標題:
A Maths,
發問:
sinA = 2cosBsinC , 其中A,B,C 是三角形ABC 的內角, 怎樣證明三角形是等腰三角形 ?
2cosBsinC=sinA =sin(180-A) =sin(B+C) =sinBcosC+sinCcosB 2cosBsinC=sinBcosC+sinCcosB cosBsinC=sinBcosC sin(B+C)=0 B+C=0 B=C As A,B,C are angles in a triangle, so that B= -C=C I hope I can help you as I am also a student 2007-02-01 20:14:04 補充： something wrong....cosBsinC=sinBcosCcosBsinC-sinBcosC=0sin(B-C)=0B-C=0B=Cdont mention that sentence= =""i always make such mistakes sorry
其他解答:
sinA=2cosBsinC sin(180-B-C)=2cosBsinC sin[180-(B+C)]=2cosBsinC sin(B+C)=sin(C+B)+sin(C-B) product to sum formula sin(C-B)=0 C-B=0 B=C 因為兩隻角一樣,所以就等腰~
A Maths,
發問:
sinA = 2cosBsinC , 其中A,B,C 是三角形ABC 的內角, 怎樣證明三角形是等腰三角形 ?
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最佳解答:2cosBsinC=sinA =sin(180-A) =sin(B+C) =sinBcosC+sinCcosB 2cosBsinC=sinBcosC+sinCcosB cosBsinC=sinBcosC sin(B+C)=0 B+C=0 B=C As A,B,C are angles in a triangle, so that B= -C=C I hope I can help you as I am also a student 2007-02-01 20:14:04 補充： something wrong....cosBsinC=sinBcosCcosBsinC-sinBcosC=0sin(B-C)=0B-C=0B=Cdont mention that sentence= =""i always make such mistakes sorry
其他解答:
sinA=2cosBsinC sin(180-B-C)=2cosBsinC sin[180-(B+C)]=2cosBsinC sin(B+C)=sin(C+B)+sin(C-B) product to sum formula sin(C-B)=0 C-B=0 B=C 因為兩隻角一樣,所以就等腰~
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