(1-cosA)/(1-cosB)=[2sin^2(A/2)]/[2sin^2(B/2)]=sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]其中sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]是怎样的转化得来的?有公式,知识点的请说明
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(1-cosA)/(1-cosB)=[2sin^2(A/2)]/[2sin^2(B/2)]=sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]其中sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]是怎样的转化得来的?有公式,知识点的请说明
(1-cosA)/(1-cosB)=[2sin^2(A/2)]/[2sin^2(B/2)]=sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]
其中sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]是怎样的转化得来的?有公式,知识点的请说明
(1-cosA)/(1-cosB)=[2sin^2(A/2)]/[2sin^2(B/2)]=sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]其中sinA/sinB=2sin(A/2)cos(A/2)/[2sin(B/2)cos(B/2)]是怎样的转化得来的?有公式,知识点的请说明
解决方案:从已知的:COSA + COSB =√2sin50
COSA * COSB =(sin50)^ 2 -1 / 2
因为0
COSA+COSB>=1,COSA
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