设△ABC的内角A，B，C所对的边分别为a,b,c 且a+c=6，b=2，cosB=7/9

余弦定理b^2=a^2+c^2-2accosB 4=a^2+c^2-2ac*7/9 a=6-c c=3 a=6-3=3 a=c 角A=角C 2角A=180-角B sin(A-B)=sin(180-3B/2)=sin(3B/2) 用三倍角公式 sin(3B/2）=3sin(B/2)-4sin^3(B/2) 用半角公式 sin(3B/2）=3sin(B/2)-4sin^3(B&#