题目

a+b+c=2π 证明sina+sinb+sinc=4sina/2sinb/2sinc/2

提问时间：2021-03-24

答案

(a+b)/2= π -c/2
sina+sinb+sinc
=2sin(a+b)cos(a-b)+2sin(c/2)cos(c/2)
=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]
=2sinc/2 *[cos(a-b)/2-cos[(a+b)/2]
=4sina/2sinb/2sinc/2
得证.