Question

[急]数学:三角形ABC,AC/AB=cosB/cosC(1)证明B=C,(2)若cosA=-1/3,求sin(4B+派/3)的值

Answer 1

AC/AB=sinB/sinC=cosB/cosC，则sinCcosB－cosCsinB=sin(B－C)=0，B=C。cosA=－cos(B＋C)=－cos(2B)，则cos2B=1/3，sin2B=2√2/3，sin4B=2sin2Bcos2B=4√2/9，cos4B=2cos²2B－1=－7/9，=(1/2)sin4B＋(√3/2)cos4B

Answer 2

1 sinBcosC-sinCcosB=0
sin(B-C)=0
B-C=0+kπ
因为ABC为三角形
B-C=0 B=C
2.cos(B+C)= - cosA=1/3
cos2B=1/3
cos4B=2(cos2B)平方-1sin4B=1-(cos4B)平方=2√2/9
sin(4B+派/3)=
写不了了= =