在△ABC中,求证:a/b-b/a=c(cosB/b-cos A/a)

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在△ABC中,求证:a/b-b/a=c(cosB/b-cos A/a)

在△ABC中,求证:a/b-b/a=c(cosB/b-cos A/a)
在△ABC中,求证:a/b-b/a=c(cosB/b-cos A/a)

在△ABC中,求证:a/b-b/a=c(cosB/b-cos A/a)
右边
=c/b·cosB-c/a·cosA
=c/b·(a^2+c^2-b^2)/(2ac)-c/a·(b^2+c^2-a^2)/(2bc)
=(a^2+c^2-b^2)/(2ab)-(b^2+c^2-a^2)/(2ab)
=(2a^2-2b^2)/(2ab)
=(a^2-b^2)/(ab)
=a/b-b/a
=左边

a/b-b/a=c(cosB/b-cos A/a) is true
iff
a^2*c-b^2*c=ac^2 cosB - bc^2 cosA
iff
a^2-b^2=ac cosB - bc cosA
iff
2(a^2-b^2)=2ac cosB - 2bc cosA
iff
[a^2-b^2 + 2bc cosA] + a^2-...

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a/b-b/a=c(cosB/b-cos A/a) is true
iff
a^2*c-b^2*c=ac^2 cosB - bc^2 cosA
iff
a^2-b^2=ac cosB - bc cosA
iff
2(a^2-b^2)=2ac cosB - 2bc cosA
iff
[a^2-b^2 + 2bc cosA] + a^2-b^2 = 2ac cosB (Note that a^2-b^2 + 2bc cosA = c^2 by Cosine Law)
iff
c^2 + a^2-b^2 = 2ac cosB which is true by Cosine Law
therefore a/b-b/a=c(cosB/b-cos A/a) is true

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正弦定理
c(cosB/b-cos A/a)
=sinC(cosB/sinB-cosA/sinA)
=sinC(sinAcosB-cosAsinB)/sinAsinB
=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)/sinAsinB
=(sin^2Acos^2B-cos^2Asin^2B)/sinAsinB
=(sin...

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正弦定理
c(cosB/b-cos A/a)
=sinC(cosB/sinB-cosA/sinA)
=sinC(sinAcosB-cosAsinB)/sinAsinB
=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)/sinAsinB
=(sin^2Acos^2B-cos^2Asin^2B)/sinAsinB
=(sin^2A(1-sin^2B)-(1-sin^2A)sin^2B)/sinAsinB
=(sin^2A-sin^2B)/sinAsinB
=sinA/sinB-sinB/sinA 正弦定理
=a/b-b/a

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