Question

三角形ABC 证明sin(A/2)*sin(B/2)*sin(C/2)

Answer 1

证明：sin(A/2)*sin(B/2)*sin(C/2)

=sin(A/2)*sin(B/2)*sin[(π-A-B)/2]

=sin(A/2)*sin(B/2)*cos[(A+B)/2]

=-0.5{cos[(A+B)/2]-cos[(A-B)/2]}*cos[(A+B)/2]

=-0.5{cos[(A+B)/2]}^2+0.5cos[(A-B)/2]cos[(A+B)/2]

可以看成是关于cos[(A+B)/2]的二次函数,显然当cos[(A+B)/2]=0.5cos[(A-B)/2]=0.5（此时cos[(A-B)/2]=1)时,sin(A/2)*sin(B/2)*sin(C/2)有最大值

-1/8+1/4=1/8

所以sin(A/2)*sin(B/2)*sin(C/2)