Question

一道数学三角函数的题，求解答过程。谢谢!

设ABC的内角A、B、C的所对的边长分别为a、b、c,且a2+b2=mc2(m为常数)，且tanC(tanA+tanB)=2tanAtanB，求m的值（答案是2）

Answer 1

tanAtanB/[（tanA+tanB）tanC] = 1/2
tanAtanB/[（tanA+tanB）tanC]
=sinAsinB/((sinAcosB+cosAsinB)tanC)
=sinAsinB/(tanCsin(A+B))
=sinAsinBcosC/(sinCsinC)
根据正弦定理得：sinA/sinC=a/c, sinB/sinC=b/c,
根据余弦定理得：cosC=(a^2+b^2-c^2)/2ab,

所以上式=ab/c^2*(a^2+b^2-c^2)/2ab
=1/2*(a^2+b^2-c^2)/ c^2
=1/2*(m-1),

根据已知：1/2*(m-1) = 1/2, m=2

Answer 2

解：∵tanC(tanA+tanB)=2tanAtanB
∴tanAtanB/[（tanA+tanB）tanC] = 1/2
∴1/2=tanAtanB/[（tanA+tanB）tanC]
=sinAsinB/((sinAcosB+cosAsinB)tanC)
=sinAsinB/(tanCsin(A+B))
=sinAsinBcosC/(sinCsinC)
=（sinA/sinC）（sinB/sinC）cosC
而sinA/sinC=a/c,
sinB/sinC=b/c,
cosC=(a^2+b^2-c^2)/2ab,
∴1/2=（a/c ）（b/c）(a^2+b^2-c^2)/2ab
=ab/c^2*(a^2+b^2-c^2)/2ab
=1/2*(a^2+b^2-c^2)/ c^2
=1/2*(m-1),
∴m=2