Question

求证:若设三角形ABC的三边长分别为a,b,c,则有:

(sinA+sinB+sinC)(cotA+cotB+cotC)=1/2(a^2+b^2+c^2)(1/ab+1/bc+1/ac)

Answer 1

由
余弦定理
,
a²
=
b²+c²-2bc·cosA,
b²
=
c²+a²-2ca·cosB,
c²
=
a²+b²-2ab·cosC.
相加得a²+b²+c²
=
2(ab·cosC+bc·cosA+ca·cosB)
=
2abc(cosA/a+cosB/b+cosC/c).
由
正弦定理
,
a
=
2R·sinA,
b
=
2R·sinB,
c
=
2R·sinC.
代入得a²+b²+c²
=
abc/R·(cosA/sinA+cosB/sinB+cosC/sinC)
=
abc/R·(cotA+cotB+cotC).
于是1/2·(a²+b²+c²)(1/(ab)+1/(bc)+1/(ca))
=
(a²+b²+c²)(a+b+c)/(2abc)
=
(a+b+c)/(2R)·(cotA+cotB+cotC)
=
(a/(2R)+b/(2R)+c/(2R))(cotA+cotB+cotC)
=
(sinA+sinB+sinC)(cotA+cotB+cotC).