Question

在三角形ABC中.求证:sinA+sinB+sinC=4*cosA/2*cosB/2*cosC/2

Answer 1

根据和差化积
sinA+sinB+sinC
=2sin[(A+B)/2]cos[(A-B)/2]+sinC
=2cos(C/2)cos[(A-B)/2]+2sin(C/2)cos(C/2)
=2cos(C/2){cos[(A-B)/2]+cos[(A+B)/2]}
=2cos(C/2)[2cos(A/2)cos(B/2)]
=4（cosA/2）（cosB/2)(cosC/2)

Answer 2

你家里有高中必修4的重难点吗？，那上面应该有，我还没学到，不好意思！

Answer 3

sinA+sinB+sinC
=2sin((A+B)/2)cos((A-B)/2)+sin(180-A-B)
=2sin((A+B)/2)cos((A-B)/2)+sin(A+B)
=2sin((A+B)/2)cos((A-B)/2)+2sin((A+B)/2)cos((A+B)/2)
=2sin((A+B)/2)(cos((A-B)/2)+cos((A+B)/2)))

Answer 4

4Cos(A/2)Cos(B/2)Cos(C/2)
=4Cos(B/2)Cos(C/2)(SinB/2·CosC/2+CosB/2·SiNC/2)
=4Sin(B/2)Cos(B/2)(Cos(C/2))^2+4Sin(C/2)Cos(C/2)(Cos(B/2))^2
=SinB(CosC+1)+SinC(CosB+1)
=Sin(B+C)+SinB+SinC=左边。