Prove the identity for triangleABC(Show your step)
root(bcsinBsinC)=[(b^2)sinC+(c^2)sinB]/b+c
Prove trigonometric identity
2008-03-22 3:41 am
回答 (1)
2008-03-22 4:34 am
✔ 最佳答案
According sine LawsinB/b=sinC/c
bsinC=csinB
So root(bcsinBsinC)=bsinC
Also
[(b^2)sinC+(c^2)sinB]/b+c
=[(bc)sinB+(bc)sinC]/(b+c)
=bc(sinB+sinC)/(b+c)
=(bc)(sinC/c)
=bsinC
So root(bcsinBsinC)=[(b^2)sinC+(c^2)sinB]/b+c
2008-03-21 20:38:25 補充：
sinC/c
=k
=k[(b+c)/(b+c)]
=(sinB+sinC)/(b+c)
收錄日期: 2021-04-25 16:59:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080321000051KK02626