正弦定理证明(在线等答案)
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正弦定理证明(在线等答案)
1.已知三角形ABC中,acosA=bcosB,试判断三角形ABC形状(详细过程)
2.在三角形ABC中.若(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),试判断三角形形状(详细过程)
1.已知三角形ABC中,acosA=bcosB,试判断三角形ABC形状(详细过程)
2.在三角形ABC中.若(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),试判断三角形形状(详细过程)
1、
正弦定理:a/sinA=b/sinB
=>
asinB=bsinA
又
acosA=bcosB
=>
sinB-cosA=sinA-cosB
=>
sinB+cosB=sinA+cosA
=>
A=B
=>
等腰三角形
2、
正弦定理:a/sinA=b/sinB
=>
[(sinA)^2+(sinB)^2]sin(A-B)=[(sinA)^2-(sinB)^2]sin(A+B)
=>
[(sinA)^2+(sinB)^2]/[(sinA)^2-(sinB)^2]=(sinAcosB+cosAsinB)/(sinAcosB-cosAsinB)
=>
[(sinA/sinB)^2+1]/[(sinA/sinB)^2-1]=[(tgA/tgB)^2+1]/[(tgA/tgB)^2-1]
=>
(sinA/sinB)^2=(tgA/tgB)^2
(sinA/sinB=-tgA/tgB时不可能,可仿以下推导,此处略)
=>
sinA/sinB=tgA/tgB
=>
sinA/sinB=(sinAcosB)/(cosAsinB)
=>
cosB/cosA=1
=>
A=B
=>
等腰三角形
正弦定理:a/sinA=b/sinB
=>
asinB=bsinA
又
acosA=bcosB
=>
sinB-cosA=sinA-cosB
=>
sinB+cosB=sinA+cosA
=>
A=B
=>
等腰三角形
2、
正弦定理:a/sinA=b/sinB
=>
[(sinA)^2+(sinB)^2]sin(A-B)=[(sinA)^2-(sinB)^2]sin(A+B)
=>
[(sinA)^2+(sinB)^2]/[(sinA)^2-(sinB)^2]=(sinAcosB+cosAsinB)/(sinAcosB-cosAsinB)
=>
[(sinA/sinB)^2+1]/[(sinA/sinB)^2-1]=[(tgA/tgB)^2+1]/[(tgA/tgB)^2-1]
=>
(sinA/sinB)^2=(tgA/tgB)^2
(sinA/sinB=-tgA/tgB时不可能,可仿以下推导,此处略)
=>
sinA/sinB=tgA/tgB
=>
sinA/sinB=(sinAcosB)/(cosAsinB)
=>
cosB/cosA=1
=>
A=B
=>
等腰三角形