∵cosA=-√2/4
∴sinA=√14/4
由正弦定理,有
a/sinA=c/sinC
则 sinC=c*sinA/a
=√2×(√14/4)÷2
=√7/4
cosC=3/4
∵sinB=sin[π-(A+C)]=sin(A+C)
∴sinB=sinA*cosC+cosA*sinC
=(√14/4)×(3/4)+(-√2/4)×(√7/4)
=√14/8
故 b=a*sinB/sinA
=2×(√14/8)÷(√14/4)
=1
∵cos(A+π/6)=cosA*cos(π/6)-sinA*sin(π/6)
=(-√2/4)×(√3/2)-(√14/4)×(1/2)
=-(√6+√14)/8
∴cos(2A+π/3)=2cos²(A+π/6)-1
=2×[-(√6+√14)/8]²-1
=(√21-3)/8cosC=3/4 ∵sinB=sin[π-(A+C)]=sin(A+C) ∴sinB=sinA*cosC+cosA*sinC =(√14/4)×(3/4)+(-√2/4)×(√7/4) =√14/8 故 b=a*sinB/sinA=2×(√14/8)÷(√14/4)=1这里我觉得不必那么麻烦直接用cosA的余弦定理 就可以了 其他的 谢谢了