Question

高一三角函数的问题

b/a + a/b =6cosC 求 tanC/tanA +tanC/tanB=
要过程

Answer 1

b/a + a/b = (b² + a²)/ab

由余弦定理
c² = a² +b² -2abcosC
(b²+ a²)/ab =( c²+2abcosC)/ab = c²/ab +2cosC = sin²C/sinAsinB +2cosC
所以 sin²C/sinAsinB = 4cosC

两边同除cosC
tanC * (sinC/sinAsinB)=4
sinC/sinAsinB = sin(A+B)/sinAsinB = (sinAcosB+cosAsinB)/sinAsinB
= cosB/sinB + cosA/sinA = 1/tanB +1/tanA

tanC * (sinC/sinAsinB) = tanC/tanA +tanC/tanB =4

Answer 2

已知式：b/a a/b=6*(a^2 b^2-c^2)/2ab[用正余弦定理边角化边] 左边通分然后化简得：2a^2 2b^2-3c^2=0 所以：c^2=2/3a^2 2/3b^2 求的式子=(sinC/cosC)/(sinA/cosA) (sinC/cosC)/(sinB/sinB)=sinCcosC/sinAcosC sinCcosB/sinBcosC=[用正余弦定理把所有角化成边，再化简](b^2 c^2-a^2)/(a^2 b^2-c^2) (a^2 c^2-b^2)/(a^2 b^2-c^2)=[把由已知式得到的式子代进去，化简](5/3b^2-1/3a^2)/(1/3a^2 1/3b^2) (5/3a^2-1/3b^2)/(1/3a^2 1/3b^2)=[4/3(a^2 b^2)]/[1/3(a^2 b^2)]=4 [爪机党好无力……………]

Answer 3

b/a+a/b=(a^2+b^2)/ab=6cosC
根据余弦定理：a^2+b^2-2abcosC=c^2
=>4ab cosC=c^2
tanC/tanA+tanC/tanB
=tanC[(sinBcosA+sinAcosB)/sinAsinB]
=tanC*sin(A+B)/(sinAsinB)
=tanC*sinC/(sinA*sinB)
=(sinC)^2/(sinA*sinB*cosC)
由正弦定理，(sinC)^2/(sinAsinB)=c^2/(ab)
所以上式 = c^2/(ab cosC)
又 4ab cosC=c^2
所以上式 = 4