cos(A)+cos(B)+cos(C)=0
sin(A)+sin(B) +sin(C)=0
求證
sin(A+B) = sinC
聽說是用力美弗定理 請問要怎樣証明捏
2006-06-12 10:19:32 · 3 個解答 · 發問者 綿羊 2 in 科學 ➔ 數學
對不起 題目抄錯
求證
sin(A+B) = sin(2C)
2006-06-12 13:13:59 · update #1
令Z1=cosA+isinA,Z2=cosB+isinB,Z3=cosC+isinCZ1,Z2,Z3的長度皆為1由題意知 Z1+Z2+Z3=0Z1+Z2=-Z3 長度也為1故Z1,Z2,Z1+Z2所構成的三角形為正三角形Z1,Z2的夾角為120度同理Z2,Z3的夾角也是120度Z1,Z3的夾角為120度於是可以假設B=C+120度A=B+120度=C+240度A+B=2C+360度sin(A+B)=sin2C
2006-06-12 17:06:17 · answer #1 · answered by ? 7 · 0⤊ 0⤋
也謝謝其他人給ㄉ意見ㄛ
2006-06-17 18:26:13 · answer #2 · answered by 綿羊 2 · 0⤊ 0⤋
sinA+sinB=-sinC => sin²A+2sinAsinB+sin²B=sin²C
cosA+cosB=-cosC => cos²A+2cosAcosB+cos²B=cos²C
相加 => 2+2(cosAcosB+sinAsinB)=1
=> cos(A-B)=-1/2 ...(*)
2006-06-15 23:34:28 補充:
sinA+sinB=-sinC
cosA+cosB=-cosC
相乘 => sinAcosA+sinAcosB+sinBcosA+sinBcosB=sinCcosC
=> sin2A+sin2B+2sin(A+B)=sin2C
=> 2sin(A+B)cos(A-B)+2sin(A+B)=sin2C [和差化積]
=> 2sin(A+B)[cos(A-B)+1]=sin2C [*]
=> sin(A+B)=sin2C
2006-06-15 19:34:20 · answer #3 · answered by Will 5 · 0⤊ 0⤋