已知三角形ABC的内角A.B.C 向量m=(cosA-2cosC,cosB),向量n=(2c-a,b),且m平行于n,求sinA/sinC当0
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已知三角形ABC的内角A.B.C 向量m=(cosA-2cosC,cosB),向量n=(2c-a,b),且m平行于n,求sinA/sinC当0
已知三角形ABC的内角A.B.C 向量m=(cosA-2cosC,cosB),向量n=(2c-a,b),且m平行于n,求sinA/sinC
当0
已知三角形ABC的内角A.B.C 向量m=(cosA-2cosC,cosB),向量n=(2c-a,b),且m平行于n,求sinA/sinC当0
两向量平行,则:
(cosA-2cosC)/(2c-a)=(cosB)/(b)
b(cosA-2cosC)=(2c-a)cosB
sinBcosA-2sinBcosC=2sinCcosB-sinAcosB
sin(A+B)=2sin(B+C)
sinC=2sinA
sinA/sinC=1/2
∵m||n
∴(cosA-2cosC)/(2c-a)=cosB/b 1
∵a/sinA=b/sinB=c/sinC
(a-2c)/(sinA-2sinC)=b/sinB 2
1式*2式得
(cosA-2cosC)/(2sinC-sinA)=cosB/sinB=1/tanB=1/tan[180-(A+C...
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∵m||n
∴(cosA-2cosC)/(2c-a)=cosB/b 1
∵a/sinA=b/sinB=c/sinC
(a-2c)/(sinA-2sinC)=b/sinB 2
1式*2式得
(cosA-2cosC)/(2sinC-sinA)=cosB/sinB=1/tanB=1/tan[180-(A+C)]
=-1/tan(A+B)=(tanAtanC-1)/(tanA+tanC)
(cosA-2cosC)*(tanA+tanC)=(tanAtanC-1)(2sinC-sinA)
cosAtanA+cosAtanC-2cosCtanA-2cosCtanC=tanAtanC2sinC-sinAtanAtanC-2sinC+sinA
sinA+cosAtanC-2cosCtanA-2sinC=tanAtanC2sinC-sinAtanAtanC-2sinC+sinA
cosAtanC-2cosCtanA=tanAtanC2sinC-sinAtanAtanC 两边乘cosAcosC
(cosA)^2sinC-2(cosC)^2sinA=2sinA(sinC)^2-(sinA)^2sinC
[1-(sinA)^2]sinC-2[1-(sinC)^2]sinA=2sinA(sinC)^2-(sinA)^2sinC
sinC-(sinA)^2sinC-2sinA+2sinA(sinC)^2=2sinA(sinC)^2-(sinA)^2sinC
sinC-2sinA=0
sinA/sinC=1/2
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