Question

已知三角形ABC中,角C＝60°,求cosA*cosB范围

Answer 1

cosA*cosB=[cos(A+B)+cos(A-B)]/2 =[cos(180度-C)+cos(A-B)]/2 =[cos(180-60度)+cos(A-B)]/2 =[-1/2+cos(A-B)]/2 =-1/4+cos(A-B)/2 又0=