Question

在三角形ABC中，求证：a/b-b/a=c(cosB/b-cosA/a)

Answer 1

利用余弦定理冲碰兆
c(cosB/b-cosA/a)
=c*[(a²+c²-b²)/(2abc)-(b²+c²-a²)/(2abc)]
=(a²+c²-b²)/(2abc)-(b²+c²吵慧-a²)/(2ab)
=(2a²-2b²)/(2ab)
=(a²-b²)/散租(ab)
=a²/(ab)-b²/(ab)
=a/b-b/a
得证

Answer 2

证姿丛明:
c(cosB/b-cosA/a)
=c{[(a^2+c^2-b^2)/2ac]/b-[(b^2+c^2-a^2)/2bc]/a}
=(a^2+c^2-b^2)/2ab-(b^2+c^2-a^2)/念姿2ab
=(2a^2-2b^2)/2ab
=(a^2-b^2)/ab
=a/b-b/a
所以a/b-b/a=c(cosB/仔册绝b-cosA/a