1.由条件=> sinA=2sinBcosC
=> sinBcosC+cosBsinC=2sinBcosC
=> sinBcosC-cosBsinC=0 => sin(B-C)=0
=> B=C
2.由条件及正弦定理=> sinA/a=sinB/b=sinC/c=cosB/b=cosC/c
=> sinB=cosB sinC=cosC => B=C=∏/4
即为等腰直角三角形
3.由三角形恒等式sin^2A+sin^2B+sin^2C=2及条件
=> sin^2B=2/(2+3+4)=2/9*3=2/3 => sinB=(√6)/3
=> B=arc sin[(√6)/3]
4.(a^2+b^2-c^2)/(2ab)=0.5=cosC => C=∏/3
sinA*sinB=sinA*sin(2∏/3-A)
=(√3)/2*sinAcosA+0.5sin^2A
=(√3)/4*sin2A-1/4*cos2A+1/4=0.5sin(2A-∏/6)+1/4=3/4 => sin(2A-∏/6)=1 => A=∏/3
又C=∏/3 所以,为正三角形