设ABC是锐角三角形,a.b.c分别是内角A.B.C所对边长,并且sin^2A=sin(60+B)sin(60-B)+sin^2B.求角A的值

题目

设ABC是锐角三角形,a.b.c分别是内角A.B.C所对边长,并且sin^2A=sin(60+B)sin(60-B)+sin^2B.求角A的值

答案

sin^2A=sin(60+B)sin(60-B)+sin^2B
sin^2A=-1/2(cos(60+B+60-B)-cos(60+B-60+B)+sin^2B
sin^2A=-1/2(-1/2-cos2B)+1/2(1-cos2B)
1/2(1-cos2A)=-1/2(-1/2-cos2B)+1/2(1-cos2B)
1-cos2A=1/2+cos2B+1-cos2B
cos2A=-1/2
2A=120度
A=60度