Question

求解数学必修五问题

在△ABC中，求证（a-c*cosB）/（b-c*cosA）=sinB/ sinA 麻烦各位学哥学姐们帮忙解一下， 顺便帮忙把过程写出来 谢谢！

Answer 1

正弦定理证明:
(a-c*cosB)/(b-c*cosA)
=(sinA-sinCcosB)/(sinB-sinCcosA)
=(sin(B+C)-sinCcosB)/(sin(A+C)-sinCcosA)
=sinBcosC/sinAcosC
=sinB/sinA

余弦定理证明:

cosB=(a^2+c^2-b^2)/2ac
cosA=(b^2+c^2-a^2)/2bc
(a-c*cosB)/(b-c*cosA)=[a-(a^2+c^2-b^2)/2a]/[b-(b^2+c^2-a^2)/2b]
=[(a^2-c^2+b^2)/2a]/[(b^2-c^2+a^2)/2b]
=b/a
=sinB/sinA(正弦定理)

Answer 2

a-ccosB
=a-c(a²+c²-b²)/2ac
=(2a²-a²-c²+b²)/2a
=(a²-c²+b²)/2a
同理b-ccosA
=(b²-c²+a²)/2b
所以左边=[(a²-c²+b²)/2a]/[(b²-c²+a²)/2b]
=(1/2a)/(1/2b)
=b/a
由正弦定理
b/a=sinB/sinA
命题得证

Answer 3

证明：
a-ccosB
=a-c(a²+c²-b²)/2ac
=(2a²-a²-c²+b²)/2a
=(a²-c²+b²)/2a
同理b-ccosA
=(b²-c²+a²)/2b
所以左边=[(a²-c²+b²)/2a]/[(b²-c²+a²)/2b]
=(1/2a)/(1/2b)
=b/a
b/a=sinB/sinA