Cho tam giác ABC thỏa mãn: sin(A/2).cos^3(B/2) = sin(B/2).cos^3(A/2). CMR tam giác ABC cân.
Bài này bạn có thể chứng minh như sau:
sin(A/2).cos^3(B/2) = sin(B/2).cos^3(A/2)
<=> sin(A/2).cos(B/2).[ 1 - sin^2(B/2)] = sin(B/2).cos(A/2)[1 -sin^2(A/2)]
<=> sin(A/2).cos(B/2) - sin(A/2).cos(B/2).sin^2(B/2) = sin(B/2).cos(A/2) - sin(B/2).cos(A/2).sin^2(A/2)
<=> sin(A/2).cos(B/2) - sin(B/2).cos(A/2) = sin(A/2).cos(B/2).sin^2(B/2) - sin(B/2).cos(A/2).sin^2(A/2)
<=> sin[(A-B)/2] = sin(A/2).sin(B/2).[cos(B/2).sin(B/2) - cos(A/2).sin(A/2)]
<=> sin[(A-B)/2] = 1/2.sin(A/2).sin(B/2).[sinB - sinA]
<=> sin[(A-B)/2] = sin(A/2).sin(B/2).cos[(A+B)/2].sin[(A-B)/2]
<=> sin[(A-B)/2].[1- sin(A/2).sin(B/2).cos[(A+B)/2] = 0 (***)
Vì [ 1 - sinA/2.sinB/2 cos(A+B)/2] >0 nên để (***) bằng 0 thì:
sin[(A-B)/2] =0
<=> A=B => tam giác ABC cân tại C