F.4 AM

2008-01-06 6:56 am
Prove the following identities for triangle ABC
a)(b^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC=0
b)cotA-cotB=b^2-a^2/absinC

回答 (1)

2008-01-06 10:51 pm
✔ 最佳答案
(a)
LHS
= (b^2 - c^2)cotA+(c^2 -a^2)cotB+(a^2 -b^2) cotC
=(b^2 - c^2)cosA/sinA + (c^2 - a^2)cosB/sinB + (a^2 - b^2)cosC/sinC
=(b^2 - c^2)(b^2 + c^2 - a^2)/2bcsinA + (c^2 - a^2)(a^2 + b^2 - c^2)/2ac (bsinA/a ) + (a^2 - b^2)(a^2 + c^2 - b^2)/2ab (csinA/a)
= 1/2bcsinA *{ (b²-c²)(b²+c²-a²) + (c²-a²)(c²+a²-b²) + (a²-b²)(a²+b²-c²) }
= 1/2bcsinA *{ (b²-c²)(b²+c²) -a²(b²+c²) + (c²-a²)(c²+a²) - b²(c²+a²) + (a²-b²)(a²+b²) -c²(a²+b²) }
= 1/2bcsinA * (b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2)
=0
=RHS

(b)
LHS
=cotA-cotB
=cosA/sinA - cosB/sinB
= cosAsinB-sinAcosB/si nAsinB
={sinB (b²+c²-a²/2bc) - sinA(a²+c²-b²/2ac)}/sinAsinB
=sinA[(b²+c²-a²)/2bc * b/a - (a²+c²-b²)/2ac]/sinAsinB
=(b²+c²-a²-a²+b²)/2acsinB
=2(b²-a²)/[2ac*(bsinC/c)]
=(b²-a²)/absinC
=RHS


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