Question

在△ABC中,求证:sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)

高一数学.解三角形.请不要用和差积化公式.谢谢.

Answer 1

证明:
∵在三角形ABC中，
∴A+B+C=180度，得SINA=SIN(B+C)
则A/2=90度-(B+C)/2,得COSA/2=SIN((B+C)/2)
左消中袭边=Sin(B+C)+SinB+SinC
则4Cos(A/2)Cos(B/2)Cos(C/2)
=4Sin((B+C)/2)Cos(B/2)Cos(C/2)
=4Cos(B/2)Cos(C/拿兄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=右培蚂边
原式成立!