在三角形中,cosAcosBcosC的最大值?

楼上不要胡写
公式都错着呢，算啥着呢
cosAcosBcosC
=-cos(B+C)cosBcosC
=-(cosBcosC-sinBsinC)cosBcosC
设cosBcosC=x，sinBsinC=y
=-(x-y)x
=-(x^2-xy)
=-(x-xy+1/4y-1/4y)
=-(x-1/2y)^2+1/4y
当x=1/2y取得最大，即1/4y
cosBcosC=1/2sinBsinC
tanBtanC=2
-tan(A+C)tanC=2
-((tanA+tanC)/(1-tanAtanC))tanC=2
(tanAtanC+tanC^2)=-2(1-tanAtanC)
设tanA=m,tanC=n
mn+n^2=-2(1-mn)
n^2-mn+2=0
n^2-mn+m^2/4-m^2/4+2=0
(n-1/2m)^2+2-1/4m^2=0
n=1/2m取最大
tanA=1/2tanC
剩下你自己看哈

设COS A COSB COSC=t

∵A+B+C=π，又因t= [cos(A+B)+cos(A－B)]cosC=－ [cos2C－cos(A－B)cosC]

∴cos＾2C－cos(A－B)cosC+2t=0

∵cosC∈R,故△≥0，即cos＾2(A－B)－8t≥0

∴t≤ cos2(A－B)/8 ≤ 1/8
故cosAcosBcosC ≤ 1/8

最大值1/8