已知三角形ABC中,角A,B,C对应的边为a,b,c,A等于2B,cosB等于3分之根号6,求sinC的值 若角A的平分...

(1)sinB=√3/3
cosA=cos2B=2(cosB)^2-1=1/3
sinA=2√2/3
sinC=sin(A+B)=sinAcosB+cosAsinB=5√3/9
(2)cos(A/2)=√6/3
sin(A/2)=√3/3
sin∠ADC=sin(A/2+C)=2√2/3
2/sinC=b/sin∠ADC
b=4√6/5