Question

在三角形ABC中,试证明等式,sinA+sinB+sinC=4cosA/2*cosB/2*cosC/2

Answer 1

主要运用了和差化积和半角公式

答案是别人的：

sinA+sinB+sinC

=2sin((A+B)/2)cos((A-B)/2)+2sin(C/2)cos(C/2)

=2sin((π-C)/2)cos((A-B)/2)+2sin(π-(A+B)/2)cos(C/2)

=2cos(C/2)cos((A-B)/2)+2cos((A+B)/2)cos(C/2)

=2cos(C/2)(cos((A-B)/2)+cos((A+B)/2))

=2cos(C/2)2cos(A/2)cos(B/2)

=4cos(A/2)cos(B/2)cos(C/2)