问题标题：
△ABC的内切圆半径为R,外接圆半径为R,则r/(4R)得知等于A.sin（A/2)sin(B/2)sin(C/2)B.cos（A/2)cos(B/2)sin(C/2)C.sin（A/2）cos(B/2)cos(C/2)D.sin（A/2）sin(B/2)cos(C/2)求详解,最好标明公式的值sinA+sinB=2sin[(A+B)/2]cos[(A
问题描述：
△ABC的内切圆半径为R,外接圆半径为R,则r/(4R)得知等于
A.sin（A/2)sin(B/2)sin(C/2)
B.cos（A/2)cos(B/2)sin(C/2)
C.sin（A/2）cos(B/2)cos(C/2)
D.sin（A/2）sin(B/2)cos(C/2)
求详解,最好标明公式
的值
sinA+sinB
=2sin[(A+B)/2]cos[(A-B)/2]是怎么出来的？
高素芳回答：
解析：∵sinA+sinB+sinC
=sinA+sinB+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]}
=4cos(A/2)cos(B/2)cos(C/2),
∵S△ABC=(a+b+c)r/2=abc/4R
∴r/4R=abc/8R^2(a+b+c)
=sinAsinB*c/2(a+b+c)
=sinAsinBsinC/2(sinA+sinB+sinC)
=8sin(a/2)cos(a/2)sin(B/2)cos(B/2)sin(C/2)cos(C/2)/8cos(A/2)cos(B/2)cos(C/2)
=sin（A/2)sin(B/2)sin(C/2)
选择A
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