Question

如何证明 在三角形abc中，cosa+cosb+cosc=1+r/r

Answer 1

r 为三角形内切圆的半径，R 为三角形外接圆的半径
证明：先证一个三角恒等式
sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
a+b+c＝2p
pr＝S＝(1/2)absinC
用正弦定理
Rr(sinA+sinB+sinC)=2RRsinAsinBsinC
再利用三角恒等式
r/R＝4sin(A/2)sin(B/2)sin(C/2)
cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)
cos(180-B-C)+cosB+cosC=1+2sin(A/2)[2sin(B/2)sin(C/2)]
cos(180-B-C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)]
-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)]
-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[cos(B/2-C/2)-cos(B/2+C/2)]
-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)cos(B/2-C/2)-2[cos(B/2+C/2)]^2
cosB+cosC=2cos(B/2+C/2)cos(B/2-C/2)