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=2sinC
tanC=
∵cos2C=
=
且角C为锐角
∴cosC=
sinC=
,
sinA=
sinC =
cosA=
∵cosB=-cos(A+B)=-(cosAcosC-sinAsinC)
∴cosB=-(
×
-
×
)=-
∴cosC=
∵cos2C=
∴cosC=
sinA=
∵cosB=-cos(A+B)=-(cosAcosC-sinAsinC)
∴cosB=-(
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