三角函数,
问题描述：
三角函数,
三角形中,若tanAtanB/(tanAtanC+tanBtanC)=-1/6,
则(a^2+b^2)/c^2=?没分了,化了个半死,还没出来
答
在三角形ABC中,由余弦定理：
c^2=a^2+b^2-2ab*cosC 两边同除以c^2
(a^2+b^2)/c^2=1+2(ab*cosC)/c^2.(1)
已知：tanAtanB/(tanAtanC+tanBtanC)=-1/6
(sinA*sinB)/[cosAcosB*tanC*(tanA+tanB)]=-1/6
(sinA*sinB)/[tanC*(sinA*cosB+cosB*sinA)]=-1/6
(sinA*sinB)/[(sinC/cosC)*sin(A+B)]=-1/6
(sinA*sinB)/[(sinC/cosC)*sin(180°-C)]=-1/6
(sinA*sinB)/[(sinC/cosC)*sinC]=-1/6
(sinA*sinB*cosC)/[(sinC)^2]=-1/6 .(2)
由正弦定理：a/sinA=b/sinB=c/sinC=2R=t
sinA=a/t,sinB=b/t,sinC=C/t
代入（2）式得：(ab*cosC)/c^2=-1/6
2(ab*cosC)/c^2=-1/3 代入（1）式得：
(a^2+b^2)/c^2=1+2(ab*cosC)/c^2
=1-1/3
=2/3
故：(a^2+b^2)/c^2=2/3.