内容
已知角B为钝角的△ABC的内角A、B、C所对应边分别为a,b,c,若a=c,cosC=sinA,则cosB= ( )
A. -
B. -
C. -
D. -
A. -
B. -
C. -
D. -
A
∵a=c,由正弦定理知,sinA=sinC
∴cosC=sinA=2sinCtanC=
∵cos2C= =且角C为锐角
∴cosC=sinC=,
sinA=sinC =cosA=
∵cosB=-cos(A+B)=-(cosAcosC-sinAsinC)
∴cosB=-(×-×)=-
∴cosC=sinA=2sinCtanC=
∵cos2C= =且角C为锐角
∴cosC=sinC=,
sinA=sinC =cosA=
∵cosB=-cos(A+B)=-(cosAcosC-sinAsinC)
∴cosB=-(×-×)=-
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