在∆ABC中,証明
sin(B-C)/sin(B+C)=[bcosC-ccosB]/[bcosC+ccosB]
a-maths~~~~~~
2007-10-18 1:51 am
回答 (2)
2007-10-19 7:03 am
✔ 最佳答案
如下:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/Crazytrigo11.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/Crazytrigo12.jpg
參考: My Maths knowledge
2007-10-18 2:34 am
sin(B-C)/sin(B+C)
=(sinBcosC-sinCcosB)/(sinBcosC+sinCcosB)
=((sinB/sinC)cosC-cosB)/((sinB/sinC)cosC+cosB)
=((b/c)cosC-cosB)/((b/c)cosC+cosB)............................(applying sin law)
=(bcosC-ccosB)/(bcosC+ccosB)
=(sinBcosC-sinCcosB)/(sinBcosC+sinCcosB)
=((sinB/sinC)cosC-cosB)/((sinB/sinC)cosC+cosB)
=((b/c)cosC-cosB)/((b/c)cosC+cosB)............................(applying sin law)
=(bcosC-ccosB)/(bcosC+ccosB)
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