Question

在三角形ABC中，角A,B,C所对的对边长分别是a,b,c.

(1)设向量x=(sinB,sinC),向量y=(cosB,cosC),向量z=(cosB,-cosC),若z//(x+y),求tanB+tanC的值； (2)若sinAcosC+3cosAsinC=0,证明：(a^2)-(c^2)=2b^2.

Answer 1

x+y=(sinB+cosB,sinC+cosC)
z//(x+y)
cosB/(sinB+cosB)=-cosC/(sinC+cosC)
1/(tanB+1)=-1/(tanC+1)
tanB+1=-(tanC+1)
tanB+tanC=-2

sinAcosC+3cosAsinC=0
sinAcosC+cosAsinC=-2cosAsinC
sin(A+C)=-2cosAsinC=sinB
-2cosA=sinB/sinC=b/c
cosA=(b^2+c^2-a^2)/(2bc)=-b/(2c)
整理得
(a^2)-(c^2)=2b^2

Answer 2

1.cosB=SINB+cosB -cosC=sinC+cosC SINB=0 SINC=-2COS

tanB+tanC=sinB/cosB+sinC/cosC=-2